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Teacher’s Solutions Manual and Answer Keys
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TEACHER’S
SOLUTIONS MANUAL
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Holt Physics
Print Solutions Manual
Teacher’s
Illustrations
All work is contributed by Holt, Rinehart and Winston and TSI Graphics.
Cover Photo: © Lawrence Manning/CORBIS
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Contents
Section I Pupil’s Edition Solutions
Chapter 1
The Science of Physics
Chapter 2
Motion in One Dimension
Chapter 3
Two-Dimensional Motion and Vectors
Chapter 4
Forces and the Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-4-1
Chapter 5
Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-5-1
Chapter 6
Momentum and Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-6-1
Chapter 7
Rotational Motion and the Law of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . I-7-1
Chapter 8
Rotational Equilibrium and Dynamics
Chapter 9
Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-9-1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-1-1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-2-1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-3-1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-8-1
Chapter 10 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-10-1
Chapter 11 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-11-1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Chapter 12 Vibrations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-12-1
Chapter 13 Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-13-1
Chapter 14 Light and Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-14-1
Chapter 15 Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-15-1
Chapter 16 Interference and Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-16-1
Chapter 17 Electric Forces and Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-17-1
Chapter 18 Electrical Energy and Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-18-1
Chapter 19 Current and Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-19-1
Chapter 20 Circuits and Circuit Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-20-1
Contents
iii
Chapter 21 Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-21-1
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Chapter 22 Induction and Alternating Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-22-1
Chapter 23 Atomic Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-23-1
Chapter 24 Modern Electronics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-24-1
Chapter 25 Subatomic Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-25-1
Section II Problem Workbook Solutions
Chapter 1
The Science of Physics
Chapter 2
Motion in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-2-1
Chapter 3
Two-Dimensional Motion and Vectors
Chapter 4
Forces and the Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-4-1
Chapter 5
Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-5-1
Chapter 6
Momentum and Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-6-1
Chapter 7
Rotational Motion and the Law of Gravity . . . . . . . . . . . . . . . . . . . . . . . . II-7-1
Chapter 8
Rotational Equilibrium and Dynamics
Chapter 9
Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-9-1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-1-1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . II-8-1
Chapter 10 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-10-1
Chapter 11 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-11-1
Chapter 12 Vibrations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-12-1
Chapter 13 Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-13-1
Chapter 14 Light and Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-14-1
Chapter 15 Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-15-1
iv
Contents
Copyright © by Holt, Rinehart and Winston. All rights reserved.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . II-3-1
Chapter 16 Interference and Diffraction
Print
Chapter 17 Electric Forces and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-17-1
Chapter 18 Electrical Energy and Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-18-1
Chapter 19 Current and Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-19-1
Chapter 20 Circuits and Circuit Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-20-1
Chapter 21 Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-21-1
Chapter 22 Induction and Alternating Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-22-1
Chapter 23 Atomic Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-23-1
Chapter 25 Subatomic Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-25-1
(The Holt Physics Problem Workbook does not contain practice problems for Chapter 24.)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II-16-1
Section III Section Review Worksheets Answers
III-1
Section IV Interactive Tutor Worksheets Answers
IV-1
Section V Problem Bank Solutions
Chapter 1
The Science of Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-1-1
Chapter 2
Motion in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-2-1
Chapter 3
Two-Dimensional Motion and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-3-1
Chapter 4
Forces and the Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-4-1
Chapter 5
Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-5-1
Chapter 6
Momentum and Collisions
Chapter 7
Rotational Motion and the Law of Gravity
Chapter 8
Rotational Equilibrium and Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-8-1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-6-1
. . . . . . . . . . . . . . . . . . . . . . . . V-7-1
Contents
v
Chapter 9
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Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-9-1
Print
Chapter 10 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-10-1
Chapter 11 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-11-1
Chapter 12 Vibrations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-12-1
Chapter 13 Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-13-1
Chapter 14 Light and Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-14-1
Chapter 15 Refraction
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-15-1
Chapter 16 Interference and Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-16-1
Chapter 17 Electric Forces and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-17-1
Chapter 18 Electrical Energy and Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-18-1
Chapter 19 Current and Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-19-1
Chapter 20 Circuits and Circuit Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-20-1
Chapter 21 Magnetism. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-21-1
Chapter 23 Atomic Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-23-1
Chapter 25 Subatomic Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-25-1
(Problem banks for the above chapters may be found on the Holt Physics One-Stop
Planner. It does not contain a problem bank for Chapter 24.)
vi
Contents
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Chapter 22 Induction and Alternating Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V-22-1
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Section
Pupil’s Edition
Solutions
I
Holt Physics
I
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The Science of Physics
Chapter
1
I
Practice 1A, p. 14
Givens
Solutions
1. diameter = 50 mm
1 × 10−6 m
50 mm ×  = 5 × 10−5 m
1 mm
2. period = 1 ms
1 × 10−6 s
1 ms ×  = 1 × 10−6 s
1 ms
3. diameter = 10 nm
1 × 10−9 m
a. 10 nm ×  = 1 × 10−8 m
1 nm
1 mm
b. 1 × 10−8 m × 
= 1 × 10−5 mm
1 × 10−3 m
1 mm
c. 1 × 10−8 m × 
= 1 × 10−2 mm
1 × 10−6 m
4. distance = 1.5 × 1011 m
1 Tm
1.5 × 1011 m ×  
= 1.5 × 10-1 Tm
1 × 1012 m
1 km
1.5 × 1011 m × 
= 1.5 × 108 km
1 × 103 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. mass = 1.440 × 106 g
1 kg
1.440 × 106 g × 
= 1.440 × 103 kg
1 × 103 g
Section Review, p. 19
2. mass = 6.20 mg
4.
1 kg
1 × 10−3 g
a. 6.20 mg ×  × 
= 6.20 × 10−6 kg
1 × 103 g
1 mg
time = 3 × 10−9 s
1 ms
b. 3 × 10−9 s × 
= 3 × 10−6 ms
1 × 10−3 s
distance = 88.0 km
1 × 103 m
c. 88.0 km ×  = 8.80 × 104 m
1 km
a. 26 × 0.02584 = 0.67184 = 0.67
b. 15.3 ÷ 1.1 = 13.90909091 = 14
c. 782.45 − 3.5328 = 778.9172 = 778.92
d. 63.258 + 734.2 = 797.458 = 797.5
Section One — Pupil’s Edition Solutions
I Ch. 1–1
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Chapter Review and Assess, pp. 27–31
I
Givens
Solutions
11. 2 dm
1 mm
1 × 10–1 m
a. 2 dm ×  × 
= 2 × 102 mm
1 × 10−3 m
1 dm
2 h 10 min
60 min
b. 2 h ×  = 120 min
1h
120 min + 10 min = 130 min
60 s
130 min ×  = 7.8 × 103 s
1 min
16 g
1 mg
c. 16 g × 
= 1.6 × 107 mg
1 × 10−6 g
0.75 km
1 cm
1 × 103 m
d. 0.75 km ×  × 
= 7.5 × 104 cm
1 × 10−2 m
1 km
0.675 mg
1 × 10−3 g
e. 0.675 mg ×  = 6.75 × 10−4 g
1 mg
462 mm
1 cm
1 × 10−6 m
f. 462 mm ×  × 
= 4.62 × 10−2 cm
1 × 10−2 m
1 mm
35 km/h
1h
35 km
1 × 103 m
g.  ×  ×  = 9.7 m/s
3600 s
h
1 km
1 dekaration
 = 1 dekaration
a. 10 rations × 
101 rations
2000 mockingbirds
1 kmockingbirds
b. 2000 mockingbirds × 
= 2 kilomockingbirds
1 × 103 mockingbirds
10−6 phones
1 mphone
 = 1 microphone
c. 10−6 phones × 
10−6 phones
10−9 goats
1 ngoat
 = 1 nanogoat
d. 10−9 goats × 
10−9 goats
1018 miners
1 Eminer
 = 1 examiner
e. 1018 miners × 
1018 miners
13. speed of light =
3.00 × 108 m/s
3.00 × 108 m 3600 s
1 km
 ×  × 1 h × 
= 1.08 × 109 km
s
1h
1 × 103 m
∆t = 1 h
14. 1 ton = 1.000 × 103 kg
mass/person = 85 kg
I Ch. 1–2
1 person
1.000 × 103 kg ×  = 11 people
85 kg
Note that the numerical answer, 11.8 people, must be rounded down to 11 people.
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
12. 10 rations
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Givens
Solutions
20.
a. 756 g + 37.2 g + 0.83 g + 2.5 g = 796.53 g = 797 g
3.2 m
b.  = 0.898119562 m/s = 0.90 m/s
3.563 s
c. 5.67 mm × p = 17.81283035 mm = 17.8 mm
I
d. 27.54 s − 3.8 s = 23.74 s = 23.7 s
21. 93.46 cm, 135.3 cm
22.
l = 38.44 m
w = 19.5 m
26. s = (a + b + c) ÷ 2
93.46 cm + 135.3 cm = 228.76 cm = 228.8 cm
38.44 m + 38.44 m + 19.5 m + 19.5 m = 115.88 m = 115.9 m
r=
(s − a)(s − b)(s − c)

s
r, a, b, c, and s all have units of L.
length =
length × length × length
=
length
(length)3  = (l
)2 = length
en
gt
h
length
Thus, the equation is dimensionally consistent.
27.
T = 2p
g
L
Substitute the proper dimensions into the equation.
time =
length
=
(t
im
e)
2 = time
[length/(time)2]
Thus, the dimensions are consistent.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
28.
(m/s)2 ≠ m/s2 × s
m2/s2 ≠ m/s
The dimensions are not consistent.
29.
Estimate one breath every 5 s.
365 days 24 h 3600 s 1 breath
70 years ×  ×  ×  ×  = 4 × 108 breaths
1 year
1 day
1h
5s
30.
Estimate one heart beat per second.
24 h 3600 s 1 beat
1 day ×  ×  ×  = 9 × 104 beats
1 day
1h
s
31.
Ages will vary.
365 days 24 h 3600 s
17 years ×  ×  ×  = 5.4 × 108 s
1 year
1 day
1h
32.
Estimate a tire’s radius to be 0.3 m.
1.609 km 103 m
1 rev
50 000 mi ×  ×  ×  = 4 × 107 rev
1 mi
1 km 2 p (0.3 m)
Section One — Pupil’s Edition Solutions
I Ch. 1–3
Givens
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Solutions
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33.
Estimate 30 balls lost per game.
30 balls
81 games × = 2 × 103 balls
1game
I
1
Estimate 4 lb per burger and 800 lb per head of cattle.
34.
0.25 lb
5 × 1010 burgers × = 1 × 1010 lb
1burger
0.25 lb 1 head
5 × 1010 burgers × ×  = 2 × 107 head of cattle
1burger 800 lb
35.
Estimate 1 tuner per 10,000 people.
1 tuner
8 × 106 people ×  = 8 × 102 tuners
10,000 people
36. diameter = 3.8 cm
l = 4 m w = 4 m h = 3m
Find the number of balls that can fit along the length and width.
1 ball
4 m ×  = 100 balls
0.038 m
Find the number that can be stacked to the ceiling.
1 ball
3 m ×  = 80 balls
0.038 m
Multiply all three figures to find the number of balls that can fit in the room.
100 balls × 100 balls × 80 balls = 8 × 105 balls
A rough estimate: divide the volume of the room by the volume of a ball.
37. r = 3.5 cm
a. C = 2pr = 2p (3.5 cm) = 22 cm
r = 4.65 cm
b. C = 2pr = 2p (4.65 cm) = 29.2 cm
A = pr 2 = p (4.65 cm)2 = 67.9 cm2
1s
1h
1 day
1 year
5 × 109 bills ×  ×  ×  ×  = 272 years
1 bill 3600 s 14 h 365 days
38.
Take the $5000. It would take 272 years to count 5 billion $1 bills.
3.786 × 10−3 m3
V = 1 quart ×  = 9.465 × 10−4 m3
4 quarts
39.
V = L3
L= 3
V = 3
9.46
5×10−4m
3 = 9.818 × 10−2 m
40. mass = 9.00 × 10−7 kg
density = 918 kg/m3
r = 41.8 cm
area = pr 2
I Ch. 1–4
mass
9.00 × 10−7 kg
volume = = 
= 9.80 × 10−10 m3
density
918 kg/m3
volume 9.80 × 10−10 m3
diameter =  = 
= 1.79 × 10−9 m
area
p (0.418 m)2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
A = pr 2 = p (3.5 cm)2 = 38 cm2
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Givens
Solutions
41.
Estimate 1 palm to be 0.08 m.
V ark = 300 cubits × 50 cubits × 30 cubits = 4.5 × 105 cubits3
63 palms3 (0.08 m)3
 = 5 × 104 m3
4.5 × 105 cubits3 × 
×
1 cubit3
13 palm3
I
Estimate the average size of a house to be 2000 ft2 and 10 ft tall.
V house = 2000 ft2 × 10 ft = 2 × 104 ft3
(1 m)3
2 × 104 ft3 × 
= 6 × 102 m3
3.2813 ft3
42. d = 1.0 × 10−6 m
l = 1.0 m
number of micrometeorites per side:
1 micrometeorite
= 1.0 × 106 micrometeorites
1.0 m × 
1.0 × 10−6 m
number of micrometeorites needed to cover the moon to a depth of 1.0 m:
(1.0 × 106 micrometeorites)3 = 1.0 × 1018 micrometeorites
1s
1h
1 day
1 year
1.0 × 1018 micrometeorites ×  ×  ×  ×  =
1 micrometeorite 3600 s 24 h 365 days
3.2 × 1010 years
Note that a rougher estimate can be made by dividing the volume of the 1.0 m3 box
by the volume of a micrometeorite.
43. V = 1.0 cm3
m = 1.0 × 10−3 kg
44. density = r = 1.0 × 103 kg/m3
1.0 × 10−3 kg
1 cm 3

× 
× 1.0 m3 = 1.0 × 103 kg
3
1.0 cm
(1 × 10−2 m)3
a.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
diameter = 1.0 mm
l = 4.0 mm
diameter = 2r = 2.0 mm
density =r = 1.0 × 103 kg/m3
45. r = 5.85 × 109 cm
m = 5.68 × 1029 g
r = 5.85 × 107 m
1.0 × 10−6 m
4
4
V = 3 pr 3 = 3 p  = 5.2 × 10−19 m3
2
3
m = rV = (1.0 × 103 kg/m3)(5.2 × 10−19 m3)(0.9) = 5 × 10−16 kg
b.
2.0 × 10−3 m
V = l pr 2 = (4.0 × 10−3m) (p)  = 1.3 × 10−8 m3
2
2
m = rV = (1.0 × 103 kg/m3)(1.3 × 10−8 m3)(0.9) = 1 × 10−5 kg
4
4
a. V = 3 pr 3 = 3p (5.85 × 109 cm)3 = 8.39 × 1029 cm3
m
5.68 × 1029g
= 0.677 g/cm3
density =  = 
V 8.39 × 1029 cm3
b. A = 4pr 2 = 4p(5.85 × 107 m)2 = 4.30 × 1016 m2
Section One — Pupil’s Edition Solutions
I Ch. 1–5
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Motion In One Dimension
Chapter
2
I
Practice 2A, p. 44
Givens
1. vavg = 0.98 m/s east
∆t = 34 min
2. ∆t = 15 min
vavg = 12.5 km/h south
3. ∆t = 9.5 min
vavg = 1.2 m/s north
4. vavg = 48.0 km/h east
∆x = 144 km east
5. vavg = 56.0 km/h east
∆x = 144 km east
6. ∆x1 = 280 km south
vavg,1 = 88 km/h south
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t2 = 24 min
vavg,2 = 0 km/h
∆x3 = 210 km south
vavg,3 = 75 km/h south
Solutions
∆x = vavg ∆t = (0.98 m/s)(34 min)(60 s/min)
= 2.0 × 103 m = 2.0 km east
1h
∆x = vavg ∆t = (12.5 km/h)(15 min) 
60 min
= 3.1 km south
∆x = vavg ∆t = (1.2 m/s) (9.5 min)(60 s/min)
= 6.8 × 102 m north
144 km
∆x
∆t =  = = 3.00 h
vavg 48.0 km/h
144 km
∆x
∆t =  = = 2.57 h
vavg 56.0 km/h
time saved = 3.00 h − 2.57 h = 0.43 h = 25.8 min
∆x1
∆x3
a. ∆ttot = ∆t1 + ∆t2 + ∆t3 = + ∆t2 + vavg,1
vavg,3
280 km
1h
210 km
∆t = + (24 min)  + 
88 km/h
60 min
75 km/h
∆t = 3.2 h + 0.40 h + 2.8 h = 6.4 h = 6 h 24 min
∆xtot ∆x1 + ∆x2 + ∆x3
b. ∆avg, tot = 
= 
∆ttot
∆t1 + ∆t2 + ∆t3
1h
∆x2 = ∆avg,2 ∆t2 = (0.km/h)(24 min)  = 0 km
60 min
280 km + 0 km + 210 km 490 km
∆avg, tot =  =  = 77.0 km/h south
6.4 h
6.4 h
Section Review, p. 47
2. v = 3.5 mm/s ∆x = 8.4 cm
∆x
8.4 cm
∆t =  =  = 24 s
v
0.35 cm/s
4. v = 1.5 m/s ∆x = 9.3 m
9.3 m
∆x
∆t =  =  = 6.2 s
1.5 m/s
v
Section One — Pupil’s Edition Solutions
I Ch. 2–1
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Givens
Solutions
5. ∆x1 = 50.0 m south
∆t1 = 20.0 s
∆x2 = 50.0 m north
I
∆t2 = 22.0 s
∆x
50.0 m
a. vavg,1 = 1 =  = 2.50 m/s south
∆t1
20.0 s
∆x
50.0 m
b. vavg,2 = 2 =  = 2.27 m/s north
∆t2
22.0 s
∆xtot = ∆x1 + ∆x2 = (50.0 m) + (−50.0 m) = 0.0 m
∆ttot = ∆t1 + ∆t2 = 20.0 s + 22.0 s = 42.0 s
∆xtot 0.0 m
vavg = 
=  = 0.0 m/s
∆ttot 42.0 s
6. v1 = 0.90 m/s
v2 = 1.90 m/s
∆x = 780 m
780 m
∆x
a. ∆t1 =  =  = 870 s
v1 0.90 m/s
780 m
∆x
∆t2 =  =  = 410 s
v2 1.90 m/s
∆t1 − ∆t2 = 870 s − 410 s = 460 s
∆t1 − ∆t2 =
(5.50 min)(60 s/min) =
3.30 × 102 s
b. ∆x1 = v1∆t1
∆x2 = v2∆t2
∆x1 = ∆x2
v1∆t1 = v2∆t2
v1[∆t2 + (3.30 × 102 s)] = v2∆t2
v1∆t2 + v1(3.30 × 102 s) = v2∆t2
∆t2 (v1 − v2) = −v1(3.30 × 102 s)
2
−(0.90 m/s)(3.30 × 102 s)
−v1(3.30 × 102 s) −(0.90 m/s)(3.30 × 10 s)
 =  = 
∆t2 = 
0.90 m/s − 1.90 m/s
−1.00 m/s
v1 − v2
∆t2 = 3.00 × 102 s
∆t1 = ∆t2 + (3.30 × 102 s) = (3.00 × 102 s) + (3.30 × 102 s) = 630 s
∆x2 = v2 ∆t2 = (1.90 m/s)(3.00 × 102 s) = 570 m
Practice 2B, p. 49
1. aavg = − 4.1 m/s2
vi = 9.0 m/s
vf – vi
∆v
– 9.00 m/s
0 m/s – 9.00 m/s
∆t =  =  = 
= 2 = 2.2 s
aavg
aavg
– 4.1 m/s
– 4.1 m/s2
vf = 0 m/s
2. aavg = 2.5 m/s2
vi = 7.0 m/s
∆v vf – vi
12.0 m/s – 7.0 m/s 5.00 m/s
∆t =  =  = 
= 2 = 2.0 s
aavg
aavg
2.5 m/s2
2.5 m/s
vf = 12.0 m/s
3. aavg = −0.50 m/s2
vi = 13.5 m/s
vf − vi 0 m/s − 13.5 m/s
−13.5
∆t =  = 
=  = 27 s
∆t
−0.50 s
− 0.50 m/s2
vf = 0 m/s
I Ch. 2–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆x1 = v1∆t1 = (0.90 m/s)(630 s) = 570 m
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Givens
4. vi = −1.2 m/s
vf = −6.5 m/s
Solutions
vf − vi −6.5 m/s − (−1.2 m/s) −5.3 m/s
aavg =  =  =  = −3.5 × 10−3 m/s2
(25 min)(60 s/min)
1500 s
∆t
∆t = 25 min
5. aavg = 4.7 × 10−3 m/s2
∆t = 5.0 min
I
a. ∆v = aavg ∆t = (4.7 × 10−3 m/s2)(5.0 min)(60 s/min) = 1.4 m/s
b. vf = ∆v + vi = 1.4 m/s + 1.7 m/s = 3.1 m/s
vi = 1.7 m/s
Practice 2C, p. 53
1. vi = 0 km/h = 0 m/s
vf = 23.7 km/h
23.7 × 103 m
1
1
∆x = 2(vi + vf )∆t = 2 0 m/s +  (6.5 s) = 21 m
(1 h)(3600 s/h)
∆t = 6.5 s
2. vi = 15.00 m/s
1
1
∆x = 2(vi + vf )∆t = 2(15.00 m/s + 0 m/s)(2.50 s) = 18.8 m
vf = 0 m/s
∆t = 2.50 s
vf = 0 m/s
vf − vi 0 m/s − 100 m/s
∆t =  = 
= 20 s
−5.0 m/s2
aavg
a = −5.0 m/s2
∆x = 2(vi + vf )∆t = 2(100 m/s + 0 m/s)(20 s) = 1000 m = 1 km
3. vi = 100 m/s
1
1
The plane cannot land at an airport with a 0.80 km-long runway. The runway must
be at least 1 km long.
4. vi = 78 km/h
∆x = 99 m
(2)(99 m)(3600 s/h)
2∆x
∆t =  = 
= 9.1 s
3
vi + vf 78 × 10 m/h + 0 m/h
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vf = 0 m/s
5. vi = 6.4 m/s
∆x = 3.2 km
(2)(3.2 × 103 m)
2∆x
vf =  − vi =  − 6.4 m/s = 3.0 × 101 m/s − 6.4 m/s = 24 m/s
(3.5 min)(60 s/min)
∆t
∆t = 3.5 min
Practice 2D, p. 55
1. vi = 23.7 km/h
vf = vi + a∆t = (23.7 × 103 m/h)(1 h/3600 s) + (0.92 m/s2)(3.6 s)
a = 0.92 m/s2
vf = 6.58 m/s + 3.3 m/s = +9.9 m/s
∆t = 3.6 s
or (9.9 × 10−3 km/s)(3600 s/h) = +36 km/h
1
∆x = vi ∆t + 2a∆t 2
1
∆x = (23.7 × 103 m/h)(1 h/3600 s)(3.6 s) + 2(0.92 m/s2)(3.6 s)2
∆x = 24 m + 6.0 m = 3.0 × 101 m = +0.03 km
Section One — Pupil’s Edition Solutions
I Ch. 2–3
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Givens
Solutions
2. vi = 4.30 m/s
a = 3.00 m/s2
∆t = 5.0 s
vf = vi + a∆t = 4.30 m/s + (3.00 m/s2)(5.0 s)
vf = 4.30 m/s + 15 m/s = 19 m/s
1
∆x = vi ∆t + 2a∆t 2
1
∆x = (4.30 m/s)(5.0 s) + 2(3.00 m/s2)(5.0 s)2
I
∆x = 22 m + 38 m = 6.0 × 101 m
3. vi = 0.0 m/s
vf = vi + a∆t = 0 m/s + (−1.5 m/s2)(5.0 s) = −7.5 m/s
1
1
∆t = 5.0 s
∆x = vi ∆t + 2a∆t 2 = (0 m/s)(5.0 s) + 2(−1.5 m/s2)(5.0 s)2 = −19 m
a = −1.5 m/s2
distance traveled = 19 m
4. vi = 15.0 m/s
a = −2.0 m/s2
vf = 10.0 m/s
vf − vi 10.0 m/s − 15.0 m/s −5.0
∆t =  = 
=  s = 2.5 s
a
−2.0 m/s2
−2.0
1
∆x = vi ∆t + 2a∆t 2
1
∆x = (15.0 m/s)(2.5 s) + 2(−2.0 m/s2)(2.5 s)2
∆x = 38 m + (−6.2 m) = 32 m
distance traveled during braking = 32 m
Practice 2E, p. 58
a = 0.500 m/s2
∆x = 6.32 m
2. vi = +7.0 m/s
a = +0.80 m/s2
∆x = 245 m
vf = vi 2+
2a∆
x
vf = (0
m/s
)2
+(2)
m/s
m2/
s2 = ±2.51 m/s
(0
.5
00
2)(6
.3
2m
) = 6.3
2
vf = +2.51 m/s
a. vf = vi 2+
2a∆
x
vf = (7
m/s
)2
+(2)
m)
.0
(0
.8
0m
/s
2)(2
45
vf = 49
m2/
s2
+390
m2/
s2 = 44
s2 = ±21 m/s
0m
2/
vf = +21 m/s
∆x = 125 m
b. vf = (7
m/s
)2
+(2)
m)
.0
(0
.8
0m
/s
2)(1
25
vf = 49
m2/
s2
+(2.
02
m2/
s2) = 25
s2
0×1
0m
2/
vf = ±16 m/s = +16m/s
∆x = 67 m
c. vf = (7
m/s
)2
+(2)
s2
+110
m2/
s2
.0
(0
.8
0m
/s
2)(6
7m
) = 49
m
2/
vf = 16
s2 = ±13 m/s = +13 m/s
0m
2/
3. vi = 0 m/s
a = 2.3 m/s2
∆x = 55 m
a. vf = vi 2+
m/s
)2
+(2)
m/s
2a∆
x = (0
(2
.3
2)(5
5m
)
vf = 25
s2 = ±16 m/s
0m
2/
car speed = 16 m/s
vf
16 m/s
b. ∆t =  = 2 = 7.0 s
a 2.3 m/s
I Ch. 2–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. vi = 0 m/s
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Givens
4. a = 0.85 m/s2
vi = 83 km/h
vf = 94 km/h
Solutions
2
1h
[(94 × 103 m/h)2 − (83 × 103 m/h)2] 
3600 s
∆x =  = 
2a
(2)(0.85 m/s2)
2
vf − vi2
2
1h
(8.8 × 109 m2/h2 − 6.9 × 109 m2/h2) 
3600 s
∆x = 
(2)(0.85 m/s2)
I
2
1h
(1.9 × 109 m2/h2) 
3600 s
1.5 × 102 m2/s2
= 88 m
∆x = 
= 
2
(2)(0.85 m/s )
1.7 m/s2
distance traveled = 88 m
5. vi = 0 km/h
vf = 120 km/h
∆x = 240 m
6. vi = 6.5 m/s
vf = 1.5 m/s
2
2
1 h 103m
[(120 km/h)2 − (0 km/h)2]  
2
2
vf − vi
3600 1 km
a =  = 
2∆x
(2)(240 m)
1100 m2/s2
a =  = 2.3 m/s2
480 m
vf 2 − vi2 (1.5 m/s)2 − (6.5 m/s)2
−40 m2/s2
∆x =  = 
= 2 = 7.4 m
2
(2)(−2.7 m/s )
−5.4 m/s
2a
a = −2.7 m/s2
Section Review, p. 59
1. a = +2.60 m/s2
vi = 88.5 km/h
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vf = 96.5 km/h
3. vi = 0 m/s
vf = 12.5 m/s
∆t = 2.5 s
vf − vi (96.5 × 103 m/h) − (88.5 × 103 m/h)(1 h/3600 s)
∆t =  = 
2.60 m/s2
a
2.20 m/s
∆t = 2 = 0.85 s
2.60 m/s
vf − vi 12.5 m/s − 0 m/s
a. a =  =  = +5.0 m/s2
∆t
2.5 s
b. ∆x = vi∆t + 2a∆t 2 = (0 m/s)(2.5 s) + 2(5.0 m/s2)(2.5 s)2 = +16 m
1
1
∆x 16 m
c. vavg =  =  = +6.4 m/s
∆t 2.5 s
Practice 2F, p. 64
1. ∆y = −239 m
vi = 0 m/s
a = −3.7 m/s2
vf − v
−42.1 m/s − 0 m/s
a. ∆t =  i = 
= 11 s
a
−3.7 m/s2
b. vf = vi 2+
m/s
)2
+(2)
2a∆
y = (0
(−
3.
7m
/s
2)(−
23
9m
)
vf = 1.
03
m2/
s2 = ±42 m/s
77
×1
vf = −42 m/s
2. ∆y = −25.0 m
vi = 0 m/s
a = −9.81 m/s2
a. vf = vi 2+
m/s
m/s
m)
2a∆
y = (0
)2+(2)
(−
9.
81
2)(−
25
.0
vf = 4.9
02
m2/
s2 = −22.1 m/s
0×1
vf − vi −22.1 m/s − 0 m/s
= 2.25 s
b. ∆t =  = 
a
−9.81 m/s2
Section One — Pupil’s Edition Solutions
I Ch. 2–5
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Givens
Solutions
2
a. vf = v
+2a⌬
m/s
)2
+(2)
m/s
m)
y = (8
.0
(−
9.
81
2)(0
i
3. vi = +8.0 m/s
a = −9.81 m/s2
vf = 64
m/s
= ±8.0 m/s = −8.0 m/s
vf − vi −8.0 m/s − 8.0 m/s −16.0 m/s
= 2 = 1.63 s
b. ∆t =  = 
a
−9.81 m/s
−9.81 m/s2
∆y = 0 m
I
vf 2 − vi 2 (0 m/s)2 − (7.5 m/s)2
−56 m2/s2
a. ∆y =  = 
= 
= 2.9 m
2
2a
(2)(−9.81 m/s2)
(2)(−9.81 m/s )
4. y1 = 0.80 m
vi = +7.5 m/s
a = −9.81 m/s2
maximum height = y2 = y1 + ∆y = 0.80 m + 2.9 m = 3.7 m
vf − v 0 m/s − 7.5m/s
b. ∆t = i = 
= 0.76 s
a
−9.81 ms2
vf = 0 m/s
5. y1 = 1.3 m
vf 2 − vi2 (0 m/s)2 − (2.3 m/s)2
−5.3 m2/s2

 = +0.27 m
a. ∆y =  = 
=
2a
(2)(−9.81 m/s2)
(2)(−9.81 m/s2)
vi = +2.3 m/s
maximum height = y2 = y1 + ∆y = 1.3 m + 0.27 m = 1.6 m
vf = 0 m/s
a = −9.81 m/s2
The apple will not reach the treehouse, which is 3.7 m farther up.
b. Setting vi = 0 m/s and ∆y = −1.6 m,
vi2+
m/s
)2
+(2)
m/s
vf = 2a∆
y = (0
(−
9.
81
2)(−
1.
6m
)
vf = 31
m2/
s2 = ±5.6 m/s = −5.6 m/s
For the entire time period, vi = +2.3 m/s and vf = −5.6 m/s.
vf − vi −5.6 m/s − 2.3 m/s
−7.9 m/s
∆t =  = 
= 2 = 0.81 s
2
a
−9.81 m/s
−9.81 m/s
vf = +1.1 m/s2
a = −9.81 m/s2
vf 2 − vi 2
(1.1 m/s)2 − (6.0 m/s)2
∆y =  = 
(2)(−9.81 m/s2)
2a
1.2m2/s2 − 36 m2/s2
−35 m2/s2
∆y = 
=  2 = +1.8 m
2
−19.6 m/s
−19.6 m/s
Section Review, p. 65
1
1
∆y = vi ∆t + 2a∆t 2 = (0 m/s)(1.5 s) + 2(−9.81 m/s2)(1.5 s)2
2. vi = 0 m/s
∆t = 1.5 s
∆y = 0 m + (−11 m) = −11 m
2
a = −9.81 m/s
the distance to the water’s surface = 11 m
Chapter Review and Assess, pp. 69–75
8. ∆t = 5.2 h
vavg = 73 km/h south
9. ∆t = 0.530 h
∆x = vavg ∆t = (73 km/h)(5.2 h) = 3.8 × 102 km south
∆x = vavg ∆t = (19.0 km/h)(0.530 h) = 10.1 km east
vavg = 19.0 km/h east
I Ch. 2–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. vi = +6.0 m/s
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Givens
Solutions
10. ∆t = 3.0 s
a. ∆x = 4.0 m + (−4.0 m) + (−2.0 m) + 0.0 m = −2.0 m
∆x −2.0 m
b. vavg =  =  = −0.67 m/s
∆t
3.0 s
11. ∆t = 2.00 h, 9.00 min, 21.0 s
vavg = 5.436 m/s
∆x = vavg ∆t = (5.436 m/s) [(2.00 h)(3600 s/h) + (9.00 min)(60 s/min) + 21.0 s]
I
∆x = (5.436 m/s)(7200 s + 540 s + 21.0 s) = (5.436 m/s)(7761 s)
∆x = 4.22 × 104 m = 4.22 × 101 km
12. ∆t = 5.00 s
a. ∆xA = +70.0 m
distance between
b. ∆xB = 70.0 m + 70.0 m = +140 m
poles = 70.0 m
∆xA 70.0 m
c. vavg,A = 
=  = +14 m/s
∆t
5.0 s
∆x
140 m
d. vavg,B = B =  = +28 m/s
∆t
5.0 s
13. v1 = 80.0 km/h
a. ∆x1 = v1∆t1 = (80.0 km/h)(30.0 min)(1 h/60 min) = 40.0 km
∆t1 = 30.0 min
∆x2 = v2 ∆t2 = (105 km/h)(12.0 min)(1 h/60 min) = 21.0 km
v2 = 105 km/h
∆x3 = v3 ∆t3 = (40.0 km/h)(45.0 min)(1 h/60 min) = 30.0 km
∆t2 = 12.0 min
∆x4 = v4 ∆t4 = (0 km/h)(15.0 min)(1 h/60 min) = 0 km
v3 = 40.0 km/h
∆xtot ∆x1 + ∆x2 + ∆x3 + ∆x4
vavg = 
= 
∆ttot
∆t1 + ∆t2 + ∆t3 + ∆t4
∆t3 = 45.0 min
v4 = 0 km/h
∆t4 = 15.0 min
40.0 km + 21.0 km + 30.0 km + 0.00 km
vavg = 
(30.0 min + 12.0 min + 45.0 min + 15.0 min)(1 h/60 min)
91.0 km
vavg =  = 53.5 km/h
(102.0 min)(1 h/60 min)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
b. ∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4
∆xtot = 40.0 km + 21.0 km + 30.0 km + 0 km = 91.0 km
14. ∆t1 = 10.0 min − 0 min
= 10.0 min
∆t2 = 20.0 min − 10.0 min
= 10.0 min
∆t3 = 30.0 min − 20.0 min
= 10.0 min
a. ∆x1 = (2.4 × 103 m) − (0 × 103 m) = +2.4 × 103 m
(2.4 × 103 m)
∆x
v1 = 1 =  = +4.0 m/s
∆t1 (10.0 min)(60 s/min)
b. ∆x2 = (3.9 × 103 m) − (2.4 × 103 m) = +1.5 × 103 m
(1.5 × 103 m)
∆x
v2 = 2 =  = +2.5 m/s
∆t2 (10.0 min)(60 s/min)
c. ∆x3 = (4.8 × 103 m) − (3.9 × 103 m) = +9.0 × 102 m
(9.0 × 102 m)
∆x
v3 = 3 =  = +1.5 m/s
∆t3 (10.0 min)(60 s/min)
d. ∆xtot = ∆x1 + ∆x2 + ∆x3 = (2.4 × 103 m) + (1.5 × 103 m) + (9.0 × 102 m)
∆xtot = +4.8 × 103 m
∆xtot ∆x1 + ∆x2 + ∆x3
(4.8 × 103 m)
vavg = 
=  = 
∆ttot ∆t1 + ∆t2 + ∆t3
(30.0 min)(60 s/min)
vavg = +2.7 m/s
Section One — Pupil’s Edition Solutions
I Ch. 2–7
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Givens
Solutions
15. vA = 9.0 km/h east
∆xA = vA ∆t = x − xi, A
= +9.0 km/h
I
∆xB = vB ∆t = x − xi, B
xi, A = 6.0 km west of
flagpole = −6.0 km
∆xA − ∆xB = (x − xi, A) − (x − xi, B) = xi, B − xi, A
vB = 8.0 km/h west
5.0 km − (−6.0 km)
11.0 km
xi, B − xi, A
 =  = 
∆t = 
9.0 km/h − (−8.0 km/h)
17.0 km/h
vA − vB
= −8.0 km/h
xi, B = 5.0 km east of
flagpole = +5.0 km
x = distance from flagpole
to point where runners’
paths cross
∆xA − ∆xB = vA ∆t − vB ∆t = (vA − vB) ∆t
∆t = 0.647 h
∆xA = vA∆t = (9.0 km/h)(0.647 h) = 5.8 km
∆xB = vB∆t = (−8.0 km/h)(0.647 h) = −5.2 km
for runner A, x = ∆xA + xi, A = 5.8 km + (−6.0 km) = −0.20 km
x = 0.20 km west of the flagpole
for runner B, x = ∆xB + xi, B = −5.2 km + (−5.0 km) = −0.20 km
x = 0.20 km west of the flagpole
20. vi = +5.0 m/s
aavg = +0.75 m/s2
vf = +8.0 m/s
21. vi = +7.0 m/s
∆v vf − v
8.0 m/s − 5.0 m/s
3.0 m/s
∆t =  = i = 
= 2
aav g
aavg
0.75 m/s2
0.75 m/s
∆t = 4.0 s
vf = a∆t + vi = (0.80 m/s2)(2.0 s) + 7.0 m/s = 1.6 m/s + 7.0 m/s = +8.6 m/s
a = +0.80 m/s2
∆t = 2.0 s
a1 = +0.50 m/s2 ∆t = 7.0 s
a2 = −0.60 m/s2 vf = 0 m/s
23. vi = +16 m/s
vf = +32 m/s
∆t = 10.0 s
a. vf = a 1∆t + vi = (0.50 m/s2)(7.0 s) + 3.0 m/s = 3.5 m/s + 3.0 m/s = +6.5 m/s
vf − vi 0 m/s − 3.0 m/s
b. ∆t =  = 
= 5.0 s
a2
−0.60 m/s2
vf − vi 32 m/s − 16 m/s 16 m/s
a. a =  =  =  = +1.6 m/s2
10.0 s
10.0 s
∆t
1
1
1
b. ∆x = 2(vi + vf )∆t = 2(16 m/s + 32 m/s)(10.0 s) = 2(48 m/s)(10.0 s)
∆x = +240 m
∆x 240 m
vavg =  =  = +24 m/s
∆t 10.0 s
c. distance traveled = +240 m (See b.)
24. vi = 0 m/s
a = 3.3 m/s2
∆t = 7.5 s
25. vi = 75.0 km/h = 21.0 m/s
vf = 0 km/h = 0 m/s
∆t = 21.0 s
I Ch. 2–8
1
∆x = vi ∆t + 2a∆t 2
1
∆x = (0 m/s)(7.5 s) + 2(3.3 m/s2)(7.5 s)2 = 0 m + 93 m = 93 m
1
1
1
∆x = 2(vi + vf) ∆t = 2(21.0 m/s + 0 m/s)(21.0 s) = 2(21.0 m/s)(21.0 s)
∆x = 220 m
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
22. vi = +3.0 m/s
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Givens
Solutions
26. vi = 0 m/s
a. vf = vi + a∆t = 0 m/s + (−3.00 m/s2)(5.0 s) = −15 m/s
a = −3.00 m/s 2
∆t = 5.0 s
1
1
b. ∆x = vi ∆t + 2a∆t 2 = (0 m/s)(5.0 s) + 2(−3.00 m/s2)(5.0 s)2 = −38 m
1
27. vi = 0 m/s
I
1
∆x = 2 (vi + vf) ∆t = 2(0 m/s + 18 m/s)(12 s) = 110 m
vf = 18 m/s
∆t = 12 s
28. vi = 0 m/s
For the first time interval,
∆t 1 = 5.0 s
vf = vi + ai ∆ti = 0 m/s + (1.5 m/s2)(5.0 s) = +7.5 m/s
a1 = +1.5 m/s2
∆x 1 = vi∆t1 + 2 a1 ∆ti 2 = (0 m/s)(5.0 s) + 2 (1.5 m/s2)(5.0 s)2 = +19 m
∆t 2 = 3.0 s
For the second time interval,
2
a2 = −2.0 m/s
1
1
vi = +7.5 m/s
vf = vi + a2∆t2 = 7.5 m/s + (−2.0 m/s2)(3.0 s) = 7.5 m/s − 6.0 m/s = +1.5 m/s
1
1
∆x 2 = vi∆t2 + 2 a2 ∆t2 = (7.5 m/s)(3.0 s) + 2 (−2.0 m/s2)(3.0 s)2 = 22 m − 9.0 m = +13 m
∆x tot = ∆x1 + ∆x2 = 19 m + 13 m = +32 m
29. a = 1.40 m/s2
vi = 0 m/s
vf 2 − vi2 (7.00 m/s)2 − (0 m/s)2 49.0 m2/s2
∆x =  = 
= 
= 17.5 m
2a
2.80 m/s2
(2)(1.40 m/s2)
vf = 7.00 m/s
30. For 0 s to 5.0 s:
vi = −6.8 m/s
vf = −6.8 m/s
For 0 s to 5.0 s,
vf − vi −6.8 m/s − (−6.8 m/s)
aavg =  =  = 0.0 m/s2
∆t
5.0 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 5.0 s
For 5.0 s to 15.0 s:
vi = −6.8 m/s
vf = +6.8 m/s
For 5.0 s to 15.0 s,
vf − vi 6.8 m/s − (−6.8 m/s) 13.6 m/s
aavg =  =  =  = +1.36 m/s2
∆t
10.0 s
10.0 s
∆t = 10.0 s
For 0 s to 20.0 s:
vi = −6.8 m/s
vf = +6.8 m/s
For 0 s to 20.0 s,
vf − vi 6.8 m/s − (−6.8 m/s) 14 m/s
aavg =  =  =  = +0.68 m/s2
∆t
20.0 s
20.0 s
∆t = 20.0 s
31. vi = +120 m/s
2
a = −6.0 m/s
vf = 0 m/s
a.
vf − vi 0 m/s − 120 m/s
∆t =  = 
= 2.0 × 101 s
a
−6.0 m/s2
b.
∆x = vi ∆t + 2a ∆t 2 = (120 m/s)(2.0 × 101 s) + 2(−6.0 m/s2)(2.0 × 101 s)2
1
1
∆x = −2400 m − 1200 m = 1200 m = +1.2 km
The carrier is not long enough for the plane to land. The plane requires 1.2 km.
Section One — Pupil’s Edition Solutions
I Ch. 2–9
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Givens
Solutions
32. vi = 0 m/s
a. vf = vi2+
m/s
)2
+(2)
m) = 12
s2 = ±11 m/s
2a∆
x = (0
(0
.2
1m
/s
2)(2
80
0m
2/
vf = 11 m/s
a = 0.21 m/s2
∆x = 280 m
vf − v 11.0 m/s − 0 m/s
b. ∆t = i = 
= 52.0 s
a
0.21 m/s2
I
33. vi = +1.20 m/s
a = −0.31 m/s2
∆x = +0.75 m
38. vi = 0 m/s
vf = vi2+
)2
+(2)
m/s
2a∆
x = (1
.2
0m
/s
(−
0.
31
2)(0
.7
5m
)
vf = 1.
m2/s2
−0.4
m2/
44
6m
2/s2 = 0.
98
s2
= ±0.99 m/s = +0.99 m/s
When vi = 0 m/s,
∆y = −80.0 m
a = −9.81 m/s2
v 2 = 2 ay
v = 2ay
v = (2
m/s
m )
)(
−9.
81
2)(−
80
.0
v = ± 15
m2/
s2
70
v = −39.6 m/s
39. vi = 0 m/s
When vi = 0 m/s,
2
a = −9.81 m/s
∆y = −76.0 m
40. vi, 1 = +25 m/s
∆t =
2∆ay = (2
−)9(.−8716
m.0/ms ) =
2
15
.5
s2 = 3.94 s
1
∆y1 = y − yi, 1 = vi, 1 ∆t + 2a∆t 2
1
vi, 2 = 0 m/s
∆y2 = y − yi, 2 = vi, 2 ∆t + 2a∆t 2
a = −9.81 m/s2
∆y1 − ∆y2 = (y − yi, 1) − (y − yi, 2) = yi, 2 − yi, 1
yi, 1 = 0 m
∆y1 − ∆y2 = (vi, 1 ∆t + 2a∆t 2) − (vi, 2 ∆t + 2a∆t 2) = (vi, 1 − vi, 2) ∆t
yi, 2 = 15 m
∆y1 − ∆y2 = yi, 2 − yi, 1 = (vi, 1 − vi, 2)∆t
y = distance from ground
to point where both balls
are at the same height
yi, 2 − yi, 1
15 m − 0 m
15 m
 =  =  = 0.60 s
∆t = 
vi, 1 − vi, 2 25 m/s − 0 m/s 25 m/s
41. vi = +25.0 m/s
yi = +2.0 m/s
2
a = −9.81 m/s
For the trip up: vf = 0 m/s
1
a. For the trip up,
vf − v
0 m/s − 25.0 m/s
∆t = i = 
= 2.55 s
a
−9.81 m/s2
1
1
∆y = vi ∆t + 2a∆t 2 = (25.0 m/s)(2.55 s) + 2(−9.81 m/s2)(2.55 s)2
∆y = 63.8 m − 31.9 m = +31.9 m
For the trip down: vi = 0 m/s
∆y = (−31.9 m − 2.0 m)
∆y = −33.9 m
I Ch. 2–10
b. For the trip down, because vi = 0 m/s,
∆t =
2∆ay = (2
−)9(.−8313
m.9/ms ) =
Holt Physics Solution Manual
2
6.
91
s2 = 2.63 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
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Givens
Solutions
42.
a. For ∆y = 0.20 m = maximum height of ball, ∆t = 0.20 s
b. For ∆y = 0.10 m = one-half maximum height of ball,
∆t = 0.060 s as ball goes up
I
∆t = 0.34 s as ball comes down
c. Estimating v from t = 0.040 s to t = 0.060 s:
∆x 0.10 m − 0.07 m 0.03 m
v =  =  =  = + 1.5 m/s
∆t 0.060 s − 0.040 s 0.020 s
Estimating v from t = 0.090 s to t = 0.11 s:
∆x 0.10 m − 0.13 m 0.030 m
v =  =  =  = +1.5 m/s
∆t
0.11 s − 0.090 s
0.020 s
Estimating v from t = 0.14 s to t = 0.16 s:
∆x 0.18 m − 0.17 m 0.010 m
v =  =  =  = + 0.50 m/s
∆t
0.16 s − 0.14 s
0.020 s
Estimating v from t = 0.19 s to t = 0.21 s:
∆x 0.20 m − 0.20 m
v =  =  = 0.00 m/s
∆t
0.21 s − 0.19 s
Graph of v vs t should be similar to the following:
Velocity (m/s)
2.0
1.5
1.0
0.5
Copyright © by Holt, Rinehart and Winston. All rights reserved.
0
0.05 0.10 0.15 0.20
Time (s)
∆v 0 m/s − 2.0 m/s − 2.0 m/s
d. a =  =  =  = −10 m/s2 ≈ −9.81 m/s2
∆t
0.20 s − 0 s
0.20 s
43. vavg = 27 800 km/h
rearth = 6380 km
∆x = 320.0 km
circumference = 2p(rearth + ∆x)
circumference 2p (6380 km + 320.0 km) 2p(6.700 × 103 km)
∆t =  =  =  = 1.51 h
vavg
27 800 km/h
27 800 km/h
Section One — Pupil’s Edition Solutions
I Ch. 2–11
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Givens1
Solutions
44. ∆xAB = ∆xBC = ∆xCD
∆xAB
2∆xAB
∆xAB
∆tAB = 
=
= 
vAB,avg (vB + 0)
vB

2
∆tAB + ∆tBC + ∆tCD =
5.00 min
∆xBC
Because the train’s velocity is constant from B to C, ∆tBC = 
.
vB
∆xCD
2∆xCD
∆xCD
∆tCD = 
=
= 
vCD,avg (0 + vB)
vB

2
∆t
∆t
 = ∆tBC = CD
, or
Because ∆xAB = ∆xBC = ∆xCD , AB
2
2
∆tAB = ∆tCD = 2∆tBC.
I
We also know that ∆tAB + ∆tBC + ∆tCD = 5.00 min.
Thus, the time intervals are as follows:
a. ∆tAB = 2.00 min
b. ∆tBC = 1.00 min
c. ∆tCD = 2.00 min
45. ∆y = −19.6 m
vi,1 = −14.7 m/s
vi,2 = +14.7 m/s
a = −9.81 m/s2
2 2a∆y = (−14.7 m/s)2 + (2)(−9.81 m/s2)(−19.6 m)
a. vf,1 = vi
,1+
vf,1 = 21
s2
+385
m2/
s2 = 60
s2 = ±24.5 m/s = −24.5 m/s
6m
2/
1m
2/
2 2a∆y = (14.7 m/s)2 + (2)(−9.81 m/s2)(−19.6 m)
vf,2 = vi
,2+
vf,2 = 21
s2
+385
m2/
s2 = 60
s2 = ±24.5 m/s = −24.5 m/s
6m
2/
1m
2/
vf,1 − vi,1 −24.5 m/s − (−14.7 m/s)
−9.8 m/s
∆t1 =  = 
= 2 = 1.0 s
a
−9.81 m/s
−9.81 m/s2
vf,2 − vi,2 −24.5 m/s − 14.7 m/s −39.2 m/s
= 2 = 4.00 s
∆t2 =  = 
a
−9.81 m/s
−9.81 m/s2
b. vf,1 = −24.5 m/s (See a.)
vf,2 = −24.5 m/s (See a.)
∆t = 0.800 s
1
1
c. ∆y1 = vi,1∆t + 2a∆t 2 = (−14.7 m/s)(0.800 s) + 2(−9.81 m/s2)(0.800 s)2
∆y1 = −11.8 m − 3.14 m = −14.9 m
1
1
∆y2 = vi,2 ∆t + 2a∆t 2 = (14.7 m/s)(0.800 s) + 2(−9.81 m/s2)(0.800 s)2
∆y2 = 11.8 m − 3.14 m = +8.7 m
distance between balls = ∆y2 − ∆y1 = 8.7 m − (−14.9 m) = 23.6 m
I Ch. 2–12
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
difference in time = ∆t2 − ∆t1 = 4.00 s − 1.0 s = 3.0 s
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46. For the first time interval:
While the rocket accelerates,
1
1
vi = 0 m/s
∆y1 = v1∆t + 2a∆t 2 = (0 m/s)(4.00 s) + 2(29.4 m/s2)(4.00 s)2 = +235 m
a = +29.4 m/s2
vf = vi + a∆t = 0 m/s + (29.4 m/s2)(4.00 s) = +118 m/s
∆t = 4.00 s
For the second time interval:
vi = +117 m/s
I
After the rocket runs out of fuel,
a = −9.81 m/s2
vf 2 − vi 2 (0 m/s)2 − (118 m/s)2
∆y2 =  = 
= +710 m
2a
(2)(−9.81 m/s2)
vf = 0 m/s
total height reached by rocket = ∆y1 + ∆y2 = 233 m + 710 m = 943 m
47. v1 = 85 km/h
v2 = 115 km/h
∆x = 16 km
16 km
∆x
a. ∆t1 =  =  = 0.19 h
v1 85 km/h
16 km
∆x
∆t2 =  =  = 0.14 h
v2 115 km/h
The faster car arrives ∆t1 − ∆t2 = 0.19 h − 0.14 h = 0.050 h earlier.
∆t1 − ∆t2 = 15 min
b. ∆x1 = ∆x2
∆t1v1 = ∆t2 v2
∆t1 = ∆t2 + (15 min)(1 h/60 min)
∆t1 = ∆t2 + 0.25 h
(∆t2 + 0.25 h)v1 = ∆t2 v2
∆t2 v1 + (0.25 h)v1 = ∆t2 v2
−(0.25 h)(v1) −(0.25 h)(85 km/h)
−(0.25 h)(85 km/h)
∆t2 = 
=  = 
v1 − v2
85 km/h − 115 km/h
−3.0 × 101 km/h
∆t2 = 0.71 h
∆x2 = ∆t2 v2 = (0.71 h)(115 km/h) = 82 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
or ∆x1 = ∆t1v1 = (∆t2 + 0.25 h)v1 = (0.71 h + 0.25 h)(85 km/h)
∆x 1 = (0.96 h)(85 km/h) = 82 km
48. vi = −1.3 m/s
a = −9.81 m/s2
∆t = 2.5 s
vf = vi + a∆t = −1.3 m/s + (−9.81 m/s2)(2.5 s)
vf = −1.3 m/s − 25 m/s = −26 m/s
1
1
∆xkit = 2(vi + vf )∆t = 2(−1.3 m/s − 26 m/s)(2.5 s)
1
∆xkit = 2(−27 m/s)(2.5 s) = −34 m
∆xclimber = (−1.3 m/s)(2.5 s) = −3.2 m
The distance between the kit and the climber is ∆xclimber − ∆xkit .
∆xclimber − ∆xkit = −3.2 m − (−34 m) = 31 m
Section One — Pupil’s Edition Solutions
I Ch. 2–13
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Givens
Solutions
49. vi = +0.50 m/s
a. vf = vi + a∆t = 0.50 m/s + (−9.81 m/s2)(2.5 s) = 0.50 m/s − 25 m/s
∆t = 2.5 s
vf = −24 m/s
2
a = −9.81 m/s
I
1
1
b. ∆xfish = 2(vi + vf )∆t = 2(0.50 m/s − 24 m/s)(2.5 s)
1
∆xfish = 2(−24 m/s)(2.5 s) = −30 m
∆xpelican = (0.50 m/s)(2.5 s) = +1.3 m
The distance between the fish and the pelican is ∆xpelican − ∆xfish .
∆xpelican − ∆xfish = 1.3 m − (−30 m) = 33 m
50. vi = 56 km/h
For the time interval during which the ranger decelerates,
vf = 0 m/s
a = −3.0 m/s2
vf − v 0 m/s − (56 × 103 m/h)(1 h/3600 s)
= 5.2 s
∆t = i = 
a
−3.0 m/s2
1
1
∆x = vi ∆t + 2a∆t 2 = (56 × 103 m/h)(1 h/3600 s)(5.2 s) + 2(−3.0 m/s2)(5.2 s)2
∆xtot = 65 m
∆x = 81 m − 41 m = 4.0 × 101 m
maximum reaction distance = ∆xtot − ∆x = 65 m − (4.0 × 101 m) = 25 m
maximum reaction distance
maximum reaction time = 
vi
25 m
maximum reaction time = 
= 1.6 s
(56 × 103 m/h)(1 h/3600 s)
51. vs = 30.0 m/s
a. ∆xs = ∆xp
vi,p = 0 m/s
∆xs = vs ∆t
2
ap = 2.44 m/s
Because vi,p = 0 m/s,
1
∆xp = 2a p ∆t 2
2v (2)(30.0 m/s)
∆t = s = 
= 24.6 s
ap
2.44 m/s2
b. ∆xs = vs ∆t = (30.0 m/s)(24.6 s) = 738 m
1
1
or ∆xp = 2a p ∆t 2 = 2(2.44 m/s2)(24.6 s)2 = 738 m
I Ch. 2–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
vs ∆t = 2a p ∆t 2
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Givens
Solutions
52. For ∆t1:
When vi = 0 m/s,
1
vi = 0 m/s
∆x1 = 2a∆t12
a = +13.0 m/s2
∆t2 = 90.0 s − ∆t1
vf = v
∆x2 = v∆t2 = v(90.0 s − ∆t1)
I
1
For ∆t2 :
a = 0 m/s2
v = constant velocity
∆xtot = +5.30 × 103 m
∆ttot = ∆t1 + ∆t2 = 90.0 s
∆xtot = ∆x1 + ∆x2 = 2a∆t12 + v(90.0 s − ∆t1)
v = vf during the first time interval = a∆t1
1
1
∆xtot = 2a∆t12 + a∆t1(90.0 s − ∆t1) = 2a∆t12 + (90.0 s)a∆t1 − a∆t12
1
∆xtot = − 2a∆t12 + (90.0 s)a∆t1
1
a∆t 2
1
2
− (90.0 s)a∆t1 + ∆xtot = 0
Using the quadratic equation,
(90.0 s)(a) ± [(90.0 s)(a)]2 − 42a(∆xtot)
∆t1 = 
1
22a
1
(90.0 s)(13.0 m/s2) ± [(90.0s)(13.0m/s2)]2 −42(13.0 m/s)(5.30 × 103 m)


∆t1 = 
1
22(13.0 m/s2)
1
.3
7×1
(1.
38
×1
1170 m/s ± (1
06
m2/
s2)−
05
m2/
s2)
∆t1 = 
13.0 m/s2
2/
1170 m/s ± 1.
23
×1
06
m
s2 1170 m/s ± 1110 m/s
60 m/s
= 
= 2 = 5.0 s
∆t1 = 
2
13.0 m/s
13.0 m/s
13.0 m/s2
∆t2 = ∆ttot − ∆t1 = 90.0 s − 5 s = 85 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v = a∆t1 = (13.0 m/s2)(5 s) = +60 m/s
53. ∆x1 = +5800 m
a = −7.0 m/s2
vi = +60 m/s
vf = 0 m/s
54. vi = +10.0 m/s
vf = −8.0 m/s
vf 2 − vi2 (0 m/s)2 − (60 m/s)2
a. ∆x2 =  = 
= +3.0 × 102 m
2a
(2)(−7.0 m/s2)
sled’s final poisition = ∆x 1 + ∆x2 = 5800 m + 300 m = 6100 m
vf − vi 0 m/s − 60 m/s
b. ∆t =  = 
= 9s
a
−7.0 m/s2
vf − vi −8.0 m/s − 10.0 m/s
aavg =  =  = −1.5 × 103 m/s2
∆t
0.012 s
∆t = 0.012 s
55. vi = −10.0 m/s
∆y = −50.0 m
a = −9.81 m/s2
(−
m/s
)2
+(2)
m/s
m)
10
.0
(−
9.
81
2)(−
50
.0
vf = (1
02
m2/s2)+
m2/s2 = 10
m2/s2 = ±32.9 m/s = −32.9 m/s
.0
0×1
981
81
a. vf = vi2+
2a∆
y =
vf − vi −32.9 m/s − (−10.0 m/s)
−22.9 m/s
∆t =  = 
= 2 = 2.33 s
−9.81 m/s2
a
−9.81 m/s
b. vf = −32.9 m/s (See a.)
Section One — Pupil’s Edition Solutions
I Ch. 2–15
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Givens
Solutions
56. ∆y = −50 m
2 2a∆y =
a. vf,1 = vi
,1+
vi,1 = +2.0 m/s
∆t1 = ∆t2 + 1.0 s
I
a = −9.81 m/s2
(2
m/s
)2
+(2)
m/s
m)
.0
(−
9.
81
2)(−
50
vf,1 = 4.
m2/s2 = 98
0m
2/s2+981
5m
2/s2 = ±31.4 m/s = −31.4 m/s
vf,1 − vi,1 −31.4 m/s − 2.0 m/s
−33.4 m/s
∆t1 =  = 
= 2 = 3.4 s
2
a
− 9.81 m/s
− 9.81 m/s
b. ∆t2 = ∆t1 − 1.0 s = 3.4 s − 1.0 s = 2.4 s
1
∆y = vi,2 ∆t2 + 2a∆t22
1
1
−50 m − 2(−9.81 m/s2)(2.4 s)2
∆y − 2a∆t22
vi,2 =  = 
2.4 s
∆t2
−50 m + 28 m −22 m
 =  = −9.2 m/s
vi,2 = 
2.4 s
2.4 s
c. vf,1 = −31 m/s (See a.)
vf,2 = vi,2 + a∆t2 = −9.2 m/s + (−9.81 m/s2)(2.4 s)
vf,2 = −9.2 m/s − 24 m/s = −33 m/s
vi,1 = +50.0 m/s
a1 = +2.00 m/s2
∆y1 = +150 m
For the second time interval:
vi,2 = +55.7 m/s
vf,2 = 0 m/s
a2 = −9.81 m/s2
2 2a ∆y = (50.0 m/s)2 + (2)(2.00 m/s2)(150 m)
a. vf,1 = vi
11
,1+
vf,1 = (2
03
m2/
s2)+
02
m2/
s2) = 3.
03
m2/
s2
.5
0×1
(6.
0×1
10
×1
vf,1 = ±55.7 m/s = +55.7 m/s
vf,22 − vi,22 (0 m/s)2 − (55.7 m/s)2
∆y2 =  = 
= +158 m
(2)(−9.81 m/s2)
2a2
maximum height = ∆y1 + ∆y2 = 150 m + 158 m = 310 m
b. For the first time interval,
(2)(150 m)
2∆y1
(2)(150 m)
∆tup,1 = 
=  =  = 2.8 s
vi,1 + vf,1 50.0 m/s + 55.7 m/s
105.7 m/s
For the second time interval,
(2)(158 m)
2∆y2
=  = 5.67 s
∆tup,2 = 
vi,2 + vf,2 55.7 m/s + 0 m/s
∆tup,tot = ∆tup,1 + ∆tup,2 = 2.8 s + 5.67 s = 8.5 s
For the trip down:
∆y = −310 m
vi = 0 m/s
c. Because vi = 0 m/s, ∆tdown =
(2)(−310 m)

=
−9.81 m/s2
∆ttot = ∆tup,tot + ∆tdown = 8.5 s + 7.9 s = 16.4 s
a = −9.81 m/s2
I Ch. 2–16
2∆y
 =
a
Holt Physics Solution Manual
2
63
s = 7.9 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
57. For the first time interval:
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Givens
Solutions
58. a1 = +5.9 m/s2
Because both cars are initially at rest,
a2 = +3.6 m/s2
∆t1 = ∆t2 − 1.0 s
1
a. ∆x1 = 2a1∆t12
1
∆x2 = 2a2 ∆t22
∆x1 = ∆x2
1
a ∆t 2
2 1 1
I
1
= 2a2 ∆t22
a1(∆t2 − 1.0 s)2 = a2∆t22
a1[∆t22 − (2.0 s)(∆t2) + 1.0 s2] = a2 ∆t22
(a1)(∆t2)2 − a1(2.0 s)∆t2 + a1(1.0 s2) = a2 ∆t22
(a1 − a2)∆t22 − a1(2.0 s)∆t2 + a1(1.0 s2) = 0
Using the quadratic equation,
2)
[
a1(
2.0
s)
]2
−4
(a
−a
)a1 (1
.0s
(a1)(2.0 s) ± 1
2
∆t2 = 
2(a1 − a2)
a1 − a2 = 5.9 m/s2 − 3.6 m/s2 = 2.3 m/s2
(5.9 m/s2)(2.0 s) ± [(
m/s
m/s
5.
9m
/s
2)(2
.0
s)]
2−(4)
(2
.3
2)(5
.9
2)(1
.0
s2)
∆t2 = 
2
(2)(2.3 m/s )
0m
2/
2/
s2
−54m
s2 12 m/s ± 90
m2/
s2
12 m/s ± 14
∆t2 = 
= 
2
2
(2)(2.3 m/s )
(2)(2.3 m/s )
21 m/s
12 m/s ± 9 m/s
= 
= 4.6 s
∆t2 = 
(2)(2.3 m/s2)
(2)(2.3 m/s2)
∆t1 = ∆t2 − 1.05 = 4.65 − 1.05 = 3.6 s
b. ∆t1 = ∆t2 − 1.0 s = 4.6 s − 1.0 s = 3.6 s
1
1
∆x1 = 2a1∆t12 = 2(5.9 m/s2)(3.6 s)2 = 38 m
1
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
or ∆x2 = 2a2∆t22 = 2(3.6 m/s2)(4.6 s)2 = 38 m
distance both cars travel = 38 m
c. v1 = a1∆t1 = (5.9 m/s2)(3.6 s) = +21 m/s
v2 = a2∆t2 = (3.6 m/s2)(4.6 s) = +17 m/s
59. vi,1 = +25 m/s
vi,2 = +35 m/s
vf,1 − vi,1 0 m/s − 25 m/s
= 13 s
a. ∆t1 =  = 
a1
−2.0 m/s2
∆x2 = ∆x1 + 45 m
a1 = −2.0 m/s2
vf,1 = 0 m/s
vf,2 = 0 m/s
1
1
b. ∆x1 = 2(vi,1 + vf,1)∆t1 = 2(25 m/s + 0 m/s)(13 s) = +163 m
∆x2 = ∆x1 + 42 m = 163 m + 45 m = +208 m
vf,22 − vi,22 (0 m/s)2 − (35 m/s)2
a2 =  =  = −2.9 m/s2
(2)(208 m)
2∆x2
vf,2 − vi,2 0 m/s − 35 m/s
c. ∆t2 =  = 
= 12 s
a2
−2.9 m/s2
Section One — Pupil’s Edition Solutions
I Ch. 2–17
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Givens
Solutions
60. ∆x1 = 50.0 m
∆x
50.0 m
∆t = 1 =  = 12.5 s
v1 4.00 m/s
v1 = 4.00 m/s
∆x2 = v2(0.500 s) + 50.0 m
I
v2(0.500 s) + 50.0 m
∆x
v2 = 2 = 
∆t
∆t
v2∆t = v2(0.500 s) + 50.0 m
v2(∆t − 0.50 s) = 50.0 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
50.0 m
50.0 m
50.0 m
v2 =  =  =  = 4.17 m/s
∆t − 0.500 s (12.5 s − 0.500 s) 12.0 s
I Ch. 2–18
Holt Physics Solution Manual
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Two-Dimensional Motion and Vectors
Chapter
3
I
Section Review, p. 87
Givens
2. ∆x1 = 85 m
d2 = 45 m
q2 = 30.0°
Solutions
Students should use graphical techniques. Their answers can be checked using the
techniques presented in Section 3-2. Answers may vary.
∆x2 = d2(cos q2) = (45 m)(cos 30.0°) = 39 m
∆y2 = d2(sin q2) = (45 m)(sin 30.0°) = 22 m
∆xtot = ∆x1 + ∆x2 = 85 m + 39 m = 124 m
∆ytot = ∆y2 = 22 m
xtot
)2
+(∆
ytot
)2 = (1
m
)2
+(22
m
)2
d = (∆
24
d = 15
m2+
m2 = 15
m2 = 126 m
400
480
900
∆ytot
22 m
= tan−1  = (1.0 × 101)° above the horizontal
q = tan−1 
∆xtot
124 m
3. vy, 1 = 2.50 × 102 km/h
Students should use graphical techniques.
v2 = 75 km/h
vx,2 = v2(cos q2) = (75 km/h)[cos (−45°)] = 53 km/h
q2 = −45°
vy,2 = v2(sin q2) = (75 km/h)[sin (−45°)] = −53 km/h
vy,tot = vy,1 + vy,2 = 2.50 × 102 km/h − 53 km/h = 197 km/h
vx,tot = vs,2 = 53 km/h
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2
2
vy
)2
+(1
)2
v = (v
x,to
(
3km
/h
97
km
/h
t)+
,to
t) = (5
v = 28
00
km
2/h2+3880
0km
2/h2 = 41
600
km
2/h2 = 204 km/h
v tot
204 km/h
q = tan−1 y,
= tan−1  = 75° north of east
vx,tot
53 km/h
2.50 × 102 km/h
4. vy,1 = 
2
= 125 km/h
vx,2 = 53 km/h
vy,2 = −53 km/h
Students should use graphical techniques.
vy,dr = vy,1 + vy,2 = 125 km/h − 53 km/h = 72 km/h
vx,dr = vx,2 = 53 km/h
2
)2
=vy,d
)2
+ (7
)2
v = (v
x,dr
3km
/h
2km
/h
r) = (5
v = 28
00
km
2/h2+520
0km
2/h2 = 8.
0
103km
2/h2 = 89 km/h
vy dr
72 km/h
q = tan−1 ,
= tan−1  = 54° north of east
vx,dr
53 km/h
Section One—Pupil’s Edition Solutions
I Ch. 3–1
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Practice 3A, p. 91
Givens
Solutions
1. ∆x1 = 8 km east
∆x1 = 8 km
I
∆x2 = 3 km west = −3km, east
a. d = ∆x1 + ∆x2 + ∆x3 = 8 km + 3 km + 12 km = 23 km
b. ∆xtot = ∆x1 + ∆x2 + ∆x3 = 8 km + (−3 km) + 12 km = 17 km east
∆x2 = 3 km
∆x3 = 12 km east
∆x3 = 12 km
∆y = 0 km
2. ∆x = 7.5 m
∆y = 45.0 m
d= ∆
x2
+
∆
y 2 = (7
m
)2
+(45
m
)2
.5
.0
d = 56
m2+
m2 = 45.6 m
202
0m
2 = 20
80
Measuring direction with respect to y (north),
∆x
7.5 m
q = tan−1  = tan−1  = 9.5° east of due north
∆y
45.0 m
3. ∆x = 6.0 m
∆y = 14.5 m
d= ∆
∆y2 = (6
m
)2
+(14
m
)2
x2+
.0
.5
d = 36
m2+
02
m2 = 24
2.1
0×1
6m
2 = 15.7 m
Measuring direction with respect to the length of the field (down the field),
∆x
6.0 m
q = tan−1  = tan−1  = 22° to the side of downfield
∆y
14.5 m
∆y = –1.4 m
d= ∆
∆y2 = (1
m)2
+–
(1.
x2+
.2
4m
)2
d = 1.
m2 = 3.
4m
2+2.0
4m
2 = 1.8 m
∆y
–1.4 m
q = tan−1  = tan−1  = –49° = 49° below the horizontal
∆x
1.2 m
Practice 3B, p. 94
1. v = 105 km/h
vx = v(cos q ) = (105 km/h)(cos 25°) = 95 km/h
q = 25°
2. v = 105 km/h
vy = v(sin q ) = (105 km/h)(sin 25°) = 44 km/h
q = 25°
3. v = 22 m/s
vx = v(cos q ) = (22 m/s)(cos 15°) = 21 m/s
q = 15°
vy = v(sin q ) = (22 m/s)(sin 15°) = 5.7 m/s
4. d = 5 m
q = 90°
I Ch. 3–2
∆x = d(cos q) = (5 m)(cos 90°) = 0 m
∆y = d(sin q) = (5 m)(sin 90°) =
Holt Physics Solution Manual
5m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. ∆x = 1.2 m
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Practice 3B, p. 94 cont.
Givens
5. d = 125 m
q = −25°
6. d = 23.0 m
q = −30.5°
7. a = 2.5 m/s2
q = −18°
Solutions
∆x = d(cos q) = (125 m)[cos (−25°)] = 1.1 × 102 m
∆y = d(sin q ) = (125 m)[sin (−25°)] = −53 m
I
∆x = d(cos q ) = (23.0 m)[cos (−30.5°)] = 19.8 m
∆y = d(sin q) = (23.0 m)[sin (−30.5°)] = −11.7 m
ax = a(cos q) = (2.5 m/s2)[cos (−18°)] = 2.4 m/s2
ay = a(sin q) = (2.5 m/s2)[sin (−18°)] = −0.77 m/s2
Practice 3C, p. 97
1. d1 = 35 m
∆x1 = d1(cos q1) = (35 m)(cos 0.0°) = 35 m
q1 = 0.0°
∆y1 = d1(sin q1) = (35 m)(sin 0.0°) = 0.0 m
d2 = 15 m
∆x2 = d2(cos q2 ) = (15 m)(cos 25°) = 14 m
q2 = 25°
∆y2 = d2(sin q2 ) = (15 m)(sin 25°) = 6.3 m
∆xtot = ∆x1 + ∆x2 = 35 m + 14 m = 49 m
∆ytot = ∆y1 + ∆y2 = 0.0m + 6.3m = 6.3 m
dtot = (∆
xtot
)2
+(∆
ytot
)2 = (4
)2
+(6.
)2
9m
3m
dtot = 24
m2+
m2 = 49 m
00
40m
2 = 24
00
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6.3 m
∆ytot
= tan−1  = 7.3° to the right of downfield
qtot = tan−1 
49 m
∆xtot
2. d1 = 2.5 km
∆x1 = d1(cos q1) = (2.5 km)(cos 35°) = 2.0 km
q1 = 35°
∆y1 = d1(sin q1) = (2.5 km)(sin 35°) = 1.4 km
d2 = 5.2 km
∆x2 = d2(cos q2) = (5.2 km)(cos 22°) = 4.8 km
q2 = 22°
∆y2 = d2(sin q2) = (5.2 km)(sin 22°) = 1.9 km
∆xtot = ∆x1 + ∆x2 = 2.0 km + 4.8 km = 6.8 km
∆ytot = ∆y1 + ∆y2 = 1.4 km + 1.9 km = 3.3 km
xt
)2
+(∆
yt
)2 = (6
)2
+(3.
)2
dtot = (∆
.8
km
3km
ot
ot
dtot = 46
11
km
2 =
km
2 = 57
km
2 = 7.5 km
∆ytot
3.3 km
qtot = tan−1 
= tan−1  = 26° above the horizontal
∆xtot
6.8 km
Section One—Pupil’s Edition Solutions
I Ch. 3–3
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Givens
Solutions
3. d1 = 8.0 m
I
Measuring direction with respect to y = (north),
q1 = 0.0°
∆x1 = d1(sin q1) = (8.0 m)(sin 0.0°) = 0.0 m
d2 = 3.5 m
∆y1 = d1(cos q1) = (8.0 m)(cos 0.0°) = 8.0 m
q2 = 35°
∆x2 = d2(sin q2) = (3.5 m)(sin 35°) = 2.0 m
d3 = 5.0 m
∆y2 = d2(cos q2) = (3.5 m)(cos 35°) = 2.9 m
q3 = 90.0°
∆x3 = d3(sin q3) = (5.0 m)(sin 90.0°) = 5.0 m
∆y3 = d3(cos q3) = (5.0 m)(cos 90.0°) = 0.0 m
∆xtot = ∆x1 + ∆x2 + ∆x3 = 0.0. m + 2.0 m + 5.0 m = 7.0 m
∆ytot = ∆y1 + ∆y2 + ∆y3 = 8.0. m + 2.9 m + 0.0 m = 10.9 m
xt
)2(∆
yt
)2 = (7
m
)2
+(10
m
)2
dtot = (∆
.0
.9
ot
ot
dtot = 49
m2+
m2 = 16
119
8m
2 = 13.0 m
7.0 m
∆xtot
qtot = tan−1 
= tan−1  = 33° east of north
10.9 m
∆ytot
4. d1 = 75 km
Measuring direction with respect to y (north),
q1 = −30.0°
∆x1 = d1(sin q1) = (75 km)(sin − 30.0°) = −38 km
d2 = 155 km
∆y1 = d1(cos q1) = (75 km)(cos − 30.0°) = 65 km
q2 = 60.0°
∆x2 = d2(sin q2) = (155 km)(sin 60.0°) = 134 km
∆y2 = d2(cos q2) = (155 km)(cos 60.0°) = 78 km
∆xtot = ∆x1 + ∆x2 = −38 km + 134 km = 96 km
∆ytot = ∆y1 + ∆y2 = 65 km + 78 km = 143 km
2
2
dtot = (∆
xt
)2
+(
∆
yt
)2 = (9
6
km
)2
+(
14
3
km
)2 = 9.
2
×1
03k
m
.0
×1
04 km
+2
ot
ot
∆xtot
96 km
q = tan−1 
= tan−1  = 34° east of north
∆ytot
143 km
Section Review, p. 97
2. vx = 3.0 m/s
vy = 5.0 m/s
2
a. v = vx2+
m/s
)2
=(5.
)2
v
.0
0m
/s
y = (3
v = 9.
s2
+25m
s2 = 34
m2/
s2 = 5.8 m/s
0m
2/
2/
vy
5.0 m/s
q = tan−1  = tan−1  = 59° downriver from its intended path
vx
3.0 m/s
vx = 1.0 m/s
vy = 6.0 m/s
2
m/s
)2
+(6.
)2
b. v = vx2+
v
.0
0m
/s
y = (1
v = 1.
s2
+36m
s2 = 37
m2/
s2 = 6.1 m/s
0m
2/
2/
v
1.0 m/s
q = tan−1 x = tan−1  = 9.5° from the direction the wave is traveling
vy
6.0 m/s
I Ch. 3–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
= 2.
92
×104km
2 = 171 km
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Givens
Solutions
3. d = 10.0 km
a. ∆x = d(cos q) = (10.0 km)(sin 45.0°) = 7.07 km
q = 45.0°
∆y = d(sin q) = (10.0 km)(cos 45.0°) = 7.07 km
a = 2.0 m/s2
q = 35°
b. ax = a(cos q ) = (2.0 m/s2)(cos 35°) = 1.6 m/s2
I
ay = a(sin q) = (2.0 m/s2)(sin 35°) = 1.1 m/s2
∆x1 = d1(cos q) = (10.0 m)(cos 55°) = 5.7 m
4. d1 = 10.0 m
∆y1 = d1(sin q) = (10.0 m)(sin 55°) = 8.2 m
q1 = 55°
∆x2 = d2(cos q 2) = (5.0 m)(cos 0.0°) = 5.0 m
d2 = 5.0 m
∆y2 = d2(sin q 2) = (5.0 m)(sin 0.0°) = 0.0 m
q2 = 0.0°
∆xtot = ∆x1 + ∆x2 = 5.7 m + 5.0 m = 10.7 m
∆ytot = ∆y1 + ∆y2 = 8.2 m + 0.0 m = 8.2 m
dtot = (∆
xt
)2
+(∆
yt
)2 = (1
)2
+(8.
)2
0.
7m
2m
ot ot
dtot = 11
4m
2+67m
2 = 18
1m
2 = 13.5 m
∆ytot
8.2 m
= tan−1  = 37° north of west
qtot = tan−1 
∆xtot
10.7 m
Practice 3D, p. 102
1. ∆y = −0.70 m
∆x = 0.25 m
x
g = 9.81 m/s2
2. ∆y = −1.0 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2∆y ∆x
−g = v 
−g
v = 
2∆y ∆x = 
(2−)9(.−801.
7m0/ms) (0.25 m) =
∆t =
∆x = 2.2 m
2
x
2∆y ∆x
−g = v 
−g
 ∆x =  (2.2 m) =
v = −
(−29)(.8−11.m0/ms)
2∆y
∆t =
x
g = 9.81 m/s2
3. ∆y = −5.4 m
∆x = 8.0 m
g = 9.81 m/s
4. vx = 7.6 m/s
∆y = −2.7 m
2
g = 9.81 m/s
2
x
7.6 m/s
2∆y ∆x
−g = v 
2∆y
(2)(−2.7 m)
∆x = 
−g v = 
−9.81
m/s (7.6 m/s) =
∆t =
4.9 m/s
2∆y ∆x
−g = v 
−g
m/s
v = 
2∆y ∆x = 
(−29)(.8−15.
4m) (8.0 m) =
x
2
2
x
∆t =
0.66 m/s
x
x
2
5.6 m
Section One—Pupil’s Edition Solutions
I Ch. 3–5
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Practice 3E, p. 104
Givens
Solutions
1. ∆x = 4.0 m
q = 15°
I
∆x = vi (cos q )∆t
vi = 5.0 m/s
∆x
4.0 m
∆t =  =  = 0.83 s
vi (cos q ) (5.0 m/s)(cos 15°)
∆ymax = −2.5 m
∆y = vi (sin q )∆t − 2 g ∆t 2 = (5.0 m/s)(sin 15°)(0.83 s) − 2 (9.81 m/s2)(0.83 s)2
g = 9.81 m/s2
∆y = 1.1 m − 3.4 m = −2.3 m
2. ∆x = 301.5 m
q = 25.0°
g = 9.81 m/s2
1
1
yes
At maximum height, vy,f = 0 m/s.
vy,f = vi (sin q ) − g∆t = 0
vi (sin q )

∆t = 
g
v2
∆x = vi (cos q )∆t = (cos q )(sin q)i
g
vi =
x
(9.81 m/s )(301.5 m)
=  = 87.9 m/s

(cos qg)∆
(
n
q ) si
(cos 25.0°)(sin 25.0°)
2
vy,f 2 = vi 2(sin q )2 − 2g∆ymax = 0
2
2
vi 2(sin q)2 (87.9 m/s) (sin 25.0°)
 = 
= 70.3 m
∆y max = 
2
(2)(9.81 m/s )
2g
q = 25°
∆x
∆x
42.0 m
∆t =  =  =  = 2.0 s
vx vi (cos q) (23.0 m/s)(cos 25°)
vi = 23.0 m/s
At maximum height, vy,f = 0 m/s.
3. ∆x = 42.0 m
g = 9.81 m/s2
vy,f 2 = vy,i2 − 2g∆ymax = 0
4. ∆x = 2.00 m
∆y = 0.55 m
q = 32.0°
g = 9.81 m/s2
∆x = vi(cos q)∆t
∆x
∆t = 
vi(cos q)
2
g∆x2

2
2(cos q) [∆x(tan q) − ∆y]
vi =
vi =

(2)(cos 32.0°) [(2.00 m)(tan 32.0°) − 0.55 m]
vi =
I Ch. 3–6
∆x
∆x
∆y = vi(sin q)∆t − 12g∆t 2 = vi(sin q)  − 12 g 
vi(cos q)
vi(cos q)
g∆x2

∆y = ∆x(tan q) − 
2vi2(cos q)2
g∆x2
∆x(tan q) − ∆y = 
2 2
2vi (cos q)
g∆x2
2vi2(cos q)2 =  
∆x(tan q) − ∆y
(9.81 m/s2)(2.00 m)2
2
(9.81 m/s2)(2.00 m)2

=
(2)(cos 32.0°)2(1.25 − 0.55 m)
(9.81 m/s2)(2.00 m)2

= 6.2 m/s
(2)(cos 32.0°)2(0.70 m)
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vy,i2 vi2(sin q)2 (23.0 m/s)2(sin 25°)2
 = 
∆ymax =  = 
= 4.8 m
(2)(9.81 m/s2)
2g
2g
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Givens
5. ∆x = 31.5 m
q = 40.0°
g = 9.81 m/s2
Solutions
∆x = vi(cos q)∆t
∆x
∆t = 
vi(cos q)
∆y = vi(sin q)∆t− 12 g∆t2 = 0
I
∆y = vi(sin q) − 12 g∆t = 0
g∆x
∆y = vi(sin q) −  = 0
2vi(cos q)
(9.81 m/s2)(31.5 m)
g∆x
vi =  =  = 17.7 m/s
(2)(cos 40.0°)(sin 40.0°)
2(cos q)(sin q)
At maximum height, vy,f = 0 m/s.
vy,f2 = vi2(sin q)2− 2g∆ymax = 0
2
2
vi2(sin q)2 (17.7 m/s) (sin 40.0°)
 = 
∆ymax = 
= 6.60 m
2
(2)(9.81 m/s )
2g
Section Review, p. 105
3. vx = 100.0 m/s
∆y = −50.0 m
g = 9.81 m/s2
4. vx = 100.0 m/s
∆x = 319 m
∆y = −50.0 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
1
∆y = − 2g∆t2
∆t =
2∆y
−g
2∆y
∆x = vx∆t = vx  = (100.0 m/s)
−g
(2)(−50.0 m)

−9.81
m/s =
2
319 m
2
vy = vy
=2g∆
m/s
)2
−(2)
)2(−
m) = ±31.3 m/s = −31.3 m/s
y = (0
(9
.8
1m
/s
50
.0
,i 2
vtot = vy2+
m/s
)2
+(−
m/s
)2 = 1.
04
m2s2+
m2/
s2
00
.0
31
.3
00
0×1
9.8
0×102
v
y = (1
vtot = 10
m2/
s2 = 104.8 m/s
980
v
−31.3 m/s
q = tan−1 y = tan−1  = −17.4°
vx
100.0 m/s
q = 17.4° below the horizontal
5. ∆y = −125 m
vx = 90.0 m/s
g = 9.81 m/s2
1
∆y = − 2g∆t 2
2∆y
∆t =  =
−g
(2)(−125 m)

−9.81
m/s =
2
5.05 s
∆x = vx ∆t = (90.0 m/s)(5.05 s) = 454 m
6. ∆t = 0.50 s
∆x = 1.5 m
q = 33°
7. ∆t = 0.35 s
q = 67°
vi = 5.0 m/s
g = 9.81 m/s2
∆x = vi (cos q)∆t
∆x
1.5 m
vi =  =  = 3.6 m/s
(cos q )∆t (cos 33°)(0.50 s)
1
∆y = vi (sin q)∆t − 2 g∆t 2
1
∆y = (5.0 m/s)(sin 67°)(0.35 s) − 2 (9.81 m/s2)(0.35 s)2
∆y = 1.6 m − 0.60 m = 1.0 m
Section One—Pupil’s Edition Solutions
I Ch. 3–7
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Practice 3F, p. 109
Givens
Solutions
1. vte = +15 m/s
vbt = −15 m/s
vbe = vbt + vte = −15 m/s + 15 m/s = 0 m/s
I
2. vaw = +18.0 m/s
vsa = −3.5 m/s
3. vfw = 2.5 m/s north
vwe = 3.0 m/s east
vsw = vsa + vaw = − 3.5 m/s 18.0 m/s
vsw = 14.5 m/s in the direction that the aircraft carrier is moving
vfe = vfw + vwe
2 v 2 = (2.5 m/s)2 + (3.0 m/s)2
vtot = vf
w
w+
e
vtot = 6.
s2
+9.0
m2/
s2 = 15
.2
m2/
s2 = 3.90 m/s
2m
2/
vfw
2.5 m/s
q = tan−1  = tan  = (4.0 × 101)° north of east
vwe
3.0 m/s
4. vtr = 25.0 m/s north
vdt = 1.75 m/s at 35.0° east
of north
vdr = vdt + vtr
vx,tot = vx,dt = (1.75 m/s)(sin 35.0°) = 1.00 m/s
vy,dt = (1.75 m/s)(cos 35.0°) = 1.43 m/s
vy,tot = vtr + vy,dt = 25.0 m/s + 1.43 m/s = 26.4 m/s
2 (v
2
vtot = (v
)2
+(26
m/s
)2
x,
to
y .0
0m
/s
.4
t)+
,to
t) = (1
vtot = 1.
m2/
s2
+697
m2/
s2 = 69
s2 = 26.4 m/s
00
8m
2/
vx, tot
1.00 m/s
 = tan−1  = 2.17° east of north
q = tan−1 
vy, tot
26.4 m/s
2. vwg = −9 m/s
vbg = 1 m/s
3. vbw = 0.15 m/s north
vwe = 1.50 m/s east
vbw = vbgvgw = vbg − vwg = (1 m/s) – (–9 m/s) = 1 m/s + 9 m/s
vbw =
10 m/s away in the oppposite direction
vbe = vbw + vwe
2 v 2 = (0.15 m/s)2 + (1.50 m/s)2
vtot = vb
we w+
vtot = 0.
s2
+2.2
s2 = 2.
m2/
s2 = 1.51 m/s
02
2m
2/
5m
2/
27
vw
0.15 m/s
= tan−1  = 5.7° north of east
q = tan−1 b
vwe
1.50 m/s
I Ch. 3–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section Review, p. 109
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Chapter Review and Assess, pp. 113–119
Givens
6. A = 3.00 units (u)
B = −4.00 units (u)
Solutions
Students should use graphical techniques.
a. A + B = A2+
B 2 = (3
)2
+(−
)2
.0
0u
4.
00
u
A + B = 9.
00
u2+16.
0u2 = 25
.0
u2 = 5.00 units
I
B
− 4.00 u
q = tan−1  = tan−1  = 53.1° below the positive x-axis
A
3.00 u
b. A − B = A2+
)2 = (3
)2
+(4.
)2
(−B
.0
0u
00
u
A − B = 9.
00
u2+16.
0u2 = 25
.0
u2 = 5.00 units
−B
4.00 u
q = tan−1  = tan−1  = 53.1° above the positive x-axis
A
3.00 u
A2+
)2 = (3
)2
+(−
)2
c. A + 2B = (2B
.0
0u
8.
00
u
A + 2B = 9.
00
u2+64.
0u2 = 73
.0
u2 = 8.54 units
2B
−8.00 u
q = tan−1  = tan−1  = 69.4° below the positive x-axis
A
3.00 u
d. B − A = B 2+
)2 = (−
)2
+(−
)2 = 5.00 units
(−A
40
0u
3.
00
u
B
− 4.00 u
q = tan−1  = tan−1  = 53.1° below the negative x-axis
−A
−3.00 u
or 127° clockwise from the positive x-axis
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. A = 3.00 m
Students should use graphical techniques.
B = 3.00 m
Ax = A(cos q) = (3.00 m)(cos 30.0°) = 2.60 m
q = 30.0°
Ay = A(sin q) = (3.00 m)(sin 30.0°) = 1.50 m
2 + (A + B)2 = (2.60 m)2 + (4.50 m)2
a. A + B = A
y x A + B = 6.
m2+
m2 = 5.20 m
76
20.
2m
2 = 27
.0
Ay + B
4.50 m
q = tan−1  = tan−1  = 60.0° above the positive x-axis
Ax
2.60 m
2 + (A − B)2 = (2.60 m)2 + (−1.50 m)2
b. A − B = A
y x A − B = 6.
m2+
m2 = 3.00 m
76
2.2
5m
2 = 9.
01
Ay − B
−1.50 m
q = tan−1  = tan−1  = 30.0° below the positive x-axis
2.60 m
Ax
Section One—Pupil’s Edition Solutions
I Ch. 3–9
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Chapter Review and Assess, pp. 113–119 continued
Givens
Solutions
2 (−A )2 = (1.50 m)2 + (−2.60 m)2
c. B − A = (B
A
−
y)+
x B − A = 3.00 m
I
1.50 m
B − Ay
q = tan−1  = tan−1  = 30.0° above the negative x-axis
−2.60 m
−Ax
or 150° counterclockwise from the positive x-axis
2 + (A − 2B)2 = (2.60)2 + (−4.50)2 = 5.20 m
d. A − 2B = A
y x Ay − 2B
− 4.50 m
q = tan−1  = tan−1  = 60.0° below the positive x-axis
Ax
2.60 m
8. ∆y1 = −3.50 m
Students should use graphical techniques.
d2 = 8.20 m
∆x2 = d2 (cos q2 ) = (8.20 m)(cos 30.0°) = 7.10 m
q2 = 30.0°
∆y2 = d2 (sin q2 ) = (8.20 m)(sin 30.0°) = 4.10 m
∆x3 = 15.0 m
∆xtot = ∆x2 + ∆x3 = 7.10 m − 15.0 m = −7.9 m
∆ytot = ∆y1 + ∆y2 = −3.50 m + 4.10 m = 0.60 m
xtot
m)2
d = (∆
)2+(∆
ytot)2 = (−
7.
9m
)2+(0.
60
d = 62
m2+
0.3
6m
2 = 62
m
2= 7.9 m
∆ytot
0.60 m
q = tan−1 
= tan−1  = 4.3° north of west
∆xtot
−7.9 m
∆y = 13.0 m
Students should use graphical techniques.
d= ∆
x2
+
∆
y 2 = (−
m
)2
+(13
m
)2
8.
00
.0
d = 64
m2+
m2 = 23
.0
169
3m
2 = 15.3 m
∆y
13.0 m
q = tan−1  = tan−1  = 58.4° south of east
∆x
−8.00 m
22. ∆x1 = 3 blocks west
= −3 blocks east
∆y = 4 blocks north
∆x2 = 6 blocks east
a. ∆xtot = ∆x1 + ∆x2 = −3 blocks + 6 blocks = 3 blocks
∆ytot = ∆y = 4 blocks
xtot
)2
+(∆
ytot
)2 = (3
d = (∆
blo
ck
s)
2+(4blo
ck
s)
2
d = 9b
s2
+16b
s2 = 25
s2 = 5 blocks
lo
ck
lo
ck
blo
ck
4 blocks
∆ytot
q = tan−1 
= tan−1  = 53° north of east
3 blocks
∆xtot
b. distance traveled = 3 blocks + 4 blocks + 6 blocks = 13 blocks
I Ch. 3–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. ∆x = −8.00 m
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Givens
Solutions
23. ∆x = 6.00 m
d= ∆
∆y2 = (6
+ (−5.
m)2
x2+
.0
0m
)2
40
∆y = −5.40 m
d = 36
m2+
m2 = 8.07 m
.0
29.
2m
2 = 65
.2
∆y
−5.40 m
q = tan−1  = tan−1  = 42.0° south of east
∆x
6.00 m
24. ∆y1 = −10.0 yards
I
∆ytot = ∆y1 + ∆y2 = −10.0 yards + 50.0 yards = 40.0 yards
∆x = 15.0 yards
∆xtot = ∆x = 15.0 yards
∆y2 = 50.0 yards
d = (∆
xt
)2
+(∆
yt
)2 = (1
5.
0ya
rd
s)
2+(40
.0
yar
ds)
2
ot
ot
d = 22
s2
+1.6
03yar
s2 = 18
s2 = 42.7 yards
5ya
rd
0×1
d
20
yar
d
25. ∆y1 = −40.0 m
Case 1: ∆ytot = ∆y1 + ∆y2 = −40.0 m − 20.0 m = −60.0 m
∆x = ±15.0 m
∆xtot = ∆x = +15.0 m
∆y2 = ±20.0 m
d = (∆
ytot
)2
+(∆
xtot
)2 = (−
m
)2
+(15
m
)2
60
.0
.0
d = 3.
m2+
m2 = 38
m2 = 61.8 m
60
×103
225
20
∆ytot
−60.0 m
q = tan−1 
= tan−1  = 76.0° south of east
∆xtot
15.0 m
Case 2: ∆ytot = ∆y1 + ∆y2 −40.0 m+ 20.0 m+ −20.0 m
∆xtot = ∆x = +15.0 m
yt
)2
+)∆
xt
)2 = (−
m
)2
+(15
m
)2
d = (∆
20
.0
.0
ot
ot
d = 4.
02
+225
m2 = (6
m
)2 = 25.0 m
00
×1
25
∆ytot
−20.0 m
∆ = tan−1 
= tan−1  = 53.1° south of east
∆tot
15.0 m
Case 3: ∆ytot = ∆y1 + ∆y2 = −40.0 m − 20.0 m = −60.0 m
∆xtot = ∆x = −15.0 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
yt
)2
+(∆
xt
)2 = (−
m
)2
+(−
m
)2
d = (∆
60
.0
15
.0
ot
ot
d = 61.8 m
∆ytot
−60.0 m
q = tan−1 
= tan−1  = 76.0° south of west
∆xtot
−15.0 m
Case 4: ∆ytot = ∆y1 + ∆y2 = −40.0 m + 20.0 m = −20.0 m
∆xtot = ∆x = −15.0 m
ytot
xtot
m)2
+(−
m)2
d = (∆
)2+(∆
)2 = (−
20
.0
15
.0
d = 25.0 m
∆ytot
−20.0 m
q = tan−1 
= tan−1  = 53.1° south of west
∆xtot
−15.0 m
26. d = 110.0 m
∆ = −10.0°
∆x = d(cos q) = (110.0 m)[cos(−10.0°)] = 108 m
∆x = d(sin q) = (110.0 m)[sin(−10.0°)] = −19.1 m
Section One—Pupil’s Edition Solutions
I Ch. 3–11
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Givens
Solutions
27. q = 25.0°
∆x = d(cos q) = (3.10 km)(cos 25.0°) = 2.81 km east
d = 3.10 km
I
28. d = 41.1 m
∆y = d(sin q) = (3.10 km)(sin 25.0°) = 1.31 km north
∆x = d(cos q) = (41.1 m)(cos 40.0°) = 31.5 m
q = 40.0°
∆y = d(sin q) = (41.1 m)(sin 40.0°) = 26.4 m
29. d1 = 100.0 m
∆x1 = d1(cos q1) = (100.0 m)(cos 0.00°) = 100.0 m
q1 = 0.00° east = 0.00°
∆y1 = d1(sin q1) = (100.0 m)(sin 0.00°) = 0.000 m
d2 = 300.0 m
∆x2 = d2(cos q2) = (300.0 m)[cos (−90.0°)] = 10.00 m
q2 = 90.0° south = −90.0°
∆y2 = d2(sin q2) = (300.0 m)[sin (−90.0°)] = −300.0 m
d3 = 150.0 m
∆x3 = d3(cos q3) = (150.0 m)[cos (−150°)] = −129.9 m
q3 = 30.0° south of west
∆y3 = d3(sin q3) = (150.0 m)[sin (−150°)] = −75.00 m
= 180.0° − 30.0° south of
east
= −150°
d4 = 200.0 m
q4 = 60.0 ° north of west
= 180° − 60.0° north of
east
= 120°
∆x4 = d4(cos q4) = (200.0 m)(cos (120°)] = −100.0 m
∆y4 = d4(sin q4) = (200.0 m)[sin (120°)] = 173.2 m
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (100.0 m) + (0.00 m) + (−129.9 m) + (−100.0 m)
= −129.9 m
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = (0.000 m) + (−300.0 m) + (−75.00 m) + (173.2 m)
= −201.8 m
d = (∆
xtot
)2
+(∆
ytot
)2 = (−
)2
+(−
)2
12
9.
9m
20
1.
8m
d = 16
m2+
m2 = 240.0 m
870
4072
0m
2 = 57
590
34. ∆y = −0.809 m
∆x = 18.3 m
2
g = 9.81 m/s
∆x
∆t =  
vx
2
−g
m/s
v = 
2∆y ∆x = 
(2−)(9−.801.8
09 m) (18.3 m) =
1
1 ∆x
∆y = − 2g∆t 2 = − 2 g 
vx
2
x
35. vi = 1.70 × 103 m/s
q = 55.0°
g = 9.81 m/s2
1
45.1 m/s, or 162 km/h
1
a. ∆y = vi(sin q)∆t − 2g∆t2 = vi(sin q ) − 2g∆t = 0
2vi(sin q) (2)(1.70 × 103 m/s)(sin 55.0°)
 = 
∆t = 
= 284 s
9.81 m/s2
g
∆x = vi(cos q)∆t = (1.70 × 103 m/s)(cos 55.0°)(284 s) = 2.77 × 105 m
b. ∆t = 284 s (See a.)
I Ch. 3–12
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆ytot
−201.8 m
q = tan−1 
= tan−1  = 57.23° south of west
∆xtot
−129.9 m
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Givens
Solutions
36. vx = 18 m/s
∆y = − 2 g∆t2
1
∆y = −52 m
g = 9.81 m/s2
∆t =
2∆y
(2)(−52 m)
−g = 
−9.81
m/s =
2
3.3 s
When the stone hits the water,
I
vy = −g∆t = (−9.81 m/s)(3.3 s) = −32 m/s
2
)2+(−
m/s
)2
vtot = vx2+
v
8m
/s
32
y = (1
vtot = 32
s2
+100
s2 = 13
m2/
s2 = 36 m/s
0m
2/
0m
2/
00
37. vx,s = 15 m/s
vx,o = 26 m/s
∆y = −5.0 m
g = 9.81 m/s2
1
∆y = − 2g∆t2
2∆y
∆t =  =
−g

−9.81
/s = 1.0 s
m
2(−5.0 m)
2
∆xs = vx,s ∆t = (15 m/s)(1.0 s) = 15 m
∆xo = vx,o ∆t = (26 m/s)(1.0 s) = 26 m
∆xo − ∆xs = 26 m − 15 m = 11 m
38. ∆x = 36.0 m
vi = 20.0 m/s
q = 53°
∆ybar = 3.05 m
g = 9.81 m/s2
a. ∆x = vi(cos q)∆t
∆x
36.0 m
∆t =  =  = 3.0 s
vi(cos q) (20.0 m/s)(cos 53°)
1
1
∆y = vi(sin q)∆t − 2g∆t2 = (20.0 m/s)(sin 53°)(3.0 s) − 2(9.81 m/s2)(3.0 s)2
∆y = 48 m − 44 m = 4 m
∆y = ∆ybar = 4 m − 3.05 m = 1 m
The ball clears the goal by 1 m.
b. vy,f = vi(sin q) − g∆t = (20.0 m/s)(sin 53°) − (9.81 m/s2)(3.0 s)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vx,f = 16 m/s − 29 m/s = −13 m/s
The velocity of the ball as it passes over the crossbar is negative; therefore, the ball
is falling.
39. vi = 25.0 m/s
q = 45.0°
∆x = 50.0 m
2
g = 9.81 m/s
∆x = vi (cos q)∆t
∆x
∆t = 
vi (cos q)
∆x
∆x
1
1
∆y = vi (sin q)∆t − 2g∆t 2 = vi (sin q)  − 2g 
vi(cos q)
vi(cos q)
2
g∆x2
(9.81 m/s2)(50.0 m)2


∆y = ∆x(tan q) − 
=
(50.0
m)(tan
45.0°)
−
2vi2(cos q)2
(2)(25.0 m/s)(cos 45.0)2
∆y = 50.0 m − 39.2 m = 10.8 m
Section One—Pupil’s Edition Solutions
I Ch. 3–13
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Givens
Solutions
40. ∆y = −1.00 m
Find the initial velocity of the water when shot at rest horizontally 1 m above the
ground.
∆x = 5.00 m
I
1
v = 2.00 m/s
∆y = − 2g∆t2
2∆ y
∆t =  
−g
∆t = 0.329 s
∆x = vx∆t
g = 9.81 m/s2
∆x
5.00 m
∆x
vx =  =  =  = 11.1 m/s
2∆y
(2)(−1.00 m)
∆t


−g
−9.81 m/s2
q = 45.0°
Find how far the water will go if it is shot horizontally 1 m above the ground while
the child is sliding down the slide.
vx, tot = vx + v(cos q)
∆x = vx, tot∆t = [vx + v(cos q)]∆t = 11.1 m/s + (2.00 m/s)(cos 45.0°)](0.329 s)
∆x = [11.1 m/s + 1.41 m/s](0.329 s) = (12.5 m/s)(0.329 s) = 4.11 m
41. ∆x1 = 2.50 × 103 m
∆x2 = 6.10 × 102 m
∆ymountain = 1.80 × 103 m
vi = 2.50 × 102 m/s
q = 75.0°
2
g = 9.81 m/s
For projectile’s full flight,
∆x
∆t = 
vi(cos q)
1
1
∆y = vi (sin q)∆t − 2g∆t 2 = vi(sin q) − 2g∆t = 0
∆x
1
vi (sin q) − 2g  = 0
vi(cos q)
2vi2(sin q)(cos q) (2)(2.50 × 102 m/s)2(sin 75.0°)(cos 75.0°)
 = 
∆x = 
= 3190 m
9.81 m/s2
g
Distance between projectile and ship = ∆x − ∆x1 − ∆x2
= 3190 m − 2.50 × 103 m − 6.10 × 102 m = 80 m
2
∆x1
∆x1
1
1
∆y = vi(sin q)∆t′ − 2g∆t′2 = vi(sin q) 
− g 
vi(cos q) 2 vi(cos q)
g∆x12
∆y = ∆x1(tan q) − 
2 2
2vi (cos q)
∆y = (2.50 × 103 m)(tan 75.0°)
(9.81 m/s2)(2.50 × 103 m)2
− 
(2)(2.50 × 102 m/s)2 (cos 75.0°)2
∆y = 9330 m − 7320 = 2010 m
distance above peak = ∆y − ∆ymountain = 2010 m − 1.80 × 103 m =
47. vap = 165 km/h south
= −165 km/h north
vpe = 145 km/ north
I Ch. 3–14
vae = vap + vpe
vae = −165 km/h + 145 km/h = −20 km/h north = 20 km/h south
Holt Physics Solution Manual
210 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
For projectile’s flight to the mountain,
∆xi
∆t′ = 
vi (cos q)
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Givens
Solutions
48. vre = 1.50 m/s east
a. vbe = vbr + vre
vbr = 10.0 m/s north
2
2
vbe vb
vr
)2
=(1.
m/s
)2
0.
0m
/s
50
r e = (1
vbe 1.
02
m2/
s2
+2.2
s2 = 10
s2 = 10.1 m/s
00
×1
5m
2/
2m
2/
v
1.50 m/s
 = tan−1  = 8.53° east of north
q = tan−1 re
vbr
10.0 m/s
I
∆x 325 m
b. ∆t =  =  = 32.5 s
vbr 10.0 m/s
∆x = 325 m
49. vwe = 50.0 km/h south
vaw = 205 km/h
vae is directed due west
∆y = vre ∆t = (1.50 m/s)(32.5 s) = 48.8 m
a. vaw = vae + (−vwe)
v
we = sin q
vaw
v
50.0 km/h
q = sin−1 we = sin−1  = 14.1° north of west
vaw
205 km/h
b. vaw2 = vae2 + vwe2
2
vae = va
)2
−(50
)2
vwe2 = (2
05
km
/h
.0
km
/h
w−
vae = 4.
04km
03km
20
×1
2/h2−2.5
0×1
2/h2
vae = 3.
04km
95
×1
2/h2 = 1.99 km/h
50. ∆x = 1.5 km
vre = 5.0 km/h
vbr = 12 km/h
51. vre = 3.75 m/s downstream
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vsr = 9.50 m/s
vse is directed across the river
The boat’s velocity in the x direction is greatest when the boat moves directly across
the river with respect to the river.
1.5 km
∆x
∆tmin =  =  = 7.5 min
(12
km/h)(1
h/60 min)
vbr
vre

= sin q
vsr
a. vsr = vse + (−vre)
3.75 m/s
q = sin−1  = 23.2° upstream from straight across
9.50 m/s
b. vsr2 = vse2 + vre2
2
vse = vs
−vre
)2
−(3.
m/s
)2
2 = (9
.5
0m
/s
75
r vse = 90
m2/
s2
−14.
s2 = 76
m2/
s2 = 8.72 m/s
.2
1m
2/
.1
vse = 8.72 m/s directly across the river
52. vbr = 12.0 m/s east
vre = 3.5 m/s south
a. vbe = vbr + vre
2
vbe = vb
+vre
)2
+(3.
)2
2 = (1
2.
0m
/s
5m
/s
r vbe = 14
4m
2/s2+12m
2/s2 = 15
6m
2/s2 = 12.5 m/s
v
3.5
 = tan−1  = 16° south of east
q = tan−1 re
vbr
12.0
∆x = 1360 m
∆x 1360 m
b. ∆t =  =  = 113 s
vbr 12.0 m/s
Section One—Pupil’s Edition Solutions
I Ch. 3–15
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Givens
Solutions
53. ∆x = 130.0 m
a. ∆x = vi(cos q)∆t
q = 35.0°
∆y = 21.0 m − 1.0 m
= 20.0 m
I
g = 9.81 m/s2
∆x
∆t = 
vi(cos q)
∆x
∆x
1
1
∆y = vi(sin q )∆t − 2 g∆t2 = vi(sin q)  − 2 g 
vi(cos q)
vi(cos q)
2
2
g∆x
∆y = ∆x(tan q) − 
2vi2(cos q)2
2vi2(cos q)2[∆x(tan q) − ∆y] = g∆x2
vi =
g∆x2

2(cos q)2[∆x(tan q) − ∆y]
(9.81 m/s2)(130.0 m)2

(2)(cos 35.0)2[(130.0 m)(tan 35.0°) − 20.0 m]
(9.81 m/s )(130.0 m)
 = 41.7 m/s
v = (2)(cos 35.0°)(91.0 m − 20.0 m)
vi =
2
2
i
130.0 m
∆x
b. ∆t =  =  = 3.81 s
vi(cos q) (41.7 m/s)(cos 35.0°)
c. vy,f = vi(sin q) − g∆t = (41.7 m/s)(sin 35.0°) − (9.81 m/s2)(3.81 s)
vy,f = 23.9 m/s − 37.4 m/s = −13.5 m/s
vx,f = vx = vi(cos q) = (41.7 m/s)(cos 35.0°) = 34.2 m/s
2
2
vf = (v
vy
)2
+(−
m/s
)2
x,
(
4.
2m
/s
13
.5
f )+
,f) = (3
vf = 11
m/s
m2/
s2 = 13
m2/
s2 = 36.7 m/s
70
2+182
50
54. ∆x = 12 m
q = 15°
2
g = 9.81 m/s
∆x
∆t = 
vi(cos q)
1
1
∆y = vi(sin q)∆t − 2 g∆t2 = vi(sin q) − 2 g∆t = 0
∆x
1
vi(sin q) − 2 g  = 0
vi(cos q)
2vi2(sin q)(cos q) = g∆x
vi =
(9.81 m/s2)(12 m)
 = 15 m/s
(2)(sin 15°)(cos 15°)
g∆x
 =
2(sin q)(cos q)
12 m
∆x
b. ∆t =  =  = 0.83 s
(15 m/s)(cos 15°)
vi(cos q)
vy,f = vi(sin q) − g∆t = (15 m/s)(sin 15°) − (9.81 m/s2)(0.83 s)
vy,f = 3.9 m/s − 8.1 m/s = −4.2 m/s
I Ch. 3–16
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
a. ∆x = vi(cos q)∆t
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Givens
Solutions
vx,f = vx = vi(cos q) = (15 m/s)(cos 15°) = 14 m/s
2
vf = (v
vy
)2
+(−
)2
x,f)2+(
4m
/s
4.
2m
/s
,f) = (1
vf = 2.
02
m2/
s2
+18m
s2 = 22
s2 = 15 m/s
0×1
2/
0m
2/
55. ∆x = 10.0 m
q = 45.0°
∆y = 3.05 m − 2.00 m
= 1.05 m
See solution to Chapter 3 Review and Assess problem 53 for a derivation of the following equation.
vi =
g = 9.81 m/s2
vi =
56. ∆t = 3.00 s
q = 30.0°
g = 9.81 m/s2
57. ∆x = 20.0 m
∆t = 50.0 s
vpe = ±0.500 m/s
g∆x2

=
2(cos q)2[∆x(tan q ) − ∆y]
I
(9.81 m/s2)(10.0 m)2

(2)(cos 45.0°)2[(10.0 m)(tan 45.0°) − 1.05 m]
(9.81 m/s2)(10.0 m)2

=
(2)(cos 45.0°)2(10.0 m − 1.05 m)
(9.81 m/s2)(10.0 m)2

= 10.5 m/s
(2)(cos 45.0°)2(8.95 m)
1
∆y = vi(sin q)∆t − 2 g∆t2 = vi(sin q) − g∆t = 0
g∆t
(9.81 m/s2)(3.00 s)
vi =  =  = 29.4 m/s
2(sin q)
(2)(sin 30.0°)
∆x 20.0 m
veg =  =  = 0.400 m/s
∆t 50.0 s
vpg = vpe + veg
a. Going up:
vpg = vpe + veg = 0.500 m/s + 0.400 m/s = 0.900 m/s
20.0 m
∆x
∆tup =  =  = 22.2 s
vpg 0.900 m/s
b. Going down:
vpg = −vpe + veg = −0.500 m/s + 0.400 m/s = −0.100 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
−20.0 m
−∆x
∆tdown =  =  = 2.00 × 102 s
vpg −0.100 m/s
58. ∆y = −1.00 m
∆x = 1.20 m
g = 9.81 m/s2
∆x
∆t = 
vx
a. ∆x = vx ∆t
1
1 ∆x
∆y = − 2 g∆t2 = − 2 g 
vx
vx =
−g∆x2
=
2y
2
g∆x2
= 
2vx2
−(9.81 m/s2)(1.20 m)2
 = 2.66 m/s
(2)(−1.00 m)
b. The ball’s velocity vector makes a 45° angle with the horizontal when vx = vy .
vx
vx = vy,f = −g∆t
∆t = 
g
vx 2
vx 2
1
1
2
∆y = − 2 g∆t = − 2 g  = − 
2g
g
(2.66 m/s)2
∆y = − 
= − 0.361 m
(2)(9.81 m/s2)
h = 1.00 m − 0.361 m = 0.64 m
Section One—Pupil’s Edition Solutions
I Ch. 3–17
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Givens
Solutions
59. v1 = 40.0 km/h
For lead car:
v2 = 60.0 km/h
∆xi = 125 m
∆xtot = v1∆t + ∆xi
For chasing car:
∆xtot = v2∆t
I
v2∆t = v1∆t + ∆xi
(125 m)(10−3 km/m)
∆xi
∆t =  = 
(60.0 km/h − 40.0 km/h)(1 h/3600 s)
v2 − v1
125 × 10−3 km
∆t =  = 22.5 s
(20.0 km/h)(1 h/3600 s)
60. q = 60.0°
d1 = v1∆t = (41.0 km/h)(3.00 h) = 123 km
v1 = 41.0 km/h
∆x1 = d1(cos q) = (123 km)(cos 60.0°) = 61.5 km
v2 = 25.0 km/h
∆y1 = d1(sin q) = (123 km)(sin 60.0°) = 107 km
∆t1 = 3.00 h
∆t2 = ∆t − ∆t1 = 4.50 h − 3.00 h = 1.50 h
∆t = 4.50 h
∆y2 = v2∆t2 = (25.0 km/h)(1.50 h) = 37.5 km
∆xtot = ∆x1 = 61.5 km
∆ytot = ∆y1 + ∆y2 = 107 km + 37.5 km = 144 km
d = (∆
xtot
)2
+(∆
ytot
)2 = (6
)2
+(14
)2
1.
5km
4km
d = 37
80
km
2+2070
0km
2 = 24
500
km
2 = 157 km
a = 4.00 m/s2
d = 50.0 m
∆y = −30.0 m
g = 9.81 m/s2
1
a. d = 2 a∆t2
∆t1 =
 = 5.00 s
a = 4.00 m/s
2d
(2)(50.0 m)
2
vi = a∆t1 = (4.00 m/s2)(5.00 s) = 20.0 m/s
2
vy,f = vi
(
si
n
q
)2
−2
g∆y = (2
]2
−(2)
m)
0.
0m
/s
2)[s
in
(−
24
.0
°)
(9
.8
1m
/s
2)(−
30
.0
vy,f = 66
m2/
s2
=589
m2/
s2 = 65
s2 = ±25.6 m/s = −25.6 m/s
.2
5m
2/
vy,f = vi(sin q) − g∆t2
vy,f − vi(sin q) −25.6 m/s − (20.0 m/s)(sin −24.0°)
∆t2 =  = 
−9.81 m/s2
−g
−25.6 m/s + 8.13 m/s −17.5 m/s
= 2 = 1.78 s
∆t2 = 
−9.81 m/s2
−9.81 m/s
∆x = vi(cos q)∆t2 = (20.0 m/s)[cos(−24.0°)](1.78 s) = 32.5 m
b. ∆t2 = 1.78 s (See a.)
62. vbw = ±7.5 m/s
vwe = 1.5 m/s
∆x d = 250 m
∆x u = −250 m
vbe = vbw + vwe
Going downstream:
vbe,d = 7.5 m/s + 1.5 m/s = 9.0 m/s
Going upstream:
vbe,u = −7.5 m/s + 1.5 m/s = −6.0 m/s
∆xd ∆xu
250 m
−250 m
+  =  +  = 28 s + 42 s = 7.0 × 101 s
∆t = 
vbe,d vbe,u 9.0 m/s − 6.0 m/s
I Ch. 3–18
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
61. q = −24.0°
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Givens
Solutions
63. q = 34°
a. ∆x = vi (cos q)∆t
∆x
∆t = 
vi (cos q)
∆x = 240 m
g = 9.81 m/s2
1
1
∆y = vi (sin q)∆t − 2 g∆t 2 = vi (sin q) − 2 g∆t = 0
∆x
g ∆x
1
vi (sin q) − 2 g  = vi2(sin q) −  = 0
vi(cos q)
2(cos q)
vi =
I
(9.81 m/s2)(240 m)
 = 5.0 × 101 m/s
(2)(cos 34°)(sin 34°)
g ∆x
 =
2(cos q )(sin q )
vy,f 2 − vy,i2
b. ∆ymax = 
−2g
Because vy,f = 0 m/s,
−vi2(sin q)2 (5.0 × 101 m/s)2(sin 34°)2
∆ymax =  = 
−2g
(2)(9.81 m/s2)
∆ymax = 4.0 × 101 m
∆x
∆x
1
vi(sin q)  − g 
2vi(cos q) 2 2vi(cos q)
∆x
g∆x

∆ymax =  (tan q) − 
2
8 vi2(cos q)2
2
2
(240 m)(tan 34°)
(9.81 m/s2)(240 m)2
∆ymax =  − 
2
(8)(5.0 × 101 m/s)2(cos 34°)
∆ymax = 81 m − 41 m = 4.0 × 101 m
64. vwe = −0.500 m/s
∆x = 0.560 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆tupstream = 0.800 s
a. vse = vsw + vwe
Going upstream:
∆x
0.560 m
vse =  =  = 0.700 m/s
∆tupstream 0.800 s
0.700 m/s = vsw + (−0.500 m/s)
vsw = 0.700 m/s + 0.500 m/s = 1.20 m/s
Going downstream:
vse = −0.500 m/s (same as the water)
−0.500 m/s = vsw + (−0.500 m/s)
vsw = −0.500 m/s + 0.500 m/s = 0.00 m/s
b. d = vsw ∆t = (1.200 m/s)(0.800 s) = 0.960 m
−∆x
− 0.560 m
c. ∆tdownstream =  =  = 1.12 s
vse
− 0.500 m/s
∆ttotal = ∆tupstream + ∆tdownstream = 0.800 s + 1.12 s = 1.92 s
d
0.960 m
vsw, avg =  =  = 0.500 m/s
∆ttotal
1.92 s
Section One—Pupil’s Edition Solutions
I Ch. 3–19
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Givens
Solutions
65. vce = 50.0 km/h east
a. vce = vrc (sin q)
v
50.0 km/h
 =  = 57.7 km/h
vrc = ce
(sin q ) (sin 60.0°)
q = 60.0°
vrc = 57.7 km/h at 60.0° west of the vertical
I
b. vre = vrc (cos q) = (57.7 km/h)(cos 60.0°) = 28.8 km/h
vre = 28.8 km/h straight down
66. ∆twalk = 30.0 s
∆tstand = 20.0 s
L
L
vpe =  = 
∆twalk 30.0 s
L
L
veg =  = 
∆tstand 20.0 s
vpg = vpe + veg
vpg = vpe + veg
L
L
2L + 3L
5L
vpg =  +  =  = 
30.0 s 20.0 s
60.0 s
60.0 s
L
5L
  = 
∆t 60.0 s
60.0 s
∆t =  = 12.0 s
5
67. ∆x Earth = 3.0 m
2
g = 9.81 m/s
1
1
∆y = vi (sin q )∆t − 2 g∆t 2 = vi (sin q) − 2 g∆t = 0
2vi (sin q )

∆t = 
g
2vi (sin q)

∆xEarth = vi (cos q)∆t = vi (cos q) 
g
2vi2(cos q)(sin q)
Because vi and q are the same for all locations,
k
∆xEarth = , where k = 2vi2(cos q)(sin q)
g
g
k = g∆xEarth =  ∆xmoon = (0.38g)∆xMars
6
∆xmoon = 6∆xEarth = (6)(3.0 m) = 18 m
∆xEarth 3.0 m
∆xMars =  =  = 7.9 m
0.38
0.38
68. vx = 10.0 m/s
q = 60.0°
g = 9.81 m/s2
The observer on the ground sees the ball rise vertically, which indicates that the
x-component of the ball’s velocity is equal and opposite the velocity of the train.
vx = vi(cos q )
10.0 m/s
vx
vi =  =  = 20.0 m/s
(cos q) (cos 60.0°)
At maximum height, vy = 0, so
vy,f 2 − vy,i2 vi2(sin q)2
∆ymax =  = 
−2g
2g
(20.0 m/s)2(sin 60.0°)2
= 15.3 m
∆ymax = 
(2)(9.81 m/s2)
I Ch. 3–20
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆xEarth = 
g
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Givens
Solutions
69. vi = 18.0 m/s
q = 35.0°
1
1
∆y = vi (sin q)∆t − 2 g∆t 2 = vi (sin q) − 2 g∆t = 0
∆xi = 18.0 m
2vi (sin q) 2(18.0 m/s)(sin 35.0°)
 = 
∆t = 
= 2.10 s
g
9.81 m/s2
g = 9.81 m/s2
∆x = vi (cos q)∆t = (18.0 m/s)(cos 35.0°)(2.10 s) = 31.0 m
I
∆xrun = ∆x − ∆xi = 31.0 m − 18.0 m = 13.0 m
∆x un 13.0 m
vrun = r
=  = 6.19 m/s downfield
∆t
2.10 s
70. q = 53°
ay = a(sin q) = (25 m/s2)(sin 53°) = 2.0 × 101 m/s2
vi = 75 m/s
∆y = vi (sin q)∆t + 2 ay ∆t 2 = (75 m/s)(sin 53°)(25 s) + 2 (2.0 × 101 m/s2)(25 s)2
∆t = 25 s
∆y = 1500 m + 6200 m = 7700 m
2
a = 25 m/s
1
1
vf = vi + a∆t = 75 m/s + (25 m/s2)(25 s) = 75 m/s + 620 m/s = 7.0 × 102 m/s
For the motion of the rocket after the boosters quit:
vi = vf = 7.0 × 102 m/s
vy,f = vi (sin q) − g∆t = 0
q = 53°
2
vi (sin q) (7.0 × 10 m/s)(sin 53°)
 = 
∆t = 
= 57 s
9.81 m/s2
g
g = 9.81 m/s2
1
1
∆y = vi (sin q)∆t − 2 g∆t 2 = (7.0 × 102 m/s)(sin 53°)(57 s) − 2 (9.81 m/s2)(57 s)2
∆y = 32 000 m − 16 000 m = 16 000 m
a. ∆ytotal = 7700 m + 16 000 m = 2.4 × 104 m
1
b. ∆y = − 2 g∆t 2
∆t =
2∆y
 = 7.0 × 10 s
−g = −9.81 m/s
(2)(−24 000 m)
2
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆ttotal = 25 s + 57 s + 7.0 × 101 s = 152 s
c. ax = a(cos q) = (25 m/s2)(cos 53°) = 15 m/s2
1
1
∆x = vi (cos q)∆t + 2 a∆t 2 = (75 m/s)(cos 53°)(25 s) + 2 (75 m/s2)(25 s)2
∆x = 1.1 × 103 m + 2.3 × 104 m = 2.4 × 104 m
vi = 7.0 × 102 m/s
After the rockets quit:
q = 53°
∆t = 57 s + 7.0 × 101 s = 127 s
∆x = vi (cos q)∆t = (7.0 × 101 m/s)(cos 53°)(127 s) = 5.4 × 104 m
∆xtot = 2.4 × 104 m + 5.4 × 104 m = 7.8 × 104 m
Section One—Pupil’s Edition Solutions
I Ch. 3–21
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Forces and the Laws of Motion
Chapter
4
I
Practice 4A, p. 133
Givens
1. F = 70.0 N
q = + 30.0°
Solutions
Fx = F(cos q) = (70.0 N)(cos 30.0°) = 60.6 N
Fy = F(sin q) = (70.0 N)(sin 30.0°) = 35.0 N
2. Fx,1 = 82 N
Fx,2 = −115 N
a. Fx,net = Fx,1 + Fx,2 = 82 N − 115 N = −33 N
Fy,1 = 565 N
b. Fy,net = Fy,1 + Fy,2 = 565 N + (−236 N) = 329 N
Fy,2 = −236 N
2
)2 = (−
N
)2
+(32
)2
c. Fnet = (F
x,
(Fy,n
et
33
9N
ne
t)+
Fnet = 1.
03
N2+
05
N2 = 1.
05
N2 = 3.30 × 102 N
09
×1
1.0
8×1
09
×1
Fy,net
329 N
q = tan−1  = tan−1  = −84° above negative x-axis
Fx,net
−33 N
θ = 180.0° − 84° = 96° counterclockwise from the positive x-axis
3. Fy = −9.25 N
Fx = 1.05 N
2
2
Fnet = F
+
F
)2
+(−
N
)2
.0
5N
9.
25
x y = (1
Fnet = 1.
N2+
N2 = 9.31 N
10
85.
6N
2 = 86
.7
1.05 N
F
q = tan−1 x = tan−1 
−9.25 N
Fy
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = −6.48° = 6.48° to the right of vertical
4. Fwind = 452 N north
Fwater = 325 N west
Fx = Fwater = −325 N
Fy = Fwind = 452 N
2
2
Fnet = F
+
F
)2
+(45
)2
32
5N
2N
x y = (−
Fnet = 1.
N2+
N2 = 3.
N2
06
×105
2.0
4×105
10
×105
Fnet = 557 N
F
452 N
q = tan−1 y = tan−1  = −54.3°
Fx
−325 N
q = 54.3° north of west, or 35.7° west of north
Section Review, p. 135
4. Fy = 130.0 N
Fx = 4500.0 N
2
2
+
F
)2
+(13
)2
Fnet = F
50
0.
0N
0.
0N
x y = (4
Fnet = 2.
07
N2+
04
N2 = 2.
07
N2 = 4502 N
02
5×1
1.6
90
×1
02
7×1
Fy
130.0 N
q = tan−1  = tan−1  = 1.655° forward of the side
Fx
4500.0 N
Section One—Pupil’s Edition Solutions
I Ch. 4–1
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Practice 4B, p. 138
Givens
Solutions
1. Fnet = 7.0 N forward
m = 3.2 kg
I
2. Fnet = 390 N north
m = 270 kg
3. Fnet = 6.75 × 103 N east
m = 1.50 × 103 kg
7.0 N
F
 =  = 2.2 m/s2 forward
a = net
m 3.2 kg
390 N
F
 =  = 1.4 m/s2 north
a = net
m 270 kg
6.75 × 103 N
F
 = 
a = net
= 4.50 m/s2 east
m 1.50 × 103 kg
1
4. m = 2.0 kg
∆x = 85 cm down the incline
∆t = 0.50 s
5. Fnet = 13.5 N to the right
2
a = 6.5 m/s to the right
∆x = 2a(∆t)2
2∆x (2)(0.85 m)
a = 2 = 
= 6.8 m/s2 down the incline
∆t
(0.50 s)2
Fnet = ma = (2.0 kg)(6.8 m/s2) = 14 N down the incline
F et Fnet 13.5 N
m = n
=  = 2 = 2.1 kg
a
6.5 m/s
a
Section Review, p. 140
a. Fnet = ma = (6.0 kg)(2.0 m/s2) = 12 N
1. m = 6.0 kg
a = 2.0 m/s2
12 N
F
 =  = 3.0 m/s2
b. a = net
4.0 kg
m
4. Fy = 390 N, north
Fx = 180 N, east
m = 270 kg
2 + F 2 = (180 N)2 + (390 N)2
Fnet = F
x y
Fnet = 3.
04
N2+1.5
05
N2 = 1.
05
N2 = 420 N
2×1
×1
8×1
Fy
390 N
q = tan−1  = tan−1 
Fx
180 N
q = 65° north of east
420 N
F
 =  = 1.6 m/s2
a = net
270 kg
m
Practice 4C, p. 145
1. Fk = 53 N
F
Fk
53 N
mk = k = 
=  = 0.23
Fn mg (24 kg)(9.81 m/s2)
m = 24 kg
g = 9.81 m/s2
2. m = 25 kg
Fs, max = 365 N
Fk = 327 N
2
g = 9.81 m/s
I Ch. 4–2
Fs,max Fs,max
365 N
 =  = 
a. ms = 
= 1.5
Fn
mg
(25 kg)(9.81 m/s2)
F
Fk
327 N
b. mk = k = 
=  = 1.3
Fn mg (25 kg)(9.81 m/s2)
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m = 4.0 kg
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Givens
Solutions
3. m = 145 kg
a. Fs,max = msFn = msmg = (0.61)(145 kg)(9.81 m/s2) = 8.7 × 102 N
ms = 0.61
Fk = mkFn = mkmg = (0.47)(145 kg)(9.81 m/s2) = 6.7 × 102 N
mk = 0.47
g = 9.81 m/s2
m = 15 kg
I
2
2
b. Fs,max = msFn = msmg = (0.74)(15 kg)(9.81 m/s ) = 1.1 × 10 N
ms = 0.74
Fk = mkFn = mkmg = (0.57)(15 kg)(9.81 m/s2) = 84 N
mk = 0.57
g = 9.81 m/s2
m = 250 kg
c. Fs,max = msFn = msmg = (0.4)(250 kg)(9.81 m/s2) = 1 × 103 N
ms = 0.4
Fk = mkFn = mkmg = (0.2)(250 kg)(9.81 m/s2) = 5 × 102 N
mk = 0.2
g = 9.81 m/s2
m = 0.55 kg
d. Fs,max = msFn = msmg = (0.9)(0.55 kg)(9.81 m/s2) = 5 N
ms = 0.9
Fk = mkFn = mkmg = (0.4)(0.55 kg)(9.81 m/s2) = 2 N
mk = 0.4
g = 9.81 m/s2
Practice 4D, p. 147
1. Fapplied = 185 N at 25.0°
above the horizontal
m = 35.0 kg
Fapplied, y = Fapplied (sin q)
Fy, net = ΣFy = Fn + Fapplied, y − Fg = 0
mk = 0.27
g = 9.81 m/s2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fapplied, x = Fapplied (cos q)
Fn = Fg − Fapplied, y = mg − Fapplied (sin q)
Fn = (35.0 kg)(9.81 m/s2) − (185 N)(sin 25.0°) = 343 N − 78.2 N = 265 N
Fk = mkFn = (0.27)(265 N) = 72 N
Fx, net = ΣFx = Fapplied, x − Fk = Fapplied(cos q) − Fk
Fx, net = (185 N)(cos 25.0°) − 72 N = 168 N − 72 N = 96 N
F
96 N
 =  = 2.7 m/s2
ax = x,net
m
35.0 kg
a = ax = 2.7 m/s2 in the positive x direction
2. q1 = 12.0°
Fg,y = mg(cos q1) = (35.0 kg)(9.81 m/s2)(cos 12.0°) = 336 N
q2 = 25.0°
Fg,x = mg(sin q1) = (35.0 kg)(9.81 m/s2)(sin 12.0°) = 71.4 N
Fapplied = 185 N
Fapplied,x = Fapplied (cos q2) = (185 N)(cos 25.0°) = 168 N
m = 35.0 kg
Fapplied,y = Fapplied (sin q2) = (185 N)(sin 25.0°) = 78.2 N
mk = 0.27
Fy,net = SFy = Fn + Fapplied,y − Fg,y = 0
2
g = 9.81 m/s
Fn = Fg,y − Fapplied,y = 336 N − 78.2 N = 258 N
Fk = mkFn = (0.27)(258 N) = 7.0 × 101 N
Fx,net = SFx = Fapplied,x − Fk − Fg,x = max
Fapplied,x − Fk − Fg,x
ax = 
m
Section One—Pupil’s Edition Solutions
I Ch. 4–3
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Givens
Solutions
168 N − 7.0 × 101 N − 71.4 N 2 7 N
ax =  =   = 0.77 m/s2
35 kg
35 N
a = ax = 0.77 m/s2 up the ramp
I
3. m = 75.0 kg
a. Fx,net = max = Fg,x − Fk
q = 25.0 °
Fg,x = mg(sin q)
2
Fk = Fg,x − max = mg(sin q) − max
2
Fk = (75.0 kg)(9.81 m/s2)(sin 25.0°) − (75.0 kg)(3.60 m/s2)
= 311 N − 2.70 × 102 N = 41 N
a = 3.60 m/s
g = 9.81 m/s
Fn = Fg,y = mg(cos q) = (75.0 kg)(9.81 m/s2)(cos 25.0°) = 667 N
F
41 N
mk = k =  = 0.061
Fn 667 N
m = 175 kg
b. Fx,net = Fg,x − Fk = mg(sin q) − mkFn
mk = 0.061
Fn = Fg,y = mg(cos q)
Fx,net = mg(sin q) − mkmg(cos q)
Fx,net = (175 kg)(9.81 m/s2)(sin 25.0°) − (0.061)(175 kg)(9.81 m/s2)(cos 25.0°)
Fx,net = 726 N − 95 N = 631 N
F
631 N
 =  = 3.61 m/s2
ax = x,net
m
175 kg
a = ax = 3.61 m/s2 down the ramp
4. Fg = 325 N
Fx,net = Fapplied,x − Fk = 0
Fapplied = 425 N
Fk = Fapplied,x = Fapplied (cos q) = (425 N)[cos(−35.2°)] = 347 N
q = −35.2°
Fy,net = Fn + Fapplied,y − Fg = 0
Fn = Fg − Fapplied,y = Fg − Fapplied (sin q)
Fn = 325 N − (425 N)[sin (−35.2°)] = 325 N + 245 N = 5.70 × 102 N
Section Review, p. 149
1
2. m = 2.26 kg
1
a. Fg = 6 mg = 6 (2.26 kg)(9.81 m/s2) = 3.70 N
g = 9.81 m/s2
3. m = 2.0 kg
b. Fg = (2.64)mg = (2.64)(2.26 kg)(9.81 m/s2) = 58.5 N
a. Fx,net = F(cos q) − mg(sin q) = 0
q = 60.0°
g = 9.81 m/s2
mg (sin q ) (2.0 kg)(9.81 m/s2)(sin 60.0°)
F =  =  = 34 N
cos q
cos 60.0°
b. Fy,net = Fn − F(sin q) − mg(cos q) = 0
Fn = F(sin q ) + mg(cos q) = (34 N)(sin 60.0°) + (2.0 kg)(9.81 m/s2)(cos 60.0°)
Fn = 29 N + 9.8 N = 39 N
4. m = 55 kg
Fs, max = 198 N
Fk = 175 N
2
g = 9.81 m/s
I Ch. 4–4
Fs,max Fs,max
198 N
 =  = 
ms = 
= 0.37
Fn
mg
(55 kg)(9.81 m/s2)
F
Fk
175 N
mk = k = 
= 
= 0.32
Fn mg (55 kg)(9.81 m/s2)
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
F
347 N
mk = k = 
= 0.609
Fn 5.70 × 102 N
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Review and Assess, pp. 151–157
Givens
Solutions
10. Fx,1 = 950 N
Fx,net = Fx,1 + Fx,2 = 950 N + (−1520 N) = −570 N
Fx,2 = −1520 N
Fy,1 = 5120 N
Fy,2 = −4050 N
Fy,net = Fy,1 + Fy,2 = 5120 N + (−4050 N) = 1070 N
2 (F
2
Fnet = (F
)2
+(10
N
)2
y,
57
0N
70
x,
ne
t)+
ne
t) = (−
Fnet = 3.
N2+
06
N2 = 1.
06
N2 = 1.21 × 103 N
2×105
1.1
4×1
46
×1
I
F net
1070 N
q = tan−1 y,
= tan−1  = −62°
Fx,net
−570 N
q = 62° above the 1520 N force
11. F1 + F2 = 334 N
−F1 + F2 = −106 N
b.
F1 + F2 = 334 N
+(−F1 + F2) = (−106 N)

2F2 = 228 N
F2 = 114 N
F1 + 114 N = 334 N
F1 = 220 N
F1 = 220 N right for the first situation and left for the second
F2 = 114 N right
12. F = 5 N
Fx = F(cos q) = (5 N)(cos 37°) = 4 N
q = 37°
Fy = F(sin q) = (5 N)(sin 37°) = 3 N
20. m = 24.3 kg
Fnet = 85.5 N
21. m = 1.5 × 107 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fnet = 7.5 × 105 N
vf = 85 km/h
vi = 0 km/h
22. F1 = 380 N
F
85.5 N
 =  = 3.52 m/s2
a = net
m
24.3 kg
F
7.5 × 105 N
 = 
a = net
= 5.0 × 10−2 m/s2
m
1.5 × 107 kg
(85 km/h − 0 km/h)(103 m/km)(1 h/3600 s)
vf − v
∆t = i = 
5.0 × 10−2 m/s2
a
∆t = 4.7 × 102 s
a. F1,x = F1(sin q1) = (380 N)(sin 30.0°) = 190 N
q1 = 30.0°
F1,y = F1(cos q1) = (380 N)(cos 30.0°) = 330 N
F2 = 450 N
F2,x = F2(sin q2) = (450 N)[sin (−10.0°)] = −78 N
q2 = −10.0°
F2,y = F2(cos q2) = (450 N)[cos (−10.0°)] = 440 N
Fy,net = F1,y + F2,y = 330 N + 440 N = 770 N
Fx,net = F1,x + F2,x = 190 N − 78 N = 110 N
2
Fnet = (F
)2 = (1
N
)2
+(77
)2
x,
(F
y,n
et
10
0N
ne
t)+
Fnet = 1.
04
N2+
05
N2 = 6.
05
N2 = 770 N
2×1
5.9
×1
0×1
F net
110 N
q = tan−1 x,
= tan−1  = 8.13° to the right of forward
Fy,net
770 N
m = 3200 kg
F
770 N
b. a =  =  = 0.241 m/s2
m 3200 kg
anet = 0.241 m/s2 at 8.13° to the right of forward
Section One—Pupil’s Edition Solutions
I Ch. 4–5
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Givens
Solutions
23. m = 3.00 kg
a. Because the ball is dropped, vi = 0 m/s
∆y = −176.4 m
Fw = 12.0 N
2
I
g = 9.81 m/s
1
∆y = − 2g∆t 2
∆t =
2∆y
(2)(−176.4 m)
−g = 
−9.81
m/s =
2
6.00 s
12.0 N
F
 =  = 4.00 m/s2
b. ax = w
3.00 kg
m
1
1
∆x = 2ax ∆t2 = 2(4.00 m/s2)(6.00 s)2 = 72.0 m
c. vy = −g∆t = −(9.81 m/s2)(6.00 s) = −58.9 m/s
vx = ax ∆t = (4.00 m/s2)(6.00 s) = 24.0 m/s
2
m/s
v = vx2+
v
4.
0m
/s
)2+(−58
.9
)2
y = (2
v = 57
m2/s2 = 63.6 m/s
6m
2/s2+347
0m
2/s2 = 40
50
a. Fg,x = mg(sin q) = (40.0 kg)(9.81 m/s2)(sin 18.5°) = 125 N
24. m = 40.0 kg
q = 18.5°
Fg,y = mg(cos q) = (40.0 kg)(9.81 m/s2)(cos 18.5°) = 372 N
Fapplied,x = 1.40 × 102 N
Fx,net = max = Fapplied,x − Fg,x = 1.40 × 102 N − 125 N = 15 N
∆x = 30.0 m
g = 9.81 m/s
F
15 N
 =  = 0.38 m/s2
ax- = x,net
m
40.0 kg
vi = 0 m/s
vf = vi 2+
ax ∆
m/s
)2
+(2)
2
x = (0
(0
.3
8m
/s
2)(3
0.
0m
)
2
vf = 4.78 m/s
∆x = 80.0 m
b. vf = vi 2
+2
ax ∆
m/s
)2
+(2)
x = (0
(0
.3
8m
/s
2)(8
0.
0m
)
vf = 7.80 m/s
F x = 12 N
∆t = 3.5 s
vi = 0 m/s
Fg,child = 85 N
g = 9.81 m/s2
27. m = 0.150 kg
F
12 N
a. a = x =  = 0.38 m/s2
m 32 kg
1
1
∆x = vi ∆t + 2a∆t 2 = (0 m/s)(3.5 s) + 2(0.38 m/s2)(3.25 s)2 = 2.3 m
85 N
Fg,child
b. mchild = 
 = 2 = 8.7 kg
g
9.81 m/s
Fx
12 N
12 N
a =  =  =  = 0.29 m/s2
m + mchild
32 kg + 8.7 kg 41 kg
∆x = vi ∆t + 12a∆t 2= (0 m/s)(3.5 s) + 12(0.29 m/s2)(3.5 s)2 = 1.8 m
a. Fnet = −mg = −(0.150 kg)(9.81 m/s2) = −1.47 N
vi = 20.0 m/s
g = 9.81 m/s2
I Ch. 4–6
b. same as part a
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
25. m = 32 kg
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Givens
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29. m = 5.5 kg
a. Fn = mg = (5.5 kg)(9.81 m/s2) = 54 N
g = 9.81 m/s2
q = 12°
b. Fn = mg(cos q) = (5.5 kg)(9.81 m/s2)(cos 12°) = 53 N
q = 25°
c. Fn = mg(cos q) = (5.5 kg)(9.81 m/s2)(cos 25°) = 49 N
q = 45°
d. Fn = mg(cos q) = (5.5 kg)(9.81 m/s2)(cos 45°) = 38 N
I
Fn = Fg = mg = (95 kg)(9.81 m/s2) = 930 N
37. m = 95 kg
Fs,max = 650 N
Fk = 560 N
g = 9.81 m/s2
Fs,max 650 N
ms =  =  = 0.70
Fn
930 N
F 560 N
mk =  k =  = 0.60
F n 930 N
Fn = mg(cos q)
38. q = 30.0°
2
mg(sin q) − Fk = max , where Fk = mkFn = mkmg(cos q)
2
mg(sin q) − mkmg(cos q) = max
a = 1.20 m/s
g = 9.81 m/s
sin q
ax
ax
mg(sin q) − ma
mk = x =  − 
= tan q − 
cos q
g(cos q)
g(cos q)
mg(cos q)
1.20 m/s2
mk = (tan 30.0°) − 
= 0.577 − 0.141 = 0.436
(9.81 m/s2)(cos 30.0°)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
39. m = 4.00 kg
Fapplied,x = Fapplied(cos q) = (85.0 N)(cos 55.0°) = 48.8 N
Fapplied = 85.0 N
Fapplied,y = Fapplied(sin q) = (85.0 N)(sin 55.0°) = 69.6 N
q = 55.0°
Fn = Fapplied,y −mg = 69.6 N − (4.00 kg)(9.81 m/s2) = 69.6 N − 39.2 N = 30.4 N
ax = 6.00 m/s2
max = Fapplied,x − Fk = Fapplied,x − mkFn
g = 9.81 m/s2
48.8 N − (4.00 kg)(6.00 m/s2)
Fapplied,x − max
 = 
mk = 
30.4 N
Fn
48.8 N − 24.0 N 24.8 N
mk =  =  = 0.816
30.4 N
30.4 N
40. m = 5.4 kg
Fn = mg(cos q) = (5.4 kg)(9.81 m/s2)(cos 15°) = 51 N
q = 15°
g = 9.81 m/s2
Section One—Pupil’s Edition Solutions
I Ch. 4–7
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Givens
Solutions
41. Fapplied = 185.0 N
Fapplied,x = Fapplied (cos q) = (185.0 N)(cos 25.0°) = 168 N
q = 25.0°
Fapplied,y = Fapplied (sin q) = (185.0 N)(sin 25.0°) = 78.2 N
m = 35.0 kg
Fy,net = Fn + Fapplied,y − mg = 0
mk = 0.450
Fn = mg − Fapplied,y = (35.0 kg)(9.81 m/s2) − 78.2 N = 343 N − 78.2 N = 265 N
g = 9.81 m/s2
Fk = mkFn = (0.450)(265 N) = 119 N
Fx,net = max = Fapplied,x − Fk
49 N
F
168 N − 119 N
 =  =  = 1.4 m/s2
ax = x,net
m
35.0 kg
35.0 kg
a = ax = 1.4 m/s2 down the aisle
42. Fg = 925 N
Fapplied,x = Fapplied (cos q) = (325 N)(cos 25.0°) = 295 N
Fapplied = 325 N
Fapplied,y = Fapplied (sin q) = (325 N)(sin 25.0°) = 137 N
q = 25.0°
Fy,net = Fn + Fapplied,y − Fg = 0
mk = 0.25
Fn = Fg − Fapplied,y = 925 N − 137 N = 788 N
2
g = 9.81 m/s
Fk = mkFn = (0.250)(788 N) = 197 N
Fx,net = max = Fapplied,x − Fk = 295 N − 197 N = 98 N
925 N
Fg
m =  = 2 = 94.3 kg
9.81 m/s
g
98 N
F
 =  = 1.0 m/s2
ax = x,net
m
94.3 kg
43. m = 2.00 kg
Fn = mg(cos q) = (2.00 kg)(9.81 m/s2)(cos 36.0°) = 15.9 N
q = 36.0°
44. m = 35 kg
ms = 0.300
g = 9.81 m/s2
45. m = 25 kg
a = 2.2 m/s2
46. m = 5.0 kg
q = 30.0°
g = 9.81 m/s2
Fs,max msmg
 =  = msg = (0.300)(9.81 m/s2) = 2.94 m/s2
b. atruck,max = 
m
m
atruck,max = 2.94 m/s2 forward
Fnet = ma = (25 kg)(2.2 m/s2) = 55 N
Fnet = 55 N to the right
Fx,net = Fx − mg(sin q) = F(cos q) − mg(sin q) = 0
(5.0 kg)(9.81 m/s2)(sin 30.0°)
mg(sin q)
F =  =  = 28 N
cos 30.0°
cos q
Fy,net = Fn − F(sin q) − mg(cos q) = 0
Fn = F(sin q) + mg(cos q) = (28 N)(sin 30.0°) + (5.0 kg)(9.81 m/s2)(cos 30.0°)
Fn = 14 N + 42 N = 56 N
I Ch. 4–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
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Givens
Solutions
47. m = 2.0 kg
Because vi = 0 m/s,
−1
∆x = 8.0 × 10
m
1
∆x = 2a∆t2
∆t = 0.50 s
2∆x (2)(0.80 m)
a = 2 = 
= 6.4 m/s2
∆t
(0.50 s)2
vi = 0 m/s
Fnet = ma= (2.0 kg)(6.4 m/s2) = 13 N
I
Fnet = 13 N down the incline
48. m = 2.26 kg
b. Fg = mg = (2.26 kg)(9.81 m/s2) = 22.2 N
∆y = −1.5 m
g = 9.81 m/s2
49. Fg = 5.30 N
a. Fg,M = (0.378)Fg = (0.378)(5.30 N) = 2.00 N
b. Fg,N = (1.14)Fg = (1.14)(5.30 N) = 6.04 N
FT − Fg = ma
50. m = 5.0 kg
2
a = 3.0 m/s
FT = ma + Fg = ma + mg
g = 9.81 m/s2
FT = (5.0 kg)(3.0 m/s2) + (5.0 kg)(9.81 m/s2) = 15 N + 49 N = 64 N
FT = 64 N upward
51. m = 3.46 kg
b. Fg = mg = (3.46 kg)(9.81 m/s2) = 33.9 N
g = 9.81 m/s2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
52. F1 = 2.10 × 103 N
a. Fnet = F1 + F2 = 2.10 × 103 N + (−1.80 × 103 N)
F2 = −1.80 × 103 N
Fnet = 3.02 × 102 N
m = 1200 kg
F
3.02 × 102 N
 =  = 0.250 m/s2
anet = net
m
1200 kg
anet = 0.250 m/s2 forward
∆t = 12 s
vi = 0 m/s
1
1
b. ∆x = vi ∆t + 2a∆t 2= (0 m/s)(12 s) + 2(0.25 m/s2)(12 s)2
∆x = 18 m
c. vf = a∆t + vi = (0.25 m/s2)(12 s) + 0 m/s
vf = 3.0 m/s
53. vi = 7.0 m/s
mk = 0.050
Fg = 645 N
g = 9.81 m/s2
vf = 0 m/s
Fk = mkFn = (0.050)(645 N) = 32 N
645 N
Fg
m =  = 2 = 65.7 kg
9.81 m/s
g
−32 N
−F
a = k =  = −0.49 m/s2
65.7 kg
m
2
2
(0 m/s)2 − (7.0 m/s)2
vf − vi
∆x = 
= 
(2)(−0.49 m/s2)
2a
1
∆x = 5.0 × 10 m
Section One—Pupil’s Edition Solutions
I Ch. 4–9
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Givens
Solutions
54. Fg = 319 N
a. Fapplied, x = Fapplied(cos q) = (485 N)[cos(−35°)] = 4.0 × 102 N
Fapplied = 485 N
Fapplied,y = Fapplied(sin q) = (485 N)[sin(−35°)] = −2.8 × 102 N
q = −35°
Fy,net = Fn + Fapplied,y − Fg = 0
mk = 0.57
Fn = Fg − Fapplied,y = 319 N − (−2.8 × 102 N) = 6.0 × 102 N
∆x = 4.00 m
Fk = mkFn = (0.57)(6.0 × 102 N) = 3.4 × 102 N
g = 9.81 m/s2
Fx,net = max = Fapplied,x − Fk = 4.0 × 102 N − 3.4 × 102 N = 6 × 101 N
vi = 0 m/s
319 N
Fg
m =  = 2 = 32.5 kg
9.81 m/s
g
Fx,net 6 × 101 N
ax =  =  = 2 m/s2
m
32.5 kg
1
∆x = vi ∆t + 2ax ∆t 2
Because vi = 0 m/s,
∆t =
2m
/s =
a = 
2∆x
(2)(4.00 m)
2
x
mk = 0.75
2s
b. Fk = mkFn = (0.75)(6.0 × 102 N) = 4.5 × 102 N
Fk > Fapplied,x ; the box will not move
55. m = 3.00 kg
1
a. ∆x = vi ∆t + 2a∆t 2
q = 30.0°
Because vi = 0 m/s,
∆x = 2.00 m
2∆x (2)(2.00 m)
2
a = 2 = 
= 1.78 m/s
∆t
(1.50 s)2
∆t = 1.50 s
g = 9.81 m/s2
vi = 0 m/s
b. Fg,x = mg(sin q) = (3.00 kg)(9.81 m/s2)(sin 30.0°) = 14.7 N
Fg,y = mg(cos q) = (3.00 kg)(9.81 m/s2)(cos 30.0°) = 25.5 N
Fn = Fg,y = 25.5 N
14.7 N − (3.00 kg)(1.78 m/s2)
Fg,x − ma
mk = x = 
25.5 N
Fn
14.7 N − 5.34 N
9.4 N
mk =  =  =
25.5 N
25.5 N
c. Fk = mkFn = (0.37)(25.5 N) =
d. vf2 = vi2+ 2ax∆x
0.37
9.4 N
vi 2+
ax ∆
m/s
)2
+(2)
vf = 2
x = (0
(1
.7
8m
/s
2)(2
.0
0m
) = 2.67 m/s
56. m = 75 kg
ma = FR − Fg = FR − mg
∆y = −25 m
FR − mg 95 N − (75 kg)(9.81 m/s2) 95 N − 740 N
 =  = 
a=
75 kg
m
75 kg
FR = 95 N
−640 N
a =  = −8.5 m/s2
75 kg
g = 9.81 m/s2
vf = vi 2
+2a∆
m/s
m) = ±21 m/s
y = (0
)2+(2)
(−
8.
5m
/s
2)(−
25
vi = 0 m/s
vf = −21 m/s
I Ch. 4–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
max = Fg,x − Fk = Fg,x − mkFn
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Givens
Solutions
57. Fg = 8820 N
Fg
8820 N
m =  = 2 = 899 kg
g 9.81 m/s
vi = 35 m/s
∆x = 1100 m
vf 2= vi 2 + 2a∆x = 0
g = 9.81 m/s2
−vi 2
−(35 m/s)2
a = 
=  = −0.56 m/s2
2∆x (2)(1100 m)
I
2
Fnet = ma = (899 kg)(−0.56 m/s2) = −5.0 × 10 N
58. mcar = 1250 kg
mtrailer = 3225 kg
2
a = 2.15 m/s
a. Fnet = mcara = (1250 kg)(2.15 m/s2) = 2690 N
Fnet = 2690 N forward
b. Fnet = mtrailera = (325 kg)(2.15 m/s2) = 699 N
Fnet = 699 N forward
Fs, max = msFn
59. m = 3.00 kg
q = 35.0°
mg(sin q) = ms[F + mg(cos q)]
ms = 0.300
2
g = 9.81 m/s
mg[sin q − ms(cos q)]
mg(sin q) − msmg(cos q)
F = 
= 
ms
ms
(3.00 kg)(9.81 m/s2)[sin 35.0° 0.300 (cos 35.0°)]
F = 
0.300
(3.00 kg)(9.81 m/s2)(0.574 0.246)
(3.00 kg)(9.81 m/s2)(0.328)
F =  
0.300
0.300
F=
60. m = 64.0 kg
32.2 N
At t = 0.00 s, v = 0.00 m/s. At t = 0.50 s, v = 0.100 m/s.
0.100 m/s − 0.00 m/s
vf − v
a = i =  = 0.20 m/s2
0.50 s
tf − ti
Copyright © by Holt, Rinehart and Winston. All rights reserved.
F = ma = (64.0 kg)(0.20 m/s2) = 13 N
At t = 0.50 s, v = 0.100 m/s. At t = 1.00 s, v = 0.200 m/s.
vf − vi 0.200 m/s − 0.100 m/s
a =  =  = 0.20 m/s2
tf − ti
0.50 s
F = ma = (64.0 kg)(0.20 m/s2) = 13 N
At t = 1.00 s, v = 0.200 m/s. At t = 1.50 s, v = 0.200 m/s.
a = 0 m/s2; therefore, F = 0 N
At t = 1.50 s, v = 0.200 m/s. At t = 2.00 s, v = 0.00 m/s.
vf − vi 0.00 m/s − 0.200 m/s
a =  =  = −0.40 m/s2
tf − ti
0.50 s
F = ma = (64.0 kg)(−0.40 m/s2) = −26 N
Section One—Pupil’s Edition Solutions
I Ch. 4–11
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Givens
Solutions
61. Fapplied = 3.00 × 102 N
Fnet = Fapplied − mg(sin q) = 0
Fg = 1.22 × 104 N
Fapplied
sin q = 
mg
m = 1.24 × 103 kg
3.00 × 102 N
Fapplied
 = sin−1 
q = sin−1 
(1.24 × 103 kg)(9.81 m/s2)
mg
g = 9.81 m/s2
62. Fg = 95.5 N
ms = 0.663
q = 0.141°
Fs,max = msFn
Fy,net = 2Fs,max + Fg = 2msFn − Fg = 0
Fg
95.5 N
Fn =  =  = 72.0 N
2ms (2)(0.663)
63. m1 = 2.00 kg
m2 = 3.00 kg
m3 = 4.00 kg
F = 180 N to the right
180 N
F
F
a. a =  =  = 
2.00
kg
+
3.00
kg + 4.00 kg
m m1 + m2 + m3
180 N
a =  = 2.0 × 101 m/s2
9.00 kg
a=
2.0 × 101 m/s2 to the right
b. F1 = m1a = (2.00 kg)(2.0 × 101 m/s2) = 4.0 × 101 N
F1 = 4.0 × 101 N to the right
F2 = m2a = (3.00 kg)(2.0 × 101 m/s2) = 6.0 × 101 N
F2 = 6.0 × 101 N to the right
F3 = m3a = (4.00 kg)(2.0 × 101 m/s2) = 8.0 × 101 N
F3 = 8.0 × 101 N to the right
c. m3 by m2: F32 = F3 =
8.0 × 101 N
m2 by m1: F21 = F2 + F32 = 6.0 × 101 N + 8.0 × 101 N
F21 = 1.40 × 102 N
m1 by m2: F12 = −1.40 × 102 N
I Ch. 4–12
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m2 by m3: F23 = −8.0 × 101 N
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Givens
Solutions
64. vi = 50.0 km/h
−F
−m mg
a. ax = k = k = −mkg = −(0.100)(9.81 m/s2) = −0.981 m/s2
m
m
mk = 0.100
g = 9.81 m/s2
vf 2 = vi 2+ 2a∆x = 0
−(50.0 km/h)2 (103 m/km)2(1 h/3600 s)2
−vi 2
∆x = 
= 
= 98.3 m
(2)(−0.981 m/s2)
2a
mk = 0.600
−(50.0 km/h)2 (103 m/km)2 (1 h/3600 s)2
−vi2
−vi2
b. ∆x = 
= 
= 
= 16.4 m
(2)(0.600)(−9.81 m/s2)
2(−mkg)
2a
65. m1 = 45.0 kg
a. Fs,1 = Fnms,1 = m1gms,1 = (45.0 kg)(9.81 m/s2)(0.60) = 260 N
m2 = 23.5 kg
Fs,2 = (m1 + m2)gms,2 = (45.0 kg + 23.5 kg)(9.81 m/s2)(0.30)
ms,1 = 0.60
Fs,2 = (68.5 kg)(9.81 m/s2)(0.30) = 2.0 × 102 N
ms,2 = 0.30
g = 9.81 m/s2
I
Fs,1 > Fs,2; thus, the slippage occurs between the 23.5 kg mass and the table.
b. (m1 + m2)ms,2 > m1ms,1
m ms,1
(45.0 kg)(0.60) (45.0 kg)(0 .60)
ms,2 > 1 
=  =  
m1 + m2 45.0 kg + 23.5 kg
68.5 kg
ms,2 > 0.39
66.
a. Apply Newton’s second law to find an expression for the acceleration of the truck.
ma = Ff = mkFn = mkmg
a = mkg
Because the acceleration of the truck does not depend on the mass of the truck,
the stopping distance will be ∆x regardless of the mass of the truck.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2
b. vf = vi
+2a∆
x = 0
−vi 2
a = 
2∆x
The acceleration will be the same regardless of the intial velocity.
a1 = a2 = mkg
−vi,12 −vi,22
 = 
2∆x
2∆x2
vi,22∆x
1
 where vi,2 = 2vi,1
∆x2 = 
vi,12
2
2vi,1 ∆x
1
∆x2 = 
=
vi,12
1

4
∆x
Section One—Pupil’s Edition Solutions
I Ch. 4–13
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Givens
Solutions
67. vi = 12.0 m/s
a. vf = vi + a∆t
∆t = 5.00 s
vf = 6.00 m/s
vf − vi 6.00 m/s − 12.0 m/s −6.0 m/s
a =  =  = 
∆t
5.00 s
5.00 s
a = −1.2 m/s2
I
b. Fk = −ma = m(1.2 m/s2)
F
m (1.2 m/s2)
mk = k = 
= 0.12
Fn m(9.81 m/s2)
1
1
c. ∆x = vi ∆t + 2 a∆t 2 = (12.0 m/s)(5.00 s) + 2 (−1.2 m/s2)(5.00 s)2
∆x = 60.0 m − 15 m = 45 m
68. Fg = 625 N
q = 37°
Fy,net = FT,1 (sin q) − Fg = 0
Fg
625 N
FT,1 =  =  = 1.0 × 103 N
sin q sin 37°
Fx,net = FT,1(cos q) − FT,2 = 0
Fg(cos q)
625 N (cos 37°)
FT,2 = FT,1(cos q) =  =  =
(sin q)
(sin 37°)
69. Fg,1 = 150 N
Fg,2 = 75 N
8.3 × 102 N
a. Fx,net = Fg,2 − Fs,max = 0
Fs,max = Fg,2 = 75 N
0.50
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fs,max Fg,2
75 N
 =  =  =
b. ms = 
Fg,1 150 N
Fn
I Ch. 4–14
Holt Physics Solution Manual
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Work and Energy
Chapter
5
I
Practice 5A, p. 170
Givens
1. Fnet = 5.00 × 103 N
Solutions
Wnet = Fnet d(cos q) = (5.00 × 103 N)(3.00 × 103 m)(cos 0°) = 1.50 × 107 J
d = 3.00 km
q = 0°
2. d = 2.00 m
Wnet = Fnet d(cos q) = (350 N)(2.00 m)(cos 0°) = 7.00 × 102 J
q = 0°
Fnet = 350 N
3. F = 35 N
W = Fd(cos q) = (35 N)(50.0 m)(cos 25°) = 1.6 × 103 J
q = 25°
d = 50.0 m
4. W = 2.0 J
m = 180 g
2.0 J
W
W
d =  =   = 
= 1.1 m
F(cos q) mg(cos q) (0.18 kg)(9.81 m/s2)(cos 0°)
g = 9.81 m/s2
q = 0°
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section Review, p. 171
5. Fg = 1.50 × 103 N
Fapplied = 345 N
q = 0°
a. W1 = Fappliedd(cos q) = (345 N)(24.0 m)(cos 0°) = 8.28 × 103 J
b. W2 = Fkd(cos q) = −Fg mkd(cos q)
W2 = −(1.50 × 103 N)(0.220)(24.0 m)(cos 0°) = −7.92 × 103 J
d = 24.0 m
mk = 0.220
6. m = 0.075 kg
c. Wnet = W1 + W2 = 8.28 × 103 J + (−7.92 × 103 J) = 3.6 × 102 J
Wnet = Fnet d(cos q) = (Fg − Fk)d(cos q ) = (mg − Fk)d(cos q )
Fk = 0.350 N
Wnet = [(0.075 kg)(9.81 m/s2) − 0.350 N](1.33 m)(cos 0°)
g = 9.81 m/s2
Wnet = (0.74 N − 0.350 N)(1.32 m) = (0.39 N)(1.33 m) = 0.519 J
d = 1.32 m
q = 0°
Practice 5B, p. 174
1. m = 8.0 × 104 kg
KE = 1.1 × 109 J
v=
2KE
 =
m
(2)(1.1 × 109 J)

= 1.7 × 102 m/s
8.0 × 104 kg
Section One—Pupil’s Edition Solutions
I Ch. 5–1
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Givens
Solutions
2. m = 0.145 kg
KE = 109 J
3. m1 = 3.0 g
I
v=
09 J)
= 38.8 m/s
2mKE = (
02.)1(41
5
kg
KE1 = 2 m1v12 = 2 (3.0 × 10−3 kg)(40.0 m/s)2 = 2.4 J
1
v1 = 40.0 m/s
m2 = 6.0 g
v2 = 40.0 m/s
4. m1 = 3.0 g
1
KE2 = 2 m2 v22 = 2 (6.0 × 10−3 kg)(40.0 m/s)2 = 4.8 J
1
1
KE2 4.8 J
2
 =  = 
KE1 2.4 J
1
KE1 = 2 m1v12 = 2 (3.0 × 10−3 kg)(40.0 m/s)2 = 2.4 J
1
v1 = 40.0 m/s
m2 = 3.0 g
v2 = 80.0 m/s
5. KE = 4.32 × 105 J
v = 23 m/s
1
KE2 = 2 m2 v2 2 = 2 (3.0 × 10−3 kg)(80.0 m/s)2 = 9.6 J
1
1
KE2 9.6 J
4
 =  = 
KE1 2.4 J
1
2 KE (2)(4.32 × 105 J)
m = 
= 
= 1.6 × 103 kg
v2
(23 m/s)2
Practice 5C, p. 176
1. Fnet = 45 N
Wnet = ∆KE = KEf − KEi
KEf = 352 J
Wnet = Fnet d(cos q)
KEi = 0 J
KEf − KEi
352 J − 0 J
352 J
d =  =  =  = 7.8 m
Fnet (cos q) (45 N)(cos 0°) 45 N
q = 0°
2. m = 2.0 × 103 kg
F1 = 1140 N
1
1
Wnet = ∆KE = KEf − KEi = 2 mvf 2 − 2 mvi 2
1
 m(v 2
f
2
− vi2)
(2.0 × 103 kg)[(2.0 m/s)2 − (0 m/s)2]
d =  = 
(2)(1140 N − 950 N)(cos 0°)
(F1 −F2)(cos q)
vf = 2.0 m/s
vi = 0 m/s
(2.0 × 103 kg)(4.0 m2/s2)
d =  = 21 m
(2)(190 N)
q = 0°
3. m = 2.1 × 103 kg
q = 20.0°
1
1
Wnet = ∆KE = KEf − KEi = 2 mvf 2 − 2 mvi 2
Wnet = Fnet d(cos q′)
3
Fnet = mg(sin q) − Fk
2
g = 9.81 m/s
[mg(sin q) − Fk]d(cos q ′) = 2 m(vf2 − vi2)
vf = 3.8 m/s
Simplify the equation by noting that
vi = 0 m/s
vi = 0 m/s and q ′ = 0.
q ′ = 0°
 mv 2
f
(2.1 × 103 kg)(3.8 m/s)2
2
d =  = 
3
mg(sin q) − Fk (2)[(2.1 × 10 kg)(9.81 m/s2)(sin 20.0°) − 4.0 × 103 N]
Fk = 4.0 × 10 N
1
1
(2.1 × 103 kg)(3.8 m/s)2
(2.1 × 103 kg)(3.8 m/s)2
d = 
= 
= 5.1 m
3
3
(2)(7.0 × 10 N − 4.0 × 10 N)
(2)(3.0 × 103 N)
I Ch. 5–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Wnet = Fnetd(cos q) = (F1 − F2)d(cos q)
F2 = 950 N
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Givens
Solutions
1
1
Wnet = ∆KE = KEf − KEi = 2 mvf 2 − 2 mvi2
4. m = 75 kg
d = 4.5 m
Wnet = Fnetd(cos q)
vf = 6.0 m/s
1
 m(v 2
f
2
− vi2)
(75 kg)[(6.0 m/s)2 − (0 m/s)2]
(75 kg)(36 m2/s2)
Fnet =  =  = 
(2)(4.5 m)(cos 0°)
(2)(4.5 m)
d(cos q)
vi = 0 m/s
q = 0°
I
2
Fnet = 3.0 × 10 N
5. m = 10.0 kg
a. Wg = Fg,x d = −mg(sin q )d
vi = 1.5 m/s
Wg = −(10.0 kg)(9.81 m/s2)(sin 15.0°)(7.5 m) = −1.9 × 102 J
Fapplied = 100.0 N
q = 15.0°
b. Wfriction = Fkd = Fn mk d = −mg(cos q)mkd
Wfriction = −(10.0 kg)(9.81 m/s2)(cos 15.0°)(0.40)(7.5 m) = −2.8 × 102 J
mk = 0.40
d = 7.5 m
2
g = 9.81 m/s
q ¢ = 0° = 0°
c. Wpuller = Fappliedd (cos q′) = (100.0 N)(7.5 m)(cos 0°) = 7.5 × 102 J
d. Wnet = ∆KE
Wnet = Wg + Wfriction + Wpuller = −1.9 × 102 J + (−2.8 × 102 J) + 7.5 × 102 J
Wnet = ∆KE = 2.8 × 102 J
1
1
e. ∆KE = KEf − KEi = 2 mvf 2 − 2 mvi 2
vf =
vf =
2∆KE
 + vi2 =
m
(2)(2.8 × 102 J)
 + (1.5 m/s)2
10.0 kg
2/s2 = 7.6 m/s = 7.6 m/s
56
m2/
s2
+2.2
m2/
s2 = 58
m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Practice 5D, p. 180
1. k = 5.2 N/m
x = 3.57 m − 2.45 m
= 1.12 m
2. k = 51.0 N/m
x = 0.150 m − 0.115 m
= 0.035 m
3. m = 40.0 kg
h1 = 2.00 m
g = 9.81 m/s2
1
1
PE = 2 kx 2 = 2 (5.2 N/m)(1.12 m)2 = 3.3 J
PE = 2 kx 2 = 2 (51.0 N/m)(0.035 m)2 = 3.1 × 10−2 J
1
1
a. PE1 = mgh1 = (40.0 kg)(9.81 m/s2)(2.00 m) = 785 J
b. h2 = (2.00 m)(1 − cos 30.0°) = (2.00 m)(1 − 0.866)
h2 = (2.00 m)(0.134) = 0.268 m
PE2 = mgh2
PE2 = (40.0 kg)(9.81 m/s2)(0.268 m) = 105 J
c. h3 = 2.00 m − 2.00 m = 0.00 m
PE3 = mgh3
PE3 = (40.0 kg)(9.81 m/s2)(0.00 m) = 0.00 J
Section One—Pupil’s Edition Solutions
I Ch. 5–3
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Section Review, p. 180
Givens
Solutions
3. v = 42 cm/s
KE = 2 mv 2 = 2 (50.0 × 10−3 kg)(0.42 m/s)2 = 4.4 × 10−3 J
1
m = 50.0 g
I
4. m = 0.75 kg
1
1
1
Wnet = ∆KE = KEf − KEi = 2 mvf2 − 2 mvi2
d = 1.2 m
Wnet = Fnetd(cos q) = Fkd (cos q) = mkmgd(cos q)
mk = 0.34
vf = 0 m/s
1
mkmgd(cos q) = 2 m(vf2 − vi2)
g = 9.81 m/s2
q = 180°
5. h = 21.0 cm
2
g = 9.81 m/s
vi =
vf 2−
2m
kgd(
co
sq) =
(0
m/s
)2
−(2)
(0
.3
4)
(9
.8
1m
/s
2)(1
.2
m)(
co
s18
0°
)
vi =
)(
(2
0.
34
)(
9.
81
m/s
2)(1
.2
m) = 2.8 m/s
PEg = mgh = (30.0 × 10−3 kg)(9.81 m/s2)(0.210 m) = 6.18 × 10−2 J
m = 30.0 g
Practice 5E, p. 185
1. vi = 18.0 m/s
PEi + KEi = KEf
1
1
mghi + 2 mvi 2 = 2 mvf 2
m = 2.00 kg
hi = 5.40 m
g = 9.81 m/s2
2
vf = 2g
m/s
m/s
)2
hi +v
)(
9.
81
2)(5
.4
0m
)+(18
.0
i = (2
2/s2 + 324 m2/s2 = 4.30 × 102 m2/s2 = 20.7 m/s = 20.7 m/s
vf = 10
6m
PEi = PEf + KEf
2. Fg = 755 N
hi = 10.0 m
1
mghi = mghf + 2 mvf 2
2
g = 9.81 m/s
2)(10.0 m) − (2)(9.81 m/s2)(5.00 m)
vf = 2g
m/s
hi −2gh
f = (2
)(
9.
81
2/s2 − 98.1 m2/s2 = 98 m2/s2 = 9.90 m/s
vf = 19
6m
diver speed at 5 m = vf = 9.90 m/s
hf = 0 m
PEi = KEf
1
mghi = 2 mvf 2
vf = 2g
m/s
hi = (2
)(
9.
81
2)(1
0.
0m
) = 14.0 m/s
diver speed at 0 m = vf = 14.0 m/s
I Ch. 5–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
hf = 5.00 m
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Givens
Solutions
3. hi = 10.0 m
PEi + KEi = KEf
g = 9.81 m/s2
vi = 2.00 m/s
1
1
mghi + 2 mvi 2 = 2 mvf 2
2
2)(10.0 m) + (2.00 m/s)2
vf = 2g
m/s
hi +v
)(
9.
81
i = (2
2/s2 + 4.00 m2/s2 = 2.00 × 102 m2/s2 = 14.1 m/s = 14.1 m/s
vf = 19
6m
4. vi = 2.2 m/s
I
KEi = PEf
2
g = 9.81 m/s
1
 mv 2
i
2
= mghf
v2
(2.2 m/s)2
hf = i = 
= 0.25 m
2g (2)(9.81 m/s2)
PEi = KEf
5. vf = 1.9 m/s
g = 9.81 m/s2
1
mghi = 2 mvf2
vf2
(1.9 m/s)2
hi =  = 
= 0.18 m
2g (2)(9.81 m/s2)
Section Review, p. 186
1. x = −8.00 cm
k = 80.0 N/m
2
g = 9.81 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m = 50.0 g
Taking the compressed spring as the zero level, h = −x = 8.00 cm.
PEelastic,i = PEg,f + KEf
1
 kx 2
2
= mgh + 2 mvf 2
vf =
(80.0 N/m)(−0.0800 m)
 −
(2)(9.81m/s)(0.0800m)
kmx−2gh = (50.0 × 10 kg)
vf =
1
2
2
−3
2
2/s2 = 2.93 m/s
10
m2/
s2
−1.5
s2 = 8.
.2
7m
2/
6m
Practice 5F, p. 189
1. melevator = 1.0 × 103 kg
m = melevator + mload = 1.0 × 103 kg + 800.0 kg = 1.8 × 103 kg
mload = 800.0 kg
P = Fv = (Fg + Fk )v = (mg + Fk )v
Fk = 4.0 × 103 N
P = [(1.8 × 103 kg)(9.81 m/s2) + 4.0 × 103 N](3.00 m/s)
v = 3.00 m/s
P = (1.8 × 104 N + 4.0 × 103 N)(3.00 m/s)
2
g = 9.81 m/s
P = (2.2 × 104 N)(3.00 m/s) = 6.6 × 104 W = 66 kW
Section One—Pupil’s Edition Solutions
I Ch. 5–5
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Givens
Solutions
2. m = 1.50 × 103 kg
vf = 18.0 m/s
vi = 0 m/s
I
∆t = 12.0 s
Fr = 400.0 N
For vi = 0 m/s,
1
∆x = 2 vf ∆t
vf
a = 
∆t
mvf
1
 + Fr 2 vf ∆t
∆t
W F∆x (ma + Fr)∆x
P =  =  =  = 
∆t
∆t
∆t
∆t
(1.50 × 103 kg)(18.0 m/s)
1
 + 400.0 N 2(18.0 m/s)(12.0 s)
12.0 s
P = 
12.0 s
(2250 N + 400.0 N)(18.0 m/s)(12.0 s)
P = 
(2)(12.0 s)
(2650 N)(18.0 m/s)
P =  = 2.38 × 104 W, or 23.8 kW
2
3. m = 2.66 × 107 kg
P = 2.00 kW
d = 2.00 km
g = 9.81 m/s
4. P = 19 kW
W = 6.8 × 107 J
W mgd (2.66 × 107 kg)(9.81 m/s2)(2.00 × 103 m)
∆t =  =  = 
= 2.61 × 108 s
2.00 × 103 W
P
P
or (2.61 × 108 s)(1 h/3600 s)(1 day/24 h)(1 year/365.25 days) = 8.27 years
W 6.8 × 107 J
∆t =  = 
= 3.6 × 103 s
P 19 × 103 W
or (3.6 × 103 s)(1 h/3600 s) = 1.0 h
vf = 10.0 m/s
1
1
1
a. Wnet = ∆KE = KEf − KEi = 2 mvf 2 − 2 mvi 2 = 2 m (vf2 − vi 2)
1
Wnet = 2 (1.50 × 103 kg)[(10.0 m/s)2 − (0 m/s)2]
vi = 0 m/s
W = Wnet = 7.50 × 104 J
∆t = 3.00 s
W 7.50 × 104 J
b. P =  =  = 2.50 × 104 W or 25.0 kW
∆t
3.00 s
Section Review, p. 189
h = 5.00 m
W mgh (50.0 kg)(9.81 m/s2)(5.00 m)
∆t =  =  =  = 12.3 s
200.0 W
P
P
P = 200.0 W
W = mgh = (50.0 kg)(9.81 m/s2)(5.00 m) = 2.45 × 103 J
2. m = 50.0 kg
g = 9.81 m/s2
3. m = 50.0 kg
h = 5.00 m
v = 1.25 m/s
P = Fv = mgv = (50.0 kg)(9.81 m/s2)(1.25 m/s) = 613 W
W = mgh = (50.0 kg)(9.81 m/s2)(5.00 m) = 2.45 × 103 J
g = 9.81 m/s
I Ch. 5–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. m = 1.50 × 103 kg
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Chapter Review and Assess, pp. 193–199
Givens
Solutions
7. m = 4.5 kg
Wperson = Fg [d1(cos q1) + d2(cos q2 )]
d1 = 1.2 m
Wperson = mg[d1(cos 0°) + d2(cos 90°)] = mgd1
d2 = 7.3 m
Wperson = (4.5 kg)(9.81 m/s2)(1.2 m) = 53 J
2
g = 9.81 m/s
I
Wgravity = −Fg [d1(cos q1) + d2(cos q2 )]
q1 = 0°
Wgravity = −mg[d1(cos 0°) + d2(cos 90°)] = −mgd1
q2 = 90°
Wgravity = −(4.5 kg)(9.81 m/s2)(1.2 m) = −53 J
8. m = 8.0 × 103 kg
anet = 1.0 m/s2
Wnet = Fnetd(cos q) = manetd(cos q) = (8.0 × 103 kg)(1.0 m/s2)(30.0 m)(cos 0°)
W = 2.4 × 105 J
d = 30.0 m
q = 0°
W = Fd (cos q) = (475 N)(0.100 m)(cos 0°) = 47.5 J
9. F = 475 N
q = 0°
d = 10.0 cm
a. W1 = Fapplied d(cos q ) = (40.0 N)(253 m)(cos 52.0°) = 6.23 × 103 J
10. Fg = 70.0 N
Fapplied = 40.0 N
d = 253 m
b. Because the bag’s speed is constant, Fk = −Fapplied (cos q).
W2 = Fk d = −Fapplied (cos q)d = −(40.0 N)(cos 52.0°)(253 m) = −6.23 × 103 J
q = 52.0°
−Fapplied (cos q) −(40.0 N)(cos 52.0°)
F
c. mk = k =  =  = 0.352
Fn
−Fg
−70.0 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
14. v1 = 5.0 m/s
1
v2 = 25.0 m/s
1
1
20. m = 0.55 g
4
KE = 7.6 × 10 J
d = 5.0 m
g = 9.81 m/s2
KEf = 0 J
q = 180°
1
KE = 2 mv 2 = 2 (1250 kg)(11 m/s)2 = 7.6 × 104 J
v = 11 m/s
Fr = 1500 N
KE2 = 2 mv2 2
2

1
KE1 2 mv1
v 2 (5.0 m/s)2
25
 =  = 12 = 2 =  = 
25
KE2 1 mv 2 v2
(25.0 m/s)
625
2
2
19. m = 1250 kg
21. m = 50.0 kg
1
KE1 = 2 mv12
v=
2KE
 =
m
(2)(7.6 × 104 J)

= 1.7 × 104 m/s
0.55 × 10−3 kg
Wnet = ∆KE = KEf − KEi = −KEi
The diver’s kinetic energy at the water’s surface equals the gravitational potential energy associated with the diver on the diving board relative to the water’s surface.
KEi = PEg = mgh
Wnet = Fnetd(cos q) = Frd(cos q)
Frd(cos q)
(1500 N)(5.0 m)(cos 180°)
h =  = 
= 15 m
−mg
−(50.0 kg)(9.81 m/s2)
total distance = h + d = 15 m + 5.0 m = 2.0 × 101 m
Section One—Pupil’s Edition Solutions
I Ch. 5–7
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Givens
Solutions
22. vi = 4.0 m/s
Wnet = ∆KE = KEf − KEi = 2 mvf2 − 2 mvi2
vf = 0 m/s
q = 25°
I
m = 20.0 kg
mk = 0.20
g = 9.81 m/s2
q ′ = 180°
1
1
Wnet = Fnetd(cos q ′)
Fnet = mg(sin q) + Fk = mg(sin q) + mkmg(cos q)
Wnet = mg[sin q + mk(cos q)]d(cos q′)
1
m(v 2 − v 2)
vf 2 + vi2
f
i
2
d =  = 
2 g[sin q + mk(cos q)](cos q′)
mg[sin q + mk(cos q)](cos q ′)
(0 m/s)2 − (4.0 m/s)2
d = 
2
(2)(9.81 m/s )[sin 25° + (0.20)(cos 25°)](cos 180°)
−16 m2/s2
8.0 m2/s2
d = 
= 
= 1.4 m
2
−(2)(9.81 m/s )(0.42 + 0.18)
(9.81 m/s2)(0.60)
23. hA = 10.0 m
m = 55 kg
hB = 0 m
hA = 0 m
a. PEA = mghA = (55 kg)(9.81 m/s2)(10.0 m) = 5.4 × 103 J
PEB = mghB = (55 kg)(9.81 m/s2)(0 m) = 0 J
∆PE = PEA − PEB = 5400 J − 0 J = 5.4 × 103 J
b. PEA = mghA = (55 kg)(9.81 m/s2)(0 m) = 0 J
hB = −10.0 m
PEB = mghB = (55 kg)(9.81 m/s2)(−10.0 m) = −5.4 × 103 J
∆PE = PEA − PEB = 0 − (−5400 J) = 5.4 × 103 J
hA = 5.0 m
hB = −5.0 m
c. PEA = mghA = (55 kg)(9.81 m/s2)(5.0 m) = 2.7 × 103 J
PEB = mghB = (55 kg)(9.81 m/s2)(−5.0 m) = −2.7 × 103 J
24. m = 2.00 kg
h1 = 1.00 m
h2 = 3.00 m
h3 = 0 m
a. PE = mg(−h1 ) = (2.00 kg)(9.81 m/s2)(−1.00 m) = −19.6 J
b. PE = mg(h2 − h1 ) = (2.00 kg)(9.81 m/s2)(3.00 m − 1.00 m)
PE = (2.00 kg)(9.81 m/s2)(2.00 m) = 39.2 J
c. PE = mgh3 = (2.00 kg)(9.81 m/s2)(0 m) = 0 J
25. k = 500.0 N/m
x1 = 4.00 cm
x2 = −3.00 cm
x3 = 0 cm
I Ch. 5–8
1
1
1
1
1
1
a. PE = 2 kx12 = 2 (500.0 N/m)(0.0400 m)2 = 0.400 J
b. PE = 2 kx22 = 2 (500.0 N/m)(−0.0300 m)2 = 0.225 J
c. PE = 2 kx32 = 2 (500.0 N/m)(0 m)2 = 0 J
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆PE = PEA − PEB = 2700 J − (−2700 J) = 5.4 × 103 J
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Givens
Solutions
33. m = 50.0 kg
PEi = KEf
h = 7.34 m
1
mgh = 2 mvf 2
g = 9.81 m/s2
vf =
34. h = 30.0 m
h =
2g
)(
(2
9.
81
m/s
2)(7
.3
4m
) = 12.0 m/s
I
a. PEi = KEf
vi = 0 m/s
1
mghi = 2 mvf 2
q = 37.0°
2)(30.0 m)(1 − cos 37.0°)
vf = 2g
m/s
hi = 2g
h(1
−cosq) = (2
)(
9.
81
vf = (2
m/s
)(
9.
81
2)(3
0.
0m
)(
1−0.7
99
)
2)(30.0 m)(0.201) = 10.9 m/s
vf = (2
m/s
)(
9.
81
b. PEi + KEi = KEf
vi = 4.00 m/s
1
1
mghi + 2 mvi 2 = 2 mvf 2
2
2)(30.0 m)(0.201) + (4.00 m/s)2
vf = 2g
m/s
hi+v
)(
9.
81
i = (2
2/s2 + 16.0 m2/s2 = 134 m2/s2 = 11.6 m/s
vf = 11
8m
35. P = 50.0 hp
5
W = 6.40 × 10 J
W
6.40 × 105 J
∆t =  =  = 17.2 s
P (50.0 hp)(746 W/hp)
1 hp = 746 W
m
36. rate of flow = 
∆t
= 1.2 × 106 kg/s
d = 50.0 m
W Fd mgd m
P =  =  =  =  gd
∆t ∆t ∆t ∆t
P = (1.2 × 106 kg/s)(9.81 m/s2)(50.0 m) = 5.90 × 108 W
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
37. m = 215 g
hA = 30.0 cm
hB = 0 cm
2
hC = 3 (30.0 cm)
g = 9.81 m/s2
a. PEA = mghA = (215 × 10−3 kg)(9.81 m/s2)(0.300 m) = 0.633 J
b. KEB = PEA = 0.633 J
1
c. KEB = 2 mv 2
v=
 =

m = 215 × 10 kg
2KEB
(2)(0.633 J)
−3
2.43 m/s
d. PEC = mghC = (215 × 10−3 kg)(9.81 m/s2)(3)(0.300 m) = 0.422 J
2
KEC = KEB − PEC = 0.633 J − 0.422 J = 0.211 J
Section One—Pupil’s Edition Solutions
I Ch. 5–9
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Givens
Solutions
38. Fg = 700.0 N
d = 25.0 cm
Fupward = (2)(355 N)
I
KEi = 0 J
1
1 Fg
Wnet = ∆KE = KEf − KEi = KEf = 2 mvf2 = 2  vf 2
g
Wnet = Fnetd(cos q) = (Fupward − Fg)d(cos q)
Fg
g vf 2 = (Fupward − Fg)d(cos q)
g = 9.81 m/s2
1

2
q = 0°

F
(2)(9.81 m/s )[(2)(355 N) − 700.0 N](0.250 m)(cos 0°)

v = 700.0 N
(2)(9.81 m/s )(7.10 × 10 N − 700.0 N)(0.250 m)

v = 700.0 N
(2)(9.81 m/s )(1.0 × 10 N)(0.250 m)
 = 0.265 m/s
v = 700.0 N
2g(Fupward − Fg)d(cos q)
vf =
g
2
f
2
2
f
2
1
f
KEi = PEf + KEf
39. m = 50.0 kg
vi = 10.0 m/s
vf = 1.0 m/s
2
g = 9.81 m/s
1
 mv 2
i
2
1
= mgh + 2 mvf 2
vi2 − vf 2 (10.0 m/s)2 − (1.0 m/s)2 1.00 × 102 m2/s2 − 1.0 m2/s2
h =  = 
= 
(2)(9.81 m/s2)
2g
(2)(9.81 m/s2)
99 m2/s2
= 5.0 m
h = 
(2)(9.81 m/s2)
Wnet = ∆KE
d = 20.0 m
Wnet = Fnet d (cos q′) = Fnet d
q = 30.0°
∆KE = Fnet d = (Fapplied − Fk − Fg,d)d = [Fapplied − mk Fg (cos q) − Fg (sin q)]d
Fapplied = 115 N
∆KE = [115 N − (0.22)(80.0 N)(cos 30.0°) − (80.0 N)(sin 30.0°)](20.0 m)
mk = 0.22
∆KE = (115 N − 15 N − 40.0 N)(20.0 m) = (6.0 × 101 N)(20.0 m)
g = 9.81 m/s2
∆KE = 1.2 × 103 J
q ′ = 0°
41. Fg = 98.0 N
d = 12.0 m
Fapplied = 40.0 N
vi = 0 m/s
g = 9.81 m/s2
q = 0°
Wnet = Fnetd(cos q) = Fappliedd(cos q)
1

2
g(v
Fg
2
2
f − vi ) = Fappliedd(cos q)
For vi = 0 m/s and q = 0°,
vf =
I Ch. 5–10
1 Fg
1 Fg
Wnet = ∆KE = KEf − KEi = 2  vf 2 − 2  vi 2
g
g
(2)(9.81 m/s2)(40.0 N)(12.0 m)
 = 9.80 m/s
98.0 N
2gFappliedd

=
Fg
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
40. Fg = 80.0 N
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Givens
Solutions
42. mtot = 130.0 kg
For the first half of the swing,
h1 = 5.0 m
PEi = KEf
q = 30.0°
mtotgh = 2 m totvf 2
1
mJ = 50.0 kg
g = 9.81 m/s2
vf = 2g
q ) = (2
m/s
m)(
h = 2g
h
−sin
)(
9.
81
2)(5
.0
1−sin
30.
0°
)
1(1
vf = (2
m/s
m)(
m/s
m)(
)(
9.
81
2)(5
.0
1−0.5
00
) = (2
)(
9.
81
2)(5
.0
0.
50
0)
I
vf = 7.0 m/s
For the second half of the swing,
1
1
1
KET = 2 m T vf 2 = 2 (mtot − mJ)vf 2 = 2 (130.0 kg − 50.0 kg)(7.0 m/s)2
1
KET = 2 (80.0 kg)(7.0 m/s)2 = 2.0 × 103 J
KET = PEf = mT ghf
KET
2.0 × 103 J
hf = 
= 
= 2.5 m
mT g (80.0 kg)(9.81 m/s2)
43. m = 0.250 kg
3
k = 5.00 × 10 N/m
x = −0.100 m
2
g = 9.81 m/s
45. m = 0.60 kg
vA = 2.0 m/s
KEB = 7.5 J
PEelastic,i = PEg,f
1 2
kx
2
= mgh
3
2
k x 2 (5.00 × 10 N/m)(−0.100 m)
h =  = 
= 10.2 m
2
(2)(0.250 kg)(9.81 m/s )
2mg
1
1
a. KEA = 2 mvA2 = 2 (0.60 kg)(2.0 m/s)2 = 1.2 J
1
b. KEB = 2 mvB2
vB =

m = 
0.60
kg =
2KEB
(2)(7.5 J)
5.0 m/s
c. Wnet = ∆KE = KEB − KEA = 7.5 J − 1.2 J = 6.3 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
46. m = 5.0 kg
d = 2.5 m
q = 30.0°
∆t = 2.0 s
g = 9.81 m/s2
q ′ = 0°
vi = 0 m/s
a. Wgravity = Fg,dd(cos q ′) = Fg,dd = mg(sin q)d
Wgravity = (5.0 kg)(9.81 m/s2)(sin 30.0°)(2.5 m) = 61 J
1
b. Wfriction = ∆ME = KEf − PEi = 2 mvf 2 − mghi
1
d = ∆x = 2 (vi + vf)∆t
2d
2d
vf =  − vi = 
∆t
∆t
hi = d(sin q)
2
2d
1
∆ME = 2 m  − mgd(sin q)
∆t
2
(2)(2.5 m)
1
∆ME = 2 (5.0 kg)  − (5.0 kg)(9.81 m/s2)(2.5 m)(sin 30.0°)
2.0 s
∆ME = Wfriction = 16 J − 61 J = −45 J
q ′ = 90°
c. Wnormal = Fnd(cos q′)mg(cos q)d(cos q ′)
cos q ′ = cos 90° = 0, so
Wnormal = 0 J
Section One—Pupil’s Edition Solutions
I Ch. 5–11
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Givens
Solutions
47. m = 70.0 kg
a. ∆ME = KEf − KEi
mk = 0.70
g = 9.81 m/s2
I
q = 180.0°
vi = 4.0 m/s
vf = 0 m/s
1
1
1
∆ME = 2 mvf 2 − 2 mvi2 = 2 m(vf 2 − vi2)
1
∆ME = 2 (70.0 kg)[(0 m/s)2 − (4.0 m/s)2] = −5.6 × 102 J
b. W = ∆ME = − 5.6 × 102 J
W
W
W
d =  =  = 
Fk(cos q) Fnmk(cos q) mgmk(cos q)
−5.6 × 102 J
d = 
= 1.2 m
(70.0 kg)(9.81 m/s2)(0.70)(cos 180°)
48. Fapplied = 150 N
m = 40.0 kg
d = 6.00 m
q = 0°
g = 9.81 m/s2
49. m = 5.00 g
vi = 600.0 m/s
vf = 0 m/s
d = 4.00 cm
q = 180°
a. W = Fapplied d(cos q) = (150 N)(6.00 m)(cos 0°) = 9.00 × 102 J
b. Fnet = Fapplied − Fk = 0
Fapplied = Fk = mkFn
Fapplied Fapplied
150 N
mk =  =  = 
= 0.382
Fn
mg
(40.0 kg)(9.81 m/s2)
1
1
a. Wnet = ∆KE = KEf − KEi = 2 mvf2 − 2 mvi2
Wnet = Fkd(cos q)
−3
2
2

− vi2)
2 (5.00 × 10 kg)[(0 m/s) − (600.0 m/s) ]
Fk = 
= 
−2
d(cos q)
(4.00 × 10 m)(cos 180°)
1
m(v 2
f
2
1
(5.00 × 10−3 kg)(−3.600 × 105 m2/s2)
Fk = 
(2)(−4.00 × 10 − 2 m)
1
b. ∆x = 2 (vi + vf)∆t = d
2d
(2)(4.00 × 10−2 m)
∆t =  =  = 1.33 × 10−4 s
vi + vf 600.0 m/s + 0 m/s
50. m = 70.0 kg
q = 35°
W = Fd(cos q ′) = Fd(cos 0°) = Fd = mg(sin q)d
W = (70.0 kg)(9.81 m/s2)(sin 35∞)(60.0 m) = 2.4 × 104 J
q′ = 0°
d = 60.0 m
g = 9.81 m/s2
I Ch. 5–12
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fk = 2.25 × 104 N
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Givens
Solutions
51. m = 2.50 × 103 kg
a. Wnet = ∆KE = KEf − KEi
Wnet = 5.0 kJ
d = 25.0 m
vi = 0 m/s
q = 0°
52. h1 = 50.0 m
h2 = 10.0 m
g = 9.81 m/s2
q = 45.0°
vi = 0 m/s
1
Because vi = 0 m/s, KEi = 0 J, and Wnet = KEf = 2 mvf 2.
vf =
(2)(5.0 × 103 J)

= 2.0 m/s
2.50 × 103 kg
2Wnet
 =
m
I
W et
5.0 × 103 J
b. Fnet = n
=  = 2.0 × 102 N
d(cos q) (25.0 m)(cos 0∞)
a. PEi = PEf + KEf
1
mgh1 = mgh2 + 2 mvf 2
vf = 2g
m/s
h1−2gh
2 = 2g
(h
1−h
)(
9.
81
2)(5
0.
0m
−10.
0m
)
2) = (2
vf = (2
m/s
)(
9.
81
2)(4
0.
0m
) = 28.0 m/s
b. At the acrobat’s highest point, vy = 0 m/s and vx = (28.0 m/s)(cos 45.0°).
PEi = PEf + KEf
1
mgh1 = mghf + 2 mvf,x2
1
1
gh1 − 2 vf,x2
(9.81 m/s2)(50.0 m) − 2[(28.0 m/s)(cos 45.0°)]2
hf = 
= 
g
9.81 m/s2
4.90 × 102 m2/s2 − 196 m2/s2 294 m2/s2
hf = 
= 2 = 30.0 m above the ground
9.81 m/s2
9.81 m/s
53. m = 75 g
MEi = 600 mJ
a. At ∆t = 6.0 s, MEf = 500 mJ
∆ME = MEf − MEi = 500 mJ − 600 mJ = −1.0 × 102 mJ
g = 9.81 m/s2
b. At ∆t = 4.5 s, KE = 350 mJ
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
KE = 2 mv 2
v=
−3
)7(535×01×0
10kgJ) = 3.1 m/s
2mKE = (2
−3
c. PEmax = 600 mJ = mghmax
PE ax
600 × 10−3 J
hmax = m
= 
= 0.82 m
mg
(75 × 10−3 kg)(9.81 m/s2)
54. q = 10.5°
d1 = 200.0 m
mk = 0.075
g = 9.81 m/s2
KE1,i = 0 J
KE2,f = 0 J
For downhill slide,
Wnet,1 = ∆KE1 = KE1,f − KE1,i = KE1,f
Wnet,1 = Fnet,1d1(cos q′)
Fnet,1 = mg(sin q) − Fk = mg(sin q − mkmg(cos q)
Because Fnet,1 is parallel to and in the forward direction to d1, q′ = 0°, and
Wnet,1 = mgd1[sin q − mk(cos q)]
Section One—Pupil’s Edition Solutions
I Ch. 5–13
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Givens
Solutions
For horizontal slide,
Wnet,2 = ∆KE2 = KE2,f − KE2,i = −KE2,i
Wnet,2 = Fnet,2d2(cos q′) = Fkd2(cos q′) = mkmgd2(cos q′)
I
Because Fnet,2 is parallel to and in the backward direction to d2, q′ = 180°,
and Wnet,2 = −mkmgd2
For the entire ride,
mgd1[sin q − mk(cos q)] = KE1,f − mkmgd2 = −KE2,i
Because KE1,f = KE2,i, mgd1[sin q − mk(cos q)] = mkmgd2
d1[sin q − mk(cos q)]
d2 = 
mk
(200.0 m)[(sin 10.5 °) − (0.075)(cos 10.5°)]
(200.0 m)(0.182 − 0.074)
d2 =  = 
0.075
0.075
(200.0 m)(0.108)
d2 =  = 2.88 × 102 m
0.075
55. m = 10.0 kg
d1 = 3.00 m
q = 30.0°
d2 = 5.00 m
vi = 0
g = 9.81 m/s2
KEf = 0 J
a. For slide down ramp,
PEi = KEf
1
mgh = 2 mvf2
vf =
h =
2g
2g
d1(s
in
q)
vf =
)(
(2
9.
81
m/s
2)(3
.0
0m
)(
si
n30.
0°
) = 5.42 m/s
b. For horizontal slide across floor,
Wnet = ∆KE = KEf − KEi = −KEi
Wnet = Fkd2(cos q′)
= mkmgd2(cos q′)
KEi of horizontal slide equals KEf = PEi of slide down ramp.
−PEi = −mgd1(sin q) = mkmgd2(cos q′)
−d1(sin q) −(3.00 m)(sin 30.0°)
 =  = 0.300
mk = 
d2(cos q ′)
(5.00 m)(cos 180°)
KEf = 0 J
c. ∆ME = KEf − PEi = −PEi = −mgh = −mgd1(sin q)
∆ME = −(10.0 kg)(9.81 m/s2)(3.00 m)(sin 30.0°) = −147 J
I Ch. 5–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q ¢ = 180°
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Givens
Solutions
56. m = 56.0 g
Wnet = ∆KE = KEf − KEi = −KEi
h = 12.0 m
The egg’s kinetic energy at the pad’s surface equals the gravitational potential energy
associated with the egg at the window.
d = 5.00 cm
∆t = 6.25 ms
KEi = PE = mgh
g = 9.81 m/s2
The egg’s deceleration by the pad causes it to travel a distance ∆x ≤ d in a time
interval ∆t.
I
1
∆x = 2 (vi + vf)∆t
vi =

2KmE = 2m
mgh =
i
h
2g
vf = 0 m/s
1
1
h + vf)∆t = 2 [
2g
∆x = 2 (
)(
(2
9.
81
m/s
2)(1
2.
0m
) + 0 m/s](6.25 × 10−3 s)
)(
(2
9.
81
m/s
2)(1
2.
0m
)](6.25 × 10−3 s)
1
∆x = 2 [
∆x = 4.80 × 10−2 m = 4.80 cm
57. m = 75 kg
a. PEi = KEf
h = 1.0 m
1
2
g = 9.81 m/s
d = 0.50 cm
KEf = 0 J
q = 180°
mgh = 2 mvf 2
v f = 2g
m/s
m) = 4.4 m/s
h = (2
)(
9.
81
2)(1
.0
b. Wnet = ∆KE = KEf − KEi = −KEi = −PE = −mgh
Wnet = Fnetd(cos q) = −Fnetd − Fnetd = −mgh
mgh (75 kg)(9.81 m/s2)(1.0 m)
Fnet =  = 
= 1.5 × 105 N
d
(0.50 × 10−2 m)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
58. m = 5.0 kg
vi = 17 m/s
h = 25.0 m
g = 9.81 m/s2
a. Wgravity = ∆PEg = mgh = (5.0 kg)(9.81 m/s2)(25.0 m) = 1.2 × 103 J
b. ∆KE = Wgravity = 1.2 × 103 J
c. ∆KE = KEf − KEi
1
1
KEf = ∆KE + KEi = ∆KE + 2 mvi 2 = 1200 J + 2 (5.0 kg)(17 m/s)2
KEf = 1200 J + 720 J = 1.9 × 103 J
59. k = 105 N/m
m = 2.00 kg
x = −0.100 m
d = 0.250 m
g = 9.81 m/s2
KEf = 0 J
1
Wnet = ∆KE = KEf − KEi = −KEi = −PEg = − 2 kx2
Wnet = Fkd(cos q) = mkmgd(cos q) − mkmgd
1
 kx 2
2
= mkmgd
kx 2
(105 N/m)(−0.100 m)2
mk =  = 
= 0.107
2mgd (2)(2.00 kg)(9.81 m/s2)(0.250 m)
q = 180°
Section One—Pupil’s Edition Solutions
I Ch. 5–15
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I
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Givens
Solutions
60. m = 5.0 kg
a. Because v is constant,
q = 30.0°
Fy, net = F(sin q) − mkFn − mg = 0
q ′ = 0°
Fx, net = F(cos q) − Fn = 0
d = 3.0 m
Fn = F(cos q )
mk = 0.30
F(sin q ) − mkFn = mg
2
g = 9.81 m/s
F(sin q ) − mkF(cos q) = mg
mg
F = 
sin q − mk(cos q)
Upward component of F is parallel and in the same direction as motion, so
mg (sin q )d
(5.0 kg)(9.81 m/s2)(sin 30.0°)(3.0 m)
W = F(sin q )d =  = 
sin q − mk(cos q)
sin 30.0° − (0.30)(cos 30.0°)
(5.0 kg)(9.81 m/s2)(sin 30.0°)(3.0 m)
(5.0 kg)(9.81 m/s2)(sin 30.0°)(3.0 m)
W =  = 
0.500 − 0.26
0.24
W = 3.1 102 J
q′ = 180°
b. Wg = Fgd(cos q′) = −mgd = −(5.0 kg)(9.81 m/s2)(3.0 m) = −150 J
mg
(5.0 kg)(9.81 m/s2)(cos 30.0°)
c. Fn = F(cos q ) =  (cos q) = 
sin q − mk(cos q)
sin 30.0° − (0.30)(cos 30.0°)
(5.0 kg)(9.81 m/s2)(cos 30.0°)
(5.0 kg)(9.81 m/s2)(cos 30.0°)
Fn =  = 
0.500 − 0.26
0.24
Fn = 1.8 102 N
a. h = L(1 − cos q) = (2.0 m)(1 − cos 30.0°) = (2.0 m)(1 − 0.866) = (2.0 m)(0.134)
L = 2.0 m
PEmax = mgh = mgL(1 − cos q) = (25 kg)(9.81 m/s2)(2.0 m)(0.134)
q = 30.0°
PEmax = 66 J
g = 9.81 m/s2
b. PEi = KEf
1
mgh = 2 mvf 2
vf = 2g
m/s
m)(
h = (2
)(
9.
81
2)(2
.0
1−cos30
.0
°)
vf = (2
m/s
m)(
)(
9.
81
2)(2
.0
0.
13
4)
= 2.3 m/s
c. ME = PEi + KEi = 66 J + 0 J = 66 J
vf = 2.00 m/s
1
1
d. ∆KE = KEf − PEi = 2 mvf 2 − PEi = 2 (25 kg)(2.00 m/s)2 − 66 J
∆KE = (5.0 × 101 J) − 66 J = −16 J
I Ch. 5–16
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
61. m = 25 kg
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Givens
Solutions
62. m = 522 g
a. PEi = PEf + KEf
h2 = 1.25 m
vi = 0 m/s
g = 9.81 m/s2
∆x = 1.00 m
1
mgh = mgh2 + 2 mv 2
v2
h = h2 + 
2g
Choosing the point where the ball leaves the track as the origin of a
coordinate system,
I
v∆t
1
∆y = − 2 g∆t 2
At ∆y = −1.25 m (ground level),
2
1 ∆x
∆y = − 2 g 
v
v=
−(9.81 m/s2)(1.00 m)2
 = 1.98 m/s
(2)(−1.25 m)
−g∆x2
 =
2∆y
v2
(1.98 m/s)2
= 1.25 m + 0.200 m
h = h2 +  = 1.25 m + 
2g
(2)(9.81 m/s2)
h = 1.45 m
b. v = 1.98 m/s (See a.)
c. KEf = PEi
1
 mv 2
f
2
= mgh
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vf = 2g
m/s
h = (2
)(
9.
81
2)(1
.4
5m
) = 5.33 m/s
Section One—Pupil’s Edition Solutions
I Ch. 5–17
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Momentum and Collisions
Chapter
6
I
Practice 6A, p. 209
Givens
1. m = 146 kg
v = 17 m/s to the right
2. m1 = 21 kg
m2 = 5.9 kg
Solutions
p = mv = (146 kg)(17 m/s) = 2.5 × 103 kg • m/s to the right
a. ptot = mtot v = (m1 + m2)v = (21 kg + 5.9 kg)(4.5 m/s)
ptot = (27 kg)(4.5 m/s) = 1.2 × 102 kg • m/s to the northwest
v = 4.5 m/s to the northwest
b. p1 = m1 v = (21 kg)(4.5 m/s) = 94 kg • m/s to the northwest
c. p2 = m2 v = (5.9 kg)(4.5 m/s) = 27 kg • m/s to the northwest
3. m = 1210 kg
p = 5.6 × 104 kg • m/s to
the east
p 5.6 × 104 kg • m/s
v =  =  = 46 m/s to the east
m
1210 kg
Practice 6B, p. 211
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. m = 0.50 kg
vi = 15 m/s to the right
mvf − mvi (0.50 kg)(0 m/s) − (0.50 kg)(15 m/s)
 = 
Fon ball = 
∆t
0.020 s
∆t = 0.020 s
Fon ball = −3.8 × 102 N = 3.8 × 102 N to the left
vf = 0 m/s
Fon receiver = −Fon ball = −(−3.8 × 102 N) = 3.8 × 102 N to the right
2. m = 82 kg
vf = ± 2a
m/s
∆
y = ± (2
)(
−9.
81
2)(−
3.
0m
) = ±7.7 m/s = −7.7 m/s
∆y = −3.0 m
∆t = 0.55 s
vi = 0 m/s
a = −9.81 m/s2
For the time the man is in the water,
vi = 7.7 m/s downward = −7.7 m/s
vf = 0 m/s
mvf − mvi (82 kg)(0 m/s) − (82 kg)(−7.7 m/s)
 =  = 1.1 × 103 N
F= 
∆t
0.55 s
F = 1.1 × 103 N upward
3. m = 0.40 kg
∆p = mvf − mvi = (0.40 kg)(−22 m/s) − (0.40 kg)(18 m/s)
vi = 18 m/s to the north
= +18 m/s
∆p = −8.8 kg • m/s − 7.2 kg • m/s = −16.0 kg • m/s
vf = 22 m/s to the south
= −22 m/s
∆p = 16.0 kg • m/s to the south
Section One—Pupil’s Edition Solutions
I Ch. 6–1
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Givens
Solutions
4. m = 0.50 kg
F1 = 3.00 N to the right
∆t1 = 1.50 s
I
F1∆t1 + mvi,1 (3.00 N)(1.50 s) + (0.50 kg)(0 m/s)
 = 
a. vf,1 = 
m
0.50 kg
vf,1 = 9.0 m/s = 9.0 m/s to the right
vi,1 = 0 m/s
F2 = 4.00 N to the left
= −4.00 N
∆t2 = 3.00 s
vi,2 = 9.0 m/s to the right
F2∆t2 + mvi,2 (−4.00 N)(3.00 s) + (0.50 kg)(9.0 m/s)
 = 
b. vf,2 = 
m
0.50 kg
−12.0 kg • m/s + 4.5 kg • m/s −7.5 kg • m/s
vf,2 =  =  = −15 m/s
0.50 kg
0.50 kg
vf,2 = 15 m/s to the left
Practice 6C, p. 213
1. m = 2250 kg
vi = 20.0 m/s to the west
= −20.0 m/s
vf = 0 m/s
F = 8.45 × 103 N to the east
= 8.4 s × 103 N
∆p mvf − mvi (2250 kg)(0 m/s) − (2250 kg)(−20.0 m/s)
 = 
a. ∆t =  = 
F
F
8.45 × 103 N
∆t = 5.33 s
1
1
b. ∆x = 2(vi + vf)∆t = 2(−20.0 m/s + 0 m/s)(5.33 s)
∆x = −53.3 m = 53.3 m to the west
vi = 20.0 m/s to the north
= +20.0 m/s
F = 6250 N to the south
= −6250 N
∆t = 2.50 s
F∆t + mv (−6250 N)(2.50 s) + (2500 kg)(20.0 m/s)
a. vf = i = 
m
2500 kg
(−1.56 × 104 kg • m/s) + (5.0 × 104 kg • m/s) 3.4 × 104 kg • m/s
vf =  = 
2500 kg
2500 kg
vf = 14 m/s = 14 m/s to the north
1
1
b. ∆x = 2(vi + vf)(∆t) = 2(20.0 m/s + 14 m/s)(2.50 s)
1
∆x = 2(34 m/s)(2.50 s) = 42 m to the north
vf = 0 m/s
3. m = 3250 kg
vi = 20.0 m/s to the west
= −20.0 m/s
vf = 0 m/s
∆t = 5.33 s
mvf − mvi (2500 kg)(0 m/s) − (2500 kg)(20.0 m/s)
 =  = 8.0 s
c. ∆t = 
F
−6250 N
mvf − mvi (3250 kg)(0 m/s) − (3250 kg)(−20.0 m/s)
 = 
a. F = 
∆t
5.33 s
F = 1.22 × 104 N = 1.22 × 104 N to the east
1
1
b. ∆x = 2(vi + vf)(∆t) = 2(−20.0 m/s + 0 m/s)(5.33 s) = −53.3 m
∆x = 53.3 m to the west
I Ch. 6–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. m = 2500 kg
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Section Review, p. 214
Givens
2. m1 = 0.145 kg
m2 = 3.00 g
v2 = 1.50 × 103 m/s
Solutions
a. m1 v1 = m2 v2
m v
(3.00 × 10−3 kg)(1.50 × 103 m/s)
v1 = 22 =  = 31.0 m/s
m1
(0.145 kg)
1
I
1
b. KE1 = 2m1v12 = 2(0.145 kg)(31.0 m/s)2 = 69.7 J
KE2 = 2m2 v22 = 2(3.00 × 10−3 kg)(1.50 × 103 m/s)2 = 3380 J
1
KE2 > KE1
5. m = 0.42 kg
vi = 12 m/s downfield
1
The bullet has greater kinetic energy.
a. ∆p = mvf − mvi = (0.42 kg)(18 m/s) − (0.42 kg)(12 m/s)
∆p = 7.6 kg • m/s − 5.0 kg • m/s = 2.6 kg • m/s downfield
vf = 18 m/s downfield
∆t = 0.020 s
∆p 2.6 kg • m/s
b. F =  =  = 1.3 × 102 N downfield
∆t
0.020 s
Practice 6D, p. 219
1. m1 = 63.0 kg
m2 = 10.0 kg
v2,i = 0 m/s
v2, f = 12.0 m/s
v1,i = 0 m/s
2. m1 = 85.0 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m2 = 135.0 kg
v1,i = 4.30 m/s to the west
= −4.30 m/s
v2,i = 0 m/s
3. m1 = 0.50 kg
v1,i = 12.0 m/s
m1v1,i + m2 v2,i − m2 v2,f
v1,f = 
m1
(63.0 kg)(0 m/s) + (10.0 kg)(0 m/s) − (10.0 kg)(12.0 m/s)
v1,f =  = −1.90 m/s
63.0 kg
astronaut speed = 1.90 m/s
m1v1,i + m2v2,i (85.0 kg)(−4.30 m/s) + (135.0 kg)(0 m/s)
 = 
vf = 
m1 + m2
85.0 kg + 135.0 kg
(85.0 kg)(−4.30 m/s)
vf =  = −1.66 m/s = 1.66 m/s to the west
220.0 kg
m1v1,i + m2v2,i − m1v1, f
a. v2,f = 
m2
v2,i = 0 m/s
m1 = m2
m2 = 0.50 kg
v2,f = v1,i + v2,i − v1, f = 12.0 m/s + 0 m/s − 0 m/s = 12.0 m/s
v1,f = 0 m/s
v1,f = 2.4 m/s
b. v2, f = v1,i + v2,i − v1, f = 12.0 m/s + 0 m/s − 2.4 m/s = 9.6 m/s
v1,f = 0.3 m/s
c. v2,f = v1,i + v2,i − v1,f = 12.0 m/s + 0 m/s − 0.3 m/s = 11.7 m/s
Section One—Pupil’s Edition Solutions
I Ch. 6–3
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Givens
Solutions
4. m1 = 2.0 kg + mb
I
mi v1, i + m2v2, i = m1v1, f + m2v2, f
m2 = 8.0 kg
(2.0 kg + mb)(0 m/s) + (8.0 kg)(0 m/s) = (2.0 kg + mb)(−0.60 m/s) + (8.0 kg)(3.0 m/s)
v2, i = 0 m/s
(2.0 kg + mb)(0.60 m/s) = (8.0 kg)(3.0 m/s)
v2, f = 3.0 m/s
v1, i = 0 m/s
24 kg • m/s − 1.2 kg • m/s 23 kg • m/s
mb =  = 
0.60 m/s
0.60 m/s
v1, f = −0.60 m/s
mb = 38 kg
Section Review, p. 221
1. m1 = 44 kg m2 = 22 kg
v1,i = 0 m/s
v2,i = 0 m/s
v1, f = 3.5 m/s backward
= −3.5 m/s
v1,i = 0 m/s
v2, i = 4.6 m/s to the right
= +4.6 m/s
3. m1 = 215 g
v1,i = 55.0 m/s
v2,i = 0 m/s
m2 = 46 g
v1,f = 42.0 m/s
m1v1,i + m2 v2,i − m1v1,f
a. v2, f = 
m2
(44 kg)(0 m/s) + (22 kg)(0 m/s) − (44 kg)(−3.5 m/s)
v2, f =  = 7.0 m/s forward
22 kg
m1v1,i + m2v2,i (44 kg)(0 m/s) + (22 kg)(4.6 m/s)
 = 
c. vf = 
m1 + m2
44 kg + 22 kg
(22 kg)(4.6 m/s)
vf =  = 1.5 m/s to the right
66 kg
m1v1,i + m2 v2,i − m1v1,f
v2,f = 
m2
(0.215 kg)(55.0 m/s) + (0.046 kg)(0 m/s) − (0.215 kg)(42.0 m/s)
v2,f = 
0.046 kg
11.8 kg • m/s − 9.03 kg • m/s 2.8 kg • m/s
v2,f =  =  = 61 m/s
0.046 kg
0.046 kg
1. m1 = 1500 kg
v1,i = 15.0 m/s to the south
= −15.0 m/s
m2 = 4500 kg
m1v1,i + m2v2,i (1500 kg)(−15.0 m/s) + (4500 kg)(0 m/s)
vf = 
= 
1500 kg + 4500 kg
m1 + m2
(1500 kg)(−15.0 m/s)
vf = 
= −3.8 m/s = 3.8 m/s to the south
6.0 × 103 kg
v2,i = 0 m/s
2. m1 = 9.0 kg
m2 = 18.0 kg
v2,i = 0 m/s
v1,i = 5.5 m/s
3. m1 = 1.50 × 104 kg
v1,i = 7.00 m/s to the north
m2 = m1
v2,i = 1.50 m/s to the north
m1v1,i + m2v2,i (9.0 kg)(5.5 m/s) + (18.0 kg)(0 m/s)
vf = 
= 
9.0 kg + 18.0 kg
m1 + m2
(9.0 kg)(5.5 m/s)
vf =  = 1.8 m/s
27.0 kg
4
4
m1v1,i + m2v2,i (1.50 × 10 kg)(7.00 m/s) + (1.50 × 10 kg)(1.50 m/s)
 = 
vf = 
1.50 × 104 kg + 1.50 × 104 kg
m1 + m2
1.05 × 105 kg • m/s + 2.25 × 104 kg • m/s 1.28 × 105 kg • m/s
vf = 
= 
3.00 × 104 kg
3.00 × 104 kg
vf = 4.27 m/s to the north
I Ch. 6–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Practice 6E, p. 224
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Givens
4. m1 = 22 kg
m2 = 9.0 kg
v2,i = 0 m/s
vf = 3.0 m/s to the right
5. m1 = 47.4 kg
v1,i = 4.20 m/s
v2,i = 0 m/s
vf = 3.95 m/s
vf = 5.00 m/s
Solutions
(m1 + m2)vf − m2v2,i (22 kg + 9.0 kg)(3.0 m/s) − (9.0 kg)(0 m/s)
= 
v1,i = 
22 kg
m1
(31 kg)(3.0 m/s)
v1,i =  = 4.2 m/s to the right
22 kg
I
m1vf − m1v1,i (47.4 kg)(3.95 m/s) − (47.4 kg)(4.20 m/s)
a. m2 =  = 
v2,i − vf
0 m/s − 3.95 m/s
187 kg • m/s − 199 kg • m/s −12 kg • m/s
m2 =  =  = 3.0 kg
−3.95 m/s
−3.95 m/s
(m1 + m2)vf − m2v2,i (47.4 kg + 3.0 kg)(5.00 m/s) − (3.0 kg)(0 m/s)
b. v1,i =  = 
47.4 kg
m1
(50.4 kg)(5.00 m/s)
v1,i =  = 5.32 m/s
47.4 kg
Practice 6F, p. 226
1. m1 = 0.25 kg
v1,i = 12 m/s to the west
= −12 m/s
m2 = 6.8 kg
m1v1,i + m2v2,i (0.25 kg)(−12 m/s) + (6.8 kg)(0 m/s)
 = 
a. vf = 
m1 + m2
0.25 kg + 6.8 kg
(0.25 kg)(−12 m/s)
vf =  = −0.43 m/s = 0.43 m/s to the west
7.0 kg
v2,i = 0 m/s
b. ∆KE = KEf − KEi
1
1
1
1
KEi = 2m1v1,i2 + 2m2 v2,i2 = 2(0.25 kg)(−12 m/s)2 + 2(6.8 kg)(0 m/s)2
KEi = 18 J + 0 J = 18 J
1
1
KEf = 2(m1 + m2)vf 2 = 2(0.25 kg + 6.8 kg)(−0.43 m/s)2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
KEf = 2(7.0 kg)(−0.43 m/s)2 = 0.65 J
∆KE = KEf − KEi = 0.65 J − 18 J = −17 J
The kinetic energy decreases by 17 J .
2. m1 = 0.40 kg
v1,i = 8.5 m/s to the south
= −8.5 m/s
m2 = 0.15 kg
m1v1,i + m2v2,i (0.40 kg)(−8.5 m/s) + (0.15 kg)(0 m/s)
 = 
a. vf = 
m1 + m2
0.40 kg + 0.15 kg
(0.40 kg)(−8.5 m/s)
vf =  = −6.2 m/s = 6.2 m/s to the south
0.55 kg
v2,i = 0 m/s
b. ∆KE = KEf − KEi
1
1
1
1
KEi = 2m1v1,i2 + 2m2 v2,i2 = 2(0.40 kg)(−8.5 m/s)2 + 2(0.15 kg)(0 m/s)2
KEi = 14 J + 0 J = 14 J
1
1
KEf = 2(m1 + m2)vf 2 = 2(0.40 kg + 0.15 kg)(−6.2 m/s)2
1
KEf = 2(0.55 kg)(−6.2 m/s)2 = 11 J
∆KE = KEf − KEi = 11 J − 14 J = −3 J
The kinetic energy decreases by 3 J .
Section One—Pupil’s Edition Solutions
I Ch. 6–5
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Givens
Solutions
3. m1 = 56 kg
v1,i = 4.0 m/s to the north
= +4.0 m/s
I
m1v1,i + m2v2,i (56 kg)(4.0 m/s) + (65 kg)(−12.0 m/s)
 = 
a. vf = 
m1 + m2
56 kg + 65 kg
m2 = 65 kg
220 kg • m/s − 780 kg • m/s −560 kg • m/s
vf =  =  = −4.6 m/s
121 kg
121 kg
v2,i = 12.0 m/s to the south
= −12.0 m/s
vf = 4.6 m/s to the south
b. ∆KE = KEf − KEi
1
1
1
1
KEi = 2m1v1,i2 + 2m2 v2,i2 = 2(56 kg)(4.0 m/s)2 + 2(65 kg)(−12.0 m/s)2
KEi = 450 J + 4700 J = 5200 J
1
1
KEf = 2(m1 + m2)vf 2 = 2(56 kg + 65 kg)(−4.6 m/s)2
1
KEf = 2(121 kg)(−4.6 m/s)2 = 1300 J
∆KE = KEf − KEi = 1300 J − 5200 J = −3900 J
The kinetic energy decreases by 3.9 × 103 J .
Practice 6G, p. 229
1. m1 = 0.015 kg
v1,i = 22.5 cm/s to the right
= +22.5 cm/s
m1v1,i + m2v2,i − m1v1,f
a. v2,f = 
m2
m1 = m2
m2 = 0.015 kg
v2,f = v1,i + v2,i − v1,f = 22.5 cm/s + (−18.0 cm/s) − (−18.0 cm/s)
v2,i = 18.0 cm/s to the left
= −18.0 cm/s
v2,f = 22.5 cm/s = 22.5 cm/s to the right
v1,f = 18.0 cm/s to the left
= −18.0 cm/s
1
1
b. KEi = 2m1v1,i2 + 2m2 v2,i2
1
1
KEi = 3.8 × 10−4 J + 2.4 × 10−4 J = 6.2 × 10−4 J
1
1
KEf = 2m1v1,f 2 + 2m2 v2,f 2
1
1
KEf = 2(0.015 kg)(−0.180 m/s)2 + 2(0.015 kg)(0.225 m/s)2
KEf = 2.4 × 10−4 J + 3.8 × 10−4 J = 6.2 × 10−4 J
2. m1 = 16.0 kg
v1,i = 12 m/s to the left
= −12 m/s
m1v1,i + m2v2,i − m2v2,f
a. v1,f = 
m1
m2 = 4.0 kg
(16.0 kg)(−12 m/s) + (4.0 kg)(6.0 m/s) − (4.0 kg)(−22.7 m/s)
v1,f = 
16.0 kg
v2,i = 6.0 m/s to the right
= +6.0 m/s
−190 kg • m/s + 24 kg • m/s + 91 kg • m/s −75 kg • m/s
v1,f =  = 
16.0 kg
16.0 kg
v2,f = 22.7 m/s to the left
= −22.7 m/s
v1,f = −4.7 m/s = 4.7 m/s to the left
I Ch. 6–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
KEi = 2(0.015 kg)(0.225 m/s)2 + 2(0.015 kg)(−0.180 m/s)2
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Givens
Solutions
1
1
b. KEi = 2m1v1,i 2 + 2m2 v2,i 2
1
1
KEi = 2(16.0 kg)(−12 m/s)2 + 2(4.0 kg)(6.0 m/s)2
KEi = 1.2 × 103 J + 72 J = 1.2 × 103 J
1
I
1
KEf = 2m1v1,f 2 + 2m2 v2,f 2
1
1
KEf = 2(16.0 kg)(−4.8 m/s)2 + 2(4.0 kg)(−22.7 m/s)2
KEf = 180 J + 1.0 × 103 J = 1.2 × 103 J
3. m1 = 4.0 kg
v1,i = 8.0 m/s to the right
m2 = 4.0 kg
v2,i = 0 m/s
v1,f = 0 m/s
m1v1,i + m2v2,i − m1v1,f
a. v2,f = 
m2
m1 = m2
v2,f = v1,i + v2,i − v1,f = 8.0 m/s + 0 m/s − 0 m/s
v2,f = 8.0 m/s = 8.0 m/s to the right
1
1
b. KEi = 2m1v1,i2 + 2m2 v2,i2
1
1
KEi = 2(4.0 kg)(8.0 m/s)2 + 2(4.0 kg)(0 m/s)2
KEi = 130 J + 0 J = 1.3 × 102 J
1
1
KEf = 2m1v1,f 2 + 2m2 v2,f 2
1
1
KEf = 2(4.0 kg)(0 m/s)2 + 2(4.0 kg)(8.0 m/s)2
KEf = 0 J + 130 J = 1.3 × 102 J
4. m1 = 25.0 kg
v1, i = 5.00 m/s to the right
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m2 = 35.0 kg
v1, f = 1.50 m/s to the right
v2, f = 4.50 m/s to the right
miv1, f + m2v2, f − m1v1, i
a. v2, i = 
m2
(25.0 kg)(1.50 m/s) + (35.0 kg)(4.50 m/s) − (25.0 kg)(5.00 m/s)
v2, i = 
35.0 kg
37.5 kg• m/s + 158 kg • m/s − 125 kg • m/s
7.0 × 101 kg• m/s
v2, i =  = 
35.0 kg
35.0 kg
v2, i = 2.0 m/s = 2.0 m/s to the right
1
1
b. KEi = 2mi v1, i2 + 2m2 v2, i2
1
1
KEi = 2(25.0 kg)(5.00 m/s)2 + 2(35.0 kg)(2.0 m/s)2
KEi = 312 J + 7.0 × 101 J = 382 J
1
1
KEf = 2m1v1, f2 + 2m2v2, f2
1
1
KEf = 2(25.0 kg)(1.50 m/s)2 + 2(35.0 kg)(4.50 m/s)2
KEf = 28.1 J + 354 J = 382 J
Section One—Pupil’s Edition Solutions
I Ch. 6–7
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Section Review, p. 230
Givens
Solutions
3. m1 = 90.0 kg
I
v1,i = 5.0 m/s to the south
= −5.0 m/s
m1v1,i + m2v2,i (90.0 kg)(−5.0 m/s) + (95.0 kg)(3.0 m/s)
= 
a. vf = 
90.0 kg + 95.0 kg
m1 + m2
m2 = 95.0 kg
−450 kg • m/s + 280 kg • m/s −170 kg • m/s
vf =  =  = −0.92 m/s
185.0 kg
185.0 kg
v2,i = 3.0 m/s to the north
= +3.0 m/s
vf = 0.92 m/s to the south
b. ∆KE = KEf − KEi
1
1
1
1
KEi = 2m1v1,i2 + 2m2 v2,i2 = 2(90.0 kg)(−5.0 m/s)2 + 2(95.0 kg)(3.0 m/s)2
KEi = 1100 J + 430 J = 1500 J
1
1
1
KEf = 2(m1 + m2)vf 2 = 2(90.0 kg + 95.0 kg)(−0.92 m/s)2 = 2(185.0 kg)(−0.92 m/s)2 = 78 J
∆KE = KEf − KEi = 78 J − 1500 J = −1400 J = −1.4 × 103 J
The kinetic energy decreases by 1.4 × 103 J .
5. m1 = m2 = 0.40 kg
v1,i = 0 m/s
m1v1,i + m2 v2,i − m2 v2,f
a. v1,f = 
m1
v2,i = 3.5 m/s
m1 = m2
v2,f = 0 m/s
v1,f = v1,i + v2,i − v2,f = 0 m/s + 3.5 m/s − 0 m/s = 3.5 m/s
1
1
b. KE1,i = 2m1v1,i 2 = 2(0.40 kg)(0 m/s)2 = 0 J
1
1
c. KE2,f = 2m2 v2,f 2 = 2(0.40 kg)(0 m/s)2 = 0 J
12. m = 1.67 × 10−27 kg
a. p = mv = (1.67 × 10−27 kg)(5.00 × 106 m/s) = 8.35 × 10−21 kg • m/s upward
v = 5.00 × 106 m/s straight up
m = 15.0 g
b. p = mv = (15.0 × 10−3 kg)(325 m/s) = 4.88 kg • m/s to the right
v = 325 m/s to the right
m = 75.0 kg
c. p = mv = (75.0 kg)(10.0 m/s) = 7.50 × 102 kg • m/s to the southwest
v = 10.0 m/s southwest
m = 5.98 × 1024 kg
d. p = mv = (5.98 × 1024 kg)(2.98 × 104 m/s) = 1.78 × 1029 kg • m/s forward
v = 2.98 × 104 m/s forward
13. m = 0.148 kg
p = mv = (0.148 kg)(35 m/s) = 5.2 kg • m/s toward home plate
v = 35 m/s toward
home plate
I Ch. 6–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter Review and Assess, pp. 232–237
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Givens
Solutions
14. m1 = 2.5 kg
mvf − mvi (2.5 kg)(7.5 m/s) − (2.5 kg)(−8.5 m/s)
 = 
F= 
∆t
0.25 s
vi = 8.5 m/s to the left
= −8.5 m/s
vf = 7.5 m/s to the right
= +7.5 m/s
19 kg • m/s + 21 kg • m/s 4.0 × 101 kg•m/s
F =  =  = 160 N to the right
0.25 s
0.25 s
I
∆t = 0.25 s
15. m = 0.55 kg
vi = 0 m/s
vf = 8.0 m/s
mvf − mvi (0.55 kg)(8.0 m/s) − (0.55 kg)(0 m/s)
4.4 kg•m/s
F =  =  = 
0.25 s
∆t
0.25 s
F = 18 N
∆t = 0.25 s
16. m = 0.15 kg
vi = 26 m/s
vf = 0 m/s
F = −390 N
mvf − mv
(0.15 kg)(0 m/s) − (0.15 kg)(26 m/s)
∆t = i = 
F
−390 N
−(0.15 kg)(26 m/s)
∆t =  = 0.010 s
−390 N
1
1
∆x = 2(vi + vf)∆t = 2(26.0 m/s + 0 m/s)(0.010 s) = 0.13 m
24. m1 = 65.0 kg
v1,i = 2.50 m/s forward
m2 = 0.150 kg
v2,i = 2.50 m/s forward
v2,f = 32.0 m/s forward
m1v1,i + m2v2,i − m2v2,f
a. v1,f = 
m1
(65.0 kg)(2.50 m/s) + (0.150 kg)(2.50 m/s) − (0.150 kg)(32.0 m/s)
v1,f = 
65.0 kg
162 kg • m/s + 0.375 kg • m/s − 4.80 kg • m/s 162 kg • m/s − 4.42 kg • m/s
v1,f =  = 
65.0 kg
65.0 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
158 kg • m/s
v1,f =  = 2.43 m/s forward
65.0 kg
m1 = 60.0 kg
v1,i = 0 m/s
m2 = 0.150 kg
v2,i = 32.0 m/s forward
25. m1 = 55 kg
m2 = 0.057 kg
v2,i = 0 m/s
v1,i = 0 m/s
v2,f = 36 m/s to the north
m1v1,i + m2v2,i (60.0 kg)(0 m/s) + (0.150 kg)(32.0 m/s)
= 
b. vf = 
60.0 kg + 0.150 kg
m1 + m2
(0.150 kg)(32.0 m/s)
vf =  = 7.97 × 10−2 m/s forward
60.2 kg
Because the initial momentum is zero,
m1v1,f = −m2v2,f
−m v2,f −(0.057 kg)(36 m/s)
v1,f = 2
=  = −0.037 m/s
m1
55 kg
v1,f = 0.037 m/s to the south
Section One—Pupil’s Edition Solutions
I Ch. 6–9
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Givens
Solutions
26. m1 = 1.5 kg
m1v1,i + m2v2,i − m1v1,f
a. v2,f = 
m2
v1,i = 3.0 m/s to the right
m2 = 1.5 kg
I
v2,i = 0 m/s
v1,f = 0.5 m/s to the right
v1,f = 0 m/s
31. m1 = 4.0 kg
m2 = 3.0 kg
v1,i = 5.0 m/s
v2,i = −4.0 m/s
32. m1 = 1.20 kg
v1,i = 5.00 m/s
m2 = 0.800 kg
v2,i = 0 m/s
33. m1 = 10.0 kg
m2 = 2.5 kg
v1,i = 6.0 m/s
v2,i = −3.0 m/s
34. m1 = 2.00 × 104 kg
v1,i = 3.00 m/s
m2 = 2m1
v2,i = 1.20 m/s
m1 = m2
v2,f = v1,i + v2,i − v1,f
v2,f = 3.0 m/s + 0 m/s − 0.5 m/s = 2.5 m/s to the right
b. v2,f = v1,i + v2,i − v1,f = 3.0 m/s + 0 m/s − 0 m/s = 3.0 m/s to the right
m1v1,i + m2v2,i (4.0 kg)(5.0 m/s) + (3.0 kg)(−4.0 m/s)
 = 
vf = 
m1 + m2
4.0 kg + 3.0 kg
2.0 × 101 kg • m/s + (−12 kg • m/s) 8 kg • m/s
vf =  =  = 1 m/s
7.0 kg
7.0 kg
m1v1,i + m2 v2,i (1.20 kg)(5.00 m/s) + (0.800 kg)(0 m/s)
 = 
vf = 
m1 + m2
1.20 kg + 0.800 kg
(1.20 kg)(5.00 m/s)
vf =  = 3.00 m/s
2.00 kg
m1v1,i + m2 v2,i (10.0 kg)(6.0 m/s) + (2.5 kg)(−3.0 m/s)
 = 
vf = 
m1 + m2
10.0 kg + 2.5 kg
6.0 × 101 kg • m/s − 7.5 kg • m/s 52 kg • m/s
vf =  =  = 4.2 m/s
12.5 kg
12.5 kg
m1v1,i + m2 v2,i (2.00 × 104 kg)(3.00 m/s) + (2)(2.00 × 104 kg)(1.20 m/s)
 = 
a. vf = 
m1 + m2
(2.00 × 104 kg) + (2)(2.00 × 104 kg)
6.00 × 104 kg • m/s + 4.80 × 104 kg • m/s
vf = 
6.00 × 104 kg
10.80 × 104 kg • m/s
vf = 
= 1.80 m/s
6.00 × 104 kg
b. ∆KE = KEf − KEi
1
1
1
1
KEi = 2m1v1,i2 + 2m2v2,i2 = 2(2.00 × 104 kg)(3.00 m/s)2 + 2(2)(2.00 × 104 kg)(1.20 m/s)2
KEi = 9.00 × 104 J + 2.88 × 104 J = 11.88 × 104 J
1
1
KEf = 2(m1 + m2)vf 2 = 2(2.00 × 104 kg + 4.00 × 104 kg)(1.80 m/s)2 = 9.72 × 104 J
∆KE = KEf − KEi = 9.72 × 104 J − 11.88 × 104 J = −2.16 × 104 J
The kinetic energy decreases by 2.16 × 104 J .
I Ch. 6–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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35. m1 = 88 kg
m1v1,i + m2v2,i (88 kg)(5.0 m/s) + (97 kg)(−3.0 m/s)
 = 
a. vf = 
m1 + m2
88 kg + 97 kg
v1,i = 5.0 m/s to the east
= +5.0 m/s
m2 = 97 kg
v2,i = 3.0 m/s to the west
= −3.0 m/s
440 kg • m/s − 290 kg • m/s 150 kg • m/s
vf =  =  = 0.81 m/s to the east
185 kg
185 kg
I
b. ∆KE = KEf − KEi
1
1
1
1
KEi = 2m1v1,i2 + 2m2v2,i2 = 2(88 kg)(5.0 m/s)2 + 2(97 kg)(−3.0 m/s)2
KEi = 1100 J + 440 J = 1.5 × 103 J
1
1
1
KEf = 2(m1 + m2)vf 2 = 2(88 kg + 97 kg)(0.81 m/s)2 = 2(185 kg)(0.81 m/s)2 = 61 J
∆KE = KEf − KEi = 61 J − 1.5 × 103 J = −1.4 × 103 J
The kinetic energy decreases by 1.4 × 103 J .
36. m1 = 5.0 g
v1,i = 25.0 cm/s to the right
= +25.0 cm/s
m2 = 15.0 g
v2,i = 0 m/s
v1,f = 12.5 cm/s to the left
= −12.5 cm/s
m1v1,i + m2v2,i − m1v1,f
a. v2,f = 
m2
(5.0 g)(25.0 cm/s) + (15.0 g)(0 m/s) − (5.0 g)(−12.5 cm/s)
v2,f = 
15.0 g
120 g • cm/s + 62 g • cm/s 180 g • cm/s
v2,f =  =  = 12 cm/s
15.0 g
15.0 g
v2,f = 12 cm/s to the right
1
1
b. ∆KE2 = KE2,f −KE2,i = 2m2 v2,f 2 − 2m2 v2,i2
∆KE2 = 2(15.0 × 10−3 kg)(0.12 m/s)2 − 2(15.0 × 10−3 kg)(0 m/s)2
1
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆KE2 = 1.1 × 10−4 J
v2,i = 0 m/s
m1v1,i + m2 v2,i − m1 v1,f
v2,f = 
m2
m1 = m2
m1 = m2
37. v1,i = 4.0 m/s
v1,f = 0 m/s
38. m1 = 25.0 g
v1,i = 20.0 cm/s to the right
m2 = 10.0 g
v2,i = 15.0 cm/s to the right
v2,f = 22.1 cm/s to the right
v2,f = v1,i + v2,i − v1,f = 4.0 m/s + 0 m/s − 0 m/s = 4.0 m/s
m1v1,i + m2v2,i − m2v2,f
v1,f = 
m1
(25.0 g)(20.0 cm/s) + (10.0 g)(15.0 cm/s) − (10.0 g)(22.1 cm/s)
v1,f = 
25.0 g
5.00 × 102 g • cm/s + 1.50 × 102 g • cm/s − 2.21 × 102 g • cm/s
v1,f = 
25.0 g
429 g • cm/s
v1,f =  = 17.2 cm/s to the right
25.0 g
Section One—Pupil’s Edition Solutions
I Ch. 6–11
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Givens
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39. m1 = 15.0 g
m1v1,i + m2v2,i − m1v1,f
v2,f = 
m2
v1,i = 20.0 cm/s to the right
= +20.0 cm/s
m2 = 20.0 g
I
v2,i = 30.0 cm/s to the left
= −30.0 cm/s
v1,f = 37.1 cm/s to the left
= −37.1 cm/s
(15.0 g)(20.0 cm/s) + (20.0 g)(−30.0 cm/s) − (15.0 g)(−37.1 cm/s)
v2,f = 
20.0 g
3.00 × 102 g • cm/s − 6.00 × 102 g • cm/s + 5.56 × 102 g • cm/s
v2,f = 
20.0 g
256 g • cm/s
v2,f =  = 12.8 cm/s to the right
20.0 g
m1v1,i + m2v2,i − m2v2,f
v1,f = 
m1
40. m1 = m2
v1,i = 0 m/s
v2,i = 5.00 m/s to the right
v2,f = 0 m/s
m1 = m2
v1,f = v1,i + v2,i − v2,f = 0 m/s + 5.00 m/s − 0 m/s
v1,f = 5.00 m/s to the right
41. m = 0.147 kg
p = 6.17 kg• m/s toward
second base
42. KE = 150 J
p 6.17 kg•m/s
v =  =  = 42.0 m/s toward second base
m
0.147 kg
1
p = 30.0 kg • m/s
KE = 2mv 2
p
m = 
v
1 p
pv
KE =   v 2 = 
2 v
2
2KE
(2)(150 J)
v =  =  = 1.0 × 101 m/s
p
30.0 kg • m/s
43. m = 0.10 kg
a. At its maximum height, v = 0 m/s.
vi = 15.0 m/s straight up
p = mv = (0.10 kg)(0 m/s) = 0 kg • m/s
2
a = −9.81 m/s
b. Halfway to its maximum height (where vf = 0 m/s),
2
2
1 vf − vi
(0 m/s)2 − (15.0 m/s)2
∆y =   = 
= 5.73 m
2a
2
(4)(−9.81 m/s2)
Now let vf represent the velocity at ∆y = 5.73 m.
vf = ± vi2+
2a∆
x = ± (1
5.
0m
/s
)2+(2)
(−
9.8
1m
/s
2)(5
.7
3m
)
vf = ± 22
s2
−112
m2/
s2 = ± 11
s2 = ±10.6 m/s
5m
2/
3m
2/
vf = 10.6 m/s upward
p = mvf = (0.10 kg)(10.6 m/s) = 1.06 kg • m/s upward
I Ch. 6–12
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
p 30.0 kg • m/s
m =  = 
= 3.0 kg
v 1.0 × 101 m/s
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Givens
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44. m1 = 3.00 kg
m1v1,i + m2 v2,i = m1vf + m2 vf
m2 (v2,i − vf ) = m1vf − m1v1,i
v2,i = 0 m/s
vf =
1
 v , or v
1,i
3 1,i
= 3vf
m1vf − m1v1,i
m2 =  , where v2,i = 0 m/s
v2,i − vf
m1vf − m1(3vf )
m2 =  = −(m1 − 3m1) = −m1 + 3m1
− vf
I
m2 = 2m1 = (2)(3.00 kg) = 6.00 kg
45. m1 = 5.5 g
m2 = 22.6 g
v2,i = 0 m/s
∆y = −1.5 m
∆x = 2.5 m
a = −9.81 m/s2
For an initial downward speed of zero,
1
∆y = 2a∆t 2
∆t =
a
2∆y
∆x
∆x 
a
vf =  = 2∆y = ∆x 

2∆y
∆t
a
−9.81 m/s2
 = 4.5 m/s
(2)(−1.5 m)
(m1 + m2)vf − m2v2, i
v1, i = 
m1
vf = (2.5 m)
(5.5 g + 22.6 g)(1 kg/103 g)(4.5 m/s) − (22.6 × 10−3 kg)(0 m/s)
v1, i = 
5.5 × 10−3 kg
(28.1 × 10−3 kg)(4.5 m/s)
v1, i = 
= 23 m/s
5.5 × 10−3 kg
730 N
46. m1 = 2
9.81m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
R = 5.0 m
m2 = 2.6 kg
v1,i = 0 m/s
v2,i = 0 m/s
Because the initial momentum is zero,
−m2v2,f −(2.6 kg)(5.0 m/s)
v1,f =  =  = −0.17 m/s = 0.17 m/s to the south
m1
730 N
2
9.81 m/s
∆x −R
−5.0 m
∆t =  =  =  = 29 s
v1,f v1,f −0.17 m/s
v2,f = 5.0 m/s to the north
47. m = 0.025 kg
vi = 18.0 m/s
∆t = 5.0 × 10−4 s
vf = 10.0 m/s
mvf − mvi (0.025 kg)(10.0 m/s) − (0.025 kg)(18.0 m/s)
F =  = 
∆t
5.0 × 10−4 s
0.25 kg • m/s − 0.45 kg • m/s −0.20 kg • m/s
F = 
= 
= −4.0 × 102 N
5.0 × 10−4 s
5.0 × 10−4 s
magnitude of the force = 4.0 × 102 N
Section One—Pupil’s Edition Solutions
I Ch. 6–13
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48. m1 = 1550 kg
(m1 + m2)vf − m1v1,i (1550 kg + 2550 kg)(5.22 m/s) − (1550 kg)(−10.0 m/s)
= 
v2,i = 
2550 kg
m2
v1,i = 10.0 m/s to the south
= −10.0 m/s
m2 = 2550 kg
I
vf = 5.22 m/s to the north
= +5.22 m/s
(4.10 × 103 kg)(5.22 m/s) − (1550 kg)(−10.0 m/s)
v2,i = 
2550 kg
4
2.14 × 104 kg • m/s + 1.55 × 104 kg • m/s 3.69 × 10 kg • m/s
v2,i =  = 
2550 kg
2550 kg
v2,i = 14.5 m/s = 14.5 m/s to the north
49. m1 = 2150 kg
v1,i = 10.0 m/s to the east
m2 = 3250 kg
vf = 5.22 m/s to the east
(m1 + m2)vf − m1v1,i
a. v2,i = 
m2
(2150 kg + 3250 kg)(5.22 m/s) − (2150 kg)(10.0 m/s)
v2,i = 
3250 kg
(5.40 × 103 kg)(5.22 m/s) − (2150 kg)(10.0 m/s)
v2,i = 
3250 kg
2.82 × 104 kg • m/s − 2.15 × 104 kg • m/s 6700 kg • m/s
v2,i =  = 
3250 kg
3250 kg
v2,i = 2.1 m/s to the east
b. ∆KE = KEf − KEi
1
1
1
1
KEi = 2m1v1,i2 + 2m2v2,i2 = 2(2150 kg)(10.0 m/s)2 + 2(3250 kg)(2.1 m/s)2
KEi = 1.08 × 105 J + 7.2 × 103 J = 1.15 × 105 J
1
1
KEf = 2(m1 + m2 )vf 2 = 2(2150 kg + 3250 kg)(5.22 m/s)2
1
KEf = 2(5.40 × 103 kg)(5.22 m/s)2 = 7.36 × 104 J
The kinetic energy decreases by 4.1 × 104 J .
50. m1 = 0.400 kg
v1,i = 3.50 cm/s to the right
= +3.50 cm/s
m2 = 0.600 kg
v2,i = 0 m/s
v1,f = 0.70 cm/s to the left
= −0.70 cm/s
∆t = 5.0 s
m1v1,i + m2v2,i − m1v1,f
v2,f = 
m2
(0.400 kg)(3.50 × 10−2 m/s) + (0.600 kg)(0 m/s) − (0.400 kg)(−0.70 × 10−2 m/s)
v2,f = 
0.600 kg
1.40 × 10−2 kg • m/s + 2.80 × 10−3 kg • m/s
1.68 × 10−2 kg • m/s
v2,f =  = 
0.600 kg
0.600 kg
= 2.80 m/s = 2.80 cm/s to the right
∆x = v2,f ∆t = (2.80 cm/s)(5.0 s) = 14 cm
I Ch. 6–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆KE = KEf − KEi = 7.36 × 104 J − 1.15 × 105 J = −4.1 × 104 J
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51. m1 = 8.0 g
KEi = PEf
1
 mv 2
f
2
m2 = 2.5 kg
= mgh
vf = 2g
m/s
m) = 1.1 m/s
h = (2
)(
9.
81
2)(0
.0
60
v2,i = 0 m/s
h = 6.0 cm
2
g = 9.81 m/s
(m1 + m2)vf − m2 v2,i (0.0080 kg + 2.5 kg)(1.1 m/s) − (2.5 kg)(0 m/s)
v1,i =  = 
0.0080 kg
m1
I
(2.5 kg)(1.1 m/s)
v1,i =  = 340 m/s
0.0080 kg
52. m1 = 52.0 g
Because the initial momentum is zero,
v1,i = 0 m/s
−m1v1,f −(52.0 g)(2.00 m/s)
v2,f =  =  = −0.680 m/s
m2
153 g
v2,i = 0 m/s
KEi = PEf
v1,f = 2.00 m/s
1
mv 2
2,f
2
m2 = 153 g
2
g = 9.81 m/s
53. m1 = 85.0 kg
m2 = 0.500 kg
v1, i = 0 m/s
v2, i = 0 m/s
v2, f = 20.0 m/s away from
ship = −20.0 m/s
= mgh
2
(− 0.680 m/s)2
v f
h = 2,
= 
= 2.36 × 10−2 m = 2.36 cm
2g
(2)(9.81 m/s2)
Because the initial momentum is zero,
−m2v2, f −(0.500 kg)(−20.0 m/s)
v1, f =  =  = 0.118 m/s toward the ship
m1
85.0 kg
∆x
30.0 m
∆t =  =  = 254 s
v1, f 0.118 m/s
∆x = 30.0 m
54. m1 = 2250 kg
v1, i = 10.0 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m2 = 2750 kg
mi v1, i + m2v2, i
(2250 kg)(10.0 m/s) + (2750 kg)(0 m/s)
vf =  = 
2250 kg + 2750 kg
m1 + m2
v2, i = 0 m/s
(2250 kg)(10.0 m/s)
vf = 
= 4.50 m/s
5.00 × 103 kg
d = 2.50 m
From the work-kinetic energy theorem,
θ = 180°
1
2
g = 9.81 m/s
1
Wnet = ∆KE = KEf − KEi = 2 (m1 + m2)(vf′)2 − 2 (m1 + m2)(vi′)2
where
vi′ = 4.50 m/s vf′ = 0 m/s
Wnet = Wfriction = Fkd(cos θ) = (m1 + m2)gµkd(cos θ)
1
(m1 + m2)gµkd(cos θ) = 2 (m1 + m2)[(vf′)2 − (vi′)2]
(vf ′)2 − (vi′)2
(0 m/s)2 − (4.50 m/s)2
−(4.50 m/s)2
µk =  = 
= 
2
2 gd(cos θ)
(2)(9.81 m/s )(2.50 m)(cos 180°)
−(2)(9.81 m/s2)(2.50 m)
µk = 0.413
Section One—Pupil’s Edition Solutions
I Ch. 6–15
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Givens
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55. F = 2.5 N to the right
F∆t + mv
(2.5 N)(0.50 s) + (1.5 kg)(0 m/s)
a. vf = i = 
m
1.5 kg
m = 1.5 kg
∆t = 0.50 s
I
vf = 0.83 m/s 0.83 m/s to the right
vi = 0 m/s
vi = 2.0 m/s to the left
= −2.0 m/s
F∆t + mv
(2.5 N)(0.50 s) + (1.5 kg)(−2.0 m/s)
b. vf = i = 
m
1.5 kg
1.2 N • s + (−3.0 kg • m/s) −1.8 kg • m/s
vf =  =  = −1.2 m/s
1.5 kg
1.5 kg
vf = 1.2 m/s to the left
56. m = 55 kg
∆y = −5.0 m
a. vf = ±
∆
2a
y = ± (2
m/s
)(
−9.
81
2)(−
5.
0m
) = ±9.9 m/s
vf = −9.9 m/s = 9.9 m/s downward
∆t = 0.30 s
vi = 0 m/s
a = −9.81 m/s2
vi = 9.9 m/s downward
= −9.9 m/s
mvf − mvi (55 kg)(0 m/s) − (55 kg)(−9.9 m/s)
 = 
b. F = 
∆t
0.30 s
F = 1.8 × 103 N = 1.8 × 103 N upward
vf = 0 m/s
∆y = −3.00 m
m2 = 5.98 × 1024 kg
v1,i = 0 m/s
v2,i = 0 m/s
a = −9.81 m/s2
58. m1 = m2
v1,i = 22 cm/s
v2,i = −22 cm/s
a. v1,f = ±
∆
2a
y = ± (2
m/s
m) = ±7.67 m/s = −7.67 m/s
)(
−9.
81
2)(−
3.
00
Because the initial momentum is zero,
m1v1,f = −m2 v2,f
−m1v1,f −(7.50 kg)(−7.67 m/s)
v2,f =  = 
= 9.62 × 10−24 m/s
m2
5.98 × 1024 kg
Because m1 = m2 , v2,f = v1,i + v2,i − v1,f .
Because kinetic energy is conserved and the two masses are equal,
1
 v 2
2 1,i
1
1
1
+ 2 v2,i2 = 2 v1,f 2 + 2 v2,f 2
v1,i2 + v2,i2 = v1,f 2 + v2,f 2
v1,i2 + v2,i2 = v1,f 2 + (v1,i + v2,i − v1,f )2
(22 cm/s)2 + (−22 cm/s)2 = v1,f 2 + (22 cm/s − 22 cm/s − v1,f )2
480 cm2/s2 + 480 cm2/s2 = 2v1,f 2
v1, f = ± 48
s2 = ±22 cm/s
0cm
2/
Because m1 cannot pass through m2, it follows that v1, f is opposite v1, i .
v1, f = −22 cm/s
v2, f = v1, i + v2, i − v1, f
v2, f = 22 cm/s + (−22 cm/s) − (−22 cm/s) = 22 cm/s
I Ch. 6–16
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
57. m1 = 7.50 kg
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59. mnuc = 17.0 × 10−27 kg
m3 = mnuc − (m1 + m2) = (17.0 × 10−27 kg) − [(5.0 × 10−27 kg) + (8.4 × 10−27 kg)]
m1 = 5.0 × 10−27 kg
m3 = 3.6 × 10−27 kg
m2 = 8.4 × 10−27 kg
p1 = m1v1,f = (5.0 × 10−27 kg)(6.0 × 106 m/s) = 3.0 × 10−20 kg • m/s
vnuc,i = 0 m/s
p2 = m2v2,f = (8.4 × 10−27 kg)(4.0 × 106 m/s) = 3.4 × 10−20 kg • m/s
v1,f = 6.0 × 106 m/s along
the positive y-axis
Because the initial momentum is zero, the final momentum must also equal zero.
v2,f = 4.0 × 106 m/s along
the positive x-axis
p3 = −(p1 + p2)
I
p1 + p2 + p3 = 0 kg • m/s
Because p1 and p2 are along the y-axis and the x-axis, respectively, the magnitude of
p3 can be found by using the Pythagorean theorem.
2
• m/s
• m/s
0−20
)2
+(3.
)2
p3 = p12+
p
.0
×1
kg
4×10−20
kg
2 = (3
p3 = (9
m2/s2)+
m2/s2)
.0
×10−40
kg2•
(1.
2×10−39
kg2•
• m2
p3 = (2
g2 s2) = 4.6 × 10−20 kg • m/s
1×10−40
k
/
p3 4.6 × 10−20 kg • m/s
=  = 1.3 × 107 m/s
v3,f = 
m3
3.6 × 10−27 kg
Because p1,2 is between the positive x-axis and the positive y-axis and because
p3 = −p1,2 , p3 must be between the negative x-axis and the negative y-axis.
p
tan q = 1
p2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
p
3.0
q = tan−1 1 = tan−1  = 41° below the negative x-axis
p2
3.4
Section One—Pupil’s Edition Solutions
I Ch. 6–17
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Rotational Motion and the Law of Gravity
Chapter
7
I
Practice 7A, p. 247
Givens
1. ∆s = +2.50 m
∆q = +1.67 rad
2. ∆q = −p rad
∆s = −1.2 m
∆s
+2.50 m
r =  =  = 1.50 m
∆q +1.67 rad
∆s −1.2 m
r =  =  = 0.38 m
∆q −p rad
p
3. ∆q = +
4
∆s = 29.8 m
∆s +29.8 m
r =  =  = 37.9 m
∆q
p
+rad
4
4. ∆s = +0.25 m
∆s + 0.25 m
a. ∆q =  =  = 2.5 rad
r
0.10 m
r = 0.10 m
r = 8.5 m
∆q = +0.75 rad
∆s = −4.2 m
r = 0.75 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Solutions
∆q = +135°
∆s = +2.6 m
b. ∆s = r∆q = (8.5 m)(+0.75 rad) = 6.4 m
∆s − 4.2 m
c. ∆q =  =  = −5.6 rad
r
0.75 m
180°
180°
∆q (deg) =  ∆q (rad) =  (−5.6 rad) = −320 °
p rad
p rad
p rad
p rad
d. ∆q (rad) =  q(deg) = ∆  (+135°) = +2.36 rad
180°
180°
∆s
+2.6 m
r =  =  = 1.1 m
∆q +2.36 rad
Practice 7B, p. 248
1. wavg = 29 rad/s
∆q = 3.5 rev
2. wavg = 2.2 rad/s
∆q = 3.3 rad
3. wavg = 7.0 rad/s
∆q = 2.3 rad
(3.5 rev)(2p rad/rev)
∆q
∆t =  =  = 0.76 s
29 rad/s
wavg
∆q
3.3 rad
∆t =  =  = 1.5 s
wavg 2.2 rad/s
∆q
2.3 rad
∆t =  =  = 0.33 s
wavg 7.0 rad/s
Section One—Pupil’s Edition Solutions
I Ch. 7–1
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Givens
Solutions
∆t = 10.0 s
∆q +2.3 rad
a. wavg =  =  = 0.23 rad/s
∆t
10.0 s
wavg = +0.75 rev/s
b. ∆q = wavg ∆t = (+0.75 rev/s)(2p rad/rev)(0.050 s) = 0.24 rad
4. ∆q = +2.3 rad
I
∆t = 0.050 s
∆q = −1.2 turns
∆t = 1.2 s
wavg = +2p rad/s
∆q = +1.5p rad
∆q (−1.2 turns)(2p rad/turn)
c. wavg =  =  = −6.3 rad/s
∆t
1.2 s
∆q
+1.5p rad
d. ∆t =  =  = 0.75 s
wavg +2p rad/s
Practice 7C, p. 250
1. w1 = +4.0p rad/s
∆t = 3.0 s
w2 − w1 8.0p rad/s − 4.0p rad/s 4.0p rad/s
 =  =  = 4.2 rad/s2
aavg = 
∆t
3.0 s
3.0 s
w2 = +8.0p rad/s
2. w1 = 8.5 rad/s
w2 = 15.4 rad/s
w2 − w1 15.4 rad/s − 8.5 rad/s 6.9 rad/s
 =  =  = 1.3 rad/s2
aavg = 
5.2 s
∆t
5.2 s
∆t = 5.2 s
∆t = 7.0 s
∆w +121.5 rad/s
a. aavg =  =  = 17 rad/s2
∆t
7.0 s
aavg = +0.75 rad/s2
b. ∆w = aavg ∆t = (+0.75 rad/s2)(0.050 s) = 0.038 rad/s
3. ∆w = +121.5 rad/s
∆w = −1.2 turns/s
∆t = 1.2 s
∆w (−1.2 turns/s)(2p rad/turn)
c. aavg =  =  = −6.3 rad/s2
∆t
1.2 s
Practice 7D, p. 252
1. ∆q = 18.0 rad
∆t = 5.00 s
wi = 2.00 rad/s
1
∆q = wi ∆t + 2 a ∆t2
2(∆q − wi ∆t) (2) [18.0 rad − (2.00 rad/s)(5.00 s)]
a = 
= 
(5.0 s)2
∆t 2
(2)(18.0 rad − 10.0 rad) (2)(8.0 rad)
a = 
= 
(5.0 s)2
25 s2
a = 0.64 rad/s2
2. ∆t = 0.50 s
wf = 4.0p rad/s
wi = 0 rad/s
I Ch. 7–2
wf = wi + a∆t
wf − w 4.0p rad/s − 0 rad/s
a = i =  = 25 rad/s2
∆t
0.50 s
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 0.050 s
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Givens
Solutions
3. wi = 1.0 rad/s
wf = wi + a ∆t
wf = 14.5 rad/s
∆t = 4.5 s
wf − w
14.5 rad/s − 1.0 rad/s 13.5 rad/s
a = i =  = 
∆t
4.5 s
4.5 s
a = 3.0 rad/s2
4. a = 22.4 rad/s2
wi = 10.8 rad/s
∆q = 3(2p) rad = 6p rad
5. wf = 31.0 rad/s
wi = 10.8 rad/s
2
a = 22.4 rad/s
I
wf 2 = wi2 + 2a∆q
2
wf = w
+2a
)2
+(2)
∆
q = (1
0.
8ra
d/s
(2
2.
4ra
d/s
2)(6
prad
)
i wf = 11
s2
+844
s2 = 96
s2 = 31.0 rad/s
7ra
d2/
rad
2/
1ra
d2/
wf = wi + a∆t
wf − wi 31.0 rad/s − 10.8 rad/s
∆t =  = 
a
22.4 rad/s2
20.2 rad/s
∆t = 2 = 0.902 s
22.4 rad/s
Section Review, p. 252
1. q(deg) = 25°
q (deg) = 35°
q (deg) = 128°
q(deg) = 270°
2. r = 5.0 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆s = −5.0 cm
1
3. ∆q = 2 (2p rad) = p rad
∆t = 10.0 s
4. w1 = 0.50 rad/s
w2 = 0.60 rad/s
180° (25°) = 0.44 rad
p rad
180° (35°) = 0.61 rad
p rad
180° (128°) = 2.23 rad
p rad
180° (270°) = 4.7 rad
p rad
a. q (rad) =  q(deg) =
180°
p rad
b. q (rad) =  q(deg) =
180°
p rad
c. q (rad) =  q(deg) =
180°
p rad
d. q (rad) =  q(deg) =
180°
p rad
∆s − 0.050 m
∆q =  =  = −1.0 rad
r
0.050 m
∆q p rad
wavg =  =  = 0.314 rad/s
∆t 10.0 s
w2 − w1 0.60 rad/s − 0.50 rad/s 0.10 rad/s
 =  =  = 0.20 rad/s2
aavg = 
∆t
0.50 s
0.50 s
∆t = 0.50 s
5. a = 0.20 rad/s2
wi = 0.50 rad/s
∆t = 1.0 s
w f = wi + a∆t = 0.50 rad/s + (0.20 rad/s2)(1.0 s)
wf = 0.50 rad/s + 0.20 rad/s = 0.70 rad/s
Practice 7E, p. 255
1. vt = 1.8 m/s
r = 0.80 m
v 1.8 m/s
w = t =  = 2.2 rad/s
r 0.80 m
Section One—Pupil’s Edition Solutions
I Ch. 7–3
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Givens
Solutions
2. vt = 6.93 m/s
r = 0.660 m
I
3. vt = 9.0 m/s
r = 0.75 m
4. w = 121.5 rad/s
r = 0.030 m
vt = 0.75 m/s
r = 0.050 m
w = 1.2 turns/s
v
6.93 m/s
w = t =  = 10.5 rad/s
r
0.660 m
v
9.0 m/s
w = t =  = 12 rad/s
r
0.75 m
a. vt = rw = (0.030 m)(121.5 rad/s) = 3.6 m/s
v
0.75 m/s
b. w = t =  = 15 rad/s
r
0.050 m
c. vt = rw = (3.8 m)(1.2 turns/s)(2p rad/turn) = 29 m/s
r = 3.8 m
vt = 2.0 p m/s
w = 1.5 p rad/s
v
2.0p m/s
d. r = t =  = 1.3 m
w 1.5 p rad/s
Practice 7F, p. 256
a = 1.0 rad/s2
2. at = 0.18 m/s2
a = 0.35 rad/s2
3. at = 9.4 × 10−2 m/s2
r = 0.15 m
1.5 m/s2
a
r = t = 2 = 1.5 m
1.0 rad/s
a
0.18 m/s2
a
r = t = 2 = 0.51 m
a 0.35 rad/s
a
9.4 × 10−2 m/s2
a = t =  = 0.63 rad/s2
r
0.15 m
Practice 7G, p. 258
1. ac = 3.0 m/s2
vt = acr = (3
m/s
m) = 2.5 m/s
.0
2)(2
.1
r = 2.1 m
2. ac = 250 m/s2
vt = acr = (2
m/s
50
2)(0
.5
0m
) = 11 m/s
r = 0.50 m
3. ac = 1.5 m/s2
r = 1.5 m
ac = rw 2
w=
I Ch. 7–4
vt = ac r = (1
m/s
m) = 1.5 m/s
.5
2)(1
.5
ar = 1.
15.5m
m/s = 1.0 rad/s
c
Holt Physics Solution Manual
2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. at = 1.5 m/s2
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Givens
4. w = 0.512 rad/s
ac = 15.4 m/s2
5. r = 0.20 m
w = 20.5 rad/s
Solutions
a
15.4 m/s2
r = c2 = 2 = 58.7 m
w
(0.512 rad/s)
ac = rw 2 = (0.20 m)(20.5 rad/s)2 = 84 m/s2
I
Section Review, p. 259
1. w = 5.0 rad/s
vt = rw = (5.0 rad/s)(5.0 m) = 25 m/s
r = 5.0 m
3. r = 9.5 m
at = ra = (9.5 m)(0.15 rad/s2) = 1.4 m/s2
a = 0.15 rad/s2
4. r = 12 m
w = 1.2 rad/s
ac = rw 2 = (12 m)(1.2 rad/s)2 = 17 m/s2
Practice 7H, p. 261
1. r = 2.10 m
vt = 2.50 m/s
r
(2.10 m)
m = Fc 2 = (88.0 N) 2 = 29.6 kg
vt
(2.50 m/s)
Fc = 88.0 N
2. vt = 13.2 m/s
r = 40.0 m
r
(40.0 m)
m = Fc 2 = (377 N) 2 = 86.5 kg
vt
(13.2 m/s)
Fc = 377 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. r = 1.50 m
w = 1.20 rad/s
Fc
40.0 N
m = 
=  = 18.5 kg
rw 2 (1.50 m)(1.20 rad/s)2
Fc = 40.0 N
4. m = 905 kg
3.25 km
r = 
2p
vt =
.25 × 10 m 2140 N
rmF = 3
2p

905kg = 35.0 m/s
3
c
Fc = 2140 N
Practice 7I, p. 265
1. m1 = m2 = 0.800 kg
Fg = 8.92 × 10−11 N
N•m2
G = 6.673 × 10−11 
kg2
r=
G m1m2
 =
Fg
r = 0.692 m
N•m2
6.673 × 10−11 
(0.800 kg)(0.800 kg)
kg2

8.92 × 10−11 N
Section One—Pupil’s Edition Solutions
I Ch. 7–5
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Givens
Solutions
2. m1 = 6.4 × 1023 kg
15
m2 = 9.6 × 10 kg
r=
15
Fg = 4.6 × 10 N
I
N•m2
G = 6.673 × 10−11 
kg2
3. m1 = 67.5 kg
Gm m

F
=
1 2
g
N•m2
6.673 × 10−11 
(6.4 × 1023 kg)(9.6 × 1015 kg)
kg2

4.6 × 1015 N
r = 9.4 × 106 m = 9.4 × 103 km
2
24
m m2
−11 N•m (67.5 kg)(5.98 × 10 kg)



a. Fg = G 1
=
6.673
×
10
= 664 N
r2
kg2
(6.37 × 106 m)2
m2 = 5.98 × 1024 kg
r = 6.37 × 106 m
N•m2
G = 6.673 × 10−11 
kg2
2
23
m m2
−11 N•m (67.5 kg)(6.34 × 10 kg)



b. Fg = G 1
=
6.673
×
10
= 243 N
2
2
6
r
kg
(3.43 × 10 m)2
m2 = 6.34 × 1023 kg
r = 3.43 × 106 m
m m2
N•m2 (67.5 kg)(5 × 1023 kg)

c. Fg = G 1
= 6.673 × 10−11 
= 1 × 104 N
2
r
kg2
(4 × 105 m)2
m2 = 5 × 1023 kg
r = 4 × 105 m
Section Review, p. 265
3. m1 = 90.0 kg
m m2
N • m2 (90.0 kg)(60.0 kg)

Fg = G 1
= 6.673 × 10−11 
= 3.60 × 10−7 N
2
r
kg2
(1.00 m)2
r = 1.00 m
m2 = 60.0 kg
N • m2
G = 6.673 × 10−11 
kg2
4. m = 90.0 kg
Fc = mrw 2 = (90.0 kg)(11.5 m)(1.15 rad/s)2 = 1.37 × 103 N
r = 11.5 m
5. m1 = 90.0 kg
(1370 N)(6.37 × 106 m)2
Fg r 2
m2 =  =  = 9.26 × 1024 kg
Gm1
N • m2
6.673 × 10−11 
(90.0 kg)
kg2
r = 6.37 × 106 m
Fg = 1370 N
2
N•m
G = 6.673 × 10−11 
kg2
Review and Assess, pp. 269–273
5. ∆q = 0.34 rad
∆s = 12 m
I Ch. 7–6
∆s
12 m
r =  =  = 35 m
∆q 0.34 rad
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
w = 1.15 rad/s
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Givens
6. ∆q1 = 35°
∆q2 = 35 rad
Solutions
∆s
∆s
∆s
r = 1 = 2 = 3
∆q1 ∆q2 ∆q3
∆s1 = 2.5 m
9.0 × 102 m
143 m
2.5 m
r =  =  = 
(35 rev)(2p rad/rev)
35 rad
p rad
(35°) 
120°
∆s2 = 143 m
r = 4.1 m = 4.1 m = 4.1 m
∆q3 = 35 rev
∆s3 = 9.0 × 102 m
7. ∆q = 4.00 rad
wavg = 2p rad/60 s
8. w1 = 33 rev/min
w2 = 11 rev/min
∆t = 2.0 s
I
r = 4.1 m
∆q 4.00 rad
∆t =  =  = 38.2 s
wavg 2p rad

60 s
w2 − w1

aavg = 
∆t
(11 rev/min − 33 rev/min)(2p rad/rev)(1 min/60 s)
aavg = 
2.0 s
(−22 rev/min)(2p rad/rev)(1 min/60 s)
aavg =  = −1.2 rad/s2
2.0 s
9. ∆w = 2.7 rad/s
∆t = 1.9 s
10. wf = 0.20 rev/s
wi = 0 rad/s
∆t = 30.0 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
11. ∆t = 3.20 s
wf = 2628 rad/s
wi = 0 rad/s
∆w 2.7 rad/s
a =  =  = 1.4 rad/s2
∆t
1.9 s
wf = wi + a∆t
wf − w (0.20 rev/s)(2p rad/rev) − 0 rad/s
a = i =  = 0.042 rad/s2
30.0 s
∆t
wf − w 2628 rad/s − 0 rad/s
a. a = i =  = 821 rad/s2
∆t
3.20 s
1
b. ∆q = w i ∆t + 2a∆t2
1
∆q = (0 rad/s)(3.20 s) + 2(821 rad/s2)(3.20 s)2 = 4.20 × 103 rad
12. ∆q = 4.7 rev
∆t = 1.2 s
w i = 0 rad/s
21. r = 32 cm
vt = 49 m/s
22. r = 12 m
vt = 2.18 × 10−2 m/s
1
∆q = w i ∆t + 2a∆t2
Because w i = 0 rad/s,
2∆q (2)(4.7 rev)(2p rad/rev)
a = 
= 
= 41 rad/s2
∆t2
(1.2 s)2
v
49 m/s
= 1.5 × 102 rad/s
w = t = 
r 32 × 10−2 m
v 2.18 × 10−2 m/s
w = t =  = 1.8 × 10−3 rad/s
r
12 m
or
w = (1.8 × 10−3 rad/s)(3600 s/h) = 6.5 rad/h
Section One—Pupil’s Edition Solutions
I Ch. 7–7
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Givens
Solutions
23. a = 1.5 rad/s2
a
48 × 10−2 m/s2
r = t = 
= 0.32 m
a
1.5 rad/s2
at = 48 cm/s2
I
a
0.086 m/s2
r = t = 
= 0.023 m = 2.3 cm
a
3.8 rad/s2
24. a = 3.8 rad/s2
at = 0.086 m/s2
25. ac = 145 m/s2
c = (0
ra
.3
4m
)(
14
5m
/s
2) = 7.0 m/s
vt =
r = 0.34 m
26. ac = 28 m/s2
vt = ra
m/s
c = (2
7×10−2m
)(
28
2) = 2.7 m/s
r = 27 cm
r
(10.0 m)
a. m = Fc 2 = (2.06 × 104 N) 2 = 515 kg
vt
(20.0 m/s)
37. vt = 20.0 m/s
Fc = 2.06 × 104 N
r = 10.0 m
b. Fc = mg
mvt2
 = mg
r
r = 15.0 m
g = 9.81 m/s2
vt = gr
= (9
.8
1m
/s
2)(1
5.
0m
) = 12.1 m/s
mvt2
Ftotal = Fc + Fg = 
+ mg
r
Ftotal ≤ Frope, max
38. r = 10.0 m
vt = 8.0 m/s
Frope, max = 1.0 × 103 N
1.0 × 103 N
Frope, max

mmax = 
=
2
vt
(8.0 m/s)2
 + g
 + 9.81 m/s2
r
10.0 m
3
1.0 × 10 N
1.0 × 103 N

=
mmax = 
6.4 m/s2 + 9.81 m/s2
16.2 m/s2
mmax = 62 kg
39. Fg = 3.20 × 10−8 N
r=
m1 = 50.0 kg
m2 = 60.0 kg
r = 2.50 m
−11
G = 6.673 × 10
2
2
N • m /kg
40. m1 = 9.11 × 10−31 kg
m2 = 1.67 × 10−27 kg
Fg = 1.0 × 10−47 N
r=
Gm1m2 (6.673 × 10−11 N • m2/kg3)(9.11 × 10−31 kg)(1.67 × 10−27 kg)
 = 
Fg
1.0 × 10−47 N
r = 1.0 × 10−10 m = 0.10 nm
G = 6.673 × 10−11 N • m2/kg2
I Ch. 7–8
(6.673 × 10−11 N • m2/kg2)(50.0 kg)(60. kg)

3.20 × 10−8 N
Gm1m2
 =
Fg
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v2
Frope, max = mmax t + g
r
g = 9.81 m/s2
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Givens
Solutions
41. ∆q = 2p rad
∆q
2p rad
wavg =  =  = 1.99 × 10−7 rad/s
∆t
(365.25 days)(24 h/day)(3600 s/h)
∆t = 365.25 days
42. wi = 0 rad/s
1
1
∆q1 = 2(wi + wf)∆t1 = 2(0 rad/s + 11p rad/s)(8.0 s) = 44p rad
wf = 11 p rad/s
∆t1 = 8.0 s
I
For stopping stage, wi = 11p rad/s and wf = 0 rad/s
1
1
∆q2 = 2(wi + wf)∆t2 = 2(11p rad/s + 0 rad/s)(12.0 s) = 66p rad
∆t2 = 12.0 s
∆qtot = ∆q1 + ∆q2 = (44p rad + 66p rad)(1 rev/2p rad) = [(1.10 × 102)p rad](1 rev/2p rad)
∆qtot = 55.0 rev
43. vt = 105 m/s
v2
a. ac = t
r
m = 80.0 kg
(105 m/s)2
vt2
r= 
= 
= 161 m
(7.00)(9.81 m/s2)
ac
ac = 7.00g
g = 9.81 m/s2
b. Fc = mac = (80.0 kg)(7.00)(9.81 m/s2) = 5.49 × 103 N
44. vt = 30.0 m/s
2
at = −2.00 m/s
r = 0.300 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
wf = 0 rad/s
v
wf = t
r
at
a = 
r
2
v
wf2 − t
2
2
r
wf − wi
∆q = 
= 
2a
at
2 
r
2
30.0 m/s
(0 rad/s)2 − 
0.300 m
∆q =  (1 rev/2p rad) = (7.50 × 102 rad)(1 rev/2p rad)
−2.00 m/s2
(2) 
0.300 m
∆q = 119 rev
2.40 cm
45. r =  = 1.20 cm
2
wi = 18.0 rad/s
wf = 0 rad/s
2
(0 rad/s)2 − (18.0 rad/s)2
wf − wi2
∆q = 
= 
= 85.3 rad
(2)(−1.90 rad/s2)
2a
∆s = r∆q = (1.20 × 10−2 m)(85.3 rad) = 1.02 m
a = − 1.90 rad/s2
46. r = 50.0 cm
∆q = 40 rev
∆t = 1.00 min
wi = 0 rad/s
1
∆q = 2(wi + wf)∆t
(2)(40 rev)(2p rad/rev)
2∆q
wf =  − wi =  − 0 rad/s
(1.00 min)(60 s/min)
∆t
wf = 8.38 rad/s
Section One—Pupil’s Edition Solutions
I Ch. 7–9
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Givens
Solutions
47. Fg = 13 500 N
2
2
v 2 (50.0 × 103 m/h) (1h/3600 s)
a. ac = t = 
= 0.965 m/s2
2
2.00 × 10 m
r
r = 2.00 × 102 m
vt = 50.0 km/h
2
I
g = 9.81 m/s
Fg
13 500 N
b. Fc = mac =  ac = 2 (0.965 m/s2) = 1.33 × 103 N
g
9.81 m/s
c. Fc = Fk = mkFn = mkFg
F
1330 N
mk = c =  = 0.0985
Fg 13 500 N
48. m = 2.00 × 103 kg
r = 20.0 m
mk = 0.70
g = 9.81 m/s2
Fk = mkFn = mkmg
v2
Fc = mt
r
Fk = Fc
v2
mkmg = m t
r
vt = rm
m/s
0.
0m
)(
0.
70
)(
9.
81
2)
kg = (2
vt = 12 m/s
r = 1.496 × 1011 m − 0.00384 × 1011 m = 1.492 × 1011 m
ms = 1.99 × 1030 kg
N • m2
G = 6.673 × 10−11 
kg2
m ms
a. Fg = G m
r2
r = 3.84 × 108 m
mEmm

b. Fg = G 
r2
mE = 5.98 × 1024 kg
mm = 7.36 × 1022 kg
mE = 5.98 × 1024 kg
ms = 1.99 × 1030 kg
r = 1.496 × 1011 m
50. requator = 6.37 × 106 m
w = 2p rad/day
rpole = 0 m
N • m2 (7.36 × 1022 kg)(1.99 × 1030 kg)

Fg = 6.673 × 10−11 
= 4.39 × 1020 N
kg2
(1.492 × 1011 m)2
N • m2 (5.98 × 1024 kg)(7.36 × 1022 kg)

Fg = 6.673 × 10−11 
= 1.99 × 1020 N
(3.84 × 108 m)2
kg2
m ms
c. Fg = G E
r2
N • m2 (5.98 × 1024 kg)(1.99 × 1030 kg)

Fg = 6.673 × 10−11 
= 3.55 × 1022 N
kg2
(1.496 × 1011 m)2
a. ac = requatorw2 = (6.37 × 106 m)[(2p rad/day)(1 day/24 h)(1 h/3600 s)]2
ac = 3.37 × 10−2 m/s2
b. ac = rpolew2 = (0 m)(2p rad/day)2 = 0 m/s2
I Ch. 7–10 Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
49. mm = 7.36 × 1022 kg
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Givens
Solutions
51. r = 30.0 cm
Fc = Fs = msFn = msFg = msmg
ms = 0.53
Fc = mrw f 2, where w f = a∆t
a = 0.50 rad/s2
Fc = mr(a∆t)2 = mra 2∆t 2
g = 9.81 m/s2
2
wi = 0 rad/s
52. mp = 0.025 kg
r = 1.0 m
m = 1.0 kg
g = 9.81 m/s2
53. r = 3.00 m
msmg
ms g
2
2
∆t =

= 

mra = 
mr
a ra
∆t =
(0.53)(9.81 m/s )
 = 8.3 s
(0.300 m)(0.50 rad/s )
Fc
2
I
2
2 2
a. Fc = Fg = mg = (1.0 kg)(9.81 m/s2) = 9.8 N
v2
b. Fc = mp t
r
Fcr
vt =  =
mp
=

0.02
5k
g (9.8 N)(1.0 m)
2.0 × 101 m/s
Fn = Fc = mrw 2
w = 5.00 rad/s
Fs = Fg
g = 9.81 m/s2
Fs = msFn = msmrw 2
msmrw 2 = mg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g
9.81 m/s2
ms = 2 = 2 = 0.131
rw
(3.00 m)(5.00 rad/s)
Section One—Pupil’s Edition Solutions
I Ch. 7–11
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Rotational Equilibrium and Dynamics
Chapter
8
I
Practice 8A, p. 282
Givens
1. F = 3.0 N
Solutions
t = Fd(sin q) = (3.0 N)(0.25 m)(sin 90.0°) = 0.75 N • m
d = 0.25 m
q = 90.0°
2. m = 3.0 kg
a. t = Fd(sin q) = mgd(sin q)
d = 2.0 m
t = (3.0 kg)(9.81 m/s2)(2.0 m)(sin 5.0°) = 5.1 N • m
q = 5.0°
g = 9.81 m/s2
q = 15.0°
3. t = 40.0 N • m
d = 30.0 cm
b. t = mgd(sin q) = (3.0 kg)(9.81 m/s2)(2.0 m)(sin 15.0°) = 15 N • m
For a given torque, the minimum force must be applied perpendicular to the lever
arm, or sin q = 1. Therefore,
t 40.0 N • m
F =  =  = 133 N
d
0.300 m
Section Review, p. 282
3. F30 = 30.0 N
q30 = 45°
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d30 = 0 m
F25 = 25.0 N
q25 = 59°
t30 = F30d30(sin q30) = (30.0 N)(0 m)(sin 45°) = 0 N • m
t25 = F25d25(sin q25) = (25.0 N)(2.0 m)(sin 59°) = 43 N • m
t10 = F10d10(sin q10) = (10.0 N)(4.0 m)(sin 23°) = 16 N • m
The bar will rotate counterclockwise because t25 > t10 .
d25 = 2.0 m
F10 = 10.0 N
q10 = 23°
d10 = 4.0 m
Section One—Pupil’s Edition Solutions
I Ch. 8–1
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Practice 8B, p. 288
Givens
Solutions
1. L = 5.00 m
I
Apply the first condition of equilibrium.
Fg,b = 315 N
x-component: Fx = Rx − FT (cos q) = 0
Fg,p = 545 N
y-component: Fy = Ry + FT (sin q) − Fg,p − Fg,b = 0
Ry = −FT (sin q) + Fg,p + Fg,b
q = 53°
d = 1.50 m
Apply the second condition of equilibrium.
L
L
L
FT (sin q )  + Fg,p  − d − Ry  = 0
2
2
2
Substitute the Ry value from the 1st condition y-component equation.
L
L
L
FT (sin q )  + Fg,b  − d + [FT (sin q) − Fg,p − Fg,b]  = 0
2
2
2
Fg,p + Fg,b = 545 N + 315 N = 8.60 × 102 N
FT (sin 53°)(2.50 m) + (545 N)(1.00 m) + [FT (sin 53°) − 8.60 × 102 N](2.50 m) = 0
FT (2.0 m) + 545 N • m + FT (2.0 m) − 2150 N • m = 0
(4.0 m)FT − 1610 N • m = 0
1610 N • m
FT =  = 4.0 × 102 N
4.0 m
Solve for R:
Ry = −FT (sin q ) + Fg,p + Fg,b
Ry = −(4.0 × 102 N)(sin 53°) + 8.60 × 102 N = −320 N + 8.60 × 102 N = 540 N
Rx = FT (cos q ) = (4.0 × 102 N)(cos 53°) = 240 N
2 + R 2 = 240 N)2 + (540)2 = 58 000 N2 + 290 000 N2
R= R
x y
2. Fg,b = 4.00 × 105 N
Apply the first condition of equilibrium.
Fp,1 + Fp,2 − Fg,b − Fg,c = 0
db = 0 m
Fg,c = 1.96 × 10 N
Fp,1 = Fg,b + Fg,c − Fp,2
dc = 10.0 m − 8.00 m = 2.0 m
Fp,1 = 4.00 × 105 N + 1.96 × 104 N − Fp,2 = 4.20 × 105 N − Fp,2
d1 = 10.0 m − 3.00 m = 7.0 m
Apply the second condition of equilibrium using the center of mass of the bridge as
the pivet point.
4
d2 = 7.0 m
Fp,1d1 + Fg,c dc − Fp,2d2 = 0
Substitute the Fp,1 value from the first-condition equation.
(4.20 × 105 N − Fp,2)d1 + Fg,c dc − Fp,2 d2 = 0
(4.20 × 105 N)(7.0 m) − Fp,2 (7.0 m) + (1.96 × 104 N)(2.0 m) − Fp,2 (7.0 m) = 0
2.9 × 106 N • m
2.9 × 106 N • m + 3.9 × 104 N • m
Fp,2 =  =  = 2.1 × 105 N
14.0 m
14.0 m
Fp,1 = 4.20 × 105 N − 2.1 × 105 N = 2.1 × 105 N
I Ch. 8–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
R = 35
000
0N
2 = 590 N
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Givens
3. Fg,w = 700.0 N
Solutions
Apply the first condition of equilibrium.
Fg,s = 200.0 N
FT,1 + FT,2 − Fg,w − Fg,s = 0
dw = 1.00 m
FT,1 = Fg,w + Fg,s − FT,2
ds = 1.50 m
d2 = 3.00 m
Choose the end of side 1 as the pivot point and apply the second condition of
equilibrium.
I
FT,2 d2 − Fg,w dw − Fg,s ds = 0
Fg,w dw + Fg,s ds
FT,2 = 
d2
(700.0 N)(1.00 m) + (200.0 N)(1.50 m)
FT,2 = 
3.00 m
7.00 102 N • m + 3.00 102 N • m
10.00 102 N • m
FT,2 =  =  = 333 N
3.00 m
3.00 m
FT,1 = Fg,w + Fg,s − FT,2 = 700.0 N + 200.0 N − 333 N = 567 N
4. Fg,1 = 400.0 N
a. Using the pivot point as the axis:
Fg,2 = 300.0 N
Fg,1d1 − Fg,2d2 = Fg,1d1 − Fg,2 (2.0 m – d1) = 0
d2 = 2.0 m − d1
(400.0 N)d1 − (300.0 N)(2.0 m) + (300.0 N)d1 = 0
(700.0 N)d1 − 6.0 102 N • m = 0
6.0 102 N • m
d1 =  = 0.86 m from the 400.0 N child
700.0 N
d1 = 0.86 m
b. Using the pivot point as the axis point:
Fg,3 = 225 N
Fg,1d1 + Fg,3d3 − Fg,2d2 − Fg,4d4 = 0
d3 = 0.86 m − 0.200 m
= 0.66 m
Fg,1d1 + Fg,3d3 − Fg,2d2
d4 = 
Fg,4
(400.0 N)(0.86 m) + (225 N)(0.66 m) − (300.0 N)(1.1 m)
d4 = 
325 N
340 N • m + 150 N • m − 330 N • m
160 N • m
d4 =  = 
325 N
325 N
d2 = 2.0 m − 0.86 m = 1.1 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fg,4 = 325 N
d4 = 0.49 m from the pivot point on the same side as the 300.0 N child
Section Review, p. 289
5. Fg ,b = 40.0 N
Fg ,1 = 510 N
Fg ,2 = 350 N
d1 = 1.50 m
a. Choose the center of mass as the pivot point and apply the second condition of
equilibrium.
Fg ,1d1 − Fg ,2 d2 = 0
Fg ,1d1
(510 N)(1.50 m)
d2 =  =  = 2.2 m from center
Fg , 2
350 N
b. Apply the first condition of equilibrium.
Fs − Fg ,b − Fg ,1 − Fg ,2 = 0
Fs = Fg ,b + Fg ,1 + Fg ,2
Fs = 40.0 N + 510 N + 350 N = 9.0 × 102 N
Section One—Pupil’s Edition Solutions
I Ch. 8–3
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Practice 8C, p. 291
Givens
Solutions
1. R = 0.50 m
I
M = 100.0 kg
(0 rev/min − 50.0 rev/min)(2p rad/rev)(1 min/60 s)
wf − w
a. α = i =  = −0.87 rad/s2
6.0 s
∆t
wi = 50.0 rev/min
b. t = Ia = (2MR2)a
1
wf = 0 rev/min
1
t = 2(100.0 kg)(0.50 m)2(−0.87 rad/s2) = −11 N • m
∆t = 6.0 s
2. R = 0.33 m
wf − w
t = Ia = (MR2) i
∆t
M = 1.5 kg
wi = 98.7 rad/s
wf = 0 rad/s
(0 rad/s − 98.7 rad/s)
t = (1.5 kg)(0.33 m)2 = −8.1 N • m
2.0 s
∆t = 2.0 s
3. R = 0.075 m
M = 0.500 kg
1
a. t = Ia = (2MR2)a
Because F is perpenddicular to d,
∆s = 4.00 m
t = Fd = (mg)(R)
m = 5.00 kg
1
MR2a
2
2
g = 9.81 m/s
= mgR
2mg (2)(5.00 kg)(9.81 m/s2)
a =  =  = 2.6 × 103 rad/s2
(0.500 kg)(0.0075 m)
MR
wi = 0 rad/s
(2)(2.6 × 10 rad/s )(4.00 m)
w = w
+ 
+2aR∆s = (0rad/s)
(0.075 m)
∆s
b. wf 2 = wi2 + 2a∆q = wi2 + 2a 
R
f
i
2
2
3
2
Practice 8D, p. 294
1. m = 80.0 kg
1
Li = Lc,i + Lm,i = Tcwi + Imwi = 2MR2wi + mri2wi
1
M = 6.50 × 102 kg
Lf = Lc,f + Lm,f = Icwf + Imwf = 2MR2wf + mrf2wf
R = 2.00 m
Li = Lf
ri = 2.0 m
2MR2 + mri2wi = 2MR2 + mrf2wf
1
1
2MR2 + mri2wi 2(6.50 × 102 kg)(2.00 m)2 + (80.0 kg)(2.0 m)2(0.30 rad/s)
rf = 1.0 m
wi = 0.30 rad/s
1
1
wf = 
= 
1
1
MR2 + mr 2
(6.50 × 102 kg)(2.00 m)2 + (80.0 kg)(1.0 m)2
f
2
2
(1.30 × 103 kg • m2 + 320 kg • m2)(0.30 rad/s)
(1620 kg • m2)(0.30 rad/s)

wf = 
=
1.30 × 103 kg • m2 + 8.0 × 101 kg • m2
1380 kg • m2
wf = 0.35 rad/s
I Ch. 8–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
wf = 5.3 × 102 rad/s
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Givens
Solutions
2. M = 2.0 kg
Li = Lw,i + Lr,i = Iwwi + Irwi = MR2wi + mri2wi
R = 0.30 m
wi = 25 rad/s
m = 0.30 kg
Lf = Lw,f + Lr,f = Iwwf + Irwf = MR2wf + mrf2wf
Li = Lf
(MR2 + mri2)wi = (MR2 + mrf2)wf
ri = 0.19 m
rf = 0.25 m
wf =
2
(MR + mri2)wi


MR2 + mrf2
I
2
2
[(2.0 kg)(0.30 m) + (0.30 kg)(0.19 m) ](25 rad/s)
= 
(2.0 kg)(0.30 m)2 + (0.30 kg)(0.25 m)2
(0.18 kg • m2 + 1.1 × 10−2 kg • m2)(25 rad/s)
wf = 
0.18 kg • m2 + 1.9 × 10−2 kg • m2
(0.19 kg • m2)(25 rad/s)
wf = 
0.20 kg • m2
wf = 24 rad/s
3. M = 10.0 kg
1
Li = Icwi = 2MR2wi
R = 1.00 m
Lf = (Ic + Ip)wf = 2MR2 + mr2wf
1
wi = 7.00 rad/s
m = 0.250 kg
r = 0.900 m
Li = Lf
1
MR2w
i
2
= 2MR2 + mr2wf
1
1
1
(10.0 kg)(1.00 m)2 (7.00 rad/s)
MR2w
i
2
2
wf = 
= 
1
1
(10.0 kg)(1.00 m)2 + (0.250 kg)(0.900 m)2
MR2 + mr2
2
2
35.0 kg • m2/s
35.0 kg • m2/s

wf = 
=
5.20 kg • m2
5.00 kg • m2 + 0.202 kg • m2
wf = 6.73 rad/s
4. ri = 8.8 × 1010 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
rf = 5.2 × 1012 m
4
vf = 5.4 × 10 m/s
Li = Lf
Iiwi = If w f
mri2wi = mrf 2wf
vf
v
wi = i and w f = , so
rf
ri
rv
(8.8 × 1010 m)(5.4 × 104 m/s)
= 9.1 × 102 m/s
vf = i i = 
rf
5.2 × 1012 m
5. ri = 0.54 m
Li = Lf
rf = 0.040 m
Iiwi = Ifwf
m = 25 g
mri2wi = mrf2wf
wi = 0.35 rev/s
ri2wi (0.54 m)2 (0.35 rev/s)(2p rad/rev)
 = 
wf = 
rf2
(0.040 m)2
wf = 4.0 × 102 rad/s
Section One—Pupil’s Edition Solutions
I Ch. 8–5
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Practice 8E, p. 297
Givens
Solutions
MEi = MEf
1. h = 2.00 m
2
g = 9.81 m/s
I
Ivf 2 1 2
mr 2 1 2 3
1
mgh = 2 mvf 2 + 
= 2 vf m + 
= 2 vf 2 m
2
r
2r 2
4 mgh 4
vf 2 =  = 3 gh
3m
vf =
2. m = 1.5 kg
r = 0.33 m
h = 14.8 m
g = 9.81 m/s2
gh = (9.81m/s)(2.00m) = 5.11 m/s
4

3
4

3
2
The sphere has the greater speed, and so would win the race.
MEi = MEf
2
= mv (1 + 1) = mv
v
1
1
1
mgh = 2 mvf2 + 2 Iwf2 = 2 mvf 2 + (mr2) f
r
1

2
2
f
f
2
vf = gh
= (9
.8
1m
/s
2)(1
4.
8m
) = 12.0 m/s
3. r = 25 cm
MEi = MEf
d = 4.0 m
mgh = 2 mvf2 + Iwf2 = 2 mvf2 + 3mr2
q = 30.0°
mgh = mgd(sin q) = 6mvf2
g = 9.81 m/s2
vi = 0 m/s
1
1
2
5
vf =
2
=
vf

r
1 + 3 = 2 mvf23
1
 mv 2
f
2
2
1
5
in q)
6g
d(s5
1
d = 2 (vi + vf)∆t
2d
2d
∆t =  = 
6g
d(sin q)
vi + vf
vi + 
5
∆t =
(2)(4.0 m)
0 m/s +
(6)(9.81 m/s2)(4.0 m)(sin 30.0°)

5
=
(2)(4.0 m)
(6)(9.81 m/s2)(4.0 m)(sin 30.0°)

5
∆t = 1.6 s
Section Review, p. 297
1. m = 3.0 kg
Li = Iiw i = (2mri 2 + I)w i
w i = 0.75 rad/s
Lf = If w f = (2mrf 2 + I)w f
ri = 1.0 m
Li = L f
rf = 0.30 m
(2 mri2 + I)wi = (2mrf2 + I)wf
I = 3.0 kg • m2
(2 mr 2 + I)wi [(2)(3.0 kg)(1.0 m)2 + 3.0 kg • m2](0.75 rad/s)
wf = i 
= 
2 mrf2 + I
(2)(3.0 kg)(0.30 m)2 + 3.0 kg • m2
(9.0 kg • m2)(0.75 rad/s)
(6.0 kg • m2 + 3.0 kg • m2)(0.75 rad/s)
wf = 
= 
2
2
3.5 kg • m2
0.54 kg • m + 3.0 kg • m
w f = 1.9 rad/s
I Ch. 8–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Givens
2. mm = 4.0 kg
Solutions
MEi = MEf
1
1
1
mw = 8.0 kg
mm gh = 2mm v f2 + 2m w v f2 + 2Iw f2
rw = 0.50 m
1
1
1 v
mm gh = 2mmv f2 + 2mw v f2 + 2I f
rw
I = 2.0 kg • m2
h = 2.0 m
g = 9.81 m/s2
vf =
v=
3. F = 40.0 N
d = 0.15 m
r = 50.0 cm
∆v = 2.25 m/s
∆t = 3.0 s
2
2mm gh

=
I
mm + mw + 2
rw
I
(2)(4.0 kg)(9.81 m/s2)(2.0 m)

.0 kg • m2
4.0 kg + 8.0 kg + 2

(0.50 m)2
(2)(4.0 kg)(9.81 m/s2)(2.0 m)
 =
12.0 kg + 8.0 kg
(2)(4.0 kg)(9.81 m/s2)(2.0 m)
 = 2.8 m/s
20.0 kg
t = Ia
Because F is perpendicular to d, t = Fd.
t Fd∆t Fdr∆t
I =  =  = 
∆w
∆v
a
(40.0 N)(0.15 m)(0.500 m)(3.0 s)
I =  = 4.0 kg • m2
2.25 m/s
Section Review, p. 301
2. eff = 0.73
din = 18.0 m
dout = 3.0 m
m = 58 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
3. Fg = 950 N
Fapplied = 350 N
W ut
eff = o
Win
Fout dout
 where Fout = mg
eff = 
Findin
mg dout
(58 kg)(9.81 m/s2)(3.0 m)
Fin =  
=  = 1.3 × 102 N
ef f din
(0.73)(18.0 m)
F ut
Fg
950 N
= 
=  = 2.7
MA = o
Fin Fapplied 350 N
Chapter Review and Assess, pp. 305–312
9. m = 54 kg
r = 0.050 m
t = Fd(sin q) = mgr(sin q)
t = (54 kg)(9.81 m/s2)(0.050 m)(sin 90°) = 26 N • m
g = 9.81 m/s2
q = 90°
10. q = 90.0° − 8.0° = 82.0°
m = 1130 kg
d = 3.05 m − 1.12 m −
0.40 m = 1.53 m
g = 9.81 m/s2
tnet = tg + tjack = 0
mgd(sin q) + tjack = 0
tjack = −mgd(sin q) = −(1130 kg)(9.81 m/s2)(1.53 m)(sin 82.0°)
tjack = −1.68 × 104 N • m = 1.68 × 104 N • m clockwise
Section One—Pupil’s Edition Solutions
I Ch. 8–7
Givens
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11. d = 15.0 m
q = 90.0° − 20.0° = 70.0°
Solutions
a. tmax = Fg,maxd(sin q) = (450 N)(15.0 m)(sin 70.0°) = 6.3 × 103 N • m
Fg,max = 450 N
I
q = 90.0° − 40.0° = 50.0°
20. Fg,s = 205 N
l
= 3.00 m
Fg,w = 675 N
t ax
6.3 × 103 N • m
b. Fg = m
=  = 5.5 × 102 N
d(sin q) (15.0 m)(sin 50.0°)
Apply the first condition of equilibrium.
FT,1 + FT,2 − Fg,w − Fg,s = 0
FT,1 = Fg,w + Fg,s − FT,2 = 675 N + 205 N − FT,2 = 8.80 × 102 N − FT,2
dw = 1.00 m
Choose the end of the scaffold closest to the person as the pivot point.
Apply the second condition of equilibrium.
FT,2l − Fg,wdw − Fg,s l = 0
2
FT,2 =
Fg,wdw + Fg,s l
2
l
3.00 m
(675 N)(1.00 m) + (205 N) 
2
FT,2 = 
3.00 m
675 N • m + 308 N • m 983 N • m
FT,2 =  =  = 328 N
3.00 m
3.00 m
FT,1 = 8.80 × 102 N − FT,2 = 8.80 × 102 N − 328 N = 552 N
a. Apply the first condition of equilibrium in the x and y direction.
g = 9.81 m/s2
Fx = Rx − FT (cos q) = 0
q = 30.0°
Fy = Ry + FT (sin q) − mg = 0
To solve for FT, apply the second condition of equilibrium, using the end of the
beam at the pole as the pivot point. Use l to represent the length of the beam.
FT(sin q)l − mg l = 0
mg l
(20.0 kg)(9.81 m/s2)
FT =  =  = 392 N
sin q
sin 30.0°
b. Substitute the value for FT into the two first-condition equations to solve for R.
Rx = FT (cos q) = (392 N)(cos 30.0°) = 339 N
Ry = mg − FT (sin q) = (20.0 kg)(9.81 m/s2) − (392)(sin 30.0°) = 196 N − 196 N
= 0N
I Ch. 8–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
21. m = 20.0 kg
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Givens
Solutions
22. Fg,b = 1200.0 N
Apply the first condition of equilibrium in the x and y directions.
Fg,m = 2000.0 N
Fx = Rx,base − FT (cos qc) = 0
qb = 65°
Fy = Ry,base − Fg,b − Fg,m + FT (sin qc) = 0
qc = 25°
To solve for FT, apply the second condition of equilibrium, using the base of the
beam as the pivot point.
I
3
L
FT L − Fg,b  (cos qb) − Fg,m L(cos qb) = 0
4
2
3
F
4 T
= 2Fg,b(cos qb) + Fg,m(cos qb) = 2Fg,b + Fg,m(cos qb)
1
1
FT = 32Fg,b + Fg,m(cos qb) = 3(1200.0 N) + 3(2000.0 N)(cos 65°)
4 1
2
4
FT = (800.00 N + 2666.7 N)(cos 65°) = (3466.7 N)(cos 65°)
FT = 1.5 × 103 N
Substitute the value for FT into the two first-condition equations to solve for Rbase.
Rx,base = FT (cos qc ) = (1.5 × 103 N)(cos 25°) = 1.4 × 103 N
Ry,base = Fg,b + Fg,m − FT (sin qc) = 1200.0 N + 2000.0 N − (1.5 × 103 N)(sin 25°)
Ry,base = 3200.0 N − 630 N = 2.6 × 103 N
23. Fg = 10.0 N
Apply the first condition of equilibrium in the x and y directions.
dg = 15 cm
Fx = FT,1(cos q) − F = 0
d1,x = 15 cm
Fy = FT,1(sin q) + FT,2 − Fg = 0
d1,y = 30.0 cm
Choose the lower left-hand corner as the pivot point and apply the second condition
of equilibrium.
q = 50.0°
−Fg dg − FT,1d1,x(cos q) + FT,1d1,y (sin q) = 0
−(10.0 N)(0.15 m) − FT,1(0.15 m)(cos 50.0°) + FT,1(0.300 m)(sin 50.0°) = 0
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1.5 N • m
1.5 N • m
FT,1 =  =  = 11 N
0.230 m − 0.096 m
0.134 m
Substitute the value for FT,1 into the two first-condition equations and solve for the
unknown.
FT,2 = Fg − FT,1(sin q) = 10.0 N − (11 N)(sin 50.0°)
FT,2 = 10.0 N − 8.4 N = 1.6 N
F = FT,1(cos q) = (11 N)(cos 50.0°) = 7.1 N
27. M = 30.0
R = 0.180
t = Ia = 2MR2a = 2(30.0 kg)(0.180 m)2(2.30 × 10−2 rad/s2) = 1.12 × 10−2 N • m
1
1
a = 2.30 × 10−2 rad/s2
28. M = 350 kg
R = 1.5 m
wf = 3.14 rad/s
wf − w
3.14 rad/s − 0 rad/s
1
1
t = Ia = 2MR2 i = 2(350 kg)(1.5 m)2  = 620 N • m
∆t
2.00 s
wi = 0 rad/s
∆t = 2.00 s
Section One—Pupil’s Edition Solutions
I Ch. 8–9
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Givens
Solutions
35. Mt = 15.0 kg
1
Mdi = 9.0 kg
1
Lf = Itwf = 2MtR2wf
R = 25 cm
I
1
Li = Lt + Ldi = Itwi + Idiwi = 2MtR2wi + 2MdiR2wi
Li = Lf
wi = 0.75 rad/s
1
(M
t
2
1
+ Mdi)R2wi = 2MtR2wf
(Mt + Mdi)wi

wf = 
Mt
(15.0 kg + 9.0 kg)(0.75 rad/s) (24.0 kg)(0.75 rad/s)
wf =  =  = 1.2 rad/s
15.0 kg
15.0 kg
Li = Lf = 0, so
It = 1.5 × 10 kg • m
Lf = Iwww + Itwt = 0
r = 2.0 m
2
−I ww −(mr2)ww −(65 kg)(2.0 m) (−0.75 rad/s)
wt = w
=  = 
1.5 × 103 kg • m2
It
It
3
2
ww = 0.75 rad/s, clockwise
= −0.75 rad/s
37. m = 35 kg
r = 13 cm
h = 3.5 m
g = 9.81 m/s2
38. Fg = 240 N
wt = 0.13 rad/s, counterclockwise
MEi = MEf
2
vf
1
1
1
12
1
2
1
7
mgh = 2mvf2 + 2Iwf2 = 2mvf2 + 25 mr2  = 2mvf21 + 5 = 2mvf25
r
10 gh
(10)(9.81 m/s2)(3.5 m)
vf =  =  = 7.0 m/s
7
7
MEi = MEf
1
1
r = 0.20 m
mgh = 2mv f2 + 2Iw f2
d = 6.0 m
mgd(sin q ) = 2m(w f r)2 + 25mr 2w f2
q = 37°
1
12
1
g = 9.81 m/s2
wf =
45. m = 75 kg
r = 0.075 m
d = 0.25 m
1
7
gd(sin q ) = 2w f2r 2 + 5r 2w f2 = 10r 2w f2
in q )
(10)(9.81 m/s )(6.0 m)(sin 37°)
 = 36 rad/s
10
gd7(rs
= (7)(0.20 m)
2
2
2
For a force perpendicular to d, t = Fd.
t mgr (75 kg)(9.81 m/s2)(0.075 m)
F =  =  =  = 2.2 × 102 N
d
d
0.25 m
g = 9.81 m/s2
46. t = 58 N • m
d = 0.35 m
q = 56°
47. d = 1.4 m
F = 1600 N
t = Fd(sin q)
58 N • m
t
F =  =  = 2.0 × 102 N
(0.35 m)(sin 56°)
d(sin q)
t = Fd(sin q) = (1600 N)(1.4 m)(sin 53.5°) = 1800 N • m
q = 53.5°
I Ch. 8–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
36. m = 65 kg
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Givens
48.
Solutions
l s = 23.0 cm
Apply the second condition of equilibrium.
d = 2.00 cm
t net = Fs(l s − d) − Fd = 0
Fs = 84.3 N
Fs(l s − d) (84.3 N)(0.230 m − 0.0200 m)
F= 
= 
0.0200 m
d
I
(84.3 N)(0.210 m)
F =  = 885 N
0.0200 m
49. wf = 220 rad/s
wi = 0 rad/s
Wnet = 3000.0 J
50. mms = 0.100 kg
1
1
1
Wnet = ∆KE = 2Iwf2 − 2Iwi2 = 2I(wf2 − wi2)
(2)(3000.0 J)
2Wnet
 = 
= 0.12 kg• m2
I= 
(220 rad/s)2 − (0 rad/s)2
wf2 − wi2
a. Apply the first condition for equilibrium.
dms = 50.0 cm
Fs − m1g − mms g − m2 g = 0
m1 = 0.700 kg
Fs − g(m1 + mms ) 19.6 N − (9.81 m/s2)(0.700 kg + 0.100 kg)
m2 =  = 
g
9.81 m/s2
d1 = 5.00 cm
Fs = 19.6 N
ds = 40.0 cm
g = 9.81 m/s2
19.6 N − (9.81 m/s2)(0.800 kg)
m2 = 
9.81 m/s2
19.6 N − 7.85 N
11.8 N
m2 = 
= 2 = 1.20 kg
9.81 m/s2
9.81 m/s
b. Choose the zero mark as the pivot point and apply the second condition of equilibrium.
m1gd1 + m2gd2 + mmsgdms − Fsds = 0
m2gd2 = Fsds − m1gd1 − mmsgdms
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fsds − m1gd1 − mmsgdms
d2 = 
m2g
(19.6 N)(0.400 m) − (0.700 kg)(9.81 m/s2)(0.0500 m) − (0.100 kg)(9.81 m/s2)(0.500 m)
d2 = 
(1.20 kg)(9.81 m/s2)
7.84 N• m − 0.343 N • m − 0.490 N• m
7.01 N • m
d2 = 
= 
(1.20 kg)(9.81 m/s2)
(1.20 kg)(9.81 m/s2)
d2 = 0.595 m
Section One—Pupil’s Edition Solutions
I Ch. 8–11
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Givens
Solutions
51. Fg,l = 200.0 N
Apply the first condition of equilibrium in the x and y directions.
L = 8.00 m
Fx = Fs − Fwall = msRy,base − Fwall = 0
ms = 0.600
Fy = Ry,base − Fg,L − Fg,p = 0
q = 50.0°
Ry,base = Fg,L + Fg,p = 200.0 N + 800.0 N = 1000.0 N
Fg,p = 800.0 N
Fwall = msRy,base = (0.600)(1000.0 N) = 6.00 × 102 N
Choose the base of the ladder as the pivot point and apply the second condition of
equilibrium.
L
Fwall L(sin q) − Fg,L  (cos q) − Fg,pd1(cos q) = 0
2
L
FwallL(sin q) − Fg,l  (cos q)
2FwallL(tan q) − Fg,l)L
2
dp =  = 
2Fg,p
Fg,p(cos q)
(2)(6.00 × 102 N)(8.00 m)(tan 50.0°) − (200.0 N)(8.00 m)
dp = 
(2)(800.0 N)
1.14 × 104 N • m − 1.60 × 103 N • m
dp = 
1.600 × 103 N
9.8 × 103 N • m
dp = 
= 6.1 m
1.600 × 103 N • m
0.0200 m
52. r =  = 0.0100 m
2
wi = 45.0 rad/s
g = 9.81 m/s2
MEi = MEf
1
1
mv 2 + Iw 2 = mgh
i
2
2 i
1
1 1
2
m(rw ) +  mr2 w 2 = mgh
i
i
2
2 2
1
1
1
3
2
2
mr w 1 +  = mr2w 2  =
i
i 2
2
2
2
3
gh = 4r2wi2
mgh
3r2wi2 (3)(0.0100 m)2(45.0 rad/s)2
h = 
= 
4g
(4)(9.81 m/s2)
53. m = 4.0 kg
r = 2.0 m
wi = 6.0 rad/s
q = 15°
g = 9.81 m/s2
I Ch. 8–12
MEi = MEf
1
1
mv 2 + Iw 2 = mgh
i
2
2 i
1
1
m(rw )2 + (mr2)w 2 = mgd(sin q)
i
i
2
2
1
mr2w 2(1 + 1) = mr2w 2 = mgd(sin
i
1
2
q)
r2wi2
(2.0 m)2(6.0 rad/s)2
d = 
= 
= 57 m
g(sin q) (9.81 m/s2)(sin 15°)
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
h = 1.55 × 10−2 m = 1.55 cm
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Givens
Solutions
54. m = 12 kg
a. Apply Newton’s second law along the incline.
r = 10.0 cm
2
mg(sin q) − FT = ma
a = 2.0 m/s
FT = mg(sin q ) − ma = (12 kg)(9.81 m/s2)(sin 37°) − (12 kg)(2.0 m/s2)
q = 37°
FT = 71 N − 24 N = 47 N
I
g = 9.81 m/s2
b. t = Ia
Because FT is perpendicular to r, t = FT r.
t
F r
F r2
(47 N)(0.100 m)2
I =  = T = T = 
= 0.24 kg • m2
a
(a/r)
a
2.0 m/s2
∆t = 2.0 s
w i = 0 rad/s
55. Fg = Fn = 700.0 N
a
c. wf = w i + a∆t = w i +  ∆t
r
(2.0 m/s2)(2.0 s)
w f = 0 rad/s +  = 4.0 × 101 rad/s
0.100 m
First, apply the second condition of equilibrium, choosing the toe as the pivot point,
q = 21.2°
TdT − RdR = 0
f = 15.0°
dT = 25.0 cm
Td T
R= 
dR
dR = 18.0 cm
Apply the first condition of equilibrium in the y direction.
Fn − R(cos f) + T(cos q) = 0
TdT(cos f)
 + I(cos q) = 0
Fg − 
dR
Fg
700.0 N
T =  = 
d
0.250 m
T (cos f) − (cos q)
 (cos 15.0°) − (cos 21.2°)
dR
0.180 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
700.0 N
700.0 N
T =  =  = 1.7 × 103 N
1.34 − 0.932
0.41
TdT
(1.7 × 103 N)(0.250 m)
R =  =  = 2.4 × 103 N
0.180 m
dR
56. M = 0.85 kg
R = 4.0 cm
tr = 1.3 N • m
a = 66 rad/s2
a. Because F is perpendicular to r, t = Fr.
Ia = FR − tr
1
mr 2a + t
Ia + t
r
F = r = 2
R
R
t 1
1.3 N • m
1
F = 2 mra + r = 2(0.85 kg)(0.040 m)(66 rad/s2) + 
R
0.040 m
F = 1.1 N + 32 N = 33 N
∆t = 0.50 s
w i = 0 rad/s
1
b. ∆q = w i ∆t + 2a∆t2
1
∆q = (0 rad/s)(0.50 s) + 2(66 rad/s2)(0.50 s)2 = 8.2 rad
∆s = r∆q = (0.040 m)(8.2 rad) = 0.33 m
Section One—Pupil’s Edition Solutions
I Ch. 8–13
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Givens
Solutions
57. t net = 36 N • m
a. t net = Ia
∆t1 = 6.0 s
t
wf - w
 where a = i
I = net
∆t1
a
w i = 0 rad/s
I
w f = 12 rad/s
∆t2 = 65 s
w i = 12 rad/s
w f = 0 rad/s
(36 N • m)(6.0 s)
tnet∆t1
 =  = 18 kg • m2
I= 
12 rad/s − 0 rad/s
wf − wi
I(wf − wi)
(18 kg • m2)(0 rad/s −12 rad/s)
b. t f = Ia =  =  = −3.3 N • m
∆t2
65 s
(w i + w f )
(0 rad/s + 12 rad/s)
c. q1 =  ∆t1 =  (6.0 s) = 36 rad
2
2
(12 rad/s + 0 rad/s)
q2 =  (65 s) = 390 rad
2
q1 + q2
430 rad
36 rad + 390 rad
 =  =  = 68 rev
N= 
2p rad/rev
2p rad/rev
2p rad/rev
58. F1 = 120.0 N
Because F is perpendicular to the pulley’s radius, t = FR.
F2 = 100.0 N
tnet = t1 − t2 = F1R − F2R = (F1 − F2)R
M = 2.1 kg
t et (F1 − F2)R
2(F1 − F2)

a = n
= 
= 
1
 MR2
I
MR
2
R = 0.81 m
(2)(120.0 N − 100.0 N)
(2)(20.0 N)
a =  =  = 24 rad/s2
(2.1 kg)(0.81 m)
(2.1 kg)(0.81 m)
1
mv 2
i
2
MEi = MEf
r = 3.0 m
w i = 3.0 rad/s
q = 20.0°
1
m(rw )2
i
2
g = 9.81 m/s2
1
+ 2(mr 2)w i 2 = mgd(sin q)
1
mr 2w 2(1
i
2
2
1
+ 2Iw i 2 = mgh
+ 1) = mr 2w i 2 = mgd(sin q)
2
r wi
d = 
g(sin q)
(3.0 m)2(3.0 rad/s)2
d = 
= 24 m
(9.81 m/s2)(sin 20.0°)
60. M = 5.00 kg
a. Apply Newton’s second law for the bucket.
R = 0.600 m
FT − mg = −ma
m = 3.00 kg
Because FT is perpendicular to R,
∆t = 4.00 s
2
g = 9.81 m/s
vi = 0 m/s
a
1
I 
2MR2a 1
Ia
t
R
= 2mp a
FT =  =  =  = 
R
R
R
R2
1
Ma
2
− mg = −ma
mg
(3.00 kg)(9.81 m/s2)
a= 
= 
1
1
M + m
(5.00 kg) + 3.00 kg
2
2
(3.00 kg)(9.81 m/s2)
a =  = 5.35 m/s2
5.50 kg
1
1
b. ∆y = vi∆t + 2 a∆t2 = (0 m/s)(4.00 s) + 2(5.35 m/s2)(4.00 s)2 = 42.8 m
a 5.35 m/s2
c. a =  =  = 8.92 rad/s2
R
0.600 m
I Ch. 8–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
59. m = 5.0 kg
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Givens
Solutions
61. Lh = 2.7 m
Consider the total mass of each hand to be at the midpoint of that hand.
Lm = 4.5 m
mh = 60.0 kg
mm = 100.0 kg
Lm
Lh
tnet = −mh g  (sin qh ) − mm g  (sin q m)
2
2
qh = 30.0° from 6:00
2.7 m
4.5 m
tnet = −(60.0 kg)(9.81 m/s2)  (sin 30.0°) − (100.0 kg)(9.81 m/s2)  (sin 60.0°)
2
2
qm = 60.0° from 6:00
tnet = −4.0 × 102 N • m − 1.9 × 103 N • m = −2.3 × 103 N • m
I
g = 9.81 m/s2
3.00 cm
62. r =  = 1.50 cm
2
q = 30.0°
wi = 60.0 rad/s
g = 9.81 m/s2
63.
MEi = MEf
1
mv 2
i
2
1
+ 2Iwi2 = mgh
1
m(rw )2
i
2
+ 22mr2wi2 = mgd(sin q)
11
1 + 2 = 2mr2wi22 = mgd(sin q)
1
mr2w 2
i
2
1
1
3
(3)(1.50 × 10−2 m)2 (60.0 rad/s)2
3r2wi2
d = 
= 
= 0.124 m = 12.4 cm
(4)(9.81 m/s2)(sin 30.0°)
4g(sin q)
2
v
1
12
1
KErot = 2Iw2 = 25 mr2  = 5mv 2
r
1
1
7
KEtot = KErot + KEtrans = 5mv2 + 2mv 2 = 10mv 2
1
2
KErot 5 mv
1 10
 =  = 57 =
7
2
KEtot
mv
10
Copyright © by Holt, Rinehart and Winston. All rights reserved.
64. Fg = 800.0 N
F is perpendicular to R, so
R = 1.5 m
t = FR = Ia
F = 50.0 N
Because wi = 0 rad/s,
wf − w wf w
a = i =  = 
∆t
∆t ∆t
∆t = 3.0 s
wi = 0 rad/s
2

7
2
2
g = 9.81 m/s
2
1
1
1 t∆t
1 FR∆t
KE = 2Iw2 = 2I(a∆t)2 = 2I  = 2I 
I
I
(FR∆t)2 (FR∆t)2
(F∆t)2


KE =  = 
=
1
2I
M
2(2MR2)
(F∆t)2g
(F∆t)2
 = 
Fg
Fg

g
[(50.0 N)(3.0 s)]2 (9.81 m/s2)
KE =  = 2.8 × 102 J
800.0 N
Section One—Pupil’s Edition Solutions
I Ch. 8–15
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Givens
Solutions
65. I = 4.00 × 10−4 kg • m2
wi = 0 rad/s
1
Because F is in the direction of d, Wnet = Fd.
F = 5.57 N
I
1
Wnet = ∆KE = KEf − KEi = 2Iwf2 − 2Iwi2
1
1
Wnet = Fd = 2Iwf2 − 2Iwi2
d = 80.0 cm
wi = 0 rad/s, so
1
Fd = 2Iwf2
wf =
66. RE = 6.37 × 106 m
24
(2)(5.57 N)(0.800 m)
 = 149 rad/s
2FId = 4.00 × 10 kg m
−4
•
2
a. L = IEw
ME = 5.98 × 10 kg
From Conceptual Question 4 on p. 305,
w = 1 rev/day
IE = 0.331 MERE2
L = 0.331 MERE2w
2p rad 1 day 1 h
L = (0.331)(5.98 × 1024 kg)(6.37 × 106 m)2(1 rev/day)   
1 rev 24 h 3600 s
L = 5.84 × 1033 kg • m2/s
r = 1.496 × 1011 m
w = 1 rev/365.25 days
b. L = Iw = MEr2w
2p rad 1 day 1 h
L = (5.98 × 1024 kg)(1.496 × 1011 m)2 (1 rev/365.25 days)   
1 rev 24 h 3600 s
L = 2.66 × 1040 kg • m2/s
67. w i = 12.0 rad/s
a. Li = Lf
2
Ii w i = If w f
2
Iw
(41 kg • m2)(12.0 rad/s)
w f = i i = 
= 14 rad/s
If
36 kg • m2
Ii = 41 kg • m
If = 36 kg • m
1
1
1
1
KEf = 2If w f 2 = 2(36 kg • m2)(14 rad/s)2 = 3.5 × 103 J
I Ch. 8–16
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
KEi = 2Ii w i 2 = 2(41 kg • m2)(12.0 rad/s)2 = 3.0 × 103 J
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Givens
Solutions
68. I = 5.0 kg • m2
a. Apply Newton’s second law for the 2.0 kg mass.
R = 0.50 m
FT,1 − m1g = m1a
m1 = 2.0 kg
FT,1 − (2.0 kg)(9.81 m/s2) = (2.0 kg)a
m2 = 5.0 kg
FT,1 − 2.0 × 101 N = (2.0 kg)a
g = 9.81 m/s2
FT,1 = (2.0 kg)a + 2.0 × 101 N
I
Apply Newton’s second law for the 5.0 kg mass.
m2 g − FT,2 = m2 a
FT,2 = m2 g − m2 a
FT,2 = (5.0 kg)(9.81 m/s2) − (5.0 kg)a
FT,2 = 49 N − (5.0 kg)a
Find an expression for the torque on the pulley, noting that Fnet is perpendicular
to the pulley’s radius.
t net = Fnetd(sin θ) = Ia
a
a
= I  = (5.0 kg m ) 
R (0.50 m) a
(FT,2 − FT,1)r = I 
R
FT,2 − FT,1
•
2
2
2
1
FT,2 − FT,1 = (2.0 × 10 kg)a
Substitute the values for FT,1 and FT,2 from the equations above.
[49 N − (5.0 kg)a] − [(2.0 kg)a + 2.0 × 101 N] = (2.0 × 101 kg)a
29 N = (2.0 × 101 kg)a + (7.0 kg)a
29 N
a =  = 1.1 m/s2
27 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m2 accelerates downward at 1.1 m/s2 .
m1 accelerates at the same rate in the opposite direction: −1.1 m/s2
b. FT,1 = (2.0 kg)a + 2.0 × 101 N = (2.0 kg)(1.1 m/s2) + 2.0 × 101 N = 2.2 N
+ 2.0 × 101 N = 22 N
FT,2 = 49 N − (5.0 kg)a = 49 N − (5.0 kg)(1.1 m/s2) = 49 N − 5.5 N = 43 N
Section One—Pupil’s Edition Solutions
I Ch. 8–17
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Givens
Solutions
69. m1 = 4.0 kg
a. Apply Newton’s second law for the 3.0 kg mass.
m2 = 3.0 kg
F2 = m2 a = (3.0 kg)a
2
I = 0.50 kg • m
I
R = 0.30 m
g = 9.81 m/s2
Apply Newton’s second law for the 4.0 kg mass.
m1g − F1 = m1a
F1 = m1g − m1a = (4.0 kg)(9.81 m/s2) − (4.0 kg)a = 39 N − (4.0 kg)a
Find an expression for the torque on the pulley, noting that Fnet is perpendicular
to the pulley’s radius.
t net = Fnetd(sin q) = Ia
a
a
F − F = I  = (0.50 kg m ) 
R (0.30 m) a
(F1 − F2 )r = I 
R
1
2
2
•
2
2
F1 − F2 = (5.6 kg)a
Substitute the values for F1 and F2 from above.
[39 N − (4.0 kg)a] − (3.0 kg)a = (5.6 kg)a
39 N = (5.6 kg)a + (7.0 kg)a = (12.6 kg)a
39 N
a =  = 3.1 m/s2
12.6 kg
b. F1 = 39 N − (4.0 kg)(3.1 m/s2) = 39 N − 12 N = 27 N
F2 = m2 a = (3.0 kg)(3.1 m/s2) = 9.3 N
M = 500.0 kg
w = 1000.0 rev/min
a. KE = 2Iw 2 = 2 2MR2w 2
1
1 1
KE = 5.48 × 106 J
P = 7457 W
W KE
b. P =  = 
∆t
∆t
KE 5.48 × 106 J
∆t =  =  = 735 s
P
7457 W
71. w1 = 2.0 rev/s
Li = Lf
r2 = 0.50 m
Mr12w1 = Mr22w 2
r1 = 1.0 m
r12w1

w2 = 
r22
M = 4m
(1.0 m)2(2.0 rev/s)
w2 = 
= 8.0 rev/s
(0.50 m)2
I Ch. 8–18
Holt Physics Solution Manual
2
1 min 2p rad
1
KE = 4(500.0 kg)(2.00 m)2 (1000.0 rev/min)  
60 s
1 rev
Copyright © by Holt, Rinehart and Winston. All rights reserved.
70. R = 2.00 m
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Givens
Solutions
72. eff = 0.64
W ut Fout dout
eff = o
= 
Win
Findin
m = 78 kg
dout = 4.0 m
din = 24 m
g = 9.81 m/s2
73. d = 2.0 m
Foutdout mgdout
 =  = (78 kg)(9.81 m/s2)(4.0 m)\(24 m)(0.64)
Fin = 
din(eff ) din(eff )
I
Fin = 2.0 × 102 N
Wout = Fg d(sin q)
q = 15°
Win = (Ff + Fg,x )d = [mkFg (cos q) + Fg (sin q)]d
mk = 0.160
Win = Fg d[mk(cos q) + (sin q )]
W ut
sin q
Fg d(sin q)
eff = o
= 
= 
Win
mk(cos q) + (sin q)
Fg d[mk(cos q) + (sin q )]
sin 15°
sin 15°
eff =  = 
0.15 + 0.26
(0.160)(cos 15°) + (sin 15°)
sin 15°
eff =  = 0.63 = 63%
0.41
74. eff = 0.875
Fin = 648 N
m = 150 kg
dout = 2.46 m
W ut Fout dout mgdout
eff = o
=  = 
Win
Findin
Findin
mgdout
(150 kg)(9.81 m/s2)(2.46 m)
din = 
=  = 6.4 m
Fin(eff )
(648 N)(0.875)
g = 9.81 m/s2
75. dout = 3.0 m
Fin = 2200 N
din = 14 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m = 750 kg
g = 9.81 m/s2
76. Fg = 250 N
l
= 6.0 m
q = 60.0°
W ut Fout dout mgdout
eff = o
=  = 
Win
Findin
Findin
(750 kg)(9.81 m/s2)(3.0 m)
eff =  = 0.72 = 72%
(2200 N)(14 m)
Apply the first condition of equilibrium.
Fx = Ff − Fwall = 0
Fwall = Ff = Fn ms
Fy = Fn − Fg = 0
Fn = Fg
Fwall = Fg m s
Choose the base of the ladder as the pivot point and apply the second condition of
equilibrium.
l
−F  (cos q) + F m l (sin q) = 0
2l −Fg  (cos q) + Fwall l (sin q) = 0
2
g
g s
1
m s (sin q) = 2(cos q)
cos q
cos 60.0°
m s =  =  = 0.289
2(sin q)
(2)(sin 60.0°)
Section One—Pupil’s Edition Solutions
I Ch. 8–19
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Givens
77.
l
Solutions
= 15.0 m
Fg,l = 520.0 N
q = 60.0°
I
Fg,f = 800.0 N
df = 4.00 m
Apply the first condition of equilibrium in the x and y directions.
Fx = Fx,E − Fx,wall = 0
a. Fy = Fy,E − Fg,f − Fg,l = 0
Fy,E = Fg,f + Fg,l = 800.0 N + 520.0 N = 1320.0 N
Choose the base of the ladder as the pivot point and apply the second condition of
equilibrium.
l −Fg,l  (cos q) − Fg,f df (cos q) + Fx,wall l (sin q) = 0
2
F 2l + F d (cos q)
g,l
g,f f
Fx,wall =  = Fx,E
l (sin q)
[(520.0 N)(7.50 m) + (800.0 N)(4.00 m)](cos 60.0°)
Fx,E = 
(15.0 m)(sin 60.0°)
(3.90 × 103 N • m + 3.20 × 103 N • m)(cos 60.0°)
Fx,E = 
(15.0 m)(sin 60.0°)
(7.10 × 103 N • m)(cos 60.0°)
Fx,E =  = 273 N
(15.0 m)(sin 60.0°)
df = 9.00 m
b. Fx,E = Fs = msFn = msFy,E
F 2l + F d (cos q)
g,l
g,f f
F ,E
ms = x
= 
Fy,E
Fy,E l (sin q)
(3.90 × 103 N • m + 7.20 × 103 N • m)(cos 60.0°)
ms = 
(1320.0 N)(15.0 m)(sin 60.0°)
(11.1 × 103 N • m)(cos 60.0°)
ms =  = 0.324
(1320.0 N)(15.0 m)(sin 60.0°)
I Ch. 8–20
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
[(520.0 N)(7.50 m) + (800.0 N)(9.00 m)](cos 60.0°)
ms = 
(1320.0 N)(15.0 m)(sin 60.0°)
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Fluid Mechanics
Chapter
9
I
Practice 9A, p. 324
Givens
1. Fg = 50.0 N
apparent weight in water
= 36.0 N
rwater = 1.00 × 103 kg/m3
apparent weight in liquid
= 41.0 N
rmetal = 3.57 × 103 kg/m3
Solutions
a. FB = Fg − apparent weight = 50.0 N − 36.0 N = 14.0 N
Fg
(50.0 N)(1.00 × 103 kg/m3)
rmetal =  rwater = 
FB
14.0 N
rmetal = 3.57 × 103 kg/m3
b. FB = Fg − apparent weight = 50.0 N − 41.0 N = 9.0 N
F
(9.0 N)(3.57 × 103 kg/m3)
rliquid = B rmetal = 
Fg
50.0 N
rliquid = 6.4 × 102 kg/m3
2. m = 2.8 kg
FB = Fg
l = 2.00 m
rwater Vg = (m + M)g
w = 0.500 m
M = rwater V − m = rwater (l wh) − m
h = 0.100 m
M = (1.00 × 103 kg/m3)(2.00 m)(0.500 m)(0.100 m) − 2.8 kg = 1.00 × 102 kg − 2.8 kg
rwater = 1.00 × 103 kg/m3
M = 97 kg
g = 9.81 m/s2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. w = 4.0 m
l
= 6.0 m
h = 4.00 cm
Fg = FB
mg = rwaterVg = rwater (wl h)g
Fg = (1.00 × 103 kg/m3)(4.0 m)(6.0 m)(0.0400 m)(9.81 m/s2) = 9.4 × 103 N
rwater = 1.00 × 103 kg/m3
g = 9.81 m/s2
4. mballoon = 0.0120 kg
rhelium = 0.179 kg/m3
r = 0.500 m
rair = 1.29 kg/m3
g = 9.81 m/s2
a. FB = rairV g = rair 3pr 3 g
4
(1.29 kg/m3)(4p)(0.500 m)3(9.81 m/s2)
FB =  = 6.63 N
3
b. mhelium = rheliumV = rhelium 3pr 3
4
(0.179 kg/m3)(4p)(0.500 m)3
mhelium =  = 0.0937 kg
3
Fg = (mballoon + mhelium )g = (0.0120 kg + 0.0937 kg)(9.81 m/s2)
Fg = (0.1057 kg)(9.81 m/s2) = 1.04 N
Fnet = FB − Fg = 6.63 N − 1.04 N = 5.59 N
Section One—Pupil’s Edition Solutions
I Ch. 9–1
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Section Review, p. 324
Givens
Solutions
3. mballoon = 650 kg
FB = rairVg
mpack = 4600 kg
I
Fg = (mballoon + mpack + mhelium )g
3
rair = 1.29 kg/m
mhelium = rheliumV
3
rhelium = 0.179 kg/m
FB = Fg
rairVg =(mballoon + mpack + rheliumV )g
mballoon + mpack
650 kg + 4600 kg
 = 
V= 
rair − rhelium
1.29 kg/m3 − 0.179 kg/m3
5200 kg
V = 3 = 4.7 × 103 m3
1.11 kg/m
4. a = 0.325 m/s2
Use Newton’s second law.
rsw = 1.025 × 10 kg/m
msa = FB − Fg = mswg − ms g
g = 9.81 m/s2
rsVa = rswVg − rsVg
3
3
rs(a + g) = rswg
g
9.81 m/s2
rs = rsw  = (1.025 × 103 kg/m3) 
a+g
0.325 m/s2 + 9.81 m/s2
9.81 m/s2
rs = (1.025 × 103 kg/m3) 2 = 992 kg/m3
10.14 m/s
Practice 9B, p. 327
r2 = 15.0 cm
F2 = 1.33 × 104 N
F
F
a. 1 = 2
A1 A2
FA
F2 p r12 F2r12
 = 
F1 = 2 1 = 
A2
p r22
r22
(1.33 × 104 N)(0.0500 m)2
F1 = 
= 1.48 × 103 N
(0.150 m)2
F
F2
1.33 × 104 N

 = 1.88 × 105 Pa
=
b. P = 2 = 
A2 p r22 (p )(0.150 m)2
2. Fg = 1025 N
w = 1.5 m
F Fg
1025 N
P =  =  =  = 2.7 × 102 Pa
A wl (1.5 m)(2.5 m)
l = 2.5 m
3. r = 0.40 cm
a. Pnet = Pb − Pt = 1.010 × 105 Pa − 0.998 × 105 Pa = 1.2 × 103 Pa
Pb = 1.010 × 105 Pa
Pt = 0.998 × 105 Pa
b. Fnet = Pnet A = Pnet pr 2
Fnet = (1.2 × 103 Pa)(p)(4.0 × 10−3 m)2 = 6.0 × 10−2 N
I Ch. 9–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. r1 = 5.00 cm
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Practice 9C, p. 330
Givens
Solutions
1. h = 11.0 km
P = P0 + rgh = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(11.0 × 103 m)
Po = 1.01 × 105 Pa
P = 1.01 × 105 Pa + 1.11 × 108 Pa = 1.11 × 108 Pa
r = 1.025 × 103 kg/m3
I
2
g = 9.81 m/s
a. P = Po + roil ghoil
2. h water = 20.0 cm
hoil = 30.0 cm
P = 1.01 × 105 Pa + (0.70 × 103 kg/m3)(9.81 m/s2)(0.300 m)
roil = 0.70 × 103 kg/m3
P = 1.01 × 105 Pa + 2.1 × 103 Pa
rwater = 1.00 × 103 kg/m3
P = 1.03 × 105 Pa
g = 9.81 m/s2
b. Pnet = P + rwater gh water
Po = 1.01 × 105 Pa
Pnet = 1.03 × 105 Pa + (1.00 × 103 kg/m3)(9.81 m/s2)(0.200 m)
Pnet = 1.03 × 105 Pa + 1.96 × 103 Pa = 1.05 × 105 Pa
3. Po = 0 Pa
P = Po + rgh
4
P = 2.7 × 10 Pa
r = 13.6 × 103 kg/m3
P−P
2.7 × 104 Pa − 0 Pa
h = o = 
rg
(13.6 × 103 kg/m3)(9.81 m/s2)
g = 9.81 m/s2
h = 0.20 m
4. P = 3 Po
P = P0 + rgh
5
Po = 1.01 × 10 Pa
3
3
r = 1.025 × 10 kg/m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
P−P
3Po − Po 2Po
 = 
h = o = 
rg
rg
rg
(2)(1.01 × 105 Pa)
h = 
= 20.1 m
(1.025 × 103 kg/m3)(9.81 m/s2)
Section Review, p. 331
1. Fg = 25 N
w = 1.5 m
Fg = 15 N
r = 1.0 m
Fg = 25 N
w = 2.0 m
Fg = 25 N
r = 1.0 m
F Fg
a. P =  = 2
A w
25 N
P = 2 = 11 Pa
(1.5 m)
Fg
F
b. P =  = 2
A pr
15 N
P = 2 = 4.8 Pa
(p)(1.0 m)
F Fg
c. P =  = 2
A w
25 N
P = 2 = 6.2 Pa
(2.0 m)
F Fg
d. P =  = 2
A pr
25 N
P = 2 = 8.0 Pa
(p)(1.0 m)
a is the largest pressure
Section One—Pupil’s Edition Solutions
I Ch. 9–3
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Givens
Solutions
2. h = 366 m
P = rgh
3
3
r = 1.00 × 10 kg/m
P = (1.00 × 103 kg/m3)(9.81 m/s2)(366 m) = 3.59 × 106 Pa
g = 9.81 m/s2
I
4. T(°C) = 11°C
T(K) = T(°C) + 273 = 11°C + 273 = 284 K
5. h = 5.0 × 102 m
P = Po + rg h = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(5.0 × 102 m)
Po = 1.01 × 105 Pa
P = 1.01 × 105 Pa + 5.0 × 106 Pa = 5.1 × 106 Pa
r = 1.025 × 103 kg/m3
P
5.1 × 106 Pa
N =  = 
= 5.0 × 101
Po 1.01 × 105 Pa
g = 9.81 m/s2
Practice 9D, p. 337
1. h2 − h1 = 16 m
flow rate = 2.5 × 10−3 m3/min
g = 9.81 m/s2
1
1
a. P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2
The top of the tank and the spigot are open to the atmosphere, so P1 = P2 = Po.
If we assume that the hole is small, then v2 ≈ 0.
1
Po + 2 rv12 + rgh1 = Po + rgh2
1
 rv 2
1
2
= rg(h2 − h1)
2
v1 = 2g(h2 − h1)
m/s
v1 = 2g
(h
h
)(
9.
81
2)(1
6m
) = 18 m/s
2 −
1) = (2
b. flow rate = Av1
flow rate
1 2 1
 = A = pr 2 = p 2D = 4pD2
v1
(4)(2.5 × 10−3 m3/min)(1 min/60 s)

(p)(18 m/s)
4(flow rate)
 =
p v1
D = 1.7 × 10−3 m = 1.7 mm
2. r = 1.65 × 103 kg/m3
2
A1 = 10.0 cm
vi = 275 cm/s
P1 = 1.20 × 105 Pa
A2 = 2.50 cm2
a. A1 v1 = A2 v2
Av
(10.0 cm2)(10−4 m2/cm2)(2.75 m/s)
v2 = 11 = 
= 11.0 m/s
A2
2.50 × 10−4 m2
1
1
b. P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2
Because h1 = h2,
1
P2 = P1 + 2 r(v12 − v22)
1
P2 = 1.20 × 105 Pa + 2 (1.65 × 103 kg/m3)[(2.75 m/s)2 − (11.0 m/s)2]
1
P2 = 1.20 × 105 Pa + 2 (1.65 × 103 kg/m3)(7.56 m2/s2 − 121 m2/s2)
1
P2 = 1.20 × 105 Pa − 2 (1.65 × 103 kg/m3)(113 m2/s2)
P2 = 1.20 × 105 Pa − 0.932 × 105 Pa = 2.7 × 104 Pa
I Ch. 9–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
D=
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Givens
Solutions
3. v1 = 15 cm/s
1
1
P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2
v2 = 2 v1
The change in pressure, ∆P, is P2 − P1.
r = 1.29 kg/m3
Because h1 = h2,
1
1
I
1
P2 − P1 = 2 r (v12 − v22) = 2 r[v12 − (2v1)2] = 2 rv12 (1 − 4)
P2 − P1 =
−3

2
(−3)(1.29 kg/m3)(0.15 m/s)2
rv12 = 
2
P2 − P1 = −4.4 × 10 −2 Pa
Section Review, p. 337
2. P1 = 3.00 × 105 Pa
3
3
r = 1.00 × 10 kg/m
v1 = 1.00 m/s
1
r2 = 4r1
a. A1v1 = A2 v2
Av
p r 2v1 r12v1
v2 = 11 = 1
=
= 16v1
1 2
A2
p r22
4r1
v2 = (16)(1.00 m/s) = 16.0 m/s
1
1
P1 + 2 rv12 +rgh1 = P2 + 2 rv22 +rgh2
Because h1 = h2 ,
1
b. P2 = P1 + 2 r(v12 − v22)
1
P2 = 3.00 × 105 Pa + 2 (1.00 × 103 kg/m3)[(1.00 m/s)2 − (16.0 m/s)2]
1
P2 = 3.00 × 105 Pa + 2 (1.00 × 103 kg/m3)(1.00 m2/s2 − 256 m2/s2)
1
P2 = 3.00 × 105 Pa + 2 (1.00 × 103 kg/m3)(−255 m2/s2)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P2 = 3.00 × 105 Pa − 1.28 × 105 Pa = 1.72 × 105 Pa
6.0 cm
3. r1 =  = 3.0 cm
2
2.0 cm
r2 =  = 1.0 cm
2
h2 − h1 = 2.00 m
r = 1.00 × 103 kg/m3
V = 2.5 × 10−2 m3
a. flow rate = A2 v2
V

∆
t
flow rate
V
v2 =  =  = 
A2
A2 pr22∆t
2.5 × 10−2 m3
v2 = 
= 2.7 m/s
(p)(0.010 m)2(30.0 s)
∆t = 30.0 s
b. A1v1 = A2 v2
A v
p r 2v2 r22v2
v1 = 22 = 2
= 
r12
A1
p r12
(0.010 m)2(2.7 m/s)
v1 = 
= 0.30 m/s
(0.030 m)2
1
1
P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2
P1 − P2 = r 2 v22 − 2 v12 + g(h2 − h1)
1
1
P1 − P2 = (1.00 × 103 kg/m3 )2 (2.7 m/s)2 − 2 (0.30 m/s)2 + (9.81 m/s2)(2.00 m)
1
1
P1 − P2 = (1.00 × 103 kg/m3)(3.6 m2/s2 − 0.045 m2/s2 + 19.6 m2/s2)
P1 − P2 = (1.00 × 103 kg/m3)(23.2 m2/s2) = 2.32 × 104 Pa
Section One—Pupil’s Edition Solutions
I Ch. 9–5
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Practice 9E, p. 341
Givens
Solutions
1. T1 = 27°C
T1 = (273 + 27)K = 3.00 × 102 K
3
V1 = 1.5 m
I
P1 = 0.20 × 105 Pa
V2 = 0.70 m3
P2 = 0.80 × 105 Pa
P1V1 P2V2
 = 
T1
T2
5
3
P2V2T1 (0.80 × 10 Pa)(0.70 m )(3.00 × 102 K)
 = 
T2 = 
= 5.6 × 102 K
5
(0.20 × 10 Pa)(1.5 m3)
P1V1
2. P1 = 1.0 × 108 Pa
T1 = (273 + 15.0)K = 288 K
T1 = 15.0°C
T2 = (273 + 65.0)K = 338 K
N2 =
1
N
2 1
At constant volume:
T2 = 65.0°C
P1
P2

= 
N1T1 N2T2

P1N2T2 P12N1T2
= 
P2 = 
T1N1
T1N1
1
(1.0 × 108 Pa)(338 K)
PT
P2 = 1 2 =  = 5.9 × 107 Pa
(2)(288 K)
2T1
P2 = Po + rgh
h = 10.0 cm
P2 = 1.01 × 105 Pa + (13.6 × 103 kg/m3)(9.81 m/s2)(0.100 m)
rmerc = 13.6 × 103 kg/m3
P2 = 1.01 × 105 Pa + 0.133 × 105 Pa = 1.14 × 105 Pa
T2 = 27°C
T1 = (273 + 37)K = 3.10 × 102 K
T1 = 37°C
T2 = (273 + 27)K = 3.00 ×102 K
P1 = Po = 1.01 × 105 Pa
2
g = 9.81 m/s
P1V1 P2V2
 = 
T1
T2
P2V2T1 (1.14 × 105 Pa)(1.0 × 10−7m3)(3.10 × 102 K)
 = 
V1 = 
= 1.2 × 10−7m3
T2 P1
(3.00 × 102 K)(1.01 × 105 Pa)
Section Review, p. 341
1
4. P2 = 2P1
T2 =
3
T
4 1
P1V1 P2V2
 = 
T1
T2
V2 P1T2 P14T1 6 3
 =  =
=  = 
V1 T1P2 T 1P
4 2
1 2 1
3
3:2
I Ch. 9–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. V = 0.10 cm3
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Givens
5. P1 = 6.0 atm
T 1 = 27°C
Solutions
a. At constant volume:
P1T2 = P2T1
Pressure triples; thus, P2 = 3P1.
P1T2 = 3P1T1
I
3P T
T2 = 11
P1
T1 = (273 + 27)K = 3.0 × 102 K
T2 = 3T1 = (3)(3.0 × 102 K) = 9.0 × 102 K
b. Pressure and volume double; thus,
P2 = 2P1 and V2 = 2V1.
P1V1 P2V2
 = 
T1
T2
P2V2T1 (2P1)(2V1)T1
 = 
T2 = 
P1V1
P1V1
T2 = 4T1 = (4)(3.0 × 102 K) = 1.2 × 103 K
Chapter Review and Assess, pp. 343–349
8. Fg = 31.5 N
apparent weight in water
= 265 N
rwater = 1.00 × 103 kg/m3
Copyright © by Holt, Rinehart and Winston. All rights reserved.
apparent weight in oil
= 269 N
ro = 6.3 × 103kg/m3
a. FB = Fg − apparent weight = 315 N − 265 N = 5.0 × 101 N
Fg
(315 N)(1.00 × 103 kg/m3)
ro =  rwater = 
FB
5.0 × 101 N
ro = 6.3 × 103 kg/m3
b. FB = Fg − apparent weight = 315 N − 269 N = 46 N
FB
(46 N)(6.3 × 103 kg/m3)
roil =  ro = 
Fg
315 N
roil = 9.2 × 102 kg/m3
9. Fg = 300.0 N
apparent weight = 200.0 N
ralcohol = 0.70 × 103 kg/m3
FB = Fg − apparent weight = 300.0 N − 200.0 N = 100.0 N
Fg
(300.0 N)(0.70 × 103 kg/m3)
ro =  ralcohol = 
FB
100.0 N
ro = 2.1 × 103 kg/m3
16. P = 2.0 × 105 Pa
Fg = 4PA = (4)(2.0 × 105 Pa)(0.024 m2) = 1.9 × 104 N
A = 0.024 m2
17. P = 5.00 × 105 Pa
4.00 mm
r =  = 2.00 mm
2
F = PA = P(pr 2)
F = (5.00 × 105 Pa)(p)(2.00 × 10−3 m)2 = 6.28 N
Section One—Pupil’s Edition Solutions
I Ch. 9–7
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Givens
I
Solutions
0.64 cm
18. rA =  = 0.32 cm
2
3.8 cm
rB =  = 1.9 cm
2
Fg,B = 500.0 N
FA Fg ,B

= 
AA AB
Fg ,B AA Fg,B(p rA2) Fg ,BrA2
FA =  = 
= 
AB
p rB2
rB2
(500.0 N)(0.0032 m)2
FA = 
= 14 N
(0.019 m)2
F = 14 N downward
19. h = 2.50 × 102 m
a. P = Po + rgh = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(2.50 × 102 m)
rw = 1.025 × 103 kg/m3
P = 1.01 × 105 Pa + 2.51 × 106 Pa = 2.61 × 106 Pa
5
Po = 1.01 × 10 Pa
g = 9.81 m/s2
30.0 cm
r =  = 15.0 cm
2
23. h2 − h1 = 0.30 m
g = 9.81 m/s2
b. F = PA = P(pr 2) = (2.61 × 106 Pa)(p)(0.150 m)2 = 1.84 × 105 N
1
1
P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2
The top of the trough and the hole are both open to the atmosphere, so P1 = P2 = Po.
Because the hole is small, we can assume that v2 ≈ 0.
1
Po + 2 rv12 + rgh1 = Po + rgh2
1
 rv 2
2 1
= rg(h2 − h1 )
2
v1 = 2g(h2 − h1)
v1 = 2g
m/s
(h
h
)(
9.
81
2)(0
.3
0m
) = 2.4 m/s
2 −
1) = (2
A2 = 1.00 × 10−8 m2
F = 2.00 N
1
1
P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2
Because the syringe is horizontal, the above equation simplifies as follows:
1
r = 1.00 × 103 kg/m3
1
P1 + 2 rv12 = P2 + 2 rv22
Also, P1 − P2 = P1 − P0 , which equals the gauge pressure in the barrel.
F
2.00 N
= 8.00 × 104 Pa
P1,gauge = P1 − P2 =  = 
A1 2.50 × 10−5 m2
Finally, assume v1 is negligible in comparison with the fluid speed inside the needle.
1
P1 − P2 = 2 rv22
v2 =
29. T1 = 325 K
2(P1 − P2)
 =
r
(2)(8.00 × 104 Pa)

= 12.6 m/s
1.00 × 103 kg/m3
At constant volume:
5
P1 = 1.22 × 10 Pa
P2 = 1.78 × 105 Pa
P1 P2
 = 
T1 T2
P2 T1 (1.78 × 105 Pa)(325 K)
T2 =  = 
= 474 K
P1
1.22 × 105 Pa
I Ch. 9–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
24. A1 = 2.50 × 10−5 m2
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Givens
Solutions
30. P1 = 7.09 × 104 Pa
T1 = (273 + 100.0)K = 373 K
T1 = 100.0°C
T2 = (273 + 0.0)K = 273 K
P2 = 5.19 × 104 Pa
At constant volume:
T2 = 0.0°C
P1
P2
P3
 =  = 
T1 T2 T3
3
P3 = 4.05 × 10 Pa
I
P3T1
(4.05 × 103 Pa)(373 K)
T3 =  = 
= 21.3 K
P1
7.09 × 104 Pa
g
31. Fg = 4.5 N
Fg
r = 13.6 × 103 kg/m3
Fg
m
V =  =  = 
gr
r
r
4.5 N
V = 
= 3.4 × 10−5 m3
(9.81 m/s2)(13.6 × 103 kg/m3)
g = 9.81 m/s2
F = PA = (1.01 × 105 Pa)(1.00 km2)(106 m2/km2) = 1.01 × 1011 Pa
32. A = 1.00 km2
P = 1.01 × 105 Pa
Fg = PA = P(4pr 2)
33. mm = 70.0 kg
Fg
(mm + mc )g
(70.0 kg + 5.0 kg)(9.81 m/s2)


P = 2 = 
=
4p v
4p r 2
(4p )(0.010 m)2
mc = 5.0 kg
r = 1.0 cm
(75.0 kg)(9.81 m/s2)
P = 
= 5.9 × 105 Pa
(4p)(0.010 m)2
g = 9.81 m/s2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
34. V1 = 8.20 × 10−4 m3
P2 = (0.95)(Po + rgh)
P1 = 0.95Po
P2 = (0.95)[(1.013 × 105 Pa) + (1.00 × 103 kg/m3)(9.81 m/s2)(10.0 m)]
h = 10.0 m
P2 = (0.95)(1.013 × 105 Pa + 0.981 × 105 Pa) = (0.95)(1.994 × 105 Pa) = 1.9 × 105 Pa
Po = 1.013 × 105 Pa
3
Using the ideal gas law, where T1 = T2:
3
r = 1.00 × 10 kg/m
P1V1 = P2V2
g = 9.81 m/s2
(0.95)(1.013 × 105 Pa)(8.20 × 10–4m3)
PV
V2 = 1 1 = 
= 4.2 × 10–4 m3
P2
1.9 × 105 Pa
35. V1 = 1.50 cm3
P1 = Po + rgh = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(100.0 m)
P1 = 1.01 × 105 Pa + 10.1 × 105 Pa = 11.1 × 105 Pa
h = 100.0 m
T1 = T2
Using the ideal gas law, where P2 = Po and T1 = T2.
Po = 1.01 × 105 Pa
P1V1 = P2V2
g = 9.81 m/s2
3
3
r = 1.025 × 10 kg/m
36. r = 1.35 × 103 kg/m3
r = 6.00 cm
P1V1 (11.1 × 105 Pa)(1.50 cm3)
V2 =  = 
= 16.5 cm3
P2
1.01 × 105 Pa
m = rV = r 23pr 3 = 3 rpr 3
14
2
(2)(1.35 × 103 kg/m3)(π)(6.00 × 10–2 m)3
m =  = 6.11 × 10–1kg
3
Section One—Pupil’s Edition Solutions
I Ch. 9–9
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Givens
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6.00 m
37. r =  = 3.00 m
2
h = 1.50 m
a. P = Po + rgh = 1.01 × 105 Pa + (1.00 × 103 kg/m3)(9.81 m/s2)(1.50 m)
P = 1.01 × 105 Pa + 1.47 × 104 Pa = 1.16 × 105 Pa
Po = 1.01 × 105 Pa
I
r = 1.00 × 103 kg/m3
g = 9.81 m/s2
Fg mg (150 kg)(9.81 m/s2)
b. P =  = 2 = 
= 52 Pa
A pr
(p)(3.00 m)2
m = 150 kg
38. v1 = 30.0 m/s
r = 1.29 kg/m3
1
1
a. P1 + 2 rv12 + rgh1 = P2 + 2 rv22+ rgh2
We assume that the difference in height between the two points is negligible, and
note that v2 = 0.
1
1
P2 − P1 = 2 rv12 = 2 (1.29 kg/m3)(30.0 m/s)2 = 5.80 × 102 Pa
A = 175 m2
b. Fnet = Pnet A = (5.80 × 102 Pa)(175 m2)
Fnet = 1.02 × 105 N upward
39. r = 1050 kg/m3
h2 − h1 = 1.00 m
g = 9.81 m/s2
40. r1 = 0.179 kg/m3
1
1
P1 + 2 rv12 + rgh1 = P2 + 2 rv22+ rgh2
Assume v1 = v2.
P1 − P2 = rg(h2 − h1 ) = (1050 kg/m3)(9.81 m/s2)(1.00 m) = 1.03 × 104 Pa
T1 = (273 + 0.0)K = 273 K
T1 = 0.0°C
T2 = (273 + 100.0)K = 373 K
T2 = 100.0°C
At constant pressure:
VT
V2 = 12
T1
m
V =  , so
r
m2 m1T2
 =  
r2 r1T1
The amount of gas remains constant, so m1 = m2.
rT
(0.179 kg/m3)(273 K)
r2 = 11 =  = 0.131 kg/m3
T2
373 K
41. Fg = 1.0 × 106 N
Fg = mg = rVg = rAhg
r = 1.025 × 103 kg/m3
Fg
1.0 × 106 N
A =  = 
rhg (1.025 × 103 kg/m3)(2.5 × 10–2m)(9.81 m/s2)
g = 9.81 m/s2
A = 4.0 × 103 m2
h = 2.5 cm
I Ch. 9–10 Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
V1 V2
 = 
T1 T2
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42. r2 = 20.0 m
P1V1 P2V2
 = 
T1
T2
3
P2 = 3.0 × 10 Pa
T2 = 200.0 K
P2V2T1

V1 = 
T2 P1
5
P1 = 1.01 × 10 Pa
I
P2 3pr23T1
= 
T2 P1
4
T1 = 300.0 K
4
pr 3
3 1
r1 =
r1 =
43. m = 1.0 kg + 2.0 kg = 3.0 kg
rf = 916 kg/m3
mb = 2.0 kg
3
3
rb = 7.86 × 10 kg/m
2
g = 9.81 m/s
3


4
3
T1
 =
pT2 P1
P pr
3 4 2 3 2
3
3
P2T1r23

T2P1
(3.0 × 103 Pa)(300.0 K)(20.0 m)3

= 7.1 m
(200.0 K)(1.01 × 105 Pa)
rf
rf
For the spring scale, apparent weight of block = Fg,b − FB = Fg,b − Fg,b  = mbg 1 − 
rb
rb
916 kg/m3
apparent weight of block = (2.0 kg)(9.81 m/s2) 1 − 
7.86 × 103 kg/m3
= (2.0 kg)(9.81 m/s2)(1 − 0.117)
apparent weight of block = (2.0 kg)(9.81 m/s2)(0.883) = 17 N
For the lower scale, the measured weight equals the weight of the beaker and oil, plus
a force equal to and opposite in direction to the buoyant force on the block. Therefore,
rf
m r
apparent weight = mg + FB = mg + Fg,b  = m + bf g
rb
rb
3
(2.0 kg)(916 kg/m )
(9.81 m/s2)
apparent weight = 3.0 kg + 
7.86 × 103 kg/m3
= (3.0 kg + 0.23 kg)(9.81 m/s2)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
apparent weight = (3.2 kg)(9.81 m/s2) = 31 N
44. rv = 600.0 kg/m3
FB = Fg,r
A = 5.7 m2
FB =rwater Vwater g = rwater(Ah)g
Vr = 0.60 m3
Fg,r = mrg = rr Vrg
rwater =1.0 × 103 kg/m3
rwater Ahg =rr Vr g
g = 9.81 m/s2
(600.0 kg/m3)(0.60 m3)
r Vr
= 
= 6.3 × 10–2 m = 6.3 cm
h = r 
rwater A
(1.0 × 103 kg/m3)(5.7 m2)
45. P1,gauge = 1.8 atm
T1 = 293 K
P2,gauge = 2.1 atm
Po = 1.0 atm
a. P1 = P1,gauge + Po = 1.8 atm + 1.0 atm = 2.8 atm
P2 = P2,gauge + Po = 2.1 atm + 1.0 atm = 3.1 atm
At constant volume:
P P
1 = 2
T1 T2
PT
(3.1 atm)(293 K)
T2 = 21 =  = 3.2 × 102 K
P1
2.8 atm
Section One—Pupil’s Edition Solutions
I Ch. 9–11
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Givens
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b. V1 = Vi
At constant temperature:
P1V1 = P2V2
I
P1
2.8 atm
Vf = V2 = V
i = Vi = 0.90Vi
P2
3.1 atm
∆V = Vf − Vi = 0.90Vi − Vi = −0.10Vi
0.10Vi should be released from the tire
46.
l 1 = 4.0 m
At 220 m down:
3.0 m
r =  = 1.5 m
2
h = 220 m
T1 = (273 + 25)K = 298 K
T1 = 25°C
T2 = (273 + 5.0)K = 278 K
P2 = Po + rgh = 1.01 × 105 Pa + (1025 kg/m3)(9.81 m/s2)(220 m)
P2 = 1.01 × 105 Pa + 2.2 × 106 Pa = 2.3 × 106 Pa
T2 = 5.0°C
rsw = 1025 kg/m3
Po = 1.01 × 105 Pa
g = 9.81 m/s2
P1V1 P2V2
 = 
T1
T2
P1V1T2

V2 = 
T1P2
P1(pr 2 l 1)T2
pr 2l 2 = 
T1P2
l2 =
(1.01 × 105 Pa)(4.0 m)(278 K)
P1 l 1T2
= 
= 0.16 m

(298 K)(2.3 × 106 Pa)
T1P2
where l 2 is height of the remaining air inside the bell.
hwater = l 1 − l 2 = 4.0 m − 0.16 m = 3.8 m
47. h = 26 cm

F
m g
a. r =  =  = 
g
y = 3.5 cm
V
Fg = 19 N
2
g = 9.81 m/s
hwy
hwyg
19 N
r = 
= 1.0 × 103 kg/m3
(0.26 m)(0.21 m)(0.035 m)(9.81 m/s2)
Fg Fg
19 N
b. P =  =  =  = 3.5 × 102 Pa
A hw (0.26 m)(0.21 m)
Fg Fg
19 N
c. P =  =  =  = 2.1 × 103 Pa
A hy (0.26 m)(0.035 m)
I Ch. 9–12
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fg
w = 21 cm
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48. ∆y = −1.0 m
Use the equations for a horizontally-launched projectile to determine the water jet’s
initial speed.
∆x
vx = 
∆t
1
∆y = Ny,i∆t − 2 g∆t 2
∆x = 0.60 m
g = 9.81 m/s2
vy,i = 0 m/s, so
I
1
∆y = − 2 g∆t 2
g
−2∆y
∆t =
∆x
vx = 
−2∆y

g
Use Bernoulli’s equation for the height of the tank, h, noting that P1 = P2 = Po, v2 ≈ 0,
and v1 = vx .
1
1
P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2
1
rg(h2 − h1) = 2 rv12
∆x2
−g∆x2



2
2∆y
−∆x2
−2∆y
1 v1


h = h2 − h1 = 2  =  =  = 
g
4∆y
2g
g
2g
2
−(0.60 m)
h =  = 9.0 × 10–2 m = 9.0 cm
(4)(−1.0 m)
49. h1 = 5.00 cm
h2 = 12.0 cm
∆x1 = ∆x2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
Designate the position of the lower hole as point 1 and the position of the higher
hole as point 2. Use the equations for horizontally-launched projectiles to obtain expressions for the initial speeds of the water streams.
∆x
v1 = 1
∆t1
∆t1 =
−g = g
2∆y1
2h1
∆x1
∆x
v1 = 1 = 
∆t1
2h1

g
Similarly,
∆x2
∆x
v2 = 2 = 
∆t2
2h2

g
∆x1 = ∆x2, so
v1 ∆t1 = v2∆t2
g h
 =v
h
2h
g
2h2
∆t
v1 = v2 2 = v2
∆t1
1
2
2
1
Section One—Pupil’s Edition Solutions
I Ch. 9–13
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Solve for v2. Apply Bernoulli’s equation, with P1 = P2 = Po.
1
1
P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2
1
 r
2
I
1

2
(v12 − v22) = rg(h2 − h1)
v22h2
 − v22 = g (h2 − h1)
h1
2g(h2 − h1) 2gh1(h2 − h1)
 = 2gh1
v22 =  = 
(h2 − h1)
h2
 − 1
h1
v2 = 2g
h1
Apply Bernoulli’s equation again, using h3 to represent the height of the tank. Note
that P3 = P2 = Po, and v3 ≈ 0.
1
1
P2 + 2 rv22 + rgh2 = P3 + 2 rv32 + rgh3
1
 rv 2
2 2
= rg(h3 − h2)
v22 2gh1
h3 − h2 =  =  = h1
2g
2g
h3 = h1 + h2 = 5.00 cm + 12.0 cm = 17.0 cm
50. Am = 6.40 cm2
F
F
1 = 2
A1 A2
Ab = 1.75 cm2
mk = 0.50
Ab
1.75 cm2
2 (44 N) = 12 N
Fb = F
=
p
Am
6.40 cm
Fp = 44 N
Fb is the normal force exerted on the brake shoe. Fk is given as follows:
Fk = mkFn = (0.50)(12 N) = 6.0 N
flow rate = 1.55 m3/s
52. h = 2.0 cm = 0.020 m
y = 1.5 cm = 0.015 m
flow rate = Av = pr 2v
1.55 m3/s
flow rate

 = 31.6 m/s
v = 
=
(p )(0.125 m)2
pr 2
Before the oil is added:
Fg,b = FB,water = rwaterVg = rwater Ayg
3
roil = 900.0 kg/m
rwater = 1.00 × 103 kg/m3
After the oil is added:
Fg,b = FB,water + FB,oil
rwater Ayg = rwater A(h − y1)g + roil Ay1g
rwatery = rwater h − rwater y1 + roil y1
1 − r  y = h − y
roil
1
water
0.020 m − 0.015m
h−y
y1 =  = 
900.0 kg/m3
roil

1 − 
1− 
1.00 × 103 kg/m3
rwater
0.0050 m
0.0050 m
y1 =  =  = 5.0 × 10−2 m = 5.0 cm
1 − 0.9000
0.1000
I Ch. 9–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
0.250 m
51. r =  = 0.125 m
2
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53. roil = 930 kg/m3
FB = Fg
h = 4.00 cm
FB,oil + FB,water = Fg,b
rb = 960 kg/m3
moil g + mwater g = mb g
rwater = 1.00 × 103 kg/m3
roilVoil + rwaterVwater = rbVb
g = 9.81 m/s2
roil A(h − y) + rwater Ay = rb Ah
I
roil (h − y) + rwater y = rbh
roil h − roil y + rwater y = rb h
y(rwater − roil ) = h(rb − roil )
h( rb − roil ) (0.0400 m)(960 kg/m3 − 930 kg/m3)
 = 
y=
rwater − roil
1.00 × 103 kg/m3 − 930 kg/m3
(0.0400 m)(30 kg/m3)
y = 
= 1.71 × 10−2 m = 1.71 cm
70 kg/m3
54. Fg,b = 50.0 N
apparent weight of sinker
= 200.0 N − Fg,b
FB,b = Fg,b − (apparent weight of block and sinker − apparent weight of sinker)
FB,b = Fg,b − [140.0 N − (200.0 N − Fg,b)]
apparent weight of block
and sinker = 140.0 N
FB,b = Fg,b + 60.0 N − Fg,b = 60.0 N
rwater = 1.00 × 103 kg/m3
Fg,b
(50.0 N)(1.00 × 103 kg/m3)
rb =  rwater =  = 833 kg/m3
FB,b
60.0 N
55. ∆t = 1.0 s
For one molecule:
A = 8.0 cm2
mvf − mvi
F
∆p
P =  =  = 
A A ∆t
A ∆t
m = 4.68 × 10−26 kg
In a perfect elastic collision with the wall, vi = vf .
v i = 300.0 m/s
mvi − m(−vi) mvi + mvi 2mv
P =  =  = i
A∆t
A∆t
A∆t
N = 5.0 × 10
Copyright © by Holt, Rinehart and Winston. All rights reserved.
FB,b = Fg,b − apparent weight of blocks
23
For all of the molecules:
2mv
(5.0 × 1023)(2)(4.68 × 10−26 kg)(300.0 m/s)
P = N i = 
A∆t
(8.0 cm2)(1 × 10−4 m2/cm2)(1.0 s)
P = 1.8 × 104 Pa
Section One—Pupil’s Edition Solutions
I Ch. 9–15
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56. h = 10.0 m
l y = l (sin q )
h2 − h1 = h − l y = h − l (sin q)
l
= 2.0 m
q = 30.0°
h2 − h1 = 10.0 m − (2.0 m)(sin 30.0°) = 10.0 m − 1.0 m = 9.0 m
2
I
g = 9.81 m/s
Apply Bernoulli’s equation to find v1. Assume that P1 = P2 = Po , v2 ≈ 0, and r is
constant.
1
1
P1 + 2 rv12 + rgh1 = P2 + 2 rv22 + rgh2
1 2
 v
2 1
+ gh1 = gh2
v1 = 2g
m/s
m) = 13 m/s
(h
2−h
)(
9.
81
2)(9
.0
1) = (2
Find the height of the water when vy,f = 0 m/s by using an equation for projectile
motion.
vy,f 2 = vi2(sin q )2 − 2g∆y = 0
2
vi(sin q )
 =
∆y = 
2g
57. r2 = 2.0 mm
= 2.2 m above the spout opening
a. At constant temperature:
r1 = 3.0 mm
P1V1 = P2V2
3
3
r = 1.025 × 10 kg/m
5
P1 = Po = 1.01 × 10 Pa
g = 9.81 m/s2
2
(13 m/s)(sin 30.0°)

(2)(9.81 m/s2)
3

P1V1 P13pr1 P1r13
P2 =  = 
= 
4
V2
r23
pr 3
3 2
4
P2 = P1 + rgh
r3
P1 13 − 1
r2
P2 − P1
 = 
h =
rg
rg
(3.0 × 10−3 m)3
−1
(1.01 × 105 Pa) 
(2.0 × 10−3 m)3

h=
(1.025 × 103 kg/m3)(9.81 m/s2)
(1.01 × 105 Pa)(2.4)
(1.01 × 105 Pa)(3.4 − 1)

h = 
=
(1.025 × 103 kg/m3)(9.81 m/s2)
(1.025 × 103 kg/m3)(9.81 m/s2)
h = 24 m
b. P = Po + rgh = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(24 m)
P = 1.01 × 105 Pa + 2.4 × 105 Pa = 3.4 × 105 Pa
or, alternatively
P1 r13
(1.01 × 105 Pa)(3.0 × 10–3m)3
P2 = 
= 3.4 × 105 Pa
3 = 
r2
(2.0 × 10–3 m)3
I Ch. 9–16
Holt Physics Solution Manual
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58. r1 = 0.30 m
flow rate1 flow rate1
0.20 m3/s
v1 =  = 
= 2 = 0.71 m/s
2
A1
πr1
(π)(0.30 m)
flow rate1 = 0.20 m3/s
r2 = 0.15 m
A1v1 flow rate1
0.20 m3/s
v2 =  = 
= 2 = 2.8 m/s
2
A2
πv2
(π)(0.15 m)
h1 − h2 = 0.60 m
Apply Bernoulli’s equation to find the gauge pressure (P2 − P1) in the lower pipe.
g = 9.81 m/s2
P1 + 2 rv12 + rgh1 = P2 + 2 r v22 + rgh2P2 − P1 = 2 r(v12 − v22) + rg(h1 − h2)
1
= r 2 (v12 − v22) + g(h1 − h2)
P1 = Po
rwater = 1.00 × 103 kg/m3
1
1
I
1
P2 − P1 = (1.00 × 103 kg/m3)2 (0.71 m/s)2 − 2 (2.8 m/s)2
+ (9.81 m/s2)(0.60 m)]
1
1
P2 − P1 = (1.00 × 103 kg/m3)(0.25 m2/s2 − 3.9 m2/s2 + 5.9 m2/s2)
P2 − P1 = (1.00 × 103 kg/m3)(2.2 m2/s2)
P2 − P1 = 2.2 × 103 Pa
59. k = 90.0 N/m
Fnet = FB − Fg,b − Fg,hel − Fspring = 0
mb = 2.00 g
rairVg − mb g − rhelVg − k∆x = 0
V = 5.00 m3
g(rairV − mb − rhelV)
∆x = 
k
g = 9.81 m/s2
rair = 1.29 kg/m3
rhel = 0.179 kg/m3
(9.81 m/s2)[(1.29 kg/m3)(5.00 m3) − 2.00 × 10−3 kg − (0.179 kg/m3)(5.00 m3)]
∆x = 
90.0 N/m
(9.81 m/s2)(6.45 kg − 2.00 × 10−3 kg − 0.895 kg)
∆x = 
90.0 N/m
(9.81 m/s2)(5.55 kg)
∆x =  = 0.605 m
90.0 N/m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
60. A = 2.0 cm2
a. flow rate = Av = (2.0 cm2)(42 cm/s) = 84 cm3/s
r = 1.0 g/cm3
In g/s:
v = 42 cm/s
flow rate = (84 cm3/s)(1.0 g/cm3) = 84 g/s
A2 = 3.0 × 103 cm2
b. Use the continuity equation.
Av
(2.0 cm2)(42 cm/s)
v2 = 11 = 
= 0.028 cm/s = 2.8 × 10−4 m/s
A2
3.0 × 103 cm2
1.6 cm
61. ra =  = 0.80 cm
2
1.0 × 10−6 m
rc = 
2
= 0.50 × 10−6 m
va = 1.0 m/s
vc = 1.0 cm/s
Use the continuity equation.
A va
Ac = a,
vc
where Ac is the total capillary cross section needed.
(p)(8.0 × 10−3 m)2(1.0 m/s)
pr 2v
Ac = aa =  = 2.0 × 10−2 m2
vc
0.010 m/s
Ac = NA
2.0 × 10−3 m2
A
Ac

=
= 2.6 × 1010 capillaries
N = c = 
A p rc2 (p )(0.50 × 10−6 m)2
Section One—Pupil’s Edition Solutions
I Ch. 9–17
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62.
l
Solutions
= 1.5 m
V
V
l wh
∆t =  =  = 
flow rate Av pr 2v
w = 65 cm
h = 45 cm
I
(1.5 m)(0.65 m)(0.45 m)
∆t = 
= 9.3 × 102 s
(p )(0.010 m)2(1.5 m/s)
2.0 cm
r =  = 1.0 cm
2
v = 1.5 m/s
Fnet = ma = FB − Fg
63. rair = 1.29 kg/m3
FB − Fg rairVg − rhelVg
a =  = 
rhelV
m
rhel = 0.179 kg/m3
g = 9.81 m/s2
g(rair − rhel)
r
 = g air − 1
a=
rhel
rhel
1.29 kg/m3
a = (9.81 m/s2) 3 − 1 = (9.81 m/s2)(7.21 − 1)
0.179 kg/m
a = (9.81 m/s2)(6.21) = 60.9 m/s2
Fnet = (m + mair)a = FB − Fg,b − Fg,a
64. m = 1.0 kg
(m + mair)a = rwaterVg − g(m + mair)
r = 0.10 m
h = 2.0 m
3
rwater = 1.00 × 10 kg/m
g = 9.81 m/s2
rair = 1.29 kg/m3
rwater 3pr 3 g
rwaterVg
rwater(4pr3)g

 −g
a =  − g = 
−
g
=
4
m + mair
3m + rair (4pr3)
m + rair 3pr 3
4
3
(1.00 × 103 kg/m3)(4p)(0.10 m)3(9.81 m/s2)
a =  − 9.81 m/s2
(3)(1.0 kg) + (1.29 kg/m3)(4p)(0.10 m)3
(1.00 × 103 kg/m3)(4π)(0.10 m3)(9.81 m/s2)
a =  − 9.81 m/s2
3.0 kg
a = 41 m/s2 − 9.81 m/s2 = 31 m/s2
Use the following equation to find the speed of the ball as it exits the water.
Note that vi = 0.
vf 2 = vi2 + 2ah = 2ah
Use the following equation to find the maximum height of the ball above the water.
Note that vi = vf for the ball leaving the water.
vf 2 = vi2 − 2g∆y = 0
v 2 2ah ah
∆y = i =  = 
2g
2g
g
(31 m/s2)(2.0 m)
∆y = 
= 6.3 m
9.81 m/s2
I Ch. 9–18
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(1.00 × 103 kg/m3)(4p)(0.10 m)3(9.81 m/s2)
a =  − 9.81 m/s2
3.0 kg + 0.016 kg
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Givens
Solutions
65. r = 0.60rwater
3
3
rwater = 1.00 × 10 kg/m
∆y = −10.0 m
First find the speed of the sphere just before impact by using the following equation.
Assume vi = 0.
vf 2 = vi2 − 2g∆y = −2g∆y
m) = 14.0 m/s
vf = −
2g
∆
y = (−
2)
(9
.8
1m
/s
2)(−
10
.0
g = 9.81 m/s2
I
This is the initial velocity of the sphere as it enters the water. Now find the net force
on the sphere to determine its acceleration underwater.
Fnet = ma = FB − Fg
0.60rwaterVa = rwaterVg − 0.60rwaterVg
2
2
g − 0.60g 9.81 m/s − (0.60)(9.81 m/s )
a =  = 
0.60
0.60
9.81 m/s2 − 5.9 m/s2 3.9 m/s2
a =  =  = 6.5 m/s2
0.60
0.60
Use the following equation to find the maximum depth:
vf 2 = vi2 − 2ah = 0
(14.0 m/s)2
v2
h = i = 
= 15 m
2a (2)(6.5 m/s2)
66. T1 = 27°C
T1 = (273 + 27) K = 3.00 × 102 K
5
P1 = 1.01 × 10 Pa
T2 = (273 + 225) K = 498 K
T2 = 225°C
At constant volume:
P1 P2
 = 
T1 T2
P1 T2
(1.01 × 105 Pa)(498 K)
P2 =  = 
= 1.68 × 105 Pa
T1
3.00 × 102 K
Copyright © by Holt, Rinehart and Winston. All rights reserved.
67. ∆t = 1.00 min
N = 150
m = 8.0 g
v = 400.0 m/s
A = 0.75 m2
For one bullet:
mvf − mvi
F
∆p
P =  =  = 
A ∆t
A A ∆t
In a perfect elastic collision with the wall, vi = − vf.
2m v
P = 
A ∆t
For all the bullets:
2m v
(150)(2)(8.0 × 10−3 kg)(400.0 m/s)
P = N  = 
= 21 Pa
A ∆t
(0.75 m2)(1.00 min)(60 s/min)
Section One—Pupil’s Edition Solutions
I Ch. 9–19
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Givens
Solutions
68. m = 4.00 kg
a. Assume that the mass of the helium in the sphere is not significant compared with
the 4.00 kg mass of the sphere.
0.200 m
r =  = 0.100 m
2
I
Fnet = ma = FB − Fg
ma = rwaterVg − mg
h = 4.00 m
3
3
rwater = 1.00 × 10 kg/m
g = 9.81 m/s2
rwater 3pr 3g − mg
4
a=
m
(1.00 × 103 kg/m3)3(p)(0.100 m)3(9.81 m/s2) − (4.00 kg)(9.81 m/s2)
a = 
4.00 kg
4
41.1 N − 39.2 N
1.9 N
a =  =  = 0.48 m/s2
4.00 kg
4.00 kg
b. Noting that vi = 0,
1
1
∆y = vi ∆t + 2a∆t 2 = 2a∆t2
2∆ay = 2(
ha−
2r)
(2)(3.80 m)
(2)(4.00 m − 0.200 m)
 = 
∆t = = 4.0 s
0.48
m/
s 0.48 m/s
∆t =
2
2
Fnet = FB − Fg − Fspring = 0
69. k = 16.0 N/m
mb = 5.00 × 10
rwaterVg − mb g − k∆x = 0
kg
3
rb = 650.0 kg/m
3
3
rwater = 1.00 × 10 kg/m
g = 9.81 m/s2
m
rwater b g − mb g − k∆x = 0
rb
m
rwater b g − mb g
rb
∆x = 
k
 − 1 m g
r
∆x = 
rwater
b
b
k
 − 1 (5.00 × 10 kg)(9.81 m/s )
650.0 kg/m
∆x = 
1.00 × 103 kg/m3
−3
2
3
16.0 N/m
(1.54 − 1)(5.00 × 10–3kg)(9.81 m/s2)
∆x = 
16.0 N/m
(0.54)(5.00 × 10−3 kg)(9.81 m/s2)
∆x =  = 1.7 × 10−3 m
16.0 N/m
I Ch. 9–20
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
−3
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Heat
Chapter 10
I
Practice 10A, p. 363
Givens
1. TF = −128.6°F
Solutions
5
5
5
TC = 9(TF − 32.0) = 9(−128.6 − 32.0)°C = 9(−160.6)°C
TC = −89.22°C
T = TC + 273.15 = (−89.22 + 273.15) K = 183.93 K
2. TF,1 = 105°F
TF,2 = −25°F
5
5
5
TC,1 = 9(TF,1 − 32.0) = 9(105 − 32.0)°C = 9(73)°C
TC,1 = 41°C
T = (TC + 273)K
T1 = (41 + 273)K = 314 K
5
5
5
TC,2 = 9(TF,2 − 32.0) = 9(−25 − 32.0)°C = 9(−57)°C
TC,2 = −32°C
T2 = (−32 + 273)K = 241 K
3. TF,1 = 98.6°F
TF,2 = 102°F
5
5
5
TC,1 = 9(TF,1 − 32.0) = 9(98.6 − 32.0)°C = 9(66.6)°C
TC,1 = 37.0°C
5
5
5
TC,2 = 9(TF,2 − 32.0) = 9(102° − 32.0)°C = 9(7.0 × 101)°C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
TC,2 = 39°C
4. TC,i = 23°C
TC,f = 78°C
T = TC + 273.15
Ti = TC,i + 273.15 = (23 + 273.15) K = 296 K
Tf = TC,f + 273.15 = (78 + 273.15) K = 351 K
∆T = Tf − Ti = 351 K − 296 K = 55 K
Alternatively, because a degree Celsius equals a kelvin,
∆T = ∆T C = TC,f − TC,i = 78°C − 23°C
∆T = 55°C = 55 K
9
TF = 5TC + 32.0
9
9
TF,i = 5TC,i + 32.0 = 5(23)°F + 32.0° F = (41 + 32.0)°F = 73°F
9
9
TF,f = 5TC,f + 32.0 = 5(78)°F + 32.0° F = (140 + 32.0)°F = 172°F
9
9
∆TF = TF,f − TF,i = 5 (78 − 23)°F = 5 (55)°F = 99°F
5. T = 77.34 K
TC = T − 273.15 = (77.34 − 273.15)°C = −195.81°C
9
9
TF = 5TC + 32.0 = 5(−195.81)°F + 32.0°F = (−352.46 + 32.0)°F
TF = −320.5°F
Section One—Pupil’s Edition Solutions
I Ch. 10–1
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Section Review, p. 364
Givens
Solutions
3. T = 90.2 K
TC = T − 273.15 = (90.2 − 273.15)°C = −183.0°C
9
9
TF = 5(TC) + 32.0 = 5(−183.0)°F + 32.0°F = (−329.4 + 32.0)°F
I
TF = −297.4°F
4. boiling point = 444.6°C
5
5
a. ∆TC = 9(∆TF) = 9(586.1)°C = 325.6°C
melting point = 586.1°F
below boiling point
melting point = 444.6°C − 325.6°C = 119.0°C
9
9
b. TF,1 = 5(TC ,1) + 32.0 = 5(119.0)°F + 32.0°F = (214.2 + 32.0)°F
TF,1 = 246.2°F
9
9
TF,2 = 5(TC ,2 ) + 32.0 = 5(444.6)°F + 32.0°F = (800.3 + 32.0)°F
TF,2 = 832.3°F
c. T1 = TC ,1 + 273.15 = (119.0 + 273.15) K = 392.2 K
T2 = TC ,2 + 273.15 = (444.6 + 273.15) K = 717.8 K
Practice 10B, p. 370
1. m = 11.5 kg
∆U = mgh = (11.5 kg)(9.81 m/s2)(6.69 m) = 755 J
g = 9.81 m/s2
h = 6.69 m
mh = 2.50 kg
vh = 65.0 m/s
3. m = 3.0 × 10−3 kg
h = 50.0 m
4. m = 2.5 kg
vi = 5.7 m/s
5
3.3 × 10 J melts 1.0 kg of ice
5. ∆U = 209.3 J
∆U = 3(KEh ) = 3 2mh vh2 = 6 (2.50 kg)(65.0 m/s)2
1
1
∆U = 1.76 × 103 J
∆U = (0.65)(PEi) = (0.65)(mgh) = (0.65)(3.0 × 10−3 kg)(9.81 m/s2)(50.0 m)
∆U = 0.96 J
1
1
∆U = KEi = 2mvi2 = 2(2.5 kg)(5.7 m/s)2 = 41 J
(41 J)(1.0 kg)
ice melted = 
= 1.2 × 10−4 kg
3.3 × 105 J
1
∆U = KE = 2mv 2
m = 0.25 kg
v=
I Ch. 10–2
1 1
9.3 J)
2∆mU = (2
)0(.2205
kg = 41 m/s
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. ms = 0.500 kg
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Section Review, p. 370
Givens
Solutions
4. Ti = 10.0°C
∆U = PEi = mgh = (505 kg)(9.81 m/s2)(50.0 m) = 2.48 × 105 J
m = 505 kg
h = 50.0 m
2
g = 9.81 m/s
4186 J/kg increases water
temperature by 1.0°C
2.48 × 105 J
 (1.0°C)
505 kg
∆T =  = 0.12°C
4186 J/kg
I
Tf = Ti + ∆T = 10.0°C + 0.12°C = 10.1°C
Practice 10C, p. 374
1. mg = 3.0 kg
cp,wmw ∆Tw = cp,gmg ∆Tg
Tg = 99°C
cp,wmw(Tf − Tw) = cp,gmg(Tg − Tf)
cp,g = 129 J/kg • °C
mw = 0.22 kg
Tw cp,wmw + Tg cp,g mg
Tf = 
cp,wmw + cp,gmg
Tw = 25°C
cp,wmw = (4186 J/kg • °C)(0.22 kg) = 920 J/°C
cp,w = 4186 J/kg • °C
Tw cp,w mw = (25°C)(4186 J/kg • °C)(0.22 kg) = 2.3 × 104 J
cp,gmg = (129 J/kg • °C)(3.0 kg) = 390 J/°C
Tg cp,g mg = (99°C)(129 J/kg • °C)(3.0 kg) = 3.8 × 104 J
(2.3 × 104 J)(3.8 × 104 J) + (99°C)(390 J/°C) 6.1 × 104 J
Tf =  = 
1310 J/°C
(920 J/°C + 390 J/°C)
Tf = 47°C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. mt = 0.225 kg
cp,wmw ∆Tw = cp,tmt ∆Tt
Tt = 97.5°C
cp,wmw(Tf − Tw) = cp,tmt(Tt − Tf)
mw = 0.115 kg
Tw = 10.0°C
[Twcp,wmw + Ttcp,tmt]
Tf = 
(cp,wmw + cp,tmt)
cp,t = 230 J/kg • °C
cp,wmw = (4186 J/kg • °C)(0.115 kg) = 481 J/°C
cp,w = 4186 J/kg • °C
Tw cp,wmw = (10.0°)(4186 J/kg • °C)(0.115 kg) = 4.81 × 103 J
cp,tmt = (230 J/kg • °C)(0.225 kg) = 52 J/°C
Tt cp,t mt = (97.5°C)(230 J/kg • °C)(0.225 kg) = 5.0 × 103 J
4.81 × 103 J + 5.0 × 103 J 9.8 × 103 J
Tf =  = 
481 J/°C + 52 J/°C
533 J/°C
Tf = 18.0° C
3. mm = 0.032 kg
cp,mmm ∆Tm = cp,c mc ∆Tc
Tm = 11°C
Because cp,m = cp,c , mm ∆Tm = mc ∆Tc .
mc = 0.16 kg
(mm )(Tf − Tm ) = (mc )(Tc − Tf )
Tc = 91°C
mmTf − m mTm = mcTc − mcTf
cp,m = cp,c = cp,w
mmTf + mcTf = mcTc + mmTm
mcTc + mmTm (0.16 kg)(91°C) + (0.032 kg)(11°C)
 = 
Tf = 
mm + mc
0.032 kg + 0.16 kg
15 kg • °C + 0.35 kg • °C 15 kg • °C
Tf =  =  = 79°C
0.19 kg
0.19 kg
Section One—Pupil’s Edition Solutions
I Ch. 10–3
Print
Givens
Solutions
4. mc = 0.75 kg
I
cp,c mc ∆Tc = cp,w mw ∆Tw
Tc = 36.5°C
∆Tc = Tc − Tf = 36.5°C − 24.4°C = 12.1°C
mw = 1.25 kg
∆Tw = Tf − Tw = 24.4°C − 20.0°C = 4.4°C
Tw = 20.0°C
Tf = 24.4°C
cp,w mw ∆Tw (4186 J/kg • °C)(1.25 kg)(4.4°C)
cp,c =  = 
mc ∆Tc
(0.75 kg)(12.1°C)
cp,w = 4186 J/kg • °C
cp,c = 2500 J/kg • °C
5. mb = 0.59 kg
cp,b mc ∆Tb = cp,w mw ∆Tw
Tb = 98.0°C
∆Tw = Tf − Tw = 6.8°C − 5.0°C = 1.8°C
mw = 2.80 kg
∆Tb = Tb − Tf = 98.0°C − 6.8°C = 91.2°C
Tw = 5.0°C
Tf = 6.8°C
cp,w mw ∆Tw (4186 J/kg • °C)(2.80 kg)(1.8°C)
cp,b =  = 
mb ∆Tb
(0.59 kg)(91.2°C)
cp,w = 4186 J/kg • °C
cp,b = 390 J/kg • °C
6. ∆Ta = 1.0°C
mw = 1.0 kg
∆Tw = 1.0°C
cp,a ma ∆Ta = cp,w mw ∆Tw
cp,w mw ∆Tw
ma = 
cp,a ∆Ta
r = 1.29 kg/m3
cp,w mw ∆Tw 1
1
m
(4186 J/kg • °C)(1.0 kg)(1.0°C)
V = a =   =  3
cp,a ∆Ta
r
1.29 kg/m
r
(1000.0 J/kg • °C)(1.0°C)
cp,w = 4186 J/kg • °C
V = 3.2 m3
cp,a = 1000.0 J/kg • °C
7. ∆Tv = 8.39°C
mw = 101 g
cp,w = 4186 J/kg • °C
∆Tc = 838°C
cp,c = 387 J/kg • °C
cp,c mc ∆Tc = cp,w mw ∆Tw
cp,w mw ∆Tw

mc = 
cp,c ∆Tc
(4186 J/kg • °C)(0.101 kg)(8.39°C)
mc = 
(387 J/kg • °C)(838°C)
mc = 1.09 × 10−2 kg = 10.9 g
Practice 10D, p. 381
1. m = 42 g
∆T1 = melting point − Ti = 0°C − (−11°C) = 11°C
Ti = −11°C (ice)
∆T2 = boiling point − melting point = 100.0°C − 0°C = 100.0°C
Tf = 111°C (steam)
∆T3 = Tf − boiling point = 111°C − 100.0°C = 11°C
Lf = 3.33 × 105 J/kg
Q1 = mcp,ice ∆T1 = (42 × 10−3 kg)(2.09 × 103 J/kg • °C)(11°C) = 970 J
Lv = 2.26 × 106 J/kg
Q2 = mLf = (42 × 10−3 kg)(3.33 × 105 J/kg) = 1.4 × 104 J
cp,ice = 2.09 × 103 J/kg • °C
Q3 = mcp,w ∆T2 = (42 × 10−3 kg)(4186 J/kg • °C)(100.0°C) = 1.8 × 104 J
cp,steam = 2.01 × 103 J/kg • °C
Q4 = mLv = (42 × 10−3 kg)(2.26 × 106 J/kg) = 9.5 × 104 J
Q5 = mcp,steam ∆T3 = (42 × 10−3 kg)(2.01 × 103 J/kg • °C)(11°C) = 930 J
Qtot = Q1 + Q2 + Q3 + Q4 + Q5
Qtot = 970 J + (1.4 × 104 J) + (1.8 × 104 J) + (9.5 × 104 J) + 930 J
Qtot = 1.29 × 105 J
I Ch. 10–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Givens
Solutions
2. m = 1.0 kg
Q = mLv = (1.0 kg)(2.01 × 105 J/kg) = 2.0 × 105 J
T = 77 K
Lv = 2.01 × 105 J/kg
∆T = melting point − Ti = 327.3°C − 27.3°C = 300.0°C
3. m = 0.225 kg
Ti = 27.3°C
Q1 = mcp,l ∆T = (0.225 kg)(1.28 × 102 J/kg • °C)(300.0°C) = 8640 J
cp,l = 1.28 × 102 J/kg • °C
Q2 = mLf = (0.225 kg)(2.45 × 104 J/kg) = 5510 J
Lf = 2.45 × 104 J/kg
Qtot = Q1 + Q2 = 8640 J + 5510 J = 1.415 × 104 J
I
melting point = 327.3°C
4. m = (14.0 g/can)(1000
cans) = 14.0 × 103 g =
14.0 kg
∆T = melting point − Ti = 660.4°C − 26.4°C = 634.0°C
Q1 = mcp,a ∆T = (14.0 kg)(8.99 × 102 J/kg • °C)(634.0°C) = 7.98 × 106 J
Q2 = mLf = (14.0 kg)(3.97 × 105 J/kg) = 5.56 × 106 J
Ti = 26.4°C
cp,a = 8.99 × 102 J/kg • °C
Qtot = Q1 + Q2 = (7.98 × 106 J) + (5.56 × 106 J) = 1.354 × 107 J
Lf = 3.97 × 105 J/kg
melting point = 660.4°C
Q1 = cp,s ms ∆Ts
5. mi = 0.011 kg
Ti = 0°C
Q2 = miLf
ms = 0.450 kg
Q3 = cp,w mi ∆Tw
Ts = 80.0°C
By the conservation of energy, Q1 = Q2 + Q3 .
cp,s = cp,w = 4186 J/kg • °C
5
Lf = 3.33 × 10 J/kg
cp,s ms ∆Ts = mi Lf + cp,w mi ∆Tw
cp,s ms (Ts − Tf ) = mi Lf + cp,w mi (Tf − Ti )
cp,s ms Ts − cp,s msTf = mi Lf + cp,w mi Tf − cp,w miTi
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Tf (cp,w mi + cp,s ms ) = cp,s msTs + cp,w m iTi − miLf
cp,s msTs + cp,w m iTi − m i Lf
Tf =  ,
cp,w mi + cp,s ms
where Ti = 0.0°C
(4186 J/kg • °C)(0.450 kg)(80.0°C) + (0 J) − (0.011 kg)(3.33 × 105 J/kg)
Tf = 
(4186 J/kg • °C)(0.011 kg) + (4186 J/kg • °C)(0.450 kg)
(1.51 × 105 J) + (0 J) − (3.7 × 103 J) 1.47 × 105 J
Tf =  =  = 76.2°C
1930 J/°C
46 J/°C + 1880 J/°C
6. mal = 25 kg
Q = mal Lf = mair cp,air ∆T
Tair = 25°C
malLf
(25 kg)(3.97 × 105 J/kg)
∆T =  = 
= 76°C
mair cp,air (130 kg)(1.0 × 103 J/kg • °C)
cp,air = 1.0 × 103 J/kg • °C
Tf = Tair + ∆T = 25°C + 76°C = 101°C
mair = 130 kg
cp,al = 8.99 × 102 J/kg • °C
Lf = 3.97 × 105 J/kg
Section One—Pupil’s Edition Solutions
I Ch. 10–5
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Section Review, p. 382
Givens
Solutions
1. mg = 47 g
I
cp,w mw ∆Tw = cp,g mg ∆Tg
Tg = 99°C
∆Tg = Tg − Tf = 99°C − 38°C = 61°C
Tw = 25°C
∆Tw = Tf − Tw = 38°C − 25°C = 13°C
Tf = 38°C
cp,g = 1.29 × 102 J/kg • °C
cp,g mg ∆Tg (1.29 × 102 J/kg • °C)(47 × 10−3 kg)(61°C)
mw =  = 
cp,w ∆Tw
(4186 J/kg • °C)(13°C)
cp,w = 4186 J/kg • °C
mw = 6.8 × 10−3 kg = 6.8 g
4. m = 15 g = 0.015 kg
Q
7.4 × 103 J
15.8 kJ − 8.37 kJ
a. cp,l =  =  =  = 2 × 103 J/kg • °C
m∆T (0.015 kg)(300°C − 80°C)
(0.015 kg)(200°C)
Q 8.37 kJ − 1.27 kJ 7.10 × 103 J
b. Lf =  =  =  = 4.7 × 105 J/kg
m
0.015 kg
0.015 kg
Q
1.27 × 103 J
1.27 kJ − 0 kJ
c. cp,s =  =  =  = 1 × 103 J/kg • °C
m∆T (0.015 kg)(80°C − 0°C) (0.015 kg)(80°C)
Q
1 × 103 J
796 kJ − 795 kJ
d. cp,v =  =  =  = 7 × 102 J/kg • °C
m∆T (0.015 kg)(400°C − 300°C) (0.015 kg)(100°C)
Q 795 kJ − 15.8 kJ 779 × 103 J
e. Lv =  =  =  = 5.2 × 107 J/kg
0.015 kg
0.015 kg
m
5. Tf = 175°C
∆T = Tf − Tp = 175°C − 21°C = 154°C
Tp = 21°C
Q = mpcp,p ∆T = (0.105 × 10−3 kg/kernel)(125 kernels)(1650 J/kg • °C)(154°C)
cp,p = 1650 J/kg • °C
Q = 3340 J
6. mw = (0.14)(95.0 g)
L v = (0.90)(2.26 × 106 J/kg)
Q = mw Lv = (0.14)(95.0 × 10−3 kg)(0.90)(2.26 × 106 J/kg)
Q = 2.7 × 104 J
Chapter Review and Assess, pp. 387–391
9. TF = 136°F
5
5
5
TC = 9 (TF − 32.0) = 9(136 − 32.0)°C = 9(104)°C = 57.8°C
T = (TC + 273.2)K = (57.8 + 273.2)K
T = 3.31 × 102 K
10. TF = 1947°F
5
5
5
TC = 9(TF − 32.0) = 9(1947 − 32.0) = 9(1915)°C = 1064°C
T = TC + 273.15 = (1064 + 273.15) K = 1337 K
I Ch. 10–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
mp = (0.105 g/kernel)
(125 kernels)
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Givens
Solutions
19. F = 315 N
W = (0.14)(Ui )
d = 35.0 m
W = (0.14)(Ui )
20. m = 0.75 kg
vi = 3.0 m/s
29. m = 23 g
W
Fd
(315 N)(35.0 m)
Ui =  =  =  = 7.9 × 104 J
0.14 0.14
0.14
I
a. ∆U = (0.85)(KE) = (0.85)2mv 2
1
1
∆U = 2(0.85)(0.75 kg)(3.0 m/s)2 = 2.9 J
Q
16.6 kJ − 12.0 kJ
4.6 × 103 J
a. cp,l =  = 
= 
−3
m∆T (23 × 10 kg)(340°C − 160°C) (23 × 10−3 kg)(180°C)
cp,l = 1.1 × 103 J/kg • °C
Q 12.0 kJ − 1.85 kJ
10.2 × 103 J

 = 4.4 × 105 J/kg
=
b. Lf =  = 
m
23 × 10−3 kg
23 × 10−3 kg
Q
1.85 kJ − 0 kJ
1.85 × 103 J
= 
c. cp,s =  = 
−3
m∆T (23 × 10 kg)(160°C − 0°C) (23 × 10−3 kg)(160°C)
cp,s = 5.0 × 102 J/kg • °C
Q
857 kJ − 855 kJ
2.0 × 103 J
= 
d. cp,v =  = 
−3
−3
m∆T (23 × 10 kg)(540°C − 340°C) (23 × 10 kg)(2.0 × 102°C)
cp,v = 4.0 × 102 J/kg • °C
Q 855 kJ − 16.6 kJ
838 × 103 J

 = 3.6 × 107 J/kg
=
e. Lv =  = 
m
23 × 10−3 kg
23 × 10−3 kg
30. mr = 25.5 g
Tr = 84.0°C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
mw = 5.00 × 10−2 kg
Qw = Qr − 0.14 kJ
cp,wmw(Tf − Tw) = cp,rmr(Tr − Tf) − 0.14 kJ
Tw = 24.0°C
[cp,r mrTr + cp,wmwTw − 0.14 kJ]
Tf = 
(cp,wmw + cp,r mr)
Qw = Qr − 0.14 kJ
cp,rmr = (234 J/kg • °C)(2.55 × 10−2 kg) = 5.97 J/°C
cp,r = 234 J/kg • °C
cp,r mrTr = (234 J/kg • °C)(2.55 × 10−2 kg)(84.0°C) = 501 J
cp,w = 4186 J/kg • °C
cp,wmw = 4186 J/kg • °C)(5.00 × 10−2 kg) = 209 J/°C
cp,w mwTw = (4186 J/kg • °C)(5.00 × 10−2 kg)(24.0°C) = 5.02 × 103 J
501 J + (5.02 × 103 J) − 140 J 5.38 × 103 J
Tf =  = 
209 J/°C + 5.97 J/°C
215 J/°C
Tf = 25.0° C
31. m1 = 1500 kg
1
∆U = ∆KE = 2 m1(vf − vi )2 = (0.5)(1500 kg)(0 m/s − 32 m/s)2 = 7.7 × 105 J
vi = 32 m/s
Q = ∆U = 7.7 × 105 J
vf = 0 m/s
Q = m2 cp,iron ∆T
cp,iron = 448 J/kg • °C
Q
7.7 × 105 J
∆T =  =  = 120°C
m2 cp,iron (4)(3.5 kg)(448 J/kg • °C)
m2 = (4)(3.5 kg)
32. T = 0.0°C
Q = mLf = (225 × 10−3 kg)(3.33 × 105 J/kg) = 7.49 × 104 J
m = 225 g
Lf = 3.33 × 105 J/kg
Section One—Pupil’s Edition Solutions
I Ch. 10–7
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Givens
Solutions
33. mw = 1.20 × 1016 kg
Q1 = mw cp,w ∆Tw = mw cp,w(Tw − Tf )
Tw = 12.0°C
Q2 = mw Lf
Tf = 0°C
Q1 = (1.20 × 1016 kg)(4186 J/kg • °C)(12.0°C − 0°C)
cp,w = 4186 J/kg • °C
Q1 = 6.03 × 1020 J
Lf = 3.33 × 105 J/kg
Q2 = (1.20 × 1016 kg)(3.33 × 105 J/kg)
Q2 = 4.00 × 1021 J
Qtot = Q1 + Q2 = 6.03 × 1020 J + 4.00 × 1021 J
Qtot = 4.60 × 1021 J
41. TR = 0°R = absolute zero
one Rankine degree =
one Fahrenheit degree
a. TR = TF − (absolute zero in TF)
9
TF = 5TC + 32.0
absolute zero in Tc = −273.15°C
9
absolute zero in TF = 5 (−273.15)°F + 32.0°F
absolute zero in TF = (−491.67 + 32.0)°F = −459.7°F
TR = TF − (−459.7)°F = TF + 459.7°F
TR = TF + 459.7, or TF = TR − 459.7
b. T = TC + 273.15
5
TC = 9(TF − 32.0)
5
T = 9(TF − 32.0) + 273.15
TF = TR − 459.7
5
T = 9(TR − 459.7 − 32.0) + 273.15
5
T = 9(TR − 491.7) + 273.15
5
5
T = 9TR − 9(491.7) + 273.15
5
5
9
T = 9TR , or TR = 5 T
42. mr = 3.0 kg
PEi = ∆U
mw = 1.0 kg
m r gh = cp,w m w ∆T
∆T = 0.10°C
cp,w = 4186 J/kg • °C
cp,w m w ∆T (4186 J/kg • °C)(1.0 kg)(0.10°C)
h =  = 
mr g
(3.0 kg)(9.81 m/s2)
g = 9.81 m/s2
h = 14 m
43. TC = −252.87°C
9
9
a. TF = 5TC + 32.0 = 5(−252.87)°F + 32.0°F
TF = (−455.17 + 32.0)°F = −423.2°F
T = TC + 273.15 = (−252.87 + 273.15) K = 20.28 K
TC = 20.5°C
9
9
b. TF = 5TC + 32.0 = 5(20.5)°F + 32.0°F
TF = (36.9 + 32.0)°F = 68.9°F
T = TC + 273.15 = (20.5 + 273.15) K = 293.6 K
I Ch. 10–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
T = 9TR − 273.2 + 273.15
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Givens
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44. freezing point = 50°TH
(0°C)
a. Set up a graph with Celsius on the x-axis and “Too Hot” on the y-axis. The equation relating the two scales can be found by graphing one scale versus the other
scale and then finding the equation of the resulting line. In this case, the two
known coordinates of the line are (0, 50) and (100, 200).
boiling point = 200°TH
(100°C)
∆y (200 − 50) 150 3
slope = a =  =  =  = 
∆x (100 − 0) 100 2
I
y = ax + b
b = y − ax = 50 − 2 0 = 50
3
3
The values y = TTH , a = 2, x = TC , and b = 50 can be substituted into the equation
for a line to find the conversion equation.
y = ax + b
3
TTH = 2TC + 50
y−b
or x = 
a
TTH − 50
TC = 
3

2
2
TC = 3(TTH − 50)
TC = absolute zero =
−273.15°C
45. TC = −40°C
3
3
b. TTH = 2(TC ) + 50 = 2(−273.15)°TH + 50°TH = (−409.72 + 50)°TH = −360°TH
9
9
TF = 5TC + 32 = 5(−40)°F + 32°F = (−72 + 32)°F
TF = −40°F
Copyright © by Holt, Rinehart and Winston. All rights reserved.
46. A = 6.0 m2
∆T = Tf − Ti = 61°C − 21°C = (4.0 × 101)°C
P/A = 550 W/m2
mw = Vw rw = (1.0 m3)(1.00 × 103 kg/m3) = 1.0 × 103 kg
Vw = 1.0 m3
Q mw c p,w ∆T
(1.0 × 103 kg)(4186 J/kg • °C)(4.0 × 101°C)
∆t =  =  = 
(P/A)(A)
P
(550 W/m2)(6.0 m2)
Ti = 21°C
Tf = 61°C
rw = 1.00 × 103 kg/m3
cp,w = 4186 J/kg • °C
47. mc = 253 g
∆t = 5.1 × 104 s
or (5.1 × 104 s)(1 h/3600 s) = 14 h
cp,a ma ∆Ta = cp,c mc ∆Tc
Tc = 85°C
∆Tc = Tc − Tf = 85°C − 25°C = (6.0 × 101)°C
Ta = 5°C
∆Ta = Tf − Ta = 25°C − 5°C = (2.0 × 101)°C
Tf = 25°C
mc ∆Tc cp,c (0.253 kg)(6.0 × 101°C)(3.87 × 102 J/kg • °C)
ma =  = 
∆Ta cp,a
(2.0 × 101°C)(8.99 × 102 J/kg • °C)
cp,a = 8.99 × 102 J/kg • °C
cp,c = 3.87 × 102 J/kg • °C
ma = 0.33 kg = 330 g
Section One—Pupil’s Edition Solutions
I Ch. 10–9
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Givens
Solutions
48.
TF = 5TC + 32.0
9
TC = T − 273.15
9
TF − 32.0 = 5TC
I
5
T
9 F
− 9 (32.0) = TC
5
5
T
9 F
− 9 (32.0) + 273.15 = T
5
T = TF
5
T
9 F
− 9 (32.0) + 273.15 = TF
5
5
T
9 F
− 17.8 + 273.15 = TF
5
T
9 F
+ 255.4 = TF
4
255.4 = 9TF
9
TF = 4 (255.4)°F = 574.6°F
TF = T
574.6°F = 574.6 K
49. ma = 250 g
mw = 850 g
Q Qa + Qw macp,a∆T + mwcp,w ∆T
∆T
 = 
 =  = (macp,a + mw cp,w ) 
∆t
∆t
∆t
∆t
∆T
 = 1.5°C/min
∆t
Q
 = [(0.250 kg)(899 J/kg • °C) + (0.850 kg)(4186 J/kg • °)](1.5°C/min)
∆t
cp,a = 899 J/kg • °C
Q
 = (225 J/°C + 3.56 × 103 J/°C)(1.5°C/min) = (3.78 × 103 J/°C)(1.5°C/min)
∆t
cp,w = 4186 J/kg • °C
Q
 = 5.7 × 103 J/min
∆t
or (5700 J/min)(1 min/60 s) = 95 J/s
∆Qtea = ∆Qmelted ice
Tice = 0°C
mtea cp,tea ∆Ttea = mice Lf + mice cp,w ∆Tice
Tf = 15°C
mtea = 180 g
mtea cp,tea ∆Ttea
mice = 
Lf + cp,w ∆Tice
mice,tot = 112 g
∆Ttea = Ttea − Tf = 32°C − 15° C = 17°C
cp,tea = cp,w = 4186 J/kg • °C
∆Tice = Tf − Tice = 15°C − 0° C = 15°C
Lf = 3.33 × 105 J/kg
(180 × 10−3 kg)(4186 J/kg • °C)(17°C)
mice = 
3.33 × 105 J/kg + (4186 J/kg • °C)(15°C)
(180 × 10−3 kg)(4186 J/kg • °C)(17°C)
mice = 
3.33 × 105 J/kg + 6.3 × 104 J/kg
(180 × 10−3 kg)(4186 J/kg • °C)(17°C)
mice = 
= 3.2 × 10−2 kg = 32 g
3.96 × 105 kg
mass of unmelted ice = mice,tot − mice = 112 g − 32 g = 8.0 × 101 g
I Ch. 10–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
50. Ttea = 32°C
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Thermodynamics
Chapter 11
I
Practice 11A, p. 405
Givens
Solutions
1. P = 1.6 × 105 Pa
a. Vf = 2Vi = (2)(4.0 m3) = 8.0 m3
Vi = 4.0 m3
W = P∆V = P(Vf − Vi ) = (1.6 × 105 Pa)(8.0 m3 − 4.0 m3)
W = (1.6 × 105 Pa)(4.0 m3) = 6.4 × 105 J
1
1
b. Vf = 4Vi = 4(4.0 m3) = 1.0 m3
W = P∆V = P(Vf − Vi ) = (1.6 × 105 Pa)(1.0 m3 − 4.0 m3)
W = (1.6 × 105 Pa)(−3.0 m3) = −4.8 × 105 J
W = P∆V = P(Vf − Vi ) = (599.5 × 103 Pa)[(2.523 × 10−4 m3) − (5.317 × 10−4 m3)]
2. P = 599.5 kPa
Vi = 5.317 × 10−4 m3
−4
Vf = 2.523 × 10
m
W = P∆V = P(Vf − Vi) = (4.3 × 105 Pa)[(9.5 × 10−4 m3) − (1.8 × 10−4 m3)]
3. P = 4.3 × 105 Pa
Vi = 1.8 × 10−4 m3
−4
Vf = 9.5 × 10
W = (4.3 × 105 Pa)(7.7 × 10−4 m3) = 3.3 × 102 J
3
m
1.6 cm
4. r =  = 0.80 cm
2
d = 2.1 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
W = (599.5 × 103 Pa)(−2.794 × 10−4 m3) = −167.5 J
3
W = 0.84 J
W = P∆V = PAd = Ppr 2d
W
0.84 J
P = 
= 2.0 × 105 Pa
2 = 
−2
p r d p(0.80 × 10 m)2 (2.1 × 10−2 m)
Section Review, p. 408
3. A = 7.4 × 10−3 m2
W = P∆V = PAd = (9.5 × 105 Pa)(7.4 × 10−3 m2)(−7.2 × 10−2 m) = −5.1 × 102 J
d = −7.2 × 10−2 m
P = 9.5 × 105 Pa
4. P = 1.5 × 103 Pa
W = P∆V = (1.5 × 103 Pa)(5.4 × 10−5 m3) = 8.1 × 10−2 J
∆V = 5.4 × 10−5 m3
Practice 11B, p. 413
1. Ui = 27 J
Uf = 34 J
Uf − Ui = Q − W
Q = W + Uf − Ui = 26 J + 34 J − 27 J = 33 J
W = 26 J
Section One—Pupil’s Edition Solutions
I Ch. 11–1
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Givens
Solutions
2. ∆U = −195 J
Q = ∆U + W = −195 J + 52.0 J = −143 J
W = 52.0 J
I
3. Ulost = 2.0 × 103 J
Q = ∆U + W + U lost = 8.0 × 103 J + 0 J + 2.0 × 103 J = 1.00 × 104 J
W=0J
∆U = 8.0 × 103 J
4. ∆U = −344 J
Q = 0 J for adiabatic processes.
W = Q − ∆U = 0 J − (−344 J) = 344 J
5. Q = 3.50 × 108 J
∆U = Q − W = 3.50 × 108 J − 1.76 × 108 J = 1.74 × 108 J
W = 1.76 × 108 J
Section Review, p. 419
4. P = 8.6 × 105 Pa
∆V = 4.05 × 10−4 m3
Q = −9.5 J
5. P = 7.07 × 105 Pa
∆V = −1.1 × 10−4 m3
a. W = P∆V = (8.6 × 105 Pa)(4.05 × 10−4 m3) = 3.5 × 102 J
b. ∆U = Q − W = −9.5 J − 350 J = −3.6 × 102 J
Q = W + ∆U = P∆V + ∆U = (7.07 × 105 Pa)(−1.1 × 10−4 m3) + 62 J
Q = −78 J + 62 J = −16 J
∆U = 62 J
7. W = −1.51 × 104 J
Qc = 7.55 × 104 J
a. For removal of energy from inside refrigerator,
∆Uc = Qc − W = (7.55 × 104 J) − (−1.51 × 104 J) = 9.06 × 104 J
No work is done on or by outside air, so all internal energy is given up as heat.
b. For cyclic processes, ∆Uref = 0 J
c. Because there is no change in the air’s volume, Wair = 0 J
d. For air inside refrigerator, Qair = −Qc and Wair = 0 J.
Qair − Wair = ∆Uair = −Qc = −7.55 × 104 J
8. Q = −15 J
∆U = Q − W = −15 J − 13 J = −28 J
W = 13 J
Practice 11C, p. 424
1. Qh = 2.254 × 104 kJ
Qc = 1.915 × 104 kJ
2. W = 45 J
Qc = 31 J
I Ch. 11–2
Q
1.915 × 104 kJ
eff = 1 − c = 1 − 
= 1 − 0.8496 = 0.1504
Qh
2.254 × 104 kJ
Qh = W + Qc = 45 J + 31 J = 76 J
W 45 J
eff =  =  = 0.59
Qh 76 J
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Qh = ∆Uh = ∆Uc = 9.06 × 104 J transferred to outside air
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3. Qh = 1.98 × 105 J
1.49 × 105 J
Q
a. eff = 1 − c = 1 − 
= 1 − 0.753 = 0.247
1.98 × 105 J
Qh
5
Qc = 1.49 × 10 J
b. W = Qh − Qc = (1.98 × 105 J) − (1.49 × 105 J) = 4.9 × 104 J
I
Q
eff = 1 − c
Qh
4. eff = 0.21
Qc = 780 J
Q
eff − 1 = − c
Qh
Q
1 − eff = c
Qh
Qc
780 J
780 J
 =  =  = 990 J
Qh = 
1 − eff
1 − 0.21 0.79
W = Qh − Qc = 990 J − 780 J = 210 J
5. W = 372 J
Qh = W + Qc
eff = 0.330
Q
Qc
W
W + Qc − Qc
eff = 1 − c = 1 − 
= 
= 
Qh
W + Qc
W + Qc
W + Qc
W
W + Qc = 
eff
W
1
1
Qc =  − W =  − 1 W =  − 1 (372 J)
eff
eff
0.330
Qc = (3.03 − 1)(372 J) = (2.03)(372 J) = 755 J
Qc
6.0 × 102 J 6.0 × 102 J
 =  =  = 8.7 × 102 J
Qh = 
1 − eff
1 − 0.31
0.69
6. Qc = 6.0 × 102 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
eff = 0.31
Section Review, p. 424
3. Qh = 75 000 J
a. W = Qh − Qc = 75 000 J − 35 000 J = 4.0 × 104 J
W 4.0 × 104 J
b. eff =  =  = 0.53
Qh
75 000 J
Qc = 35 000 J
Chapter Review and Assess, pp. 431–435
10. Vi = 35.25 × 10−3 m3
Vf = 39.47 × 10−3 m3
W = P∆V = P(Vf − Vi ) = (2.55 × 105 Pa)[(39.47 × 10−3 m3) − (35.25 × 10−3 m3)]
W = (2.55 × 105 Pa)(4.22 × 10−3 m3) = 1.08 × 103 J
5
P = 2.55 × 10 Pa
W = P∆V = P(Vf − Vi )
11. P = 2.52 × 105 Pa
−4
Vi = 1.1 × 10
3
m
Vf = 1.50 × 10−3 m3
W = (2.52 × 105 Pa)(1.50 × 10−3 m3 − 1.1 × 10−4 m3)
W = (2.52 × 105 Pa)(1.4 × 10−3 m3)
W = 3.5 × 102 J
Section One—Pupil’s Edition Solutions
I Ch. 11–3
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Givens
Solutions
19. P = 1055 MW
W
Qh = 
eff
eff = 0.330
river flow = 1.1 × 106 kg/s
I
W
P
Qh
1055 MW
 =  =  =  = 3.20 × 103 MW
∆t eff ∆t eff
0.330
Qc = Qh − W
Qc Qh W Qh
 =  −  =  − P = 3.20 × 103 MW − 1055 MW
∆t ∆t ∆t ∆t
Qc
 = 2140 MW = 2140 MJ/s = 2.14 × 109 J/s
∆t
20. ∆U = 604 × 103 J
Q = ∆U + W = (604 × 103 J) + (43.0 × 103 J) = 647 × 103 J
3
W = 43.0 × 10 J
m2 = 6050 kg
Ti = 22°C
Tf = 47°C
d = 5.5 mm
cp = 448 J/kg • °C
g = 9.81 m/s2
28. Qh = 525 J
Qc = 415 J
29. Qh = 9.5 × 1012 J
Qc = 6.5 × 1012 J
a. Q = m1cp ∆T = m1cp (Tf − Ti ) = (150 kg)(448 J/kg • °C)(47°C − 22°C)
Q = (150 kg)(448 J/kg • °C)(25°C) = 1.7 × 106 J
b. W = Fd = m2 gd = (6050 kg)(9.81 m/s2)(5.5 × 10−3 m)
W = 3.3 × 102 J
c. ∆U = Q − W = (1.7 × 106 J) − (3.3 × 102 J) = 1.7 × 106 J
Q
415 J
eff = 1 − c = 1 −  = 1 − 0.790 = 0.210
Qh
525 J
6.5 × 1012 J
Q
eff = 1 − c = 1 − 
= 1 − 0.68 = 0.32
9.5 × 1012 J
Qh
Qc = 5.0 × 102 J
Q
5.0 × 102 J
eff = 1 − c = 1 −  = 1 − 0.59 = 0.41
Qh
850 J
W = 3.5 × 102 J
Alternatively,
30. Qh = 850 J
W 3.5 × 102 J
eff =  =  = 0.41
Qh
850 J
39. Q1 = 606 J
W1 = 418 J
W2 = 1212 J
a. ∆U1 = Q1 − W1 = 606 J − 418 J = 188 J
b. Because ∆U2 = ∆U1, Q 2 = ∆U1 + W2
Q 2 = 188 J + 1212 J = 1.40 × 103 J
40. Q = 5175 J
I Ch. 11–4
d. Because ∆V = 0, W = 0 J and ∆U = Q = 5175 J .
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
21. m1 = 150 kg
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Vibrations and Waves
Chapter 12
I
Practice 12A, p. 441
Givens
Solutions
1. x = −36 cm
a. Fg + Felastic = 0
m = 0.55 kg
Fg = −mg
2
g = 9.81 m/s
Felastic = −kx
−mg − kx = 0
−mg −(0.55 kg)(9.81 m/s2)
k =  =  = 15 N/m
x
−0.36 m
2. Fg = −45 N
Fg + Felastic = 0
x = −0.14 m
Fg + (−kx) = 0
Fg
−45 N
k =  =  = 3.2 × 102 N/m
x − 0.14 m
3. F1 = 32 N
x1 = −1.2 cm
4. x2 = −3.0 cm
−32 N
−F
k = 1 =  = 2.7 × 103 N/m
−0.012 m
x1
F2 = −kx2
F2 = −(2.7 × 103 N/m)(−0.030 m) = 81 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section Review, p. 445
F = −kx = −(13 N/m)(−0.040 m)
2. x = −4.0 cm
k = 13 N/m
F = 0.52 N
Practice 12B, p. 449
L
T = 2p 
g
1. T = 24 s
g = 9.81 m/s2
2
T
L = g 
2p
2
24 s
= (9.81 m/s2)  = 1.4 × 102 m
2p
L
T = 2p 
g
2. T = 1.0 s
2
g = 9.81 m/s
2
T
L = g 
2p
2
1.0 s
= (9.81 m/s2)  = 0.25 m = 25 cm
2p
Section One—Pupil’s Edition Solutions
I Ch. 12–1
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Givens
Solutions
L
T = 2p 
g
3. T = 3.8 s
g = 9.81 m/s2
I
2
2
T
L = g 
2p
4. L = 3.500 m
2
g1 = 9.832 m/s
3.8 s
= (9.81 m/s2)  = 3.6 m
2p
L
3.500 m
a. T1 = 2p  = 2p 2 = 3.749 s
g1
9.832 m/s
1
1
f1 =  =  = 0.2667 Hz
T1 3.749 s
g2 = 9.803 m/s2
L
3.500 m
b. T2 = 2p  = 2p 2 = 3.754 s
g2
9.803 m/s
1
1
f2 =  =  = 0.2664 Hz
T2 3.754 s
g3 = 9.782 m/s2
L
3.500 m
c. T3 = 2p  = 2p 2 = 3.758 s
g3
9.782 m/s
1
1
f3 =  =  = 0.2661 Hz
T3 3.758 s
Practice 12C, p. 451
m = 0.30 kg
2. m = 25 g = 0.025 kg
20 vibrations
f =  = 5.0 Hz
4.0 s
3. F = 125 N
g = 9.81 m/s2
T = 3.56 s
m
T = 2p 
k
4p 2m
4p 2 (0.30 kg)
2
k = 
2 =   = 2.1 × 10 N/m
T
(0.24 s)
1
m
T =  = 2p 
f
k
k = 4p 2mf 2 = 4p 2(0.025 kg)(5.0 Hz)2 = 25 N/m
m
F
T = 2p  = 2p 
k
gk
4p 2 F
(4p 2)(125 N)


k = 
=
g T2
(9.81 m/s2)(3.56 s)2
k = 39.7 N/m
4. mp = 255 kg
mc = 1275 kg
k = 2.00 × 104 N/m
I Ch. 12–2
mp + mc 255 kg + 1275 kg 1.530 × 103 kg
m =  =  =  = 382.5 kg
4
4
4
m
382.5 kg
T = 2p  = 2p 
= 0.869 s
k
2.00 × 104 N/m
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. T = 0.24 s
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Givens
Solutions
5. k = 30.0 N/m
m1 = 2.3 kg
m
2.3 kg
a. T1 = 2p 1 = 2p 
k
30.0 N/m
T1 = 1.7 s
1
1
f1 =  =  = 0.59 Hz
T1 1.7 s
I
m
0.015 kg
b. T2 = 2p 2 = 2p 
k
30.0 N/m
m2 = 15 g
T2 = 0.14 s
1
1
f2 =  =  = 7.1 Hz
T2 0.14 s
m
1.9 kg
c. T3 = 2p 3 = 2p 
k
30.0 N/m
m3 = 1.9 kg
T3 = 1.6 s
1
1
f3 =  =  = 0.62 Hz
T3 1.6 s
Section Review, p. 451
2. L = 2.5 m
2
g = 9.81 m/s
L
2.5 m
a. T = 2p  = 2p 2
g
9.81 m/s
T = 3.2 s
1
1
b. f =  =  = 0.31 Hz
T 3.2 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. m = 0.75 kg
x = −0.30 m
g = 9.81 m/s2
a. −kx − mg = 0
−mg −(0.75 kg)(9.81 m/s2)
k =  = 
x
−0.30 m
k = 25 N/m
k = 2p
25N
/m
b. T = 2p
m
0.75 kg
T = 1.1 s
4. f = 180 oscillations/min
f = (180 oscillations/min)(1 min/60 s)
f = 3.00 Hz
1
1
T =  =  = 0.330 s
3.00 Hz
f
Practice 12D, p. 457
1. f1 = 28 Hz
f2 = 4200 Hz
v = 340 m/s
v 340 m/s
l1 =  =  = 12 m
28 Hz
f1
v 340 m/s
l2 =  =  = 0.081 m
f2 4200 Hz
Section One—Pupil’s Edition Solutions
I Ch. 12–3
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Givens
Solutions
2. v = 3.00 × 108 m/s
f1 = 88.0 MHz
v 3.00 × 108 m/s
a. l1 =  = 
f1 8.80 × 107 Hz
l1 = 3.41 m
I
f2 = 6.0 × 108 MHz
v 3.00 × 108 m/s
b. l2 =  = 
f2
6.0 × 1014 Hz
l2 = 5.0 × 10−7 m
f3 = 3.0 × 1012 MHz
v 3.00 × 108 m/s
c. l3 =  = 
f3
3.0 × 1018 Hz
l3 = 1.0 × 10−10 m
3. l = 633 nm
= 6.33 × 10−7 m
v = 3.00 × 108 m/s
4. f = 256 Hz
v 3.00 × 108 m/s
f =  = 
l 6.33 × 10−7 m
f = 4.74 × 1014 Hz
a. vair = lair f = (1.35 m)(256 Hz)
lair = 1.35 m
vair = 346 m/s
vwater = 1500 m/s
vwater 1500 m/s
 = 
b. lwater = 
256 Hz
f
lwater = 5.86 m
5. l = 0.57 cm = 5.7 × 10−3 m
v = 340 m/s
v
340 m/s
f =  = 
l 5.7 × 10−3 m
f = 6.0 × 104 Hz
Chapter Review and Assess, pp. 469–473
8. m = 0.40 kg
x = −3.0 cm
g = 9.81 m/s2
−kx − mg = 0
−mg −(0.40 kg)(9.81 m/s2)
k =  = 
x
−0.030 m
k = 1.3 × 102 N/m
9. x = −0.40 m
F = 230 N
19. f = 0.16 Hz
g = 9.81 m/s2
−F −230 N
k =  =  = 580 N/m
− 0.40 m
x
1
L
T =  = 2p 
g
f
g
(9.81 m/s2)

L =  2 = 
(2p f )
(4p 2)(0.16 Hz)2
L = 9.7 m
I Ch. 12–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section Review, p. 458
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Givens
Solutions
20. T = 9.49 s
2
g = 9.81 m/s
L
T = 2p 
g
2
T g (9.49 s)2(9.81 m/s2)
L = 2 = 
4p
4p 2
L = 22.4 m
21.
I
a. Because a pendulum passes through its equilibrium position twice each cycle,
T = (2)(1.000 s) = 2.000 s .
L1 = 0.9942 m
4p 2L1 (4p 2)(0.9942 m)
b. g1 = 
= 
T2
(2.000 s)2
g1 = 9.812 m/s2
L2 = 0.9927 m
4p 2L2 (4p 2)(0.9927 m)
= 
c. g2 = 
T2
(2.000 s)2
g2 = 9.798 m/s2
22. k = 1.8 × 102 N/m
m = 1.5 kg
1.5 kg
m
a. T = 2p  = 2p 
1.8 × 102 N/m
k
T = 0.57 s
1
1
b. f =  =  = 1.8 Hz
T 0.57 s
35. f = 25.0 Hz
1
l
2
= 10.0 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2(amplitude) = 18 cm
18 cm
a. amplitude =  = 9.0 cm
2
b. l = (2)(10.0 cm) = 20.0 cm
1
1
c. T =  =  = 0.0400 s
f
25.0 Hz
d. v = lf = (0.200 m)(25.0 Hz) = 5.00 m/s
36. v = 3.00 × 108 m/s
f = 9.00 × 109 Hz
46. k = 230 N/m
x = −6.0 cm
v 3.00 × 108 m/s
l =  = 
= 0.0333 m
f
9.00 × 109 Hz
F = −kx
F = −(230 N/m)(−0.060 m)
F = 14 N
47. x = −2.0 cm
k = 85 N/m
F = −kx = −(85 N/m)(−0.020 m)
F = 1.7 N
Section One—Pupil’s Edition Solutions
I Ch. 12–5
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Givens
Solutions
48. l = 5.20 × 10−7 m
v
3.00 × 108 m/s
f =  = 
= 5.77 × 1014 Hz
l
5.20 × 10−7 m
v = 3.00 × 108 m/s
1
1
T =  = 
= 1.73 × 10−15 s
f
5.77 × 1014 Hz
I
49. l = 0.15 m
Because a wave is generated twice each second, f = 2.0 Hz .
1
1
T =  =  = 0.50 s
f
2.0 Hz
v = lf = (0.15 m)(2.0 Hz) = 0.30 m/s
50. v = 343 m/s
v∆t (343 m/s)(2.60 s)
∆x =  =  = 446 m
2
2
∆t = 2.60 s
51. f1 = 196 Hz
f2 = 2637 Hz
v 340 m/s
l1 =  =  = 1.73 m
f1 196 Hz
v 340 m/s
l2 =  =  = 0.129 m
f2 2637 Hz
v = 340 m/s
52. L = 0.850 m
L
T = 2p 
g
T = 1.86 s
4p 2L (4p 2)(0.850 m)
g = 
= 
= 9.70 m/s2
T2
(1.86 s)2
53. amplitude = 3.5 cm
k = 250 N/m
m = 0.50 kg
a. At maximum displacement:
1
ME = PEelastic = 2kx 2
x = −maximum displacement = −amplitude
F −kx −(250 N/m)(−0.035 m)
b. a =  =  = 
m
m
0.50 kg
a = 18 m/s2
54. L = 2.00 m
2
g = 9.80 m/s
∆t = 5.00 min
∆t
∆t
oscillations =  = 
T
L
2p 
g
(5.00 min)(60 s/min)
oscillations = 
2.00 m
(2p) 2
9.80 m/s
oscillations = 106
55. v = 1.97 × 108 m/s
l = 3.81 × 10−7 m
I Ch. 12–6
v 1.97 × 108 m/s
= 5.17 × 1014 Hz
f =  = 
l 3.81 × 10−7 m
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
ME = 2(250 N/m)(−0.035 m)2 = 0.15 J
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Givens
Solutions
56. m = 0.40 kg
k = 160 N/m
x = −0.15 m
a. F = −kx = −(160 N/m)(−0.15 m) = 24 N
F
24 N
b. a =  =  = 6.0 × 101 m/s2
m 0.40 kg
I
57. l = 1.20 m
8
v = lf = (1.20 m)  = 0.800 m/s
12.0 s
8
f = 
12.0 s
58. gmoon = 1.63 m/s2
∆t = 24 h
gearth = 9.81 m/s2
L
Tearth = 2p 
gearth
L
Tmoon = 2p 
gmoon
Tmoon
 =
Tearth
m/s
= 2.45
gg = 9
1..8613
m
/s
earth
moon
2
2
The clock on the moon runs slower than the same clock on Earth by a factor of 2.45.
Thus, after 24.0 h Earth time, the clock on the moon will have advanced by
24.0 h
 = 9.80 h = 9 h + (0.80 h)(60 min/h) = 9 h, 48 min
2.45
Thus, the clock will read 9:48 A.M.
40.0
59. f =  Hz
30.0
Copyright © by Holt, Rinehart and Winston. All rights reserved.
425 cm
4.25
v =  =  m/s
10.0 s
10.0
4.25
 m/s
10.0
v
l =  =  = 0.319 m
f
40.0
 Hz
30.0
Section One—Pupil’s Edition Solutions
I Ch. 12–7
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Sound
Chapter 13
I
Practice 13A, p. 488
Givens
Solutions
1. r = 5.0 m
P1
0.25 W
a. intensity = 
= 2 = 8.0 × 10−4 W/m2
2
4p r
(4p )(5.0 m)
P1 = 0.25 W
P2 = 0.50 W
P2
0.50 W
=  = 1.6 × 10−3 W/m2
b. intensity = 
4p r 2 (4p )(5.0 m)2
P3 = 2.0 W
P3
2.0 W
=  = 6.4 × 10−3 W/m2
c. intensity = 
4p r 2 (4p )(5.0 m)2
2. P = 70.0 W
P
70.0 W
intensity = 2 = 2 = 8.91 × 10−3 W/m2
4p r
(4p )(25.0 m)
r = 25.0 m
3. intensity = 4.6 × 10−7 W/m2
P = (intensity)(4pr 2) = (4.6 × 10−7 W/m2)(4p)(2.0 m)2 = 2.3 × 10−5 W
r = 2.0 m
4. intensity = 1.6 × 10−3 W/m2
P = (intensity)(4pr 2) = (1.6 × 10−3 W/m2)(4p)(15 m)2 = 4.5 W
r = 15 m
5. P = 0.35 W
−3
Copyright © by Holt, Rinehart and Winston. All rights reserved.
intensity = 1.2 × 10
2
r=
W/m
0.35 W
 = 4.8 m

ntensPi
)(4p) = (i
ty
(1.2 × 10 W/m )(4p)
−3
2
Section Review, p. 493
5. decibel level = 10 dB
intensity at 10 dB = 1.0 × 10−11 W/m2 (See Table 13-2 on page 490 of text.)
P = 0.050 W
r=
 =

ntensi
)(4p
) = (i
ty
(1.0 × 10 W/m )(4p)
P
0.050 W
−11
2
2.0 × 104 m
Practice 13B, p. 499
1. L = 0.20 m
v = 352 m/s
2. L = 66.0 cm
v = 340 m/s
nv (1)(352 m/s)
f1 =  =  = 440 Hz
4L
(4)(0.20 m)
n v (1)(340 m/s)
f1 =  =  = 260 Hz
2L (2)(0.660 m)
f2 = 2f1 = (2)(260 Hz) = 520 Hz
f3 = 3f1 = (3)(260 Hz) = 780 Hz
Section One—Pupil’s Edition Solutions
I Ch. 13–1
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Givens
Solutions
3. v = 115 m/s
L1 = 70.0 cm
nv (1)(115 m/s)
a. f1 =  =  = 82.1 Hz
2L1 (2)(0.700 m)
L2 = 50.0 cm
n v (1)(115 m/s)
b. f1 =  =  = 115 Hz
2L2 (2)(0.500 m)
L3 = 40.0 cm
n v (1)(115 m/s)
c. f1 =  =  = 144 Hz
2L3 (2)(0.400 m)
I
4. L = 50.0 cm
f1 = 440 Hz
f12L (440 Hz)(2)(0.500 m)
 =  = 440 m/s
v=
n
1
Section Review, p. 503
1. f1 = 264 Hz
f2 = 2f1 = (2)(264 Hz) = 528 Hz
2. f1 = 264 Hz
f12L (264 Hz)(2)(0.660 m)
 =  = 348 m/s
v=
n
1
L = 66.0 cm
Chapter Review and Assess, pp. 507–510
r = 5.0 m
28. P = 100.0 W
r = 10.0 m
39. L = 31.0 cm
v = 274.4 m/s
P
3.1 × 10−3 W
= 1.0 × 10−5 W/m2 = 70 dB
Intensity = 2 = 
4p r
4p (5.0 m)2
P
100.0 W
intensity = 2 = 2 = 7.96 × 10−2 W/m2
4p r
(4p )(10.0 m)
n v (1)(274.4 m/s)
f1 =  =  = 443 Hz
2L
(2)(0.310 m)
f2 = 2f1 = (2)(443 Hz) = 886 Hz
f3 = 3f1 = (3)(443 Hz) = 1330 Hz
40. L = 2.8 cm
v = 340 m/s
41. f1 = 320 Hz
v = 331 m/s
n v (1)(340 m/s)
f1 =  =  = 3.0 × 103 Hz
4L (4)(0.028 m)
n v (1)(331 m/s)
a. L =  =  = 0.52 m = 52.0 cm
2f1 (2)(320 Hz)
b. f2 = 2f1 = (2)(320 Hz) = 640 Hz
f3 = 3f1 = (3)(320 Hz) = 960 Hz
v = 367 m/s
I Ch. 13–2
n v (1)(367 m/s)
c. f1 =  =  = 350 Hz
2L (2)(0.52 m)
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
27. P = 3.1 × 10−3 W
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Givens
Solutions
42. A = 5.0 × 10−5 m2
a. P1 = (intensity1)(A) = (1.0 × 10
−12
W/m2)(5.0 × 10−5 m2) = 5.0 × 10−17 W
intensity1 =
1.0 × 10−12 W/m2
intensity2 =
1.0 × 100 W/m2
43. L = 9.00 cm
v = 345 m/s
0
2
−5
b. P2 = (intensity2)(A) = (1.0 × 10 W/m )(5.0 × 10
m2) = 5.0 × 10−5 W
I
nv
(1)(345 m/s)
a. f1 =  =  = 958 Hz
4L (4)(0.0900 m)
b. fundamental wavelength = l = 4L = (4)(9.00 cm) = 36.0 cm
4
For the second resonant position, l = 3L.
3
3
L = 4l = 4(36.0 cm) = 27.0 cm
4
c. For the third resonant position, l = 5L.
5
5
L = 4l = 4(36.0 cm) = 45.0 cm
44. f1 = 132 Hz
number of beats each second = f2 − f1 = 137 Hz − 132 Hz = 5 Hz
f2 = 137 Hz
45. v = 343 m/s
f1 = 20 Hz
f2 = 20 000 Hz
46. ∆x = (150 m)(2)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v = 1530 m/s
47. L = 2.46 m
v = 345 m/s
v 343 m/s
l1 =  =  = 20 m
f1
20 Hz
343 m/s
v
l2 =  =  = 2 × 10−2 m
f2 20 000 Hz
∆x (150 m)(2)
∆t =  =  = 0.200 s
v
1530 m/s
nv (1)(345 m/s)
a. f1 =  =  = 70.1 Hz
2L (2)(2.46 m)
nv
b.  ≤ 20 000 Hz
2L
(20 000 Hz)(2)(L) (20 000 Hz)(2)(2.46 m)
n ≤  =  = 285
v
345 m/s
48. v = 1.0 × 104 m/s
f = 2.0 × 1010 Hz
v 1.0 × 104 m/s
l =  = 
= 5.0 × 10−7 m
2.0 × 1010 Hz
f
Section One—Pupil’s Edition Solutions
I Ch. 13–3
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I
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Givens
Solutions
49. f1 = 250 Hz
f3 = 3f1 = (3)(250 Hz) = 750 Hz
50. decibel level = 70 dB
intensity at 70 dB = 1.0 × 10−5 W/m2 (See Table 13-2 on page 490 of text.)
r1 = 1.0 m
P = (intensity)(4pr12) = (1.0 × 10−5 W/m2)(4p)(1.0 m)2 = 1.3 × 10−4 W
intensity at the threshold of hearing = 1.0 × 10−12 W/m2
r2 =
51. f1,open = 261.6 Hz
f3,closed = 261.6 Hz
1.3 × 10−4 W

= 3.2 × 103 m
(1.0 × 10−12 W/m2)(4p)
P
 =
(intensity)(4p)
v
Lopen = 
(f1,open)(2)
3v
Lclosed = 
(f3,closed)(4)
3v

(3)( f1,open )(2)
(f3,closed )(4)
Lclosed
 = 
= 
v
( f3,closed )(4)
Lopen

(f1,open)(2)
Lclosed 6 1.5
 =  = 
Because f1,open = f3,closed , 
Lopen 4
1
(Lclosed) = 1.5(Lopen )
52. decibel level = 40 dB
intensity at 40 dB = 1.0 × 10−8 W/m2 (See Table 13-2 on page 490 of text.)
N = 1 mosquito
intensity at 50 dB = 1.0 × 10−7 W/m2 (See Table 13-2 on page 490 of text.)
intensity at 50 dB
1.0 × 10−7 W/m2
 = −
= 10
intensity at 40 dB
1.0 × 10 8 W/m2
Because the intensity at 50 dB is 10 times greater than the intensity at 40 dB, increas-
53. f = 2.0 × 104 Hz
v = 378 m/s
54. decibel level = 130 dB
r = 20.0 m
diameter = 1.9 × 10−2 m
378 m/s
v
l =  = 
= 1.9 × 10−2 m
2.0 × 104 Hz
f
a. intensity at 30 dB = 1.0 × 101 W/m2 (See Table 13-2 on page 490 of text.)
P = (intensity)(4pr 2) = (1.0 × 101 W/m2)(4p)(20.0 m)2 = 5.0 × 104 W
diameter
b. area = p 
2
2
1.9 × 10−2 m
= p 
2
2
1.9 × 10−2 m
P = (intensity)(area) = (1.0 × 101 W/m2)(p) 
2
P = 2.8 × 10−3 W
I Ch. 13–4
Holt Physics Solution Manual
2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ing the decibel level to 50 dB would require (1 mosquito)(10) = 10 mosquitoes .
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Light and Reflection
Chapter 14
I
Practice 14A, p. 523
Givens
Solutions
1. f = 3.0 × 1021 Hz
c = 3.00 × 108 m/s
2. f1 = 88 MHz
f2 = 108 MHz
c = 3.00 × 108 m/s
3. f1 = 3.50 MHz
f2 = 29.7 MHz
c = 3.00 × 108 m/s
4. l = 1.0 km
8
c = 3.00 × 10 m/s
5. l = 560 nm
c = 3.00 × 108 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. l = 125 nm
c = 3.00 × 108 m/s
c 3.00 × 108 m/s
l =  = 
= 1.0 × 10−13 m
f
3.0 × 1021 Hz
c 3.00 × 108 m/s
= 3.4 m
l1 =  = 
f1
8.8 × 107 Hz
c 3.00 × 108 m/s
l2 =  = 
= 2.78 m
f2 1.08 × 108 Hz
c 3.00 × 108 m/s
= 85.7 m
l1 =  = 
f1 3.50 × 106 Hz
c 3.00 × 108 m/s
l2 =  = 
= 10.1 m
f2 2.97 × 107 m/s
c 3.00 × 108 m/s
f =  = 
= 3.0 × 105 Hz
1.0 × 103 m
l
c 3.00 × 108 m/s
f =  = 
= 5.4 × 1014 Hz
5.6 × 10−7 m
l
c 3.00 × 108 m/s
f =  = 
= 2.40 × 1015 Hz
l 1.25 × 10−7 m
Section Review, p. 525
2. f = 7.57 × 1014 Hz
c = 3.00 × 108 m/s
c 3.00 × 108 m/s
l =  = 
= 3.96 × 10−7 m
f 7.57 × 1014 Hz
Section Review, p. 529
3.
The normal to the surface of the mirror is halfway between 12 o’clock and 5 o’clock.
360°
1
q = 2(5.0 h)  = 75°
12.0 h
q ⬘ = q = 75°
Section One—Pupil’s Edition Solutions
I Ch. 14–1
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Practice 14B, p. 536
Givens
Solutions
1. p1 = 10.0 cm
p2 = 5.00 cm
I
f = 10.0 cm
1
1
1
1 1
 =  −  =  −  = 0 no image (infinite q)
q1 f p1 10.0 cm 10.0 cm
1 1 1
1
1
 =  −  =  − 
q2 f p2 10.0 cm 5.00 cm
1 0.100 0.200 − 0.100
 =  −  = 
q2 1 cm 1 cm
1 cm
q2 = −10.0 cm
q
−10.0 cm
M = −  = −  = 2.00
p
5.00 cm
2. f = 33 cm
p = 93 cm
virtual, upright image
1 1 1
1
1
 =  −  =  − 
q f
p 33 cm 93 cm
1 0.030 0.011 0.019
 =  −  = 
q 1 cm
1 cm
1 cm
q = 53 cm
53 cm
q
M = −  = −  = −0.57
93 cm
p
3. p = 25.0 cm
q = −50.0 cm
real, inverted image
2 1 1
 =  + 
R p q
1 1
1
1
1
1
1
 =  +  =  +  =  − 
R 2p 2q (2)(25.0 cm) (2)(−50.0 cm) 50.0 cm 1.00 × 102 cm
R = 1.00 × 102 cm
q
−50.0 cm
M = −  = −  = 2.00
p
25.0 cm
4. p = 11.0 cm
q = 13.2 cm
1 1 1
1
1
 =  +  =  + 
f
p q 11.0 cm 13.2 cm
1 0.0909 0.0758 0.1667
 =  +  = 
f
1 cm
1 cm
1 cm
f = 5.999 cm
13.2 cm
q
M = −  = −  = −1.20
p
11.0 cm
I Ch. 14–2
Holt Physics Solution Manual
virtual image
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1 0.0200 0.0100 0.0100
 =  −  = 
R 1 cm
1 cm
1 cm
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Givens
p = 27.0 cm
Solutions
1 1 1
1
1
 =  −  =  − 
q f
p 5.999 cm 27.0 cm
1 0.1667 0.0370 0.1297
 =  −  = 
q
1 cm
1 cm
1 cm
I
q = 7.710 cm
q
7.710 cm
M = −  = −  = −0.286
p
27.0 cm
real image
Practice 14C, p. 540
1. q = −23.0 cm
f = −46.0 cm′
h⬘ = 1.70 cm
1 1 1
1
1
 =  −  =  − 
p f
q −46.0 cm −23.0 cm
1
0.0217 0.0435 0.0218
 = −  +  = 
1 cm
1 cm
1 cm
p
p = 45.9 cm
−23.0 cm
q
M = −  = −  = 0.501
45.9 cm
p
virtual, upright image
−ph⬘ −(45.9 cm)(1.70 cm)
h =  = 
−23.0 cm
q
h = 3.39 cm
2. f = −0.25 m
h⬘ = 0.080 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = −0.24 m
1 1 1
1
1
 =  −  =  − 
p f
q −0.25 m −0.24 m
1 − 4.0
−4.2 0.20
 =  −  = 
p
1m
1m
1m
p = 5.0 m
q
−0.24 m
M = −  = −  = 0.048
p
5.0 m
virtual, upright image
−ph⬘
−(5.0 m)(0.080 m)
h =  =  = 1.7 m
q
−0.24 m
3. f = −33 cm
q = −19 cm
h⬘ = 7.0 cm
1 1 1
1
1
 =  −  =  − 
p f
q −33 cm − 19 cm
1
−0.030
−0.053 0.023
 =  −  = 
p
1 cm
1 cm
1 cm
p = 43 cm
q
−19 cm
M = −  = −  = 0.44
p
43 cm
virtual, upright page
−ph⬘ −(43 cm)(7.0 cm)
h =  =  = 16 cm
q
−19 cm
Section One—Pupil’s Edition Solutions
I Ch. 14–3
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Givens
Solutions
4. R = −0.550 m
p = 3.1 m
1 1 2
 +  = 
p q R
1 2
1
2
1
 =  −  =  − 
q R
p −0.550 m 3.1 m
I
1 −3.64 0.32
4.0
 =  −  = 
q
1m
1m
1m
q = −0.25 m
q
− 0.25 m
M = −  = −  = 0.081
p
3.1 m
−6.00 cm
5. R =  = −3.00 cm
2
p = 10.5 cm
virtual, upright image
1 2 1
2
1
 =  −  =  − 
q R p −3.00 cm 10.5 cm
1 − 0.667 0.0952
− 0.762
 =  −  = 
q
1 cm
1 cm
1 cm
q = −1.31 cm
q −1.31 cm
M = −  =  = 0.125
p −10.5 cm
6. p = 49 cm
f = −35 cm
virtual, upright image
1
1
1 1 1
 =  −  =  − 
p −35 cm 49 cm
q f
1 − 0.029 0.020 − 0.049
 =  −  = 
q
1 cm
1 cm
1 cm
q = −2.0 × 101 cm
virtual, upright image
Section Review, p. 542
1. R = −1.5 cm
p = 1.1 cm
1 2 1
2
1
 =  −  =  − 
q R p −1.5 cm 1.1 cm
1 −1.3 0.91
−2.2
 =  −  = 
q 1 cm 1 cm 1 cm
q = −0.45 cm
q
− 0.45 cm
M = −  = −  = 0.41
p
1.1 cm
3. M = −95
q = 13 m
I Ch. 14–4
13 m
q
p = −  = −  = 0.14 m
−95
M
Holt Physics Solution Manual
virtual, upright image
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q
−2.0 × 101 cm
M = −  = −  = 0.41
p
49 cm
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Givens
5. R = 265.0 m
Solutions
1 1 2
 +  = 
p q R
Because objects in space are so far away,
1
 = 0; therefore,
p
I
1 2
2
7.547 × 10−3
 =  =  = 
q R 265.0 m
1m
q = 132.5 m
Chapter Review and Assess, pp. 550–555
10. f1 = 7.5 × 1014 Hz
f2 = 1.0 × 1015 Hz
c = 3.00 × 108 m/s
11. c = 3.00 × 108 m/s
f = 3 × 1014 Hz
12. f = 99.5 MHz
c = 3.00 × 108 m/s
13. f = 33 GHz
c = 3.00 × 108 m/s
34. R = 25.0 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
p = 45.0 cm
c
3.00 × 108 m/s
l1 =  = 
= 4.0 × 10−7 m
f1
7.5 × 1014 m
c
3.00 × 108 m/s
= 3.0 × 10−7 m
l2 =  = 
f2
1.0 × 1015 m
c
3.00 × 108 m/s
= 1 × 10−6 m
l =  = 
f
3 × 1014 Hz
c 3.00 × 108 m/s
l =  = 
= 3.02 m
f
9.95 × 107 Hz
c 3.00 × 108 m/s
l =  = 
= 9.1 × 10−3 m
f
3.3 × 1010 Hz
1 1 2
a.  +  = 
p q R
2
1
1 2 1
 =  −  =  − 
q R p 25.0 cm 45.0 cm
1 0.0800 0.0222
0.0578
 =  −  = 
q
1 cm
1 cm
1 cm
q = 17.3 cm
q
17.3 cm
M = −  = −  = −0.384
p
45.0 cm
p = 25.0 cm
real, inverted image
2
1
1 2 1
b.  =  −  =  − 
q R p 25.0 cm 25.0 cm
1 0.0800 0.0400 0.0400
 =  −  = 
q
1 cm
1 cm
1 cm
q = 25.0 cm
q
25.0 cm
M = −  = −  = −1.00
p
25.0 cm
real, inverted image
Section One—Pupil’s Edition Solutions
I Ch. 14–5
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Givens
Solutions
p = 5.00 cm
1 2 1
2
1
c.  =  −  =  − 
q R p 25.0 cm 5.00 cm
1 0.0800 0.200 − 0.120
 =  −  = 
q
1 cm
1 cm
1 cm
I
q = −8.33 cm
q
−8.33 cm
M = −  = −  = 1.67
p
5.00 cm
35. f = 8.5 cm
q = 2p
virtual, upright image
1 1 1 1 1
3
 =  +  =  +  = 
f p q p 2p 2p
3f (3)(8.5 cm)
p =  =  = 13 cm
2
2
q = 2p = (2)(13 cm) = 26 cm
q
26 cm
M = −  = −  = −2.0
13 cm
p
36. R = −45.0 cm
h⬘ = 1.70 cm
q = −15.8 cm
real, inverted image
1 1 2
 +  = 
p q R
1 2 1
2
1
 =  −  =  − 
p R q −45.00 cm
−15.8 cm
−0.0444
−0.0633 0.0189
1
 =  −  = 
1 cm
1 cm
1 cm
p
P = 52.9 cm
virtual, upright image
−p h⬘ −(52.9 cm)(1.70 cm)
h =  =  = 5.69 cm
−15.8 cm
q
46. M = −0.085
q = 35 cm
q
35 cm
p = −  = −  = 4.1 × 102 cm
M
− 0.085
1
1
1 1 1
 =  +  =  + 
p q 410 cm 35 cm
f
1 0.0024 0.029 0.031
 =  +  = 
f
1 cm
1 cm 1 cm
f = 32 cm
R = 2f = (2)(32 cm) = 64 cm
I Ch. 14–6
Holt Physics Solution Manual
real, inverted image
Copyright © by Holt, Rinehart and Winston. All rights reserved.
h⬘
q −15.8 cm
M =  = −  =  = 0.299
h
52.9 cm
p
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Givens
Solutions
47. M = −0.75
q
4.6 cm
p = −  = −  = 6.1 cm
M
− 0.75
q = 4.6 cm
1 1 1
1
1
 =  +  =  + 
f
p q 6.1 cm 4.6 cm
I
1 0.16 0.22 0.38
 =  +  = 
f 1 cm 1 cm 1 cm
f = 2.6 cm
48. p = 15.5 cm
1
M = 
2
real, inverted image
q 1
M = −  = 
p 2
p
q = − 
2
2 1 1 1 2
1
 =  +  =  −  = − 
R p q p p
p
R = −2p = (−2)(15.5 cm) = −31.0 cm
49. p1 = 15 cm
q1 = 8.5 cm
p2 = 25 cm
1
1
1
1
1
 =  +  =  + 
p1 q1 15 cm 8.5 cm
f
0.12
0.19
1 0.067
 =  +  = 
1 cm
1 cm
f
1 cm
1 1
1
0.19
1
 =  −  =  − 
q2 f
p2 1 cm 25 cm
1
0.19
0.040 0.15
 =  −  = 
q2 1 cm
1 cm 1 cm
q2 = 6.7 cm
real image
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q
8.5 cm
M1 = − 1 = −  = −0.57
p1
15 cm
q
6.7 cm
M2 = − 2 = −  = −0.27
p2
25 cm
50. p = 195 cm
q = −12.8 cm
both images are inverted
1 1 1
1
1
 =  +  =  + 
f
p q 195 cm −12.8 cm
1
5.13 × 10−3
−7.81 × 10−2
−7.30 × 10−2
 =  +  = 
f
1 cm
1 cm
1 cm
f = −13.7 cm
q
−12.8 cm
M = −  = −  = 0.0656
p
195 cm
51. R = −11.3 cm
1
M = 
3
virtual, upright image
q 1
M = −  = 
p 3
p
q = − 
3
2 1 1 1 3
2
 =  +  =  −  = − 
R p q p p
p
p = −R = 11.3 cm
Section One—Pupil’s Edition Solutions
I Ch. 14–7
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Givens
Solutions
52. p = 10.0 cm
q
2.00 × 102 cm
M = −  = −  = −20.0
p
10.0 cm
q = 2.00 m
I
53.
Triangle 1
f
f
real, inverted image
Let q1 and q2 be the angles of incidence at the first and second surfaces, respectively.
Triangle 1:
2q1 + 2q2 + f = 180°
θ2
θ1
θ
f = 180° − 2(q1 + q2 )
Triangle 2:
Triangle 2
(90° − q1) + (90° − q2 ) + q = 180°
q = q 1 + q2
Substitute this q value into the equation for f.
f = 180° − 2q
54. R = ∞
1 1 2
 +  = 
p q R
1 1 2
 +  = 
p q ∞
1 1
 +  = 0
p q
1
1
 = − 
q
p
q = −p
= 50.0 cm
q = 10.0 cm − 20.0 cm
= −10.0 cm
2 1 1
 =  + 
R p q
1
1
1
1
1
 =  +  =  + 
R 2p 2q (2)(50.0 cm) (2)(10.0 cm)
1 0.0100 0.0500 −0.0400
 =  −  = 
R
1 cm
1 cm
1 cm
R = −25.0 cm
56. q1 = −30.0 cm
q2 = −10.0 cm
R1 = −R2
a. R1 = −R2
2
2
 = − 
R1
R2
1
1
1
1
 +  = −  + 
p1 q1
p2 q2
where p1 = p2 = p
2
1
1
1
1
 = −  −  = −  + 
p
q1
q2
−30.0 cm 10.0 cm
2 0.0333
 =  +
1 cm
p
2
p =  cm
0.133
I Ch. 14–8
Holt Physics Solution Manual
0.100 0.133
 = 
1 cm
1 cm
= 15.0 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
55. p = 70.0 cm − 20.0 cm
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Givens
Solutions
2 1 1
1
1
b.  =  +  =  + 
R p q1 15.0 cm −30.0 cm
2 0.0667 0.0333 0.0334
 =  −  = 
R 1 cm
1 cm
1 cm
I
2
R =  cm = 59.9 cm
0.0334
q
−30.0 cm
c. Mconcave = − 1 = −  = 2.00
p1
15.0 cm
q
−10.0 cm
Mconvex = − 2 = −  = 0.667
p2
15.0 cm
57. h = 2.70 cm
p = 12.0 cm
q h ⬘ 5.40 cm
M = − =  =  = 2.00
p h 2.70 cm
h⬘ = 5.40 cm
q = −Mp = −(2.00)(12.0 cm) = −24.0 cm
virtual image
2 1 1
1
1
 =  +  =  + 
R p q 12.0 cm −24.0 cm
2 0.0833 0.0417 0.0416
 =  −  = 
1 cm
1 cm
1 cm
R
2
R =  cm = 48.1 cm
0.0416
58. f = 7.5 cm
p1 = 7.5 cm
Locate the image of the coins formed by the upper mirror:
1 1
1
1
1
 =  −  =  −  = 0
q1 f
p1 7.5 cm 7.5 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Thus, q1 = ∞.
Now, locate the final image, realizing that parallel rays are reflected toward the lower
mirror by the upper mirror; thus, p2 = ∞.
1 1
1
1
1
1
 =  −  =  −  = 
q2 f
p2 7.5 cm ∞ 7.5 cm
q2 = 7.5 cm
q
q
M = M1M2 = − 1 − 2
p1 p2
where q1 = p2
q
7.5 cm
M = 2 =  = 1.0
p1 7.5 cm
Section One—Pupil’s Edition Solutions
I Ch. 14–9
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Givens
Solutions
59.
For a convex mirror:
f < 0, or f = −| f |
1
1
1
p + |f |
 =  −  = − 
q −| f | p
p| f |
I
p|f |
q = − 
p + |f |
|f |
q
1
p| f |
M = −  = −  −  = 
p
p p + |f | p + |f |
For a real object (p > 0), 1 > M > 0; this means that the image is an upright, virtual
image.
For a spherical mirror, the image is virtual and upright if p | f | and f 0. For f 0:
f > 0, or f = + | f |
1
1 p − |f |
1
 =  −  = 
q |f | p
p|f |
p|f |
q = 
p − |f |
|f |
q
1 p|f |
M = −  = −   = 
p
p p − |f | |f | − p
When p < | f |, M > 1; thus the image is enlarged, upright, and virtual.
60.
h
tan q = 
p
−h⬘
tan q ⬘ = 
q
q = q ⬘; thus, tan q = tan q⬘.
Cross multiply to find the magnification equation.
q
h⬘
M =  = − 
p
h
From two other triangles in the figure, we note that:
h
hⴕ
tan a =  and tan a = − 
p−R
R−q
so that
R−q
q
h⬘
 = −  = − 
p−R
p
h
Cross multiply to find the following:
p(R − q) = q(p − R)
pR − pq = qp − qR
R(p + q) = 2pq
2 p+q 1 1
 =  =  + 
R
pq
q p
I Ch. 14–10 Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
h
h⬘
 = − 
p
q
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Refraction
Chapter 15
I
Practice 15A, p. 567
Givens
1. ni = 1.00
nr = 1.333
qi = 25.0°
2. ni = 1.66
nr = 1.52
Solutions
ni(sin q i ) = nr(sin q r)
ni
1.00
qr = sin−1 (sin
qi ) = sin−1  (sin 25.0°) = 18.5°
nr
1.333
ni(sin q i ) = nr(sin q r)
qi = 25.0°
ni
1.66
a. qr = sin−1 (sin
qi) = sin−1 (sin 25.0°) = 27.5°
nr
1.52
ni = 1.00
ni (sin qi)
1.00(sin 14.5°)
 =  = 1.47 glycerine
b. nr = 
sin 9.80°
sin qr
qi = 14.5°
qr = 9.80°
ni = 1.00
ni
1.00
c. qr = sin−1 (sin
qi) = sin−1  (sin 31.6°) = 12.5°
nr
2.419
nr = 2.419
qi = 31.6°
3. qi = 40.0°
Copyright © by Holt, Rinehart and Winston. All rights reserved.
qr = 26.0°
ni = 1.00
ni(sin q i ) = nr(sin q r)
ni (sin qi)
(1.00)(sin 40.0°)
 =  = 1.47
nr = 
sinqr
sin 26.0°
Section Review, p. 567
3. ni = 1.00
nr = 2.419
qi = 15.0°
4. ni = 1.00
nr = 1.333
qi = 22.5°
ni(sin q i ) = nr(sin q r)
ni
1.00
qr = sin−1 (sin
qi) = sin−1  (sin 15.0°) = 6.14°
nr
2.419
ni(sin q i ) = nr(sin q r)
ni
1.00
qr = sin−1 (sin
qi ) = sin−1  (sin 22.5°) = 16.7°
nr
1.333
Section One—Pupil’s Edition Solutions
I Ch. 15–1
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Practice 15B, p. 576
Givens
Solutions
1. p = 20.0 cm
f = 10.0 cm
1 1 1
 +  = 
p q f
1
1
1 1 1
 =  −  =  − 
p
10.0 cm
20.0 cm
q f
I
1 0.100
0.0500 0.0500
 =  −  = 
q 1 cm
1 cm
1 cm
q = 20.0 cm
q
20.0 cm
M = − = −  = −1.00
p
20.0 cm
2. f = 15.0 cm
p = 10.0 cm
real, inverted image
1
1
1
1 1
 =  −  =  − 
p
15.0 cm
10.0 cm
q
f
1 0.0667
0.100 −0.0333
 =  −  = 
q
1 cm
1 cm
1 cm
q = −30.0 cm
q
−30.0 cm
M = −  = −  = 3.00
p
10.0 cm
3. p = 20.0 cm
f = −10.0 cm
virtual, upright image
1
1 1
1
1
 =  −  =  − 
q
f
p −10.0 cm 20.0 cm
1 − 0.100 0.0500
− 0.150
 =  −  = 
q
1 cm
1 cm
1 cm
q
−6.67 cm
M = −  = −  = 0.333
20.0 cm
p
4. f = 6.0 cm
q = −3.0 cm
virtual, upright image
1
1
1
1
1
a.  =  −  =  − 
p
6.0 cm
−3.0 cm
f
q
1 0.17
0.50
−0.33
 =  −  = 
p 1 cm
1 cm
1 cm
p = 2.0 cm
q
−3.0 cm
M = −  = −  = 1.5
p
2.0 cm
f = 2.9 cm
q = 7.0 cm
1
1
1
1
1
b.  =  −  =  − 
p
f
2.9 cm
7.0 cm
q
1 0.34
0.14
0.20
 =  −  = 
p 1 cm
1 cm
1 cm
p = 5.0 cm
q
7.0 cm
M = −  = −  = −1.4
p
5.0 cm
I Ch. 15–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = −6.67 cm
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Givens
f = −6.0 cm
p = 4.0 cm
Solutions
1
1
1
1
1
c.  =  −  =  − 
q
f
p −6.0 cm
4.0 cm
1 − 0.17
0.25
− 0.42
 =  −  = 
q
1 cm
1 cm
1 cm
I
q = −2.4 cm
q
−2.4 cm
M = −  = −  = 0.60
p
4.0 cm
p = 5.0 cm
d. q = −Mp = −(0.50)(5.0 cm) = −2.5 cm
M = 0.50
1 1 1
1
1
 =  +  =  + 
f p q 5.0 cm −2.5 cm
1 0.20
−0.40 − 0.20
 =  +  = 
f
1 cm
1 cm
1 cm
f = −5.0 cm
Section Review, p. 579
3. f = 4.0 cm
p = 3.0 cm + 4.0 cm = 7.0 cm
1 1 1
 +  = 
f
p q
1
1 1
1
1
 =  −  =  − 
q
f
p 4.0 cm 7.0 cm
1 0.25
0.14
0.11
 =  −  = 
q 1 cm 1 cm 1 cm
q = 9.1 cm
4. p = 7.0 cm
q = 9.1 cm
q
9.1 cm
M = −  = −  = −1.3
p
7.0 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Practice 15C, p. 582
1. ni = 1.473
nr = 1.00
2. ni = 1.473
nr = 1.333
3. ni = 1.309
nr = 1.00
4. ni = 2.419
nr = 1.00
ni = 2.20
nr = 1.00
n
sin qc = r
ni
n
1.00
qc = sin−1 r = sin−1  = 42.8°
ni
1.473
n
1.333
qc = sin−1 r = sin−1  = 64.82°
ni
1.473
n
1.00
qc = sin−1 r = sin−1  = 49.8°
ni
1.309
n
1.00
diamond: qc = sin−1 r = sin−1  = 24.4°
ni
2.419
n
1.00
cubic zirconia: qc = sin−1 r = sin−1  = 27.0°
ni
2.20
Section One—Pupil’s Edition Solutions
I Ch. 15–3
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Section Review, p. 585
Givens
Solutions
1. ni = 1.333
nr = 1.309
I
n
sin qc = r
ni
n
1.309
qc = sin−1 r = sin−1  = 79.11°
ni
1.333
Chapter Review and Assess, pp. 587–592
10. ni = 1.00
nr = 1.333
qi = 42.3°
11. ni = 1.00
nr = 1.333
qi = 36°
12. ni = 1.00
nr = 1.333
qi = 35.0°
13. qi,1 = 30.0°
ni,1 = 1.00
nr,1 = 1.50
ni (sin qi ) = nr (sin qr)
ni
1.00
qr = sin−1 (sin
qi ) = sin−1  (sin 42.3°) = 30.3°
nr
1.333
ni (sin qi ) = nr (sin qr)
ni
1.00
qr = sin−1 (sin
qi ) = sin−1  (sin 36°) = 26°
nr
1.333
ni (sin qi ) = nr (sin qr)
ni
1.00
qr = sin−1 (sin
qi) = sin−1  (sin 35.0°) = 25.5°
nr
1.333
ni (sin qi ) = nr (sin qr)
air to glass:
1.00
n 1
qr,1 = sin−1 i,(sin
qi,1) = sin−1 (sin 30.0°) = 19.5°
1.50
nr,1
qi,1 = 30.0° and qr,1 = 19.5°
qi,2 = qr,1 because they are alternate interior angles. qr,2 = qi,1 because of the
reversibility of refraction; thus, qi,2 = 19.5° and qr,2 = 30.0°
14. qr = 20.0°
ni = 1.00
ni (sin qi ) = nr (sin qr)
nr = 1.48
nr
1.48
qr) = sin−1 (sin 20.0°) = 30.4°
air to oil: q1 = qi = sin−1 (sin
ni
1.00
qi = 20.0°
ni
1.48
qi) = sin−1  (sin 20.0°) = 22.3°
oil to water: q2 = qr = sin−1 (sin
nr
1.333
ni = 1.48
nr = 1.333
I Ch. 15–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
glass to air:
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Givens
Solutions
24. f = −20.0 cm
1 1 1
a.  +  = 
p q
f
1
1
1 1 1
 =  −  =  − 
q f p −20.0 cm 40.0 cm
1 −0.0500
0.0250
−0.0750
 =  −  = 
q
1 cm
1 cm
1 cm
p = 40.0 cm
I
q = −13.3 cm
q
−13.3 cm
M = −  = −  = 0.332
p
40.0 cm
p = 20.0 cm
virtual, upright image
1
1
1 1 1
b.  =  −  =  − 
q f p −20.0 cm 20.0 cm
1 −0.0500
0.0500
−0.100
 =  −  = 
q
1 cm
1 cm
1 cm
q = −10.0 cm
q
−10.0 cm
M = − = −  = 0.500
p
20.0 cm
p = 10.0 cm
virtual, upright image
1
1
1 1 1
c.  =  −  =  − 
q f p −20.0 cm 10.0 cm
1 −0.0500
0.100 − 0.150
 =  −  = 
q
1 cm
1 cm
1 cm
q = −6.67 cm
q
−6.67 cm
M = − = −  = 0.667
p
10.0 cm
25. f = 12.5 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = −30.0 cm
virtual, upright image
1 1 1
1
1
 =  −  =  − 
p f
q 12.5 cm −30.0 cm
1 0.0800 −0.0333
0.1133
 =  −  = 
p
1 cm
1 cm
1 cm
p = 8.826 cm
q
−30.0 cm
M = −  = −  = 3.40
p
8.826 cm
26. f = 20.0 cm
p = 40.0 cm
upright image
1 1 1
1
1
a.  =  −  =  − 
q f p 20.0 cm
40.0 cm
1 0.0500 0.0250 0.0250
 =  −  = 
q
1 cm
1 cm
1 cm
q = 40.0 cm
q
40.0 cm
M = −  = −  = −1.00
p
40.0 cm
p = 10.0 cm
real, inverted image
1 1 1
1
1
b.  =  −  =  − 
q f p 20.0 cm 10.0 cm
1 0.0500
0.100
−0.0500
 =  −  = 
q
1 cm
1 cm
1 cm
q = −20.0 cm
q
−20.0 cm
M = −  = −  = 2.00
p
10.0 cm
virtual, upright image
Section One—Pupil’s Edition Solutions
I Ch. 15–5
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Givens
Solutions
36. ni = 1.473
n
sin qc = r
ni
nr = 1.00
n
1.00
qc = sin−1 r = sin−1  = 42.8°
ni
1.473
I
ni = 1.923
n
1.00
a. qc = sin−1 r = sin−1  = 31.3°
ni
1.923
ni = 1.434
n
1.00
b. qc = sin−1 r = sin−1  = 44.2°
ni
1.434
ni = 1.309
n
1.00
c. qc = sin−1 r = sin−1  = 49.8°
ni
1.309
38. ni = 1.52
nr = 1.00
qr = 45∞
39. ni = 1.00
qi = 63.5°
qr = 42.9°
40. ni = 1.00
nr = 1.333
qr = 36.2°
41. v = 1.85 × 108 m/s
c = 3.00 × 108 m/s
42. ni = 1.66
nr = 1.333
n
1.00
qc = sin−1 r = sin−1  = 41.1°
ni
1.52
qr > qc , therefore the ray will be totally internally reflected.
ni (sin qi ) = nr (sin qr)
ni (sin qi) (1.00)(sin 63.5°)
 =  = 1.31
nr = 
sin qr
sin 42.9°
ni (sin qi ) = nr (sin qr)
nr
1.333
qi = sin−1 (sin
qr) = sin−1  (sin 36.2°) = 51.9°
ni
1.00
c 3.00 × 108 m/s
= 1.62
n =  = 
v 1.85 × 108 m/s
carbon disulfide
ni (sin qi ) = nr (sin qr)
n
1.333
n (sin q ) = sin 1.66 (sin 90.0°) = 53.4°
qi = 28.7°
n
1.66
a. qr = sin−1 i (sin qi ) = sin−1  (sin 28.7°) = 36.7°
nr
1.333
qr = 90.0°
b. qi = sin−1
r
r
i
43. f = 15.0 cm
M = +2.00
p
M = − 
q
q = −Mp = −2.00p
1 1 1 1
1
1
 =  +  =  −  = 
f p q p 2.00p 2.00p
1
1
 = 
15.0 cm 2.00 p
15.0 cm
p =  = 7.50 cm
2.00
I Ch. 15–6
Holt Physics Solution Manual
−1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
37. nr = 1.00
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Givens
Solutions
44. M = 1.50
q
M = −  = 1.50
p
p = 2.84 cm
q = −1.50p = −(1.50)(2.84 cm) = −4.26 cm
1 1 1
1
1
 =  +  =  + 
f p q 2.84 cm −4.26 cm
I
1 0.352 −0.235 0.117
 =  +  = 
f
1 cm
1 cm
1 cm
f = 8.55 cm
45. M = 2.00
f = 12.0 cm
q
a. M = −  = 2.00
p
q = −2.00p
1 1 1 1
1
1
 =  +  =  −  = 
f p q p 2.00p 2.00p
f
12.0 cm
p =  =  = 6.00 cm
M
2.00
46. p = 80.0 cm
q = −40.0 cm
1
1
1 1 1
 =  +  =  + 
f p q 80.0 cm −40.0 cm
1 0.0125
−0.0250
−0.0125
 =  +  = 
f
1 cm
1 cm
1 cm
f = −80.0 cm
47. f = 2.44 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 12.9 cm
1 1 1
1
1
a.  =  −  =  + 
p f
q 2.44 cm 12.9 cm
1 0.410
0.0775
0.332
 =  +  = 
p 1 cm
1 cm
1 cm
p = 3.01 cm
q = −12.9 cm
1 1 1
1
1
b.  =  −  =  − 
p f
q 2.44 cm −12.9 cm
1 0.410
−0.0775
0.488
 =  −  = 
p 1 cm
1 cm
1 cm
p = 2.05 cm
48. q = −30.0 cm
f = −40.0 cm
1 1 1
1
1
 =  −  =  − 
p f q −40.0 cm −30.0 cm
1 −0.0250 −0.0333 0.0083
 =  −  = 
p
1 cm
1 cm
1 cm
p = 1.20 × 102 cm
q
−30.0 cm
M = −  = −  = 0.25
p
120 cm
Section One—Pupil’s Edition Solutions
I Ch. 15–7
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Givens
Solutions
49. nr = 1.331
ni (sin qi ) = nr (sin qr)
nb = 1.340
qi = 83.0°
I
ni = 1.00
50. qi = 23.1°
v = 2.17 × 108 m/s
ni = 1.00
c = 3.00 × 108 m/s
51. qi = 30.0°
qr = 22.0°
ni = 1.00
ni
1.00
blue: qr = sin−1 (sin
qi) = sin−1  (sin 83.0°) = 47.8°
nr
1.340
ni
1.00
qr) = sin−1  (sin 83.0°) = 48.2°
red: qr = sin−1 (sin
nr
1.331
c 3.00 × 108 m/s
= 1.38
nr =  = 
v 2.17 × 108 m/s
ni (sin qi ) = nr (sin qr)
ni
1.00
qr = sin−1 (sin
qi ) = sin−1 (sin 23.1°) = 16.5°
nr
1.38
ni (sin qi ) = nr (sin qr)
sin q
sin 30.0°
nr = ni i = (1.00)  = 1.33
sin qr
sin 22.0°
n
sin qc = r
ni
ni is now nr , and nr is now ni , so
n
1.00
qc = sin−1 i = sin−1  = 48.8°
nr
1.33
52. va = 340 m/s
vw = 1510 m/s
qi = 12.0°
ni (sin qi ) = nr (sin qr)
v
1510 m/s
 (sin 12.0°) = 67°
v (sin q ) = sin 
340 m/s
n
(c/va)
qr = sin−1 i (sin qi) = sin−1 
(sin qi)
nr
(c/vw)
qr = sin−1
w
−1
i
53. ni = 1.333
nr = 1.00
∆y = 2.00 m
n
sin qc = r
ni
n
1.00
qc = sin−1 r = sin−1  = 48.6°
ni
1.333
(d/2)
d
tan qc =  =  where d is the diameter of the piece of wood
∆y
2∆y
d = 2∆y(tan qc ) = (2)(2.00 m)(tan 48.6°) = 4.54 m
54. ni = 1.00
nr = 1.309
qi = 40.0°
ni (sin qi ) = nr (sin qr)
q = 180.0° − 29.4° = 110.6°
I Ch. 15–8
ni
1.00
qr = sin−1 (sin
qi ) = sin−1  (sin 40.0°) = 29.4°
nr
1.309
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
a
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Givens
Solutions
55. p = 10f
1 1 1
 +  = 
p q f
1 1 1 1
1
9
 =  −  =  −  = 
q f p f 10f
10f
I
10
q = f
9
n
sin qc = r
ni
56. ni = 1.53
nr = 1.00
n
1.00
a. qc = sin−1 r = sin−1  = 40.8°
ni
1.53
nr = 1.333
n
1.333
b. qc = sin−1 r = sin−1  = 60.6∞
ni
1.53
57. nr = 1.50
a. ni (sin qi ) = nr (sin qr)
ni = 1.00
qi = 90° − 30° = 60°
35.3°
ni
1.00
qr = sin−1 (sin
qi) = sin−1 (sin 60°) = 35.3°
nr
1.50
Using the figure at left, the angle of incidence of the ray at the bottom of the
prism can be found as follows.
qi = 180° − 35.3° − 30° − 90° = 24.7°
θi
30°
ni = 1.50
nr = 1.00
Copyright © by Holt, Rinehart and Winston. All rights reserved.
qi = 24.7°
n
1.00
q = sin  = sin  = 41.8°
n 1.50 n
b. sin qc = r
ni
−1
c
−1
r
i
The ray will pass through the bottom surface because qi < qc .
58. ni = 1.8
nr = ni (sin qc ) = (1.8)(sin 45°) = 1.3
qc = 45°
59. ni = 1.60
nr = ni (sin qc ) = (1.60)(sin 59.5°) = 1.38
qc = 59.5°
60. ni = 1.333
nr = 1.00
∆x
2.00 m
qi = tan−1  = tan−1  = 26.6°
∆y
4.00 m
∆y = 4.00 m
ni (sin qi ) = nr (sin qr)
∆x = 2.00 m
ni
1.333
qr = sin−1 (sin
qi ) = sin−1  (sin 26.6°) = 36.6°
nr
1.00
Section One—Pupil’s Edition Solutions
I Ch. 15–9
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Givens
Solutions
61. ni = 1.333
nr = 1.00
∆x
120 m
qi = tan−1  = tan−1  = 46.2°
∆y
115 m
∆x = 325 m − 205 m = 120 m
ni (sin qi ) = nr (sin qr)
∆y = 115 m
ni
1.333
qr = sin−1 (sin
qi ) = sin−1  (sin 46.2°) = 74.2°
nr
1.00
205 m
h =  = 58.0 m
(tan 74.2°)
62. ni = 1.00
nr = 1.48
qi = 50.0°
w = 3.1 mm
l
= 42 cm
ni (sin qi ) = nr (sin qr)
ni
1.00
qr = sin−1 (sin
qi ) = sin−1 (sin 50.0°) = 31.2°
nr
1.48
w
3.1 mm
d =  =  = 2.6 mm = 0.26 cm
2(tan qr) 2(tan 31.2°)
The ray travels a horizontal distance of 2d per reflection.
42 cm
l
# of reflections =  =  = 81 reflections
2d 2(0.26 cm)
63. f = 4.80 cm
= 4.80 × 10−2 m
p = 10.0 m
1
1 1
1
1
a.  =  −  = 
− 
q
f p 4.80 × 10−2 m 10.0 m
1 20.8
0.100
20.7
 =  −  = 
q 1m
1m
1m
q = 4.83 × 10−2 m = 4.83 cm
p = 1.75 m
1 1 1
1
1
b.  =  −  = 
− 
−2
q f
p 4.80 × 10 m 1.75 m
q = 4.95 × 10−2 m = 4.95 cm
∆q = 4.95 cm − 4.83 cm = 0.12 cm
64. q = 1.90 cm
p = 35.0 cm
1
1
1 1 1
 =  +  =  + 
p q 35.0 cm
1.90 cm
f
1 0.0286 0.526
0.555
 =  +  = 
f
1 cm
1 cm
1 cm
f = 1.80 cm
65. q = 1.90 cm
= 1.90 × 10−2 m
p = 15.0 m
1 1 1
1
1
 =  +  =  + 
f
p q 15.0 m 1.90 × 10−2 m
1 0.0667 52.6 52.7
 =  +  = 
f
1m
1m
1m
f = 1.90 × 10−2 m = 1.90 cm
I Ch. 15–10 Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1 20.8 0.571 20.2
 =  −  = 
q
1m
1m
1m
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Interference and Diffraction
Chapter 16
I
Practice 16A, p. 603
Givens
1. d = 0.50 mm
q = 0.059°
Solutions
d(sin q) (5.0 × 10−4 m)(sin 0.059°)
l =  =  = 5.1 × 10−7 m = 5.1 × 102 nm
m
1
m =1
2. d = 2.02 × 10−6 m
q = 16.5°
m =1
3. d = 0.250 mm
l = 546.1 nm
d(sin q) (2.02 × 10−6 m)(sin 16.5°)
l =  = 
m
1
l = 574 nm
lm
(5.461 × 10−7 m)(1)
q = sin−1  = sin−1 
= 0.125°
d
2.50 × 10−4 m
m =1
4. d = 0.250 mm
l = 546.1 nm
m =1
−1
q = sin
m + 2l
1

d
(1.5)(5.461 × 10−7 m)
q = sin−1 
= 0.188°
2.50 × 10−4 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section Review, p. 603
3. d = 0.0550 mm
m = 1 and m = 2
l = 605 nm
lm
(6.05 × 10−7 m)(1)
m = 1: q1 = sin−1  = sin−1 
= 0.630°
d
5.50 × 10−5 m
lm
(6.05 × 10−7 m)(2)
= 1.26°
m = 2: q2 = sin−1  = sin−1 
d
5.50 × 10−5 m
q = q2 − q1 = 1.26° − 0.630° = 0.630°
4. m = 2
q = 1.28°
lm
(3.35 m)(2)
d =  =  = 3.00 × 102 m
sin q
sin 1.28°
l = 3.35 m
Section One—Pupil’s Edition Solutions
I Ch. 16–1
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Practice 16B, p. 610
Givens
Solutions
1. 2.500 × 103 lines/cm
l1 = 588.995 nm
I
ml
m = 1: q1 = sin−1 1 = sin−1
d
l2 = 589.592 nm
−1
q2 = sin
(1)(5.88995 × 10−7 m)

1
= 8.46752°
5 m
2.500 × 10
(1)(5.89592 × 10−7 m)

ml2
−1
1
 = sin
= 8.47616°
5 m
d
2.500 × 10
q = q2 − q1 = 8.476° − 8.468° = 0.008°
m = 2: q1 = sin−1
(2)(5.88995 × 10−7 m)

ml1
1
 = sin−1
= 17.1274°
5 m
d
2.500 × 10
ml
q2 = sin−1 2 = sin−1
d
(2)(5.89592 × 10−7 m)

1
= 17.1453°
5 m
2.500 × 10
q = q2 − q1 = 17.15° − 17.13° = 0.02°
m = 3: q1 = sin−1
q2 = sin−1
(3)(5.88995 × 10−7 m)

ml1
1
 = sin−1
= 26.2153°
5 m
d
2.500 × 10
(3)(5.89592 × 10−7 m)

ml2
1
 = sin−1
= 26.2439°
5 m
d
2.500 × 10
2. 4525 lines/cm
blue: q = sin−1
m =1
lb = 422 nm
lr = 655 nm
ml
red: q = sin−1 r = sin−1
d
3. 1555 lines/cm
l = 565 nm
(1)(4.22 × 10−7 m)

mlb
1
 = sin−1
= 11.0°
 m
d
452 500
m = 11: q = sin−1
q < 90°
m = 12: q = sin−1
(1)(6.55 × 10−7 m)

1
= 17.2°
 m
452 500
(11)(5.65 × 10−7 m)

ml
1
 = sin−1
= 75.1°
 m
d
155 500
(12)(5.65 × 10−7 m)

ml
1
 = sin−1
=∞
 m
d
155 500
Therefore, 11 is the highest-order number that can be observed.
I Ch. 16–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = q2 − q1 = 26.24° − 26.22° = 0.02000°
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Givens
Solutions
4. 15 550 lines/cm
l = 565 nm
m = 1: q = sin−1
q < 90°
(2)(5.65 × 10−7 m)

ml
1
 = sin−1
=∞
 m
d
1 555 000
m = 2: q = sin−1
(1)(5.65 × 10−7 m)

ml
1
 = sin−1
= 61.5°
 m
d
1 555 000
I
Therefore, 1 is the highest-order number that can be observed.
5. q = 21.2°
l = 546.1 nm
m =1
ml
(1)(5.461 × 10−7 m)
d =  =  = 1.51 × 10−6 m = 1.51 × 10−4 cm
sin q
sin 21.2°
1
Lines/cm = 
= 6.62 × 103 lines/cm
1.51 × 10−4 cm
Section Review, p. 612
5. 3550 lines/cm
q = 12.07°
m =1
m =2
 m (sin 12.07°)
d(sin q) 355 000 

a. l =
=  = 5.890 × 10
1
−7
m
1
(2)(5.89 × 10−7 m)

ml
1
 = sin−1
= 24.72°
 m
d
355 000
b. q = sin−1
m = 589.0 nm
Chapter Review and Assess, pp. 620–622
9. d = 0.33 mm = 3.3 × 10−4 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 0.080°
m =1
10. d = 0.3096 mm
q = 0.218°
d(sin q) (3.3 × 10−4 m)(sin 0.080°)
l =  = 
1
m
l = 4.6 × 10−7 m = 4.6 102 nm
d(sin q) (3.096 × 10−4 m)(sin 0.218°)
a. l =  =  = 5.89 × 10−7 m = 589 nm
m
2
m =2
m =3
ml
(3)(5.89 × 10−7 m)
= 0.327°
b. q = sin−1  = sin−1 
d
3.096 × 10−4 m
m =4
ml
(4)(5.89 × 10−7 m)
= 0.436°
c. q = sin−1  = sin−1 
d
3.096 × 10−4 m
11. d = 3.2 cm = 3.2 × 10−2 m
m=2
d(sin q) (3.2 × 10−2 m)(sin 0.56°)
l =  = 
2
m
q = 0.56°
l = 1.6 × 10−4 m = 1.6 × 102 µm
Section One—Pupil’s Edition Solutions
I Ch. 16–3
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Givens
Solutions
19. 795 slits/cm
−5
l = 707 nm = 7.07 × 10 cm
7.07 × 10−5 cm
ml
m = 1:q 1 = sin−1  = sin−1 
1
d
 cm
795
q 1 = 3.22°
I
20. 795 slits/cm
l = 353 nm = 3.53 × 10−5 cm
21. 3661 slits/cm
l 1 = 478.5 nm
l 2 = 647.4 nm
l 3 = 696.4 nm
(2)(3.53 × 10−5 cm)
ml
m = 2:q 2 = sin−1  = sin−1 
1
d
 cm
795
q 2 = 3.22°
a. m = 1:
4.785 × 10−5 cm
ml
q1 = sin−1  = sin−1 
= 10.09°
1
d

3661 cm
6.474 × 10−5 cm
ml
q2 = sin−1  = sin−1 
= 13.71°
1
d

3661 cm
6.964 × 10−5 cm
ml
q3 = sin−1  = sin−1 
= 14.77°
1
d

3661 cm
b. m = 2:
(2)(6.474 × 10−5 cm)
ml
q2 = sin−1  = sin−1 
= 28.30°
1
d

3661 cm
(2)(6.964 × 10−5 cm)
ml
q3 = sin−1  = sin−1 
= 30.66°
1
d

3661 cm
26. l = 546.1 nm
q = 81.0°
sin 81.0°
sin q
lines/mm =  = 
= 603 lines/mm
(3)(5.461 × 10−4 mm)
ml
m =3
28. l = 486 nm
m =5
ml
(5)(4.86 × 10−7 m)
d =  =  = 2.41 × 10−4 m
sin q
(sin 0.578°)
q = 0.578°
I Ch. 16–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(2)(4.785 × 10−5 cm)
ml
q1 = sin−1  = sin−1 
= 20.51°
1
d

3661 cm
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Givens
Solutions
29. l1 = 540.0 nm
4l1 = 5l2
4l
4(540.0 nm)
l2 = 1 =  = 432.0 nm
5
5
30. m = 2
l = 4.000 10–7 m
q = 90.00°
31. l = 643 nm
q = 0.737°
I
ml
(2)(4.000 × 10−7 m)
d =  =  = 8.000 × 10−7 m
sin q
(sin 90.00°)
path difference = d(sin q) = (0.150 mm)(sin 0.737°) = 1.93 × 10−3 mm = 3l
a maximum
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 0.150 mm
d is at a maximum when l = 400.0 nm (red light).
Section One—Pupil’s Edition Solutions
I Ch. 16–5
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Electric Forces and Fields
Chapter 17
I
Section Review, p. 633
Givens
3. q = 10.0 C
Solutions
q
10.0 C
N = 
= 
= 6.25 × 1019 electrons
−19
1.60 × 10 C/electron
1.60 × 10−19 C/electron
Practice 17A, p. 636
1. q1 = −8.0 mC
q2 = 8.0 mC
r = 5.0 cm
2. r = 0.30 m
q1 = 12 × 10−9 C
(8.99 × 109 N • m2/C2)(8.0 × 10−6 C)2
kC q1q2

= 
F=
2
r
(0.050 m)2
F = 230 N
kC q1q2
(8.99 × 109 N • m2/C2)(12 × 10−9 C)(18 × 10−9 C)

a. F = 
= 
2
r
(0.30 m)2
q2 = −18 × 10−9 C
q1 = −3.0 × 10−9 C
q2 = −3.0 × 10−9 C
3. q1 = 6.0 mC
Copyright © by Holt, Rinehart and Winston. All rights reserved.
r = 0.12 m
F = 2.2 × 10−5 N
9
2 2
−9
2
kC q1q2 (8.99 × 10 N • m /C )(3.0 × 10 C)


=
= 9.0 × 10−7 N
b. F = 
r2
(0.30 m)2
kC q1q2 (8.99 × 109 N • m2/C2)(6.0 × 10−6 C)(4.3 × 10−6 C)
 = 
a. F = 
r2
(0.12 m)2
F = 16 N
q2 = −4.3 mC
q2
4.3 × 10−6 C
c. N = 
= 
= 2.7 × 1013 electrons
−19
1.60 × 10 C/electron 1.60 × 10−19 C/electron
6.0 × 10−6 C
q1
d. N = 
= 
= 3.8 × 1013 electrons
−19
1.60 × 10 C/electron 1.60 × 10−19 C/electron
4. q1 = 60.0 mC
q2 = 50.0 mC
F = 175 N
r=
(8.99 × 109 N • m2/C2)(60.0 × 10−6 C)(50.0 × 10−6 C)

k
qFq = 175 N
C 1 2
r = 0.393 m = 39.3 cm
Section One—Pupil’s Edition Solutions
I Ch. 17–1
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Practice 17B, p. 639
Givens
Solutions
1. q1 = 6.0 mC at x = 0 cm
q2 = 1.5 mC at x = 3.0 cm
I
q3 = −2.0 mC at x = 5.0 cm
9
2 2
−6
−6
kC q1q2 (8.99 × 10 N • m /C )(6.0 × 10 C)(1.5 × 10 C)


=
= 9.0 × 101 N
F1,2 = 
(r1,2)2
(0.030 m)2
9
2 2
−6
−6
kC q2q3 (8.99 × 10 N • m /C )(1.5 × 10 C)(2.0 × 10 C)


F2,3 = 
=
= 67 N
(r2,3)2
(0.020 m)2
9
2 2
−6
−6
kC q1q3 (8.99 × 10 N • m /C )(6.0 × 10 C)(2.0 × 10 C)

F1,3 = 
= 
= 43 N
2
2
(r1,3)
(0.050 m)
F1,tot = F1,2 + F1,3 = −(9.0 × 101 N) + (43 N) = −47 N
F1,tot = 47 N, along the negative x-axis
F2,tot = F1,2 + F2,3 = (9.0 × 101 N) + (67 N) = 157 N
F2,tot = 157 N, along the positive x-axis
F3,tot = F2,3 + F1,3 = −(67 N) − (43 N) = −11.0 × 101 N
F3,tot = 11.0 × 101 N, along the negative x-axis
2. q1 = 3.0 mC
)2
+(15
)2 = 22
a. r1,4 = r2,3 = (1
5cm
cm
0cm
2+220
cm
2 = 44
0cm
2
q2 = −6.0 mC
r1,4 = r2,3 = 21 cm
q3 = −2.4 mC
9
2 2
−6
−6
kC q1q2 (8.99 × 10 N • m /C )(3.0 × 10 C)(6.0 × 10 C)


F1,2 = 
=
= 7.2 N
(r1,2)2
(0.15 m)2
q4 = −9.0 mC
r1,2 = r2,4 = r3,4 = r1,3 =
15 cm
9
2 2
−6
−6
kC q1q3 (8.99 × 10 N • m /C )(3.0 × 10 C)(2.4 × 10 C)


F1,3 = 
=
= 2.9 N
(r1,3)2
(0.15 m)2
9
2 2
−6
−6
kC q1q4 (8.99 × 10 N • m /C )(3.0 × 10 C)(9.0 × 10 C)

F1,4 = 
= 
= 5.5 N
2
2
(r1,4)
(0.21 m)
F1,y = −(2.9 N) − (5.5 N)(sin 45°) = −2.9 N − 3.9 N = −6.8 N
2 (F )2 = (11.1 N)2 + (6.8 N)2 = 123 N2 + 46 N2
F1,tot = (F
1,
1,
x)+
y F1,tot = 16
9N
2 = 13.0 N
6.8
q = tan−1  = 31°
11.1
F1,tot = 13.0 N, 31° below the positive x-axis
b. F2,1 = 7.2 N (See a.)
9
2 2
−6
−6
kC q2q3 (8.99 × 10 N • m /C )(6.0 × 10 C)(2.4 × 10 C)

F2,3 = 
= 
= 2.9 N
2
2
(r2,3)
(0.21 m)
9
2 2
−6
−6
kC q2q4 (8.99 × 10 N • m /C )(6.0 × 10 C)(9.0 × 10 C)


F2,4 = 
=
= 22 N
(r2,4)2
(0.15 m)2
I Ch. 17–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
F1,x = (7.2 N) + (5.5 N)(cos 45°) = 7.2 N + 3.9 N = 11.1 N
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Givens
Solutions
F2,x = −(7.2 N) + (2.9 N)(cos 45°) = −7.2 N + 2.1 N = −5.1 N
F2,y = (22 N) + (2.9 N)(sin 45°) = 22 N + 2.1 N = 24 N
2 (F )2 = (5.1 N)2 + (24 N)2 = 26 N2 + 580 N2
F2,tot = (F
2,
2,
x)+
y I
F2,tot = 61
0N
2 = 25 N
24
q = tan−1  = 78°
5.1
F2,tot = 25 N, 78° above the negative x-axis
c. F4,1 = 5.5 N (See a.)
F4,2 = 22 N (See b.)
9
2 2
−6
−6
kC q4q3 (8.99 × 10 N • m /C )(9.0 × 10 C)(2.4 × 10 C)


F4,3 = 
=
= 8.6 N
(r4,3)2
(0.15 m)2
F4,x = −(5.5 N)(cos 45°) + (8.6 N) = −3.9 N + 8.6 N = 4.7 N
F4,y = (5.5 N)(sin 45°) − (22 N) = 3.9 N − 22 N = −18 N
2 (F )2 = (4.7 N)2 + (18 N)2 = 22 N2 + 320 N2
F4,tot = (F
4,
4,
x)+
y F4,tot = 34
0N
2 = 18 N
18
q = tan−1  = 75°
4.7
F4,tot = 18 N, 75° below the positive x-axis
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Practice 17C, p. 641
1. q1 = 2.00 × 10−9 C at the
origin
q2 = 4.00 × 10−9 C at
x = 1.5 m
q3 = 3.00 × 10−9 C
kC q1q3 kC q 2q3

= 
(r1,3)2
(r2,3)2
q1
q2

= 
(r1,3)2 (r2,3)2
2.00 × 10−9 C 4.00 × 10−9 C

= 
P2
(1.5 m − P )2
(4.00 × 10−9 C)(P 2) = (2.00 × 10−9 C)(1.5 m − P)2
P=
2.00 × 10−9 C
 (1.5 m − P)
4.00 × 10−9 C
P = 1.1 m − (0.707)(P)
(1.707)(P) = 1.1 m
P = 0.64 m from q1, or x = 0.64 m
Section One—Pupil’s Edition Solutions
I Ch. 17–3
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Givens
Solutions
2. q1 = −5.00 × 10−9 C
−9
q2 = −2.00 × 10
C
r1,2 = 40.0 cm
−9
I
q3 = 15.0 × 10
kC q1q3 kC q2q3

= 
(r1,3)2
(r2,3)2
q1
q2

= 
(r1,3)2 (r2,3)2
C
5.00 × 10−9 C
2.00 × 10−9 C



=
P2
(0.400 m − P)2
(2.00 × 10−9 C)(P 2) = (5.00 × 10−9 C)(0.400 m − P)2
P=
5.00 × 10−9 C

(0.400 m − P)
2.00 × 10−9 C
P = 0.632 m − (1.58)(P)
(2.58)(P) = 0.632 m
P = 0.245 m = 24.5 cm from q1
or (40.0 cm − 24.5 cm) = 15.5 cm from q2
3. q1 = q 2 = 1.60 × 10−19 C
−31
me = 9.109 × 10
kg
Felectric = Fg
kC q1q2

= me g
r2
r=
(8.99 × 109 N • m2/C2)(1.60 × 10−19 C)2

= 5.07 m
(9.109 × 10−31 kg)(9.81 m/s2)
kC q1q2
 =
me g
Section Review, p. 642
r = 12 cm
q2 = −3.5 mC
9
2 2
−6
−6
kC q1q2 (8.99 × 10 N • m /C )(2.0 × 10 C)(3.5 × 10 C)

a. F = 
= 
2
2
r
(0.12 m)
F = 4.4 N
2.0 × 10−6 C
q1

c. N = 
=
= 1.2 × 1013 electrons
1.60 × 10−19 C/electron 1.60 × 10−19 C/electron
3. q1 = 2.2 × 10−9 C at x = 1.5 m
q2 = 5.4 × 10−9 C at x = 2.0 m
q3 = 3.5 × 10−9 C at the
origin
9
2 2
−9
−9
kC q1q3 (8.99 × 10 N • m /C )(2.2 × 10 C)(3.5 × 10 C)


=
F1,3 = 
(r1,3)2
(1.5 m)2
F1,3 = 3.1 × 10−8 N
9
2 2
−9
−9
kC q2q3 (8.99 × 10 N • m /C )(5.4 × 10 C)(3.5 × 10 C)

F2,3 = 
= 
2
2
(r2,3)
(2.0 m)
F2,3 = 4.2 × 10−8 N
F3,tot = −(3.1 × 10−8 N) − (4.2 × 10−8 N) = −7.3 × 10−8 N
F3,tot = 7.3 × 10−8 N, along the negative x-axis
I Ch. 17–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. q1 = 2.0 mC
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Givens
Solutions
4. q1 = −6.00 × 10−9 C
−9
q2 = −3.00 × 10
C
r1,2 = 60.0 cm
kC q1q3 kC q2q3

= 
(r1,3)2
(r2,3)2
q1
q2

= 
(r1,3)2 (r2,3)2
6.00 × 10−9 C
3.00 × 10−9 C



=
P2
(0.600 m − P)2
I
(3.00 × 10−9 C)(P 2) = (6.00 × 10−9 C)(0.600 m − P)2
P=
6.00 × 10−9 C

(0.600 m − P)
3.00 × 10−9 C
P = 0.849 m − (1.41)(P)
(2.41)(P) = 0.849 m
P = 0.352 m from q1 = 35.2 cm from q1
or (60.0 cm − 35.2 cm) = 24.8 cm from q2
Practice 17D, p. 647
1. q1 = 5.00 mC at the origin
q2 = −3.00 mC at
x = 0.800 m
For the point y = 0.500 m on the y-axis,
9
2 2
−6
k q1 (8.99 × 10 N • m /C )(5.00 × 10 C)

=
= 1.80 × 105 N/C
E1 = C
r12
(0.500 m)2
k q2 (8.99 × 109 N • m2/C2)(3.00 × 10−6 C)
E2 = C
= 
r22
(0.8
m
)2
+(0.
)2 2
00
50
0m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(8.99 × 109 N • m2/C2)(3.00 × 10−6 C)
E2 = 
0.
640
m2
+0
.2
50
m22
(8.99 × 109 N • m2/C2)(3.00 × 10−6 C)
= 3.03 × 104 N/C
E2 = 
0.
89
0
m22
0.800
q = tan−1  = 58.0°
0.500
Ey = (1.80 × 105 N/C) − (3.03 × 104 N/C)(cos 58.0°)
Ey = (1.80 × 105 N/C) − (1.61 × 104 N/C) = 1.64 × 105 N/C
Ex = (3.03 × 104 N/C)(sin 58.0°) = 2.57 × 104 N/C
2 (E )2 = (1.64 × 105 N/C)2 + (2.57 × 104 N/C)2
Etot = (E
y)+
x Etot = (2
N2/C
.6
9×1010N
2/C
2)+(6.
60
×108
2)
Etot = 2.
010N
76
×1
2/C
2 = 1.66 × 105 N/C
1.64 × 105
2.57 × 10
j = tan−1 4 = 81.1°
Etot = 1.66 × 105 N/C, 81.1° above the positive x-axis
Section One—Pupil’s Edition Solutions
I Ch. 17–5
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Givens
Solutions
2. r = 5.3 × 10−11 m
q = 1.60 × 10−19 C
kC q (8.99 × 109 N • m2/C2)(1.60 × 10−19 C)
= 
= 5.1 × 1011 N/C
E = 
r2
(5.3 × 10−11 m)2
E = 5.1 × 1011 N/C, away from the proton
I
3. E = 2.0 × 104 N/C, along the
positive x-axis
a. F = Eqe = (2.0 × 104 N/C)(1.60 × 10−19 C)
F = 3.2 × 10−15 N, along the negative x-axis
qe = qp = 1.60 × 10−19 C
b. F = Eqp = (2.0 × 104 N/C)(1.60 × 10−19 C)
F = 3.2 × 10−15 N, along the positive x-axis
Section Review, p. 652
1. q1 = 40.0 × 10−9 C
q2 = 60.0 × 10−9 C
30.0 cm
r =  = 15.0 cm
2
9
2 2
−9
k q1 (8.99 × 10 N • m /C )(40.0 × 10 C)
= 
= 1.60 × 104 N/C
E1 = C
2
2
r
(0.150 m)
9
2 2
−9
k q2 (8.99 × 10 N • m /C )(60.0 × 10 C)

=
= 2.40 × 104 N/C
E2 = C
r2
(0.150 m)2
Etot = E1 + E2 = (1.60 × 104 N/C) − (2.40 × 104 N/C) = −0.80 × 104 N/C
Etot = 8.0 × 103 N/C toward the 40.0 × 10−9 C charge
Chapter Review and Assess, pp. 654–659
18. q1 = q2 = (46)(1.60 × 10−19 C)
r = (2)(5.9 × 10−15 m)
3.5 × 10−6 C
q

N = 
=
= 2.2 × 1013 electrons
1.60 × 10−19 C/electron 1.60 × 10−19 C/electron
kC q1q2 (8.99 × 109 N • m2/C2)[(46)(1.60 × 10−19 C)]2
 = 
F=
r2
[(2)(5.9 × 10−15 m)]2
F = 3.5 × 103 N
19. q1 = 2.5 mC
q2 = −5.0 mC
r = 5.0 cm
9
2 2
−6
−6
kC q1q2 (8.99 × 10 N • m /C )(2.5 × 10 C)(5.0 × 10 C)

F=
= 
2
2
r
(0.050 m)
F = 45 N
q2 = 79e
9
2 2
−19
2
kC q1q2 (8.99 × 10 N • m /C )(2.0)(79)(1.60 × 10 C)


=
F=
r2
(2.0 × 10−14 m)2
r = 2.0 × 10−14 m
F = 91 N
20. q1 = 2.0e
e = 1.60 × 10−19 C
I Ch. 17–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. q = 3.5 mC
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Givens
Solutions
21. q1 = 3.0 nC
r1,2 = r2,3 = (1
m
)2
+(1.
)2 = 2.
.0
0m
0m
2 = 1.4 m
q3 = 2.0 nC
9
2 2
−9
−9
kC q1q2 (8.99 × 10 N • m /C )(3.0 × 10 C)(6.0 × 10 C)


F1,2 = 
=
(r1,2)2
(1.4 m)2
r1,2 = r2,3 =
F1,2 = 8.3 × 10−8 N
q2 = 6.0 nC
(1
m
)2
+(1.
)2
.0
0m
F2,3
9
2 2
−9
−9
kC q2q3 (8.99 × 10 N • m /C )(6.0 × 10 C)(2.0 × 10 C)


=
=
(r2,3)2
(1.4 m)2
I
F2,3 = 5.5 × 10−8 N
Fx = (8.3 × 10−8 N)(cos 45°) + (5.5 × 10−8 N)(cos 45°)
Fx = (5.9 × 10−8 N) + (3.9 × 10−8 N) = 9.8 × 10−8 N
Fy = −(8.3 × 10−8 N)(sin 45°) + (5.5 × 10−8 N)(sin 45°)
Fy = −(5.9 × 10−8 N) + (3.9 × 10−8 N) = −2.0 × 10−8 N
2 (F )2 = (9.8 × 10−8 N)2 + (2.0 × 10−8 N)2
Ftot = (F
x )+
y Ftot = (9
N2)
+(4.
0−16
N2) = 1.
N2
.6
×10−15
0×1
00
×10−14
Ftot = 1.00 × 10−7 N
2.0
q = tan−1  = 12°
9.8
Ftot = 1.00 × 10−7 N, 12° below the positive x-axis
22. q1 = q2 = 2.5 × 10−9 C
q3 = 3.0 × 10−9 C
r3,1 = r3,2 = (0
)2
+(0.
m
)2 = 0.86 m
.5
0m
70
r2,1 = 1.0 m
9
2 2
−9
−9
kCq3q1 (8.99 × 10 N • m /C )(3.0 × 10 C)(2.5 × 10 C)
2 = 
F3,1 = F3,2 = 
(r3,1)
(0.86 m)2
r3,1 = r3,2
F3,1 = F3,2 = 9.1 × 10−8 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fx = F3,1 cos q + F3,2 cos q
0.70 m
0.70 m
Fx = (9.1 × 10−8 N)  + (9.1 × 10−8 N) 
0.86 m
0.86 m
Fx = 7.4 × 10−8 N + 7.4 × 10−8 N = 14.8 × 10−8 N
Fy = F3,1 sin q + F3,2 sin q
0.50 m
−0.50 m
Fx = (9.1 × 10−8 N)  + (9.1 × 10−8 N)  = 0 N
0.86 m
0.86 m
Ftot
2
2
= (F
4.8×
)2 = 14.8 × 10−8 N
(F
10−8N
1)+
4) = (1
0N
=0
q = tan−1 
14.8 × 10−8 N
Ftot = 1.48 × 10−7 N along the +x-axis
Section One—Pupil’s Edition Solutions
I Ch. 17–7
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Givens
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23. q1 = −9.0 mC at y = 6.0 m
kC q1q3 kC q2q3

= 
(r1,3)2
(r2,3)2
q2 = −8.0 mC at y = −4.0 m
9.0 × 10−6 C
8.0 × 10−6 C



=
P2
(10.0 m − P )2
(8.0 × 10−6 C)(P 2) = (9.0 × 10−6 C)(10.0 m − P)2
I
P=
9.0 × 10− 6 C

(10.0 m −P)
8.0 × 10− 6 C
P = 11 m − (1.1)(P)
(2.1)(P) = 11 m
P = 5.2 m below q1, or y = 6.0 m − 5.2 m = 0.8 m
q3 is located at y = 0.8 m
24. q1 = 3.5 nC
q2 = 5.0 nC
r = 40.0 cm
q3 = −6.0 nC
kC q1q3 kC q2q3

= 
(r1,3)2
(r2,3)2
5.0 × 10−9 C
3.5 × 10−9 C

= 2
2
(0.400 m − P )
P
(5.0 × 10−9 C)(P 2) = (3.5 × 10−9 C)(0.400 m − P)2
P=
3.5 × 10− 9 C

(0.400 m − P)
5.0 × 10− 9 C
P = 0.33 m − (0.84)(P)
(1.84)(P) = 0.33 m
38. q1 = 30.0 × 10−9 C
q2 = 60.0 × 10−9 C
9
2 2
−9
k q1 (8.99 × 10 N • m /C )(30.0 × 10 C)

E1 = C
=
= 1.20 × 104 N/C
r2
(0.150 m)2
30.0 cm
r =  = 15.0 cm
2
9
2 2
−9
k q2 (8.99 × 10 N • m /C )(60.0 × 10 C)

E2 = C
=
= 2.40 × 104 N/C
r2
(0.150 m)2
Etot = (1.20 × 104 N/C) − (2.40 × 104 N/C) = −12.0 × 103 N/C
Etot = 12.0 × 103 N/C toward the 30.0 × 10−9 C charge
I Ch. 17–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P = 0.18 m = 18 cm from q1
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39. q1 = 5.7 mC at x = −3.0 m
For E at y = 2.0 m on the y-axis,
q2 = 2.0 mC at x = 1.0 m
r1 = (2
m
)2
+(3.
)2 = 4.
m2 = 13
m2 = 3.61 m
.0
0m
0m
2+9.0
.0
r2 = (2
m
)2
+(1.
)2 = 4.
m2 = 5.
.0
0m
0m
2+1.0
0m
2 = 2.2 m
9
2 2
−6
k q1 (8.99 × 10 N • m /C )(5.7 × 10 C)

E1 = C
=
= 3.9 × 103 N/C
r12
(3.61 m)2
I
9
2 2
−6
k q2 (8.99 × 10 N • m /C )(2.0 × 10 C)

E2 = C
=
= 3.7 × 103 N/C
r22
(2.2 m)2
2.0
q = tan  = 63°
1.0 2.0
q1 = tan−1  = 34°
3.0
−1
2
Ex = (3.9 × 103 N/C)(cos 34°) − (3.7 × 103 N/C)(cos 63°)
Ex = (3.2 × 103 N/C) − (1.7 × 103 N/C) = 1.5 × 103 N/C
Ey = (3.9 × 103 N/C)(sin 34°) + (3.7 × 103 N/C)(sin 63°)
Ey = (2.2 × 103 N/C) + (3.3 × 103 N/C) = 5.5 × 103 N/C
2 (E )2 = (1.5 × 103 N/C)2 + (5.5 × 103 N/C)2
Etot = (E
x )+
y Etot = (2
06
N2/C
07
N2/C
07
N2/C
.2
×1
2)+(3.
0×1
2) = (3
.2
×1
2) = 5.7 × 103 N/C
5.5
q = tan−1  = 75°
1.5
Etot = 5.7 × 103 N/C, 75° above the positive x-axis
Copyright © by Holt, Rinehart and Winston. All rights reserved.
40. q1 = (7.0 × 1013 protons)(e)
Qnet = q1 + q2 = [(7.0 × 1013) − (4.0 × 1013)](e) = (3.0 × 1013)(e)
q2 = (4.0 × 1013 electrons)(e)
Qnet = (3.0 × 1013)(1.60 × 10−19 C)
e = 1.60 × 10−19 C
Qnet = 4.8 × 10−6 C
41. a = 6.3 × 103 m/s2
me = 9.109 × 10−31 kg
−19
q = 1.60 × 10
C
42. 1.00 g of Cu has 9.48 × 1021
atoms.
1 Cu atom has 29 electrons.
a. F = me a = (9.109 × 10−31 kg)(6.3 × 103 m/s2) = 5.7 × 10−27 N
F = 5.7 × 10−27 N, in a direction opposite E
F 5.7 × 10−27 N
= 3.6 × 10−8 N/C
b. E =  = 
q 1.60 × 10−19 C
a. 1.00 g of Cu has (9.48 × 1021 atoms)(29 electrons/atom) = 2.75 × 1023 electrons
b. qtot = (2.75 × 1023 electrons)(1.60 × 10−19 C/electron) = 4.40 × 104 C
Section One—Pupil’s Edition Solutions
I Ch. 17–9
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Givens
Solutions
43. q1 = 6.0 mC
a. E at 1.0 cm left of q2 = E1 + E2 + E3
q2 = 1.5 mC
r1 = r1,2 − 1.0 cm = 3.0 cm − 1.0 cm = 2.0 cm
q3 = −2.0 mC
r2 = 1.0 cm
r1,2 = 3.0 cm
r3 = r2,3 + 1.0 cm = 2.0 cm + 1.0 cm = 3.0 cm
r2,3 = 2.0 cm
9
2 2
−6
k q1 (8.99 × 10 N • m /C )(6.0 × 10 C)

E1 = C
=
= 1.3 × 108 N/C
r12
(0.020 m)2
9
2 2
−6
k q2 (8.99 × 10 N • m /C )(1.5 × 10 C)

E2 = C
=
= 1.3 × 108 N/C
r22
(0.010 m)2
9
2 2
−6
k q3 (8.99 × 10 N • m /C )(2.0 × 10 C)

E3 = C
=
= 2.0 × 107 N/C
r32
(0.030 m)2
Etot = (1.3 × 108 N/C) − (1.3 × 108 N/C) + (2.0 × 107 N/C)
Etot = 2.0 × 107 N/C along the positive x-axis
q4 = −2.0 mC
b. F = q4 E = (2.0 × 10−6 C)(2.0 × 107 N/C) = 4.0 × 101 N
44. q1 = 5.0 nC
9
2 2
−9
−9
kC q1q2 (8.99 × 10 N • m /C )(5.0 × 10 C)(6.0 × 10 C)

a. F1,2 = 
= 
2
2
(r1,2)
(0.30 m)
q2 = 6.0 nC
q3 = −3.0 nC
F1,2 = 3.0 × 10−6 N
r1,2 = 0.30 m
9
2 2
−9
−9
kC q1q3 (8.99 × 10 N • m /C )(5.0 × 10 C)(3.0 × 10 C)

F1,3 = 
= 
2
2
(r1,3)
(0.10 m)
r1,3 = 0.10 m
F1,3 = 1.3 × 10−5 N
2 (F )2 = (3.0 × 10−6 N)2 + (1.3 × 10−5 N)2
(F
1,
1,
2)+
3 = (9
N2)
+(1.
N2) = 1.
N2 = 1.3 × 10−5 N
.0
×10−12
7×10−10
8×10−10
F1,tot
13
q = tan−1  = 77°
3.0
F1,tot = 1.3 × 10−5 N, 77° below the negative x-axis
F 1.3 × 10−5 N
b. E =  = 
= 2.6 × 103 N/C, 77° below the negative x-axis
q1 5.0 × 10−9 C
45. q1 = (6.02 × 1023)(e)
q2 = (6.02 × 1023)(e)
r = (2)(6.38 × 106 m)
−19
e = 1.60 × 10
9
2 2
23
−19
2
kC q1q2 (8.99 × 10 N • m /C )[(6.02 × 10 )(1.60 × 10 C)]



=
F=
r2
[(2)(6.38 × 106 m)]2
F = 5.12 × 105 N
C
47. m1 = 7.36 × 1022 kg
m2 = 5.98 × 1024 kg
Fg = Felectric
Gm1m2 kC q2

= 
r2
r2
q=
q = 5.72 × 1013 C
I Ch. 17–10
(6.673 × 10−11N • m2/kg2)(7.36 × 1022 kg)(5.98 × 1024 kg)

8.99 × 109 N • m2/C2
Gm1m2
 =
kC
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
F1,tot =
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48. m1 = m2 = 0.20 g
ΣFy = 0 N, so Fg = FT,y = FT (cos 5.0°)
q = 5.0°
ΣFx = 0 N, so Felectric = FT,x = FT (sin 5.0°)
L = 30.0 cm
Felectric FT (sin 5.0°)
  =  = tan 5.0°
Fg
FT (cos 5.0°)
I
kC q 2

 = tan 5.0°
r 2mg
r = (2)(0.300 m)(sin 5.0°)
[(2)(0.300 m)(sin 5.0°)] (0.20 × 10 kg)(9.81 m/s )(tan 5.0°)

q = 8.99 × 10 N m /C
q=
r 2mg(tan 5.0°)

kC
−3
2
9
•
2
2
2
q = 7.2 × 10−9 C
49. me = 9.109 × 10−31 kg
mp = 1.673 × 10−27 kg
q = 1.60 × 10−19 C
a. F = Eq = mg
−31
2
m g (9.109 × 10 kg)(9.81 m/s )
= 5.58 × 10−11 N/C
Ee = e = 
−19
q
1.60 × 10 C
Ee = 5.58 × 10−11 N/C, downward
mp g (1.673 × 10−27 kg)(9.81 m/s2)
b. Ep =  = 
= 1.03 × 10−7 N/C
q
1.60 × 10−19 C
Ep = 1.03 × 10−7 N/C upward
50. q1 = 3.0 nC
q2 = 6.0 nC
r1,3 = r2,4 = 0.60 m
(8.99 × 109 N • m2/C2)(6.0 × 10−9 C)
k q2

E2 = C
=
= 150 N/C
(r2,4 )2
(0.60 m)2
r2,3 = r1,4 = 0.20 m
r3,4 = (0
)2
+(0.
m
)2 =
.2
0m
60
q3 = 5.0 nC
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(8.99 × 109 N • m2/C2)(3.0 × 10−9 C)
k q1

=
= 670 N/C
E1 = C
(r1,4)2
(0.20 m)2
0.
04
0m
2+0.3
6m
2
r3,4 = 0.
m2 = 0.63 m
40
(8.99 × 109 N • m2/C2)(5.0 × 10−9 C)
k q3

E3 = C
=
= 110 N/C
(0.63 m)2
(r3,4)2
0.20
q = tan−1  = 18°
0.60
Ex = −(150 N/C) − (110 N/C)(cos 18°) = −150 N/C − (1.0 × 102 N/C)
Ex = −250 N/C
Ey = (670 N/C) + (110 N/C)(sin 18°) = 670 N/C + 34 N/C
Ey = 7.0 × 102 N/C
2 (E )2 = (250 N/C)2 + (7.0 × 102 N/C)2
Etot = (E
x )+
y Etot = (6
04
N2/C
05
N2/C
05
N2/C
.2
×1
2)+(4.
9×1
2) = 5.
5×1
2 = 7.4 × 102 N/C
7.0
j = tan−1  = (7.0 × 101)°
2.5
Etot = 7.4 × 102 N/C, (7.0 × 101)° above the negative x-axis
Section One—Pupil’s Edition Solutions
I Ch. 17–11
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51. E = 520 N/C
∆t = 48 ns
vi = 0 m/s
I
me = 9.109 × 10−31 kg
mp = 1.673 × 10−27 kg
q = 1.60 × 10−19 C
F qE
a =  = 
m m
qE
vf = a∆t =  ∆t
m
For the electron,
−19
−9
qE∆t (1.60 ×10 C)(520 N/C)(48 × 10 s)
vf,e =  = 
= 4.4 × 106 m/s
−31
me
9.109 × 10 kg
For the proton,
−19
−9
qE∆t (1.60 ×10 C)(520 N/C)(48 × 10 s)
vf,p =  = 
= 2.4 × 103 m/s
−27
mp
1.673 × 10 kg
52. q = 2.0 × 10−4 C
a. Because the dome is a closed conducting surface,
E = 0.0 N/C inside the dome.
r = 1.0 m
9
2
−4
k q (8.99 × 10 N • m2/C )(2.0 × 10 C)

=
= 1.8 × 106 N/C
b. E = C
r2
(1.0 m)2
(8.99 × 109 N • m2/C2)(2.0 × 10−4 C)
k q
= 1.1 × 105 N/C
c. E = C2 = 
(4r)
(4.0 m)2
r = 2.0 m
54. E = 3.0 × 104 N/C
q = 1.60 × 10−19 C
mp = 1.673 × 10−27 kg
55. E = 3.4 × 105 N/C
6
2
Er 2 (3.0 × 10 N/C)(2.0 m)
q =  = 
= 1.3 × 10−3 C
kC
(8.99 × 109 N • m2/C2)
a. F = qE = (1.60 × 10−19 C)(3.0 × 104 N/C) = 4.8 × 10−15 N
4.8 × 10−15 N
F
= 2.9 × 1012 m/s2
b. a =  = 
mp 1.673 × 10−27 kg
F = qE = (1.60 × 10−19 C)(3.4 × 105 N/C) = 5.4 × 10−14 N
q = 1.60 × 10−19 C
56. q = 5.0 mC
r = 2.0 m
9
2
−6
k q (8.99 × 10 N • m2/C )(5.0 × 10 C)
E1 = E2 = E3 = C
= 
= 1.1 × 104 N/C
2
2
r
(2.0 m)
Ex = (1.1 × 104 N/C)(sin 60°) − (1.1 × 104 N)(sin 60°) = 0.0 N/C
Ey = (1.1 × 104 N/C) − (1.1 × 104 N/C)(cos 60°) − (1.1 × 104 N/C)(cos 60°)
Ey = (1.1 × 104 N/C) − (5.5 × 103 N/C) − (5.5 × 103 N/C) = 0.0 N/C
N/C
)2
+(0.
)2 = 0.0 N/C
Etot = (0
.0
0N
/C
I Ch. 17–12
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
53. E = 3.0 × 106 N/C
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57. q = 24 mC
Felectric = Fg
E = 610 N/C
qE = mg
−6
qE (24 × 10 C)(610 N/C)
m =  = 
= 1.5 × 10−3 kg
g
9.81 m/s2
58. E = 640 N/C
vi = 0 m/s
vf = 1.20 × 106 m/s
mp = 1.673 × 10−27 kg
q = 1.60 × 10−19 C
I
−19
F
qE (1.60 × 10 C)(640 N/C)
a. a =  =  = 
= 6.1 × 1010 m/s2
mp mp
1.673 × 10−27 kg
1.20 × 106 m/s
vf
= 2.0 × 10−5 s
b. ∆t =  = 
a 6.1 × 1010 m/s2
c. ∆x = 2a∆t 2 = (0.5)(6.1 × 1010 m/s2)(2.0 × 10−5 s)2
1
∆x = 12 m
d. KEf = 2mp vf 2 = (0.5)(1.673 × 10−27 kg)(1.20 × 106 m/s)2
1
KEf = 1.20 × 10−15 J
ΣFx = 0 N = Felectric − FT,x
59. m = 0.10 kg
L = 30.0 cm
FT,x = Felectric = FT(sin 45°)
q = 45°
ΣFy = 0 N = FT,y − Fg
FT,y = Fg = FT(cos 45°)
Felectric FT (sin 45°)
  =  = tan 45°
Fg
FT (cos 45°)
4kC q 2 + kC q 2
kC q 2
kC q 2
5kC q 2
2 + 
2 = 



Felectric = 
=
4L2(sin2q )
(L sin q )
(2L sin q)
4L2(sin2q)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fg = mg
Felectric
5kC q 2
 = 
 = tan 45°
Fg
4L2(sin2q )mg
5kC q 2 = 4L 2(sin2q)mg(tan 45°)
m
g(t5akn
45°)
(0.10 kg)(9.81 m/s )(tan 45°)

q = (2)(0.300 m)(sin 45°)
(5)(8.99 × 10 N m /C )
q=
4L2(sin2q)mg(tan 45°)
 = 2L(sin q)
5kC
C
2
9
•
2
2
q = 2.0 × 10−6 C
60. m = 2.0 g
b. FT, y = Fg = mg
L = 20.0 cm
4
E = 1.0 × 10 N/C
q = 15°
F ,y
mg
FT = T
= 
cos 15° cos 15°
mg (sin 15°)
qE = FT,x = FT (sin 15°) =  = mg(tan 15°)
cos 15°
−3
2
mg(tan 15°) (2.0 × 10 kg)(9.81 m/s )(tan 15°)
q =  = 
= 5.3 × 10−7 C
E
1.0 × 104 N/C
Section One—Pupil’s Edition Solutions
I Ch. 17–13
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Givens
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+y
−q
61.
−q
I
−q
E2
E3
72°
72°
36°
36°
72° E1
72°
E4
E5
−q
+x
Because each charge is the same size and all are the same distance from the center,
k q
E1 = E2 = E3 = E4 = E5 = C
r2
E1,y = 0 N/C
E5,y = −E2,y = −E(sin 72°)
E4,y = −E3,y = −E(sin 36°)
Ey = E1,y + E2,y + E3,y + E4,y + E5,y
Ey = 0 N/C + E(sin 72°) + E(sin 36°) − E(sin 36°) − E(sin 72°) = 0 N/C
−q
E1,x = E
E2,x = E5,x = E(cos 72°)
E3,x = E4,x = −E(cos 36°)
Ex = E1,x + E2,x + E3,x + E4,x + E5,x
Ex = E + E(cos 72°) − E(cos 36°) − E(cos 36°) + E(cos 72°)
Ex = E + 2E(cos 72°) − 2E(cos 36°) = E(1 + 0.62 − 1.62)
Ex = 0 N/C
2 (E )2 = (0 N/C)2 + (0 N/C)2 = 0 N/C
E = (E
x )+
y −19
6
F
qE (1.60 × 10 C)(3.0 × 10 N/C)
= 5.3 × 1017 m/s2
a. a =  =  = 
−31
me me
9.109 × 10 kg
62. E = 3.0 × 106 N/C
q = 1.60 × 10−19 C
me = 9.109 × 10−31 kg
b. vf 2 = 2a∆x
8
vf = (0.100)(3.00 × 10 m/s)
mp = 1.673 × 10−27 kg
63. r1 = 2.17 mm
q1 = 1.60 × 10−19 C
q2 = −1.60 × 10−19 C
r2 = (0.0100)(2.17 mm)
vf 2 [(0.100)(3.00 × 108 m/s)]2
∆x =  = 
= 8.5 × 10−4 m
2a
(2)(5.3 × 1017 m/s2)
−19
6
F
qE (1.60 × 10 C)(3.0 × 10 N/C)
c. a =  =  = 
= 2.9 × 1014 m/s2
−27
mp mp
1.673 × 10 kg
Felectric = Felastic
kC q1q2

= kr2
r12
9
2
−19
2
kC q1q 2 (8.99 × 10 N • m2/C )(1.60 × 10 C)


k=
=
r12r2
(2.17 × 10−6 m)3(0.0100)
k = 2.25 × 10−9 N/m
64. E = 370.0 N/C
∆t = 1.00 ms
me = 9.109 × 10−31 kg
mp = 1.673 × 10−27 kg
q = 1.60 × 10−19 C
F
qE (1.60 × 10−19 C)(370.0 N/C)
ae =  =  = 
= 6.50 × 1013 m/s2
me me
9.109 × 10−31 kg
∆xe = 2 ae ∆t 2 = (0.5)(6.50 × 1013 m/s2)(1.00 × 10−6 s)2 = 32.5 m
−19
F
qE (1.60 × 10 C)(370.0 N/C)
= 3.54 × 1010 m/s2
ap =  =  = 
mp mp
1.673 × 10−27 kg
1
∆xp = 2 ap ∆t 2 = (0.5)(3.54 × 1010 m/s2)(1.00 × 10−6 s)2
1
∆xp = 1.77 × 10−2 m
∆xtot = ∆xe + ∆xp = 32.5 m + (1.77 × 10−2 m) = 32.5 m
I Ch. 17–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vi = 0 m/s
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Givens
Solutions
65. E = 300.0 N/C
−19
F
qE (1.60 × 10 C)(300.0 N/C)
a. a =  =  = 
= 5.27 × 1013 m/s2
me me
9.109 × 10−31 kg
me = 9.109 × 10−31 kg
q = 1.60 × 10−19 C
b. vf = a∆t = (5.27 × 1013 m/s2)(1.00 × 10−8 s) = 5.27 × 105 m/s
∆t = 1.00 × 10−8 s
66. E = 2.0 × 103 N/C along the
positive x-axis
q = 1.60 × 10−19 C
a. F = qE = (1.60 × 10−19 C)(2.0 × 103 N/C) = 3.2 × 10−16 N
F = 3.2 × 10−16 N, along the positive x-axis
mp = 1.673 × 10−27 kg
F
3.2 × 10−16 N
= 1.9 × 1011 m/s2
b. a =  = 
mp 1.673 × 10−27 kg
vf = 1.00 × 106 m/s
vf 1.00 × 106 m/s
= 5.3 × 10−6 s
c. ∆t =  = 
a 1.9 × 1011 m/s2
67. vf,1 = (0.010)(3.00 × 108 m/s)
∆x1 = 2.0 mm
q = 1.60 × 10−19 C
me = 9.109 × 10−31 kg
I
vf,12
a. a = 
2∆x1
2
m a me vf ,1
E = e = 
2∆x1q
q
(9.109 × 10−31 kg)[(0.010)(3.00 × 108 m/s)]2
E = 
(2)(2.0 × 10−3 m)(1.60 × 10−19 C)
E = 1.3 × 104 N/C
vf,12 [(0.010)(3.00 × 108 m/s)]2
= 2.2 × 1015 m/s2
b. a =  = 
2∆x1
(2)(2.0 × 10−3 m)
∆x2 = 4.0 mm
vf,22 = 2a∆x2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vf,2 = 2a
015m
∆
x2 = (2
)(
2.
2×1
/s
2)(4
.0
×10−3m
) = 4.2 × 106 m/s
68. KE = 3.25 × 10−15 J
1
KE = 2 mp vi2
∆x = 1.25 m
vi =
vf = 0 m/s
mp = 1.673 × 10
2KE
p
q = 1.60 × 10−19 C
−27
m
2
kg
vf − vi 2
a = 
2∆x
F = qE = mpa
m m v − 2KE
m a (m )(v
E =  =  =  = 
f
2
− vi2)
p
p
q
(q)(2∆x)
−27
mp vf 2 − mp
2KE
p
p f
2
(q)(2∆x)
(q )(2∆x)
2
2
−15
(1.673 × 10 kg)(0 m/s) − (2)(3.25 × 10 J)
E = 
= −1.62 × 104 N/C
(1.60 × 10−19 C)(2)(1.25 m)
E = 1.62 × 104 N/C opposite the proton’s velocity
Section One—Pupil’s Edition Solutions
I Ch. 17–15
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Electrical Energy and Capacitance
Chapter 18
I
Practice 18A, p. 669
Givens
Solutions
1. PEelectric = 6.32 × 10−19 J
q1 = q2 = 2qp + 2qn = (2)1.60 × 10−19 C + (2)(0) = 3.20 × 10−19 C
kc q1q2
(8.99 × 109 N• m2/C2)(3.20 × 10−19 C)2
 = 
r= 
PEe lectric
6.32 × 10−19 J
r = 1.46 × 10−9 m
2. q1 = 6.4 µC = 6.4 × 10−6 C
q2 = −3.2 µC = −3.2 × 10−6 C
(8.99 × 109 N• m2/C2)(6.4 × 10−6 C)(−3.2 × 10−6 C)
kc q1q2
 = 
r= 
−4.1 × 10−2 J
PEe lectric
PEelectric = −4.1 × 10−2 J
r = 4.5 m
q1 = Nqe = (1013)(−1.60 × 10−19 C) = −1.60 × 10−6 C
3. N = 103
qe = −1.60 × 10−9
−2
PEelectric = −7.2 × 10
J
q2 = −Nqe = (−1013)(−1.60 × 10−19 C) = 1.60 × 10−6 C
(8.99 × 109 N• m2/C2)(−1.60 × 10−6 C)(1.60 × 10−6 C)
kc q1q2
 = 
r= 
−7.2 × 10−2 J
PEe lectric
r = 0.32 m
4. d = 2.0 cm = 2.0 × 10−2 m
E = 215 N/C
(−6.9 × 10−19 J)
∆PE
q = −  = − 
= 1.60 × 10−19 C
(215 N/C)(2.0 × 10−2 m)
Ed
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆PEelectric = −6.9 × 10−19 J
Section Review, p. 669
5. E = 250 N/C, in the positive
x direction
q1 = 12 mC
q1 moves from the origin to
(20.0 cm, 50.0 cm).
6. q = 35 C
d = 2.0 km
The displacement in the direction of the field (d) is 20.0 cm.
∆PE = −qEd = −(12 × 10−6 C)(250 N/C)(20.0 × 10−2 m)
∆PE = −6.0 × 10−4 J
∆PE = −qEd = −(35 C)(1.0 × 106 N/C)(2.0 × 103 m) = −7.0 × 1010 J
E = 1.0 × 106 N/C
Section One—Pupil’s Edition Solutions
I Ch. 18–1
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Practice 18B, p. 673
Givens
Solutions
1. r = 1.0 cm
q = 1.60 × 10−19 C
I
2. q1 = 5.0 nC
q2 = −3.0 nC
r = 35.0 cm
k q (8.99 × 109 N • m2/C2)(1.60 × 10−19 C)
= 1.4 × 10−7 V
∆V = C = 
r
1.0 × 10−2 m
k q
(8.99 × 109 N • m2/C2)(5.0 × 10−9 C)
∆V1 = C1 =  = 260 V
r1
(0.350 m/2)
k q
(8.99 × 109 N • m2/C2)(−3.0 × 10−9 C)
∆V2 = C2 =  = −150 V
r2
(0.350 m/2)
∆Vtot = ∆V1 + ∆V2 = 260 V − 150 V = 110 V
3. q1 = 5.0 mC
q2 = 3.0 mC
q3 = 3.0 mC
q4 = −5.0 mC
Each charge is at the corner
of a 2.0 m × 2.0 m square.
2
diagonal
(2
.0
m
)2
+(
2.0
m
)2
4.0
m
+4
.0
m2
r =  =  = 
2
2
2
2
8.
0m
r =  = 1.4 m
2
k q
(8.99 × 109 N • m2/C2)(5.0 × 10−6 C)
∆V1 = C1 =  = 3.2 × 104 V
r
1.4 m
k q
(8.99 × 109 N • m2/C2)(3.0 × 10−6 C)
∆V2 = C2 =  = 1.9 × 104 V
r
1.4 m
k q
(8.99 × 109 N • m2/C2)(3.0 × 10−6 C)
∆V3 = C3 =  = 1.9 × 104 V
r
1.4 m
k q
(8.99 × 109 N • m2/C2)(−5.0 × 10−6 C)
∆V4 = C4 =  = −3.2 × 104 V
r
1.4 m
∆Vtot = ∆V1 + ∆V2 + ∆V3 + ∆V4
∆Vtot = (3.2 × 104 V) + (1.9 × 104 V) + (1.9 × 104 V) + (−3.2 × 104 V)
∆Vtot = 3.8 × 104 V
1. ∆d = 0.060 cm
E = 3.0 × 106 V/m
5. E = 8.0 × 104 V/m
∆d = 0.50 m
q = 1.60 × 10−19 C
6. E = 1.0 × 106 V/m
∆d = 1.60 km
I Ch. 18–2
∆V = −E∆d = −(3.0 × 106 V/m)(0.060 × 10−2 m) = −1.8 × 103 V
∆V = 1.8 × 103 V
a. ∆V = −E ∆d = −(8.0 × 104 V/m)(0.50 m) = −4.0 × 104 V
b. ∆PE = −qEd = −(1.60 × 10−19 C)(8.0 × 104 V/m)(0.50 m)
∆PE = −6.4 × 10−15 J
∆V = −E∆d = −(1.0 × 106 V/m)(1.60 × 103 m) = −1.6 × 109 V
∆V = 1.6 × 109 V
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section Review, p. 675
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Practice 18C, p. 681
Givens
1. C = 4.00 mF
Solutions
a. Q = C∆V1 = (4.00 × 10−6 F)(12.0 V) = 4.80 × 10−5 C
∆V1 = 12.0 V
∆V2 = 1.50 V
b. PE = 2C(∆V2)2 = (0.5)(4.00 × 10−6 F)(1.50 V)2 = 4.50 × 10−6 J
1
∆V1 = 1.25 V
Q
6.0 × 10−6 C
a. C =  =  = 4.8 × 10−6 F
∆V1
1.25 V
∆V2 = 1.50 V
b. PE = 2C(∆V2 )2 = (0.5)(4.8 × 10−6 F)(1.50 V)2 = 5.4 × 10−6 J
2. Q = 6.0 mC
1
Q = 18.0 pC
Q 18.0 × 10−12 C
= 9.00 V
a. ∆V1 =  = 
C 2.00 × 10−12 F
∆V2 = 2.5 V
b. Q = C∆V2 = (2.00 × 10−12 F)(2.5 V) = 5.0 × 10−12 C
3. C = 2.00 pF
4. C = 1.00 F
d = 1.00 mm
I
Cd (1.00 F)(1.00 × 10−3 m)
A =  = 
= 1.13 × 108 m2
8.85 × 10−12 C2/N • m2
e0
Section Review, p. 681
d = 2.0 mm
e A (8.85 × 10−12 C2/N • m2)(2.0 × 10−4 m2)
a. C = 0 = 
= 8.8 × 10−13 F
d
2.0 × 10−3 m
∆V = 6.0 V
b. Q = C∆V = (8.8 × 10−13 F)(6.0 V) = 5.3 × 10−12 C
2. A = 2.0 cm2
3. C = 1.35 pF
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆V = 12.0 V
4. d = 800.0 m
A = 1.00 × 106 m2
E = 2.0 × 106 N/C
PE = 2C(∆V)2 = (0.5)(1.35 × 10−12 F)(12.0 V)2
1
PE = 9.72 × 10−11 J
e A (8.85 × 10−12 C2/N • m2)(1.00 × 106 m2)
a. C = 0 =  = 1.11 × 10−8 F
d
800.0 m
b. ∆V = −E∆d
Q = C∆V = C(−E∆d) = (1.11 × 10−8 F)(−2.0 × 106 N/C)(800.0 m) = −18 C
Q = ±18 C
Chapter Review and Assess, pp. 683–687
4. q1 = 9.00 × 10−9 C at the
origin
q2 = 3.00 × 10−9 C
∆PE = 8.09 × 10−2 J
∆PE = PEf − PEi
PEi = 0 J because ri = ∞
kCq1q2 (8.99 × 109 N • m2/C2)(9.00 × 10−9 C)(3.00 × 10−9 C)
 = 
rf = 
= 0.300 m = 30.0 cm
PEf
8.09 × 10−7 J
Section One—Pupil’s Edition Solutions
I Ch. 18–3
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Givens
Solutions
5. ri = 55 cm
∆PE = 2.1 × 10−28 J
q1 = 1.60 × 10−19 C
I
q2 = −1.60 × 10−19 C
kC q1q2 (8.99 × 109 N • m2/C2)(1.60 × 10−19 C)(−1.60 × 10−19 C)
 = 
PEi = 
ri
(55 × 10−2 m)
PEi = −4.2 × 10−28 J
PE f = ∆PE + PEi = (2.1 × 10−28 J) + (−4.2 × 10−28 J) = −2.1 × 10−28 J
kCq1q2 (8.99 × 109 N • m2/C2)(1.60 × 10−19 C)(−1.60 × 10−19 C)
 = 
rf = 
PEf
−2.1 × 10−28 J
rf = 1.1 m
12. E = 1.7 × 106 N/C
∆d = 1.5 cm
13. F = 4.30 × 10−2 N
q = 56.0 mC
∆V = −E∆d = −(1.7 × 106 N/C)(1.5 × 10−2 m) = −2.6 × 104 V
∆V = 2.6 × 104 V
−F∆d −(4.30 × 10−2 N)(0.200 m)
∆V = −E∆d =  = 
= −154 V
q
56.0 × 10−6 C
∆d = 20.0 cm
14. q1 = +8.0 mC
q2 = −8.0 mC
q3 = −12 mC
r1,P = 0.35 m
r2,P = 0.20 m
k q
(8.99 × 109 N • m2/C2)(8.0 × 10−6 C)
V1 = C1 =  = 2.1 × 105 V
r1,P
0.35 m
k q
(8.99 × 109 N • m2/C2)(−8.0 × 10−6 C)
V2 = C2 =  = −3.6 × 105 V
r2,P
0.20 m
(r1,P )2 + (r2,P )2 = (r3,P )2
2 (r )2 =
r3,P = (r
1,
2,P
P )+
(0
m
)2
.3
5m
)2+(0.
20
(8.99 × 109 N • m2/C2)(−12 × 10−6 C)
V3 = 
0.
m2+
m2
12
0.0
40
(8.99 × 109 N • m2/C2)(−12 × 10−6 C)
= −2.7 × 105 V
V3 = 
2
0.
16
m
Vtot = V1 + V2 + V3 = (2.1 × 105 V) + (−3.6 × 105 V) + (−2.7 × 105 V)
Vtot = −4.2 × 105 V
26. ∆V = 12.0 V
C = 6.0 pF
27. C = 0.20 mF
Q = C∆V = (6.0 × 10−12 F)(12.0 V) = 7.2 × 10−11 C
Q = ±7.2 × 10−11 C
a. Q = C∆V = (0.20 × 10−6 F)(6500 V) = 1.3 × 10−3 C
∆V = 6500 V
b. PE = 2C(∆V )2 = (0.5)(0.20 × 10−6 F)(6500 V)2 = 4.2 J
1
I Ch. 18–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
k q
(8.99 × 109 N • m2/C2)(−12 × 10−6 C)
V3 = C3 = 
r3,P
(0
)2
+(0.
m
)2
.3
5m
20
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Givens
Solutions
28. C1 = 25 mF
PE1 = 2C 1(∆V )2 = (0.5)(25 × 10−6 F)(120 V)2 = 0.18 J
C2 = 5.0 mF
∆V = 120 V
29. ∆V = 600.0 V
E = 200.0 N/C
1
PE2 = 2C 2(∆V )2 = (0.5)(5.0 × 10−6 F)(120 V)2 = 3.6 × 10−2 J
1
PEtot = PE1 + PE2 = 0.18 J + (3.6 × 10−2 J) = 0.22 J
I
kC q

∆V
r
 =  = r
kC q
E

r2
600.0 V
∆V
r =  =  = 3.000 m
200.0 N/C
E
∆Vr
(600.0 V)(3.000 m)
q =  = 
= 2.00 × 10−7 C
kC
8.99 × 109 N • m2/C2
30. d = 3.0 mm
E = 3.0 × 106 N/C
Q = −1.0 mC
Q
e A Q
C = 0 =  = 
d
∆V −E ∆d
−e0 AE∆d = Qd
−Q
A = 
e0 E
A = pr 2
r=
−6
−(−1.0 × 10 C)

Ap = e−EQp = (8.85 × 10 C /N m )(3.0 × 10 N/C)(p)
0
−12
2
•
2
6
r = 0.11 m
31. q1 = 8.0 mC
q2 = 2.0 mC
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q3 = 4.0 mC
r1,2 = 3.0 cm
)2
+(6.
)2
r1,3 = (3
.0
cm
0cm
kC q1q2 (8.99 × 109 N • m2/C2)(8.0 × 10−6 C)(2.0 × 10−6 C)
 = 
PE1,2 = 
= 4.8 J
r1,2
3.0 × 10−2 m
kC q1q3 (8.99 × 109 N • m2/C2)(8.0 × 10−6 C)(4.0 × 10−6 C)
 = 
PE1,3 = 
r1,3
(0
m)2
+(0.
.0
30
06
0m
)2
(8.99 × 109 N • m2/C2)(8.0 × 10−6 C)(4.0 × 10−6 C)
PE1,3 = 
(9
.0
×1
0−4
m2)
+(
3.6
×10−3
m2)
(8.99 × 109 N • m2/C2)(8.0 × 10−6 C)(4.0 × 10−6 C)
= 4.3 J
PE1,3 = 
4.5
×10−3
m2
PE1,tot = PE1,2 + PE1,3 = 4.8 J + 4.3 J = 9.1 J
32. ∆V = 12 V
∆d = 0.30 cm
∆V
12 V
= 4.0 × 103 V/m
E =  = 
∆d 0.30 × 10−2 m
Section One—Pupil’s Edition Solutions
I Ch. 18–5
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Givens
Solutions
33. A = 5.00 cm2
Q
a. ∆V = 
C
e0 A
C = 
d
Qd
Q
(400.0 × 10−12 C)(1.00 × 10−3 m)
∆V =  =  = 
= 90.4 V
ε0 A
e0 A (8.85 × 10−12 C2/N • m2)(5.00 × 10−4 m2)

d
d = 1.00 mm
Q = 400.0 pC
I
90.4 V
∆V
b. E =  = 
= 9.04 × 104 V/m
∆d 1.00 × 10−3 m
e A (8.85 × 10−12 C2/N • m2)(175 × 10−4 m2)
= 3.87 × 10−9 F
a. C = 0 = 
d
(0.0400 × 10−3 m)
34. A = 175 cm2
d = 0.0400 mm
Q 500.0 × 10−12 C
= 0.129 V
b. ∆V =  = 
C
3.87 × 10−9 F
Q = 500.0 pC
35. KE = 1.00 × 10−19 J
me = 9.109 × 10−31 kg
mp = 1.673 × 10−27 kg
36. ∆V = 25 700 V
(2)(1.00 × 10−19 J)

= 4.69 × 105 m/s
9.109 × 10−31 kg
a. ve =
b. vp =
2KE
 =
me
2KE
 =
mp
(2)(1.00 × 10−19 J)

= 1.09 × 104 m/s
1.673 × 10−27 kg
a. KEf = ∆PE = q∆V = (1.60 × 10−19 C)(25 700 V) = 4.11 × 10−15 J
vi = 0 m/s
q = 1.60 × 10−19 C
mp = 1.673 × 10−27 kg
b. vf =
(2)(4.11 × 10−15 J)

= 2.22 × 106 m/s
1.673 × 10−27 kg
2KEf
 =
mp
1
m v 2
2 p f
vi = 0 m/s
−27
mp = 1.673 × 10
−19
q = 1.60 × 10
kg
C
38. ∆d = 5.33 mm
vf =
= q∆V
−19
(2)(1.60 × 10 C)(120 V)
 = 1.5 × 10 m/s
2qm∆V = 1.673 × 10 kg
p
−27
5
∆V = 600.0 V
600.0 V
∆V
a. E =  = 
= 1.13 × 105 V/m
∆d 5.33 × 10−3 m
q = −1.60 × 10−19 C
b. F = qE = (−1.60 × 10−19 C)(1.13 × 105 V/m) = −1.81 × 10−14 N
F = 1.81 × 10−14 N
∆d = (5.33 mm − 2.90 mm)
= 2.43 mm
I Ch. 18–6
c. ∆PE = −qEd = −(−1.60 × 10−19 C)(1.13 × 105 V/m)(2.43 × 10−3 m)
∆PE = 4.39 × 10−17 J
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
KEf = ∆PE = q∆V
37. ∆V = 120 V
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Givens
Solutions
39. q1 = 5.0 × 10−9 C
r12 + (0.010 m)2 = (0.040 m)2
q2 = −5.0 × 10−9 C
q3 = −5.0 × 10−9 C
r1,2 = r1,3 = 4.0 cm
r2,3 = 2.0 cm
r1 = (0
m)2
−(0.
.0
40
01
0m
)2
(8.99 × 109 N • m2/C2)(5.0 × 10−9 C)
k q
V1 = C1 = 
r1
(0.0
m
)2
−(0.
)2
40
01
0m
I
(8.99 × 109 N • m2/C2)(5.0 × 10−9 C)
V1 = 
(1
.6
×10−3m
2)−(1.
0×10−4m
2)
(8.99 × 109 N • m2/C2)(5.0 × 10−9 C)
V1 = 
= 1200 V
1.
5
×10−3
m2
0.020 m
r2 = r3 =  = 0.010 m
2
kC q2 (8.99 × 109 N • m2/C2)(−5.0 × 10−9 C)
V2 =  =  = −4500 V
r2
0.010 m
k q
(8.99 × 109 N • m2/C2)(−5.0 × 10−9 C)
V3 = C3 =  = −4500 V
r3
0.010 m
Vtot = V1 + V2 + V3 = (1200 V) + (−4500 V) + (−4500 V) = −7800 V
40. q1 = −3.00 × 10−9 C at the
origin
−9
q2 = 8.00 × 10 C at
x = 2.00 m, y = 0.00 m
For the location between the two charges,
Vtot = V1 + V2 = 0 V
k q
V1 = C1
P
V1 = −V2
k q2
V2 = C
(2.00 m − P)
kC q1
−k q2
 = C
P
(2.00 m − P)
−Pq 2 = (2.00 m − P)(q1)
−Pq 2 = (2.00 m)(q1) − Pq1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P(q1 − q2 ) = (2.00 m)(q1)
(2.00 m)(q1)
(2.00 m)(−3.00 × 10−9 C)
P = 
= 
= 0.545 m
q1 − q 2
(−3.00 × 10−9 C) − (8.00 × 10−9 C)
P is 0.545 m to the right of the origin, at x = 0.545 m .
For the location to the left of the y-axis,
k q
V1 = C1
P
k q2
V2 = C
(2.00 m + P)
−k q2
kC q1
 = C
P
(2.00 m + P)
−Pq 2 = (2.00 m + P)(q1)
−Pq 2 = (2.00 m)(q1) + Pq1
P(q1 + q2) = −(2.00 m)(q1)
−(2.00 m)(q1)
−(2.00 m)(−3.00 × 10−9 C)
P = 
= 
= 1.20 m
q1 + q 2
(−3.00 × 10−9 C) + (8.00 × 10−9 C)
P is 1.20 m to the left of the origin, at x = −1.20 m .
Section One—Pupil’s Edition Solutions
I Ch. 18–7
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Givens
Solutions
41. d = 5.0 cm
F qE q∆V
a. a =  =  = 
m m
md
∆V = 550 V
1
∆x = 2a∆t 2
vi,e = 0 m/s
I
vi,p = 0 m/s
−27
mp = 1.673 × 10
−31
me = 9.109 × 10
md
2∆ax = 2∆
qx∆
V
2∆x m d
∆t = 
q∆
V
e
∆te =
kg
kg
q = 1.60 × 10−19 C
e e
p p
p
∆te = ∆tp
md

= 
2∆
qx∆
V q∆V 2∆xpmpd
e e
∆xeme = ∆xpmp
∆xe + ∆xp = d
∆xe = d − ∆xp
(d − ∆xp)me = ∆xpmp
dme − ∆xpme = ∆xpmp
dme = ∆xpmp + ∆xpme
dme
∆xp = 
mp + me
∆te = ∆tp =
2∆xpmpd
 =
q∆V
dme
mpd
2 
mp + me
 =
q∆V
2d 2memp

(mp + me)q∆V
mp + me = (1.673 × 10−27 kg) + (9.109 × 10−31 kg) = 1.674 × 10−27 kg
∆te = ∆tp =
−2
−31
−27
(2)(5.0 × 10 m) (9.109 × 10 kg)(1.673 × 10 kg)

(1.674 × 10 kg)(1.60 × 10 C)(550 V)
2
−27
−19
q∆V
(1.60 × 10−19 C)(550 V)(7.2 × 10−9 s)
= 1.4 × 107 m/s
b. ve = ae ∆te =  ∆te = 
me d
(9.109 × 10−31 kg)(5.0 × 10−2 m)
q∆V
(1.60 × 10 C)(550 V)(7.2 × 10 s)
v = a ∆t =  ∆t =  = 7.6 × 10 m/s
m d (1.673 × 10 kg)(5.0 × 10 m)
−19
p
p
p
p
p
c. ∆tp,tot =
−9
−27
−2

q∆V
2∆xtot mpd
∆xtot = d
∆tp,tot =
2mpd 2
 =
q∆V
−27
−19
∆tp,tot = 3.1 × 10−7 s
42. ∆V = 60.0 V
∆PE = 1.92 × 10−17 J
I Ch. 18–8
∆PE 1.92 × 10−17 J
q =  =  = 3.20 × 10−19 C
∆V
60.0 V
Holt Physics Solution Manual
−2
(2)(1.673 × 10 kg)(5.0 × 10 m)

(1.60 × 10 C)(550 V)
2
3
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆te = ∆tp = 7.2 × 10−9 s
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Givens
Solutions
43. ∆V = 100.0 V
Q 400.0 × 10−6 C
C =  =  = 4.000 × 10−6 F
∆V
100.0 V
Q = 400.0 mC
44. ∆V = 4.5 × 106 V
a. KEf = ∆PE = ∆Vq = (4.5 × 106 V)(1.60 × 10−19 C) = 7.2 × 10−13 J
vi = 0 m/s
I
q = 1.60 × 10−19 C
mp = 1.673 × 10−27 kg
b. vf =
−13
(2)(7.2 × 10 J)
m = 
1.673×1
0kg = 2.9 × 10 m/s
2KEf
p
KE = ∆PE
45. vf,positron = 9.0 × 107 m/s
−31
mpositron = 9.109 × 10
−19
q = 1.60 × 10
7
−27
kg
C
mproton = 1.673 × 10−27 kg
1
mv 2
2
= ∆Vq
mpositron(vf,positron)2 (9.109 × 10−31 kg)(9.0 × 107 m/s)2
∆V =  = 
2q
(2)(1.60 × 10−19 C)
∆V = 2.3 × 104 V
vf,proton =
−19
(2)(2.3 × 10 V)(1.60 × 10 C)


m2∆Vq = 1.673 × 10 kg
4
−27
proton
vf,proton = 2.1 × 106 m/s
46. vf = (0.600)(3.00 × 108 m/s)
−31
me = mp = 9.109 × 10
qe = −1.60 × 10−19 C
qp = 1.60 × 10−19 C
kg
a. ∆PE = KEf
1
∆Vq = 2mvf 2
me vf 2 (9.109 × 10−31 kg)[(0.600)(3.00 × 108 m/s)]2
∆Ve =  = 
(2)(−1.60 × 10−19 C)
2qe
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆Ve = −9.22 × 104 V
mp vf 2 (9.109 × 10−31 kg)[(0.600)(3.00 × 108 m/s)]2
b. ∆Vp =  = 
2qp
(2)(1.60 × 10−19 C)
∆Vp = 9.22 × 104 V
a. KEf = ∆PE
47. ∆V = 2200 V
−19
q = 1.60 × 10
1
mv 2
f
2
C
−31
me = 9.109 × 10
−27
mp = 1.673 × 10
kg
kg
vf,e =
= ∆Vq
−19
(2)(2200 V)(1.60 × 10 C)
 = 2.8 × 10 m/s
2∆mVq = 9.109 × 10 kg
e
b. vf,p =
(2)(2200 V)(1.60 × 10−19 C)
Q = 1.75 × 10−8 C
d = 6.50 × 10−4 m
7
 = 6.5 × 10 m/s
2∆mVq = 1.673 × 10 kg
p
48. C = 3750 pF
−31
−27
5
Q
1.75 × 10−8 C
a. ∆V =  = 
= 4.67 V
C 3750 × 10−12 F
∆V
4.67 V
= 7180 V/m
b. E =  = 
d
6.50 × 10−4 m
Section One—Pupil’s Edition Solutions
I Ch. 18–9
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Givens
Solutions
2.50 × 10−3 m
49. r = 
2
d = 1.40 × 10−4 m
e A e0p r 2 (8.85 × 10−12 C2/N • m2)(p)(2.50 × 10−3 m/2)2
 = 
a. C = 0 = 
d
d
1.40 × 10−4 m
C = 3.10 × 10−13 F
I
∆V1 = 0.12 V
b. Q = C∆V1 = (3.10 × 10−13 F)(0.12 V) = 3.7 × 10−14 C
c. PEelectric = 2Q∆V1 = (0.5)(3.7 × 10−14 C)(0.12 V) = 2.2 × 10−15 J
1
∆d1 = 1.40 × 10−4 m
d. ∆V1 = E∆d1
∆V
0.12 V
E = 1 = 
= 860 V/m
∆d1 1.40 × 10−4 m
1.40 × 10−4 m
∆d2 = 1.10 × 10−4 m −  = 4.00 × 10−5 m
2
∆V2 = E∆d2 = (860 V/m)(4.00 × 10−5 m) = 3.4 × 10−2 V
Q2 = (0.707)Q
Q
e. ∆V3 = 2
C
Because the capacitance has not changed, ∆V3 = (0.707)(∆V1).
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆V3 = (0.707)(∆V1) = (0.707)(0.12 V) = 8.5 × 10−2 V
I Ch. 18–10
Holt Physics Solution Manual
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Current and Resistance
Chapter 19
I
Practice 19A, p. 695
Givens
Solutions
1. I = 5.00 × 10−3 A
∆Q = 2.00 C
2. I = 60.0 × 10−6 A
N = 3.75 × 1014 electrons
∆Q
I = 
∆t
∆Q
2.00 C
∆t =  =  = 4.00 × 102 s
I
5.00 × 10−3 A
N(1.60 × 10−19 C/electron)
∆t = 
I
(3.75 × 1014 electrons)(1.60 × 10−19 C/electron)
∆t = 
6.00 × 10−5 A
∆t = 1.00 s
3. I = 8.00 × 10−2 A
20
N = 3.00 × 10 electrons
N(1.60 × 10−19 C/electron)
∆t = 
I
(3.00 × 1020 electrons)(1.60 × 10−19 C/electron)
∆t = 
8.00 × 10−2 A
∆t = 6.00 s × 102 s
4. I = 40.0 A
∆Q = I∆t = (40.0 A)(0.50 s) = 2.0 × 101 C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 0.50 s
5. ∆Q1 = 9.0 mC
∆t1 = 3.5 s
∆Q
9.0 × 10−3 C
a. I = 1 =  = 2.6 × 10−3 A
∆t1
3.5 s
∆t2 = 10.0 s
I∆t2
(2.6 × 10−3 A)(10.0 s)
b. N = 
= 
= 1.6 × 1017 electrons
−19
1.60 × 10 C/electron 1.60 × 10−19 C/electron
∆Q2 = 2∆Q1
∆Q
2∆Q
2(9.0 × 10−3 C)
c. I = 2 = 1 =  = 5.1 × 10−3 A
∆t1
∆t1
3.5 s
Section Review, p. 699
2. ∆t1 = 5.00 s
∆Q = 3.0 C
∆t2 = 1.0 min
∆Q 3.0 C
a. I =  =  = 0.60 A
∆t1 5.00 s
I∆t2
(0.60 A)(1.0 min)(60 s/min)
b. N = 
= 
−19
1.60 × 10 C/electron
1.60 × 10−19 C/electrons
N = 2.2 × 1020 electrons
Section One—Pupil’s Edition Solutions
I Ch. 19–1
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Practice 19B, p. 703
Givens
Solutions
1. ∆V = 1.5 V
R = 3.5 Ω
I
2. ∆V = 120 V
R = 65 Ω
3. ∆V = 120 V
R1 = 48 Ω
R2 = 20.0 Ω
4. I = 6.25 A
∆V 1.5 V
I =  =  = 0.43 A
R
3.5 Ω
∆V 120 V
I =  =  = 1.8 A
R
65 Ω
∆V 120 V
a. I1 =  =  = 2.5 A
R1
48 Ω
∆V 120 V
b. I2 =  =  = 6.0 A
R2 20.0 Ω
∆V = IR = (6.25 A)(17.6 Ω) = 1.10 × 102 V
R = 17.6 Ω
5. I = 2.5 A
∆V = 115 V
6. I1 = 0.50 A
∆V1 = 110 V
∆V 115 V
R =  =  = 46 Ω
I
2.5 A
∆V
110 V
a. R = 1 =  = 220 Ω
I1
0.50 A
∆V
90.0 V
I2 = 2 =  = 0.41 A
R
220 Ω
∆V2 = 90.0 V
∆V3 = 130 V
∆V
130 V
b. I3 = 3 =  = 0.59 A
R
220 Ω
1. R = 10.2 Ω
∆V = 120 V
2. I = 2.5 A
∆V = 9.0 V
7. R1 = 75 Ω
∆V = 115 V
R2 = 47 Ω
∆V 120 V
I =  =  = 12 A
R
10.2 Ω
∆V 9.0 V
R =  =  = 3.6 Ω
I
2.5 A
∆V 115 V
I1 =  =  = 1.5 A
R1
75 Ω
∆V 115 V
I2 =  =  = 2.4 A
R2
47 Ω
Practice 19C, p. 710
1. P = 1050 W
∆V = 120 V
(∆V )2
P = 
R
(∆V)2 (120 V)2
R =  =  = 14 Ω
P
1050 W
I Ch. 19–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section Review, p. 707
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Givens
2. P = 0.25 W
∆V = 120 V
3. P = 0.10 W
∆V = 1.50 V
4. ∆V = 50.0 V
R = 8.00 Ω
Solutions
(∆V)2 (120 V)2
R =  =  = 5.8 × 104 Ω
P
0.25 W
(∆V)2 (1.50 V)2
R =  =  = 22 Ω
P
0.10 W
I
∆V 50.0 V
I =  =  = 6.25 A
R
8.00 Ω
(∆V )2 (50.0 V)2
P =  =  = 312 W
R
8.00 Ω
Practice 19D, p. 712
1. cost of energy =
$0.080/kW • h
a. cost = (P∆t)(cost/kW • h)
cost = (75.0 × 10−3 kW)(24 h)($0.080/kW • h) = $0.14
∆t = 24 h
∆V = 115 V
P = 75.0 W
I = 20.0 A
b. P = I∆V = (20.0 A)(115 V) = 2.30 × 103 W
cost = (2.30 kW)(24 h)($0.080/kW • h) = $4.40
R = 60.0 Ω
(∆V )2 (115 V)2
c. P =  =  = 2.20 × 102 W
R
60.0 Ω
cost = (0.220 kW)(24 h)($0.080/kW • h) = $0.42
2. P1 = 75.0 W
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 24 h
a. E1 = P1∆t = (75.0 W)(24 h)(3600 s/h) = 6.5 × 106 J
b. P = I∆V = (20.0 A)(115 V) = 2.3 × 103 W
E2 = P2 ∆t = (2.30 × 103 W)(24 h)(3600 s/h) = 2.0 × 108 J
∆V 2 (115 V)2
c. P =  =  = 2.20 × 102 W
R
60.0 Ω
E3 = P3 ∆t = (2.20 × 102 W)(24 h)(3600 s/h) = 1.9 × 107 J
Section Review, p. 713
3. ∆V = 70 mV
P = I∆V = (200 × 10−6 A)(70 × 10−3 V) = 1 × 10−5 W
I = 200 mA
4. ∆t = 21 h
P = 90.0 W
cost = (P∆t)(cost/kW • h)
cost = (90.0 × 10−3 kW)(21 h)($0.070/kW • h) = $0.13
cost of energy =
$0.070/kW • h
Section One—Pupil’s Edition Solutions
I Ch. 19–3
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Chapter Review and Assess, pp. 717–721
Givens
Solutions
17. ∆Q = 10.0 C
∆Q 10.0 C
∆t =  =  = 2.0 s
I
5.0 A
I = 5.0 A
I
18. I = 9.1 A
∆Q = 1.9 × 103 C
∆Q 1.9 × 103 C
a. ∆t =  =  = 2.1 × 102 s = 3.5 min
I
9.1 A
∆Q
1.9 × 103 C

b. N = 
=
= 1.2 × 1022 electrons
1.60 × 10−19 C/electron 1.60 × 10−19 C/electron
∆Q 5.0 C
∆t =  =  = 1.0 s
I
5.0 A
19. I = 5.0 A
∆Q = 5.0 C
28. R = 15 Ω
∆V 3.0 V
I =  =  = 0.20 A
R
15 Ω
∆V = 3.0 V
29. R = 35 Ω
∆V 120 V
I =  =  = 3.4 A
R
35 Ω
∆V = 120 V
30. ∆V = 9.0 V
∆V 9.0 V
a. I1 =  =  = 1.8 A
R1 5.0 Ω
R1 = 5.0 Ω
R2 = 2.0 Ω
∆V 9.0 V
b. I2 =  =  = 4.5 A
R2 2.0 Ω
R3 = 20.0 Ω
∆V
9.0 V
c. I3 =  =  = 0.45 A
R3 20.0 Ω
34. P = 50.0 W
38. P/clock = 2.5 W
E = P∆t = (P/clock)N∆t
N = 2.5 × 10 clocks
E = (2.5 W)(2.5 × 108)(1.0 year)(365.25 d/year)(24 h/d)(3600 s/h)
∆t = 1.0 year
E = 2.0 × 1016 J
8
40. ∆V = 110 V
P = 130 W
I Ch. 19–4
(∆V)2 (110 V)2
R =  =  = 93 Ω
P
130 W
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
E = P∆t = (50.0 W)(1.00 s) = 50.0 J
∆t = 1.00 s
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Givens
Solutions
41. ∆V = 120 V
P
I =  =
∆V
P
R = 2 =
T
P = 75 W
42. P = 325 W
∆t = 3.0 h
75 W
 = 0.62 A
120 V
75 W
2 = 190 Ω
(0.62 A)
cost = (P∆t)(cost/kW • h)
I
cost = (325 × 10−3 kW)(3.0 h)($0.08/kW • h) = $0.08
cost of energy =
$0.08/kW • h
43. P = 75 W
∆t = 30 days
cost = (P∆t)(cost/kW • h)
cost = (75 × 10−3 kW)(30 days)(24 h/day)($0.15/kW • h) = $8.10
cost of energy =
$0.15/kW • h
44. ∆Q = 45 mC
∆t1 = 15 s
∆t2 = 1.0 min
∆Q 45 × 10−3 C
a. I =  =  = 3.0 × 10−3 A
∆t1
15 s
(3.0 × 10−3 A)(1.0 min)(60 s/min)
I∆t2
b. N = 
= 
−19
1.60 × 10 C/electron
1.60 × 10−19 C/electron
N = 1.1 × 1018 electrons
45. ∆V = 12 V
I = 0.40 A
46. I = 2.0 × 105 A
∆V 12 V
R =  =  = 3.0 × 101 Ω
I 0.40 A
∆Q = I∆t = (2.0 × 105 A)(0.50 s) = 1.0 × 105 C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 0.50 s
47. I = 80.0 mA
a. ∆V1 = IR1 = (80.0 × 10−6 A)(4.0 × 105 Ω) = 32 V
R1 = 4.0 × 105 Ω
R2 = 2.0 × 103 Ω
48. I = 7.0 A
b. ∆V2 = IR2 = (80.0 × 10−6 A)(2.0 × 103 Ω) = 0.16 V
P = I∆V = (7.0 A)(115 V) = 8.0 × 102 W
∆V = 115 V
49. P = 325 W
∆V = 120 V
50. ∆V = 4.0 MV
P
325 W
I =  =  = 2.7 A
∆V 120 V
P = I∆V = (25 × 10−3 A)(4.0 × 106 V) = 1.0 × 105 W
I = 25 mA
Section One—Pupil’s Edition Solutions
I Ch. 19–5
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Givens
Solutions
51. I = 6.0 A
a. P = I∆V = (6.0 A)(120 V) = 720 W
∆V = 120 V
∆t = 20.0 min
I
b. E = P∆t = (720 W)(20.0 min)(60 s/min) = 8.6 × 105 J
cost of energy =
$0.010/kW • h
c. cost = (P∆t)(cost/kW • h)
cost = (720 × 10−3 kW)(20.0 min)(1 h/60 min)($0.010/kW • h) = $0.0024
a. energy saved = (P2 − P1)(∆t) = (40.0 W − 11.0 W)(100.0 h)(3600 s/h)
52. P1 = 11.0 W
energy saved = (29.0 W)(100.0 h)(3600 s/h) = 1.04 × 107 J
P2 = 40.0 W
∆t = 100.0 h
b. money saved = (energy saved)(cost of energy)
cost of energy =
$0.080/kW • h
(1.04 × 107 J)($0.080/kW • h)
money saved = 
= $0.23
3.6 × 106 J/kW • h
53. E = 471 kW • h
a. E = (471 kW • h)(3.60 × 106 J/kW • h) = 1.70 × 109 J
E 471 kW• h
b.  =  = 14.3 kW • h/day
∆t
33 days
∆t = 33 days
E 1.70 × 109 J
 =  = 5.15 × 107 J/day
∆t
33 days
cost of energy = $0.15/kW • h
original total cost = $16.72
c. new total cost = (E)(cost/kW • h) = (471 kW • h)($0.15/kW • h) = $71
increase = new total cost − original total cost
54. matom = 3.27 × 10−25 kg
mtot = 1.25 kg
∆t = 2.78 h
−19
Qatom = 1.60 × 10
m ot
(Qatom)
∆Q = (number of atoms)(Q atom ) = t
matom
 (Q
)
∆Q m
I =  = 
mtot
C
atom
atom
∆t
∆t
 (1.60 × 10 C)

3.27 × 10 kg
I =  = 61.1 A
1.25 kg
−25
−19
(2.78 h)(3600 s/h)
55. P = 90.0 W
E = P∆t = (90.0 W)(1.0 h)(3600 s/h) = 3.2 × 105 J
∆V = 120 V
∆t = 1.0 h
56. I = 2.5 A
∆V = 120 V
E
E
3.2 × 105 J
∆t =  =  =  = 1.1 × 103 s
P I∆V (2.5 A)(120 V)
E = 3.2 × 105 J
or (1.1 × 103 s)(1 min/60 s) = 18 min
I Ch. 19–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
increase = $71 − $16.72 = $54
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Givens
Solutions
57. P = 80.0 W
Q Q∆V (90.0 A• h)(12.0 V)
∆t =  =  =  = 13.5 h
I
P
80.0 W
∆V = 12.0 V
Q = 90.0 A• h
58. ∆t = 5.0 s
a. ∆Q = I∆t = (2 A)(2 s) + (4 A)(1 s) + (6 A)(1 s) + (4 A)(1 s)
I
∆Q = 4 C + 4 C + 6 C + 4 C = 18 C
∆Q 18 C
b. I =  =  = 3.6 A
∆t 5.0 s
59. I = 50.0 A
∆V = IR = (50.0 A)(1.12 × 10−5 Ω/m)(4.0 × 10−2 m) = 2.2 × 10−5 V
R/d = 1.12 × 10−5 Ω/m
d = 4.0 cm
60. ∆V = 12 V
E = 2.0 × 107 J
P = 8.0 kW
v = 20.0 m/s
P
8.0 × 103 W
a. I =  =  = 670 A
∆V
12 V
E
b. ∆t = 
P
Copyright © by Holt, Rinehart and Winston. All rights reserved.
E
(20.0 m/s)(2.0 × 107 J)
∆x = v∆t = v  = 
= 5.0 × 104 m
P
8.0 × 103 W
Section One—Pupil’s Edition Solutions
I Ch. 19–7
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Circuits and Circuit Elements
Chapter 20
I
Practice 20A, p. 739
Givens
1. R1 = 6.75 Ω
R2 = 15.3 Ω
R3 = 21.6 Ω
∆V = 12.0 V
2. R1 = 4.0 Ω
R2 = 8.0 Ω
R3 = 12.0 Ω
∆V = 24.0 V
Solutions
a. Req = R1 + R2 + R3
Req = 6.75 Ω + 15.3 Ω + 21.6 Ω = 43.6 Ω
∆V 12.0 V
b. I =  =  = 0.275 A
Req 43.6 Ω
a. Req = R1 + R2 + R3 = 4.0 Ω + 8.0 Ω + 12.0 Ω = 24.0 Ω
∆V 24.0 V
b. I =  =  = 1.00 A
Req 24.0 Ω
c. I = 1.00 A
3. I = 0.50 A
R1 = 2.0 Ω
R2 = 4.0 Ω
R3 = 5.0 Ω
R4 = 7.0 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. ∆V = 9.00 V
R1 = 7.25 Ω
R2 = 4.03 Ω
∆V1 = IR1 = (0.50 A)(2.0 Ω) = 1.0 V
∆V2 = IR2 = (0.50 A)(4.0 Ω) = 2.0 V
∆V3 = IR3 = (0.50 A)(5.0 Ω) = 2.5 V
∆V4 = IR4 = (0.50 A)(7.0 Ω) = 3.5 V
a. Req = R1 + R2 = 7.25 Ω + 4.03 Ω = 11.28 Ω
∆V
9.00 V
I =  =  = 0.798 A
Req 11.28 Ω
b. ∆V1 = IR1 = (0.798 A)(7.25 Ω) = 5.79 V
∆V2 = IR2 = (0.798 A)(4.03 Ω) = 3.22 V
5. R1 = 7.0 Ω
∆V = 4.5 V
I = 0.60 A
∆V
Req = R1 + R2 = 
I
∆V
4.5 V
R2 =  − R1 =  − 7.0 Ω
I
0.60 A
R2 = 7.5 Ω − 7.0 Ω = 0.5 Ω
6. ∆V = 115 V
I = 1.70 A
∆V 115 V
a. Req =  =  = 67.6 Ω
I
1.70 A
R = 1.50 Ω
b. NR = Req
Req 67.6 Ω
N =  =  = 45 bulbs
R
1.50 Ω
Section One—Pupil’s Edition Solutions
I Ch. 20–1
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Practice 20B, p. 744
Givens
Solutions
1. ∆V = 9.0 V
R1 = 2.0 Ω
I
R2 = 4.0 Ω
R3 = 5.0 Ω
R4 = 7.0 Ω
∆V 9.0 V
I1 =  =  = 4.5 A
R1 2.0 Ω
∆V 9.0 V
I2 =  =  = 2.2 A
R2 4.0 Ω
∆V 9.0 V
I3 =  =  = 1.8 A
R3 5.0 Ω
∆V 9.0 V
I4 =  =  = 1.3 A
R4 7.0 Ω
2. Req = 2.00 Ω
−1
−1
1 1 1 1 1
Parallel: Req =  +  +  +  + 
R R R R R
5
= 
R
R = 5Req = 5(2.00 Ω) = 10.0 Ω
Series: Req = 5R = 5(10.0 Ω) = 50.0 Ω
3. R1 = 4.0 Ω
R2 = 8.0 Ω
R3 = 12.0 Ω
∆V = 24.0 V
−1
1
1
1
a. Req =  +  + 
R1 R2 R3
−1
1
1
1
=  +  + 
4.0 Ω 8.0 Ω 12.0 Ω
−1
1
1
1
Req = 0.25  + 0.12  + 0.0833 
Ω
Ω
Ω
−1
1
= 0.45 
Ω
Req = 2.2 Ω
∆V 24.0 V
b. I1 =  =  = 6.0 A
R1
4.0 Ω
∆V 24.0 V
I3 =  =  = 2.00 A
R3 12.0 Ω
4. R1 = 18.0 Ω
R2 = 9.00 Ω
R3 = 6.00 Ω
I2 = 4.00 A
−1
1
1
1
a. Req =  +  + 
R1 R2 R3
−1
1
1
1
Req = 0.0555  + 0.111  + 0.167 
Ω
Ω
Ω
Req = 2.99 Ω
b. ∆V = I2 R2 = (4.00 A)(9.00 Ω) = 36.0 V
∆V 36.0 V
c. I1 =  =  = 2.00 A
R1 18.0 Ω
∆V 36.0 V
I3 =  =  = 6.00 A
R3 6.00 Ω
I Ch. 20–2
Holt Physics Solution Manual
−1
1
1
1
=  +  + 
18.0 Ω 9.00 Ω 6.00 Ω
−1
1
= 0.334 
Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆V 24.0 V
I2 =  =  = 3.0 A
R2
8.0 Ω
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Section Review, p. 745
Givens
4. R1 = 2.0 Ω
R2 = 4.0 Ω
∆V = 12 V
Solutions
a. Req = R1 + R2 = 2.0 Ω + 4.0 Ω = 6.0 Ω
∆V 12 V
I1 = I2 = I =  =  = 2.0 A
Req 6.0 Ω
I
∆V1 = I1R1 = (2.0 A)(2.0 Ω) = 4.0 V
∆V2 = I2 R2 = (2.0 A)(4.0 Ω) = 8.0 V
R1 = 2.0 Ω
R2 = 4.0 Ω
∆V = 12 V
−1
1
1
b. Req =  + 
R1 R2
−1
1
1
=  + 
2.0 Ω 4.0 Ω
−1
−1
= 0.75 Ω
1
1
Req = 0.50  + 0.25 
Ω
Ω
1
= 1.3 Ω
∆V1 = ∆V2 = ∆V = 12 V
12 V
∆V
I1 = 1 =  = 6.0 A
2.0 Ω
R1
12 V
∆V
I2 = 2 =  = 3.0 A
4.0 Ω
R2
R1 = 4.0 Ω
c. Req = R1 + R2 = 4.0 Ω + 12.0 Ω = 16.0 Ω
R2 = 12.0 Ω
∆V
4.0 V
I1 = I2 = I =  =  = 0.25 A
Req 16.0 Ω
∆V = 4.0 V
∆V1 = I1R1 = (0.25 A)(4.0 Ω) = 1.0 V
∆V2 = I2 R2 = (0.25 A)(12.0 Ω) = 3.0 V
Copyright © by Holt, Rinehart and Winston. All rights reserved.
R1 = 4.0 Ω
R2 = 12.0 Ω
∆V = 4.0 V
−1
1
1
  + 
= 
4.0 Ω 12.0 Ω
−1
Req
−1
1
1
1
= 0.25  + 0.0833  = 0.33 
Ω
Ω
Ω
1
1
d. Req =  + 
R1 R2
−1
= 3.0 Ω
∆V1 = ∆V2 = ∆V = 4.0 V
4.0 V
∆V
I1 = 1 =  = 1.0 A
4.0 Ω
R1
4.0 V
∆V
I2 = 2 =  = 0.33 A
12.0 Ω
R2
Section One—Pupil’s Edition Solutions
I Ch. 20–3
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Givens
Solutions
5. R1 = 150 Ω
R2 = 180 Ω
∆V = 12 V
a. Req = R1 + R2 = 150 Ω + 180 Ω = 330 Ω
∆V
12 V
I1 = I2 = I =  =  = 0.036 A
Req 330 Ω
∆V1 = I1R1 = (0.036 A)(150 Ω) = 5.4 V
I
∆V2 = I2 R2 = (0.036 A)(180 Ω) = 6.5 V
R1 = 150 Ω
R2 = 180 Ω
∆V = 12 V
b. ∆V1 = ∆V2 = ∆V = 12 V
∆V
12 V
I1 = 1 =  = 0.080 A
R1
150 Ω
∆V
12 V
I2 = 2 =  = 0.067 A
R2
180 Ω
6. I = 0.20 A
∆V = 120.0 V
N = 35 bulbs
∆V 120.0 V
Req =  =  = 600 Ω
I
0.20 A
Req 600 Ω
R =  =  = 17 Ω
N
35
Practice 20C, p. 748
1. Ra = 25.0 Ω
Rb = 3.0 Ω
Rc = 40.0 Ω
−1
−1
1
1
1
= 0.33  + 0.0250  = 0.36 
Ω
Ω
Ω
1
1
a. Rbc =  + 
Rb Rc
1
1
=  + 
3.0 Ω 40.0 Ω
−1
Rbc
−1
Rbc = 2.8 Ω
Ra = 12.0 Ω
Rb = 35.0 Ω
Rc = 25.0 Ω
−1
−1
 + 
= 
35.0 Ω 25.0 Ω
1
1
1
= 0.0286  + 0.0400  = 0.0686 
Ω
Ω
Ω
1
1
b. Rbc =  + 
Rb Rc
1
1
−1
Rbc
Rbc = 14.6 Ω
Req = Ra + Rbc = 12.0 Ω + 14.6 Ω = 26.6 Ω
I Ch. 20–4
Holt Physics Solution Manual
−1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Req = Ra + Rbc = 25.0 Ω + 2.8 Ω = 27.8 Ω
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Givens
Ra = 15.0 Ω
Rb = 28.0 Ω
Rc = 12.0 Ω
Solutions
−1
−1
1
1
1
= 0.0357  + 0.0833  = 0.0119 
Ω
Ω
Ω
1
1
c. Rbc =  + 
Rb Rc
1
1
=  + 
28.0 Ω 12.0 Ω
−1
Rbc
−1
I
Rbc = 8.40 Ω
Req = Ra + Rbc = 15.0 Ω + 8.40 Ω = 23.4 Ω
2. Ra = 25.0 Ω
Rb = 3.0 Ω
Rc = 40.0 Ω
Rd = 15.0 Ω
Re = 18.0 Ω
−1
−1
1
1
1
= 0.0400  + 0.33  = 0.37 
Ω Ω Ω
1
1
a. Rab =  + 
Ra Rb
1
1
=  + 
25.0 Ω 3.0 Ω
−1
Rab
−1
Rab = 2.7 Ω
−1
−1
 + 
= 
15.0 Ω 18.0 Ω
1
1
1
= 0.0667  + 0.0556  = 0.1223 
Ω
Ω Ω
1
1
Rde =  + 
Rd Re
1
1
−1
Rde
−1
Rde = 8.177 Ω
Req = Rab + Rc + Rde = 2.7 Ω + 40.0 Ω + 8.177 Ω = 50.9 Ω
Ra = 12.0 Ω
Rb = 35.0 Ω
Rc = 25.0 Ω
Rd = 50.0 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Re = 45.0 Ω
−1
−1
1
1
1
= 0.0833  + 0.0286  = 0.1119 
Ω
Ω
Ω
1
1
b. Rab =  + 
Ra Rb
1
1
=  + 
12.0 Ω 35.0 Ω
−1
Rab
−1
Rab = 8.937 Ω
−1
1
1
−1
Rde
−1
 + 
= 
50.0 Ω 45.0 Ω
1
1
1
= 0.0200  + 0.0222  = 0.0422 
Ω
Ω Ω
1
1
Rde =  + 
Rd Re
−1
Rde = 23.7 Ω
Req = Rab + Rc + Rde = 8.937 Ω + 25.0 Ω + 23.7 Ω = 57.6 Ω
Section One—Pupil’s Edition Solutions
I Ch. 20–5
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Practice 20D, p. 751
I
Givens
Solutions
Ra = 5.0 Ω
Rab = Ra + Rb = 5.0 Ω + 7.0 Ω = 12.0 Ω
Rb = 7.0 Ω
Rc = 4.0 Ω
1
1
Rabc =  + 
Rab Rc
Rd = 4.0 Ω
Rabc
Re = 4.0 Ω
Rf = 2.0 Ω
−1
−1
 + 
= 
12.0 Ω 4.0 Ω
1
1
1
= 0.0833  + 0.25  = 0.33  = 3.0 Ω
Ω Ω Ω
1
1
1
1
=  +  =  + 
R R 4.0 Ω 4.0 Ω
1
1
1
= 0.25  + 0.25  = 0.50  = 2.0 Ω
Ω Ω Ω
1
1
−1
−1
Rde
d
∆V = 14.0 V
−1
−1
e
−1
Rde
−1
Req = Rabc + Rde + Rf = 3.0 Ω + 2.0 Ω + 2.0 Ω = 7.0 Ω
∆V 14.0 V
I =  =  = 2.0 A
Req 7.0 Ω
∆Vabc = IRabc = (2.0 A)(3.0 Ω) = 6.0 V
6.0 V
∆Vabc
Iab = 
=  = 0.50 A
12.0 Ω
Rab
Ra: Ia = Iab = 0.50 A
∆Va = IaRa = (0.50 A)(5.0 Ω) = 2.5 V
Rb : Ib = Iab = 0.50 A
∆Vb = Ib Rb = (0.50 A)(7.0 Ω) = 3.5 V
Rc : ∆Vc = ∆Vabc = 6.0 V
6.0 V
∆V
Ic = c =  = 1.5 A
4.0 Ω
Rc
Rd: ∆Vd = ∆Vde = 4.0 V
4.0 V
∆V
Id = d =  = 1.0 A
4.0 Ω
Rd
Re : ∆Ve = ∆Vde = 4.0 V
4.0 V
∆V
Ie = e =  = 1.0 A
4.0 Ω
Re
Rf : If = I = 2.0 A
∆Vf = If Rf = (2.0 A)(2.0 Ω) = 4.0 V
I Ch. 20–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆Vde = IRde = (2.0 A)(2.0 Ω) = 4.0 V
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Section Review, p. 752
Givens
1. R1 = 5.0 Ω
R2 = 5.0 Ω
R3 = 5.0 Ω
R4 = 5.0 Ω
R5 = 1.5 Ω
Solutions
R23 = R2 + R3 = 5.0 Ω + 5.0 Ω = 10.0 Ω
−1
−1
 + 
= 
10.0 Ω 5.0 Ω
1
1
1
= 0.100  + 0.20  = 0.30 
Ω Ω Ω
1
1
R234 =  + 
R23 R4
1
1
−1
R234
−1
I
= 3.3 Ω
Req = R1 + R234 + R5 = 5.0 Ω + 3.3 Ω + 1.5 Ω = 9.8 Ω
R1
18.0 V
R2
R4
R5
2. Req = 9.8 Ω
∆V = 18.0 V
3. I5 = 1.8 A
R3
∆V 18.0 V
I5 = I =  =  = 1.8 A
Req 9.8 Ω
∆V5 = I5 R5 = (1.8 A)(1.5 Ω) = 2.7 V
R5 = 1.5 Ω
5. ∆V = 120 V
RT = 16.9 Ω
RM = 8.0 Ω
RP = 10.0 Ω
RC = 0.01 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. ∆V = 120 V
−1
1
1
1
=  +  + 
16.9 Ω 8.0 Ω 10.0 Ω
−1
RTMP
Req = 3.6 Ω
RTMP = 3.6 Ω
∆V T = ∆V TMP = IRTMP = (33 A)(3.6 Ω) = 120 V
RT = 16.9 Ω
∆V
120 V
IT = T =  = 7.1 A
RT
16.9 Ω
∆V = 120.0 V
N = 35 bulbs
n = 3 strands
8. ∆V = 120.0 V
Req, strand = NR = (35)(15.0 Ω) = 525 Ω
−1
−1
Req
−1
= 525Ω + 525Ω + 525Ω
1
1
1
1
= 0.0019  + 0.0019  + 0.0019  = 0.0057  = 175 Ω
Ω
Ω
Ω
Ω
1
1
1
Req =  +  + 
Req, strand Req, strand Req, strand
1
1
1
−1
∆Vstrand = ∆V = 120.0 V
Req, strand = 520 Ω
∆V rand 120.0 V
Istrand = st
=  = 0.23 A
Req, strand
520 Ω
R = 15 Ω
∆V = IstrandR = (0.23 A)(15 Ω) = 3.4 V
Req = 170 Ω
−1
Req = RTMP + RC = 3.6 Ω + 0.01 Ω = 3.6 Ω
∆V 120 V
I =  =  = 33 A
Req 3.6 Ω
7. R = 15.0 Ω
−1
1
1
1
1
= 0.0592  + 0.12  + 0.100  = 0.28  = 3.6 Ω
Ω Ω Ω Ω
1
1
1
RTMP =  +  + 
RT RM RP
Section One—Pupil’s Edition Solutions
I Ch. 20–7
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Givens
Solutions
9. Req, strand = 520 Ω
R = 15 Ω
∆V rand 120.0 V
Istrand = st
=  = 0.23 A
Req, strand
520 Ω
n = 2 strands
∆V = IstrandR = (0.23 A)(15 Ω) = 3.4 V
∆V = 120.0 V
I
∆Vstrand = ∆V = 120.0 V
N = 35 bulbs
Chapter Review and Assess, pp. 754–759
16. R = 0.15 Ω
Req = 5R = 5(0.15 Ω) = 0.75 Ω
17. R1 = 4.0 Ω
a. Req = R1 + R2 + R3 = 4.0 Ω + 8.0 Ω + 12 Ω = 24 Ω
R2 = 8.0 Ω
R3 = 12 Ω
∆V = 24 V
18. R1 = 4.0 Ω
R2 = 8.0 Ω
R3 = 12 Ω
∆V = 24 V
∆V 24 V
b. I =  =  = 1.0 A
Req 24 Ω
−1
1
1
1
=  +  + 
4.0 Ω 8.0 Ω 12 Ω
−1
Req
−1
1
1
1
1
= 0.25  + 0.12  + 0.083  = 0.45  = 2.2 Ω
Ω Ω Ω Ω
1
1
1
a. Req =  +  + 
R1 R2 R3
−1
∆V 24 V
b. I =  =  = 11 A
Req 2.2 Ω
R2 = 9.00 Ω
R3 = 6.00 Ω
∆V = 12 V
−1
1
1
1
−1
Req
−1
 +  + 
= 
18.0 Ω 9.00 Ω 6.00 Ω
1
1
1
1
= 0.0556  + 0.111  + 0.167  = 0.334 
Ω Ω Ω Ω
1
1
1
a. Req =  +  + 
R1 R2 R3
−1
= 2.99 Ω
∆V
12 V
b. I =  =  = 4.0 A
Req 2.99 Ω
23. R1 = 12 Ω
R2 = 18 Ω
R3 = 9.0 Ω
R4 = 6.0 Ω
∆V = 30.0 V
24. R1 = 7.0 Ω
R2 = 7.0 Ω
R3 = 7.0 Ω
R4 = 7.0 Ω
R5 = 1.5 Ω
1
1
1
−1
R234
−1
−1
= 2.9 Ω
Req = R1 + R234 = 12 Ω + 2.9 Ω = 15 Ω
R34 = R3 + R4 = 7.0 Ω + 7.0 Ω = 14.0 Ω
−1
1
1
R234 =  + 
R2 R34
−1
1
1
=  + 
7.0 Ω 14.0 Ω
Holt Physics Solution Manual
−1
1
1
= 0.14  + 0.0714 
Ω
Ω
Req = R1 + R234 + R5 = 7.0 Ω + 4.8 Ω + 1.5 Ω = 13.3 Ω
∆V = 12.0 V
I Ch. 20–8
−1
= 18Ω + 9.0Ω + 6.0Ω
1
1
1
1
= 0.056  + 0.11  + 0.17  = 0.34 
Ω Ω Ω Ω
1
1
1
R234 =  +  + 
R2 R3 R4
= 4.8 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
19. R1 = 18.0 Ω
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Givens
Solutions
25. R1 = 6.0 Ω
Current:
R2 = 9.0 Ω
∆V12 = IR12 = (1.8 A)(3.6 Ω) = 6.5 V
R3 = 3.0 Ω
∆V12 6.5 V
I1 = 
=  = 1.1 A
6.0 Ω
R1
∆V = 12 V
I
∆V12 6.5 V
I2 = 
=  = 0.72 A
9.0 Ω
R2
−1
−1
1
1
1
= 0.17  + 0.11  = 0.28 
Ω Ω Ω
1
1
R12 =  + 
R1 R2
1
1
=  + 
6.0 Ω 9.0 Ω
−1
R12
−1
= 3.6 Ω
Req = R12 + R3 = 3.6 Ω + 3.0 Ω = 6.6 Ω
∆V 12 V
I =  =  = 1.8 A
Req 6.6 Ω
I3 = 1.8 A
Potential difference:
∆V1 = ∆V2 = ∆V12 = 6.5 V
∆V3 = I3R3 = (1.8 A)(3.0 Ω) = 5.4 V
26. R1 = 3.0 Ω
R2 = 3.0 Ω
−1
−1
1
1
=  + 
6.0 Ω 6.0 Ω
R234 = R2 + R34 = 3.0 Ω + 2.9 Ω = 5.9 Ω
R4 = 6.0 Ω
R5 = 4.0 Ω
1
1
R56 =  + 
R5 R6
R6 = 12.0 Ω
R56
∆V = 18.0 V
−1
−1
1
= 0.34 
Ω
= 2.9 Ω
−1
1
1
1
= 0.25  + 0.0833  = 0.33 
Ω
Ω
Ω
1
1
=  + 
4.0 Ω 12.0 Ω
−1
1
1
= 0.17  + 0.17 
Ω
Ω
R3 = 6.0 Ω
R7 = 2.0 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
1
a. R34 =  + 
R3 R4
−1
−1
= 3.0 Ω
R567 = R56 + R7 = 3.0 Ω + 2.0 Ω = 5.0 Ω
−1
1
1
=  + 
5.9 Ω 5.0 Ω
−1
R234567
−1
1
1
1
= 0.17  + 0.20  = 0.37  = 2.7 Ω
Ω Ω Ω
1
1
R234567 =  + 
R234 R567
−1
Req = R1 + R234567 = 3.0 Ω + 2.7 Ω = 5.7 Ω
∆V 18.0 V
I =  =  = 3.2 A
Req 5.7 Ω
∆V234567 = IR234567 = (3.2 A)(2.7 Ω) = 8.6 V
∆V234567 8.6 V
 =  = 1.7 A
I7 = I567 = 
R567
5.0 Ω
Section One—Pupil’s Edition Solutions
I Ch. 20–9
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Givens
Solutions
b. ∆V7 = I7R7 = (1.7 A)(2.0 Ω) = 3.4 V
c. ∆V56 = I567R56 = (1.7 A)(3.0 Ω) = 5.1 V
∆V6 = ∆V56 = 5.1 V
I
5.1 V
∆V
d. I6 = 6 =  = 0.42 A
12.0 Ω
R6
27. R1 = 8.0 Ω
R2 = 6.0 Ω
∆V2 = 12 V
28. R1 = 9.0 Ω
Req = R1 + R2 = 8.0 Ω + 6.0 Ω = 14.0 Ω
12 V
∆V
I2 = 2 =  = 2.0 A
6.0 Ω
R2
∆V = I2Req = (2.0 A)(14.0 Ω) = 28 V
∆V1 = I1R1 = (0.25 A)(9.0 Ω) = 2.2 V
R2 = 6.0 Ω
I1 = 0.25 A
29. R1 = 9.0 Ω
Req = R1 + R2 = 9.0 Ω + 6.0 Ω = 15.0 Ω
R2 = 6.0 Ω
∆V = I1Req = (0.25 A)(15.0 Ω) = 3.8 V
I1 = 0.25 A
R2 = 6.0 Ω
∆V
12 V
I = 2 =  = 2.0 A
R2
6.0 Ω
∆V2 = 12 V
Req = R1 + R2 = 9.0 Ω + 6.0 Ω = 15.0 Ω
30. R1 = 9.0 Ω
∆V = IReq = (2.0 A)(15.0 Ω) = 3.0 × 101 V
R2 = 9.00 Ω
R3 = 6.00 Ω
I2 = 4.00 A
a. Req = R1 + R2 + R3 = 18.0 Ω + 9.00 Ω + 6.00 Ω = 33.0 Ω
b. I = I2 = 4.00 A
∆V = IReq = (4.00 A)(33.0 Ω) = 132 V
c. I1 = I3 = I2 = 4.00 A
33. R1 = 90.0 Ω
R12 = R1 + R2 = 90.0 Ω + 10.0 Ω = 100.0 Ω
R2 = 10.0 Ω
R34 = R3 + R4 = 10.0 Ω + 90.0 Ω = 100.0 Ω
R3 = 10.0 Ω
R4 = 90.0 Ω
1
1
R1234 =  + 
R12 R34
Req = 60.0 Ω
R1234
−1
1
= 0.020000  = 50.00 Ω
Ω
−1
1
1
=  + 
100.0 Ω 100.0 Ω
−1
Req = R + R1234
R = Req − R1234 = 60.0 Ω − 50.00 Ω = 10.0 Ω
I Ch. 20–10 Holt Physics Solution Manual
−1
1
1
= 0.01000  + 0.01000 
Ω
Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
31. R1 = 18.0 Ω
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Givens
Solutions
34. Req = 150.0 Ω
N
Req, string = 
R
∆V = 120.0 V
N = 25
−1
−1
25
= 
R
R
= 
25
R
R 2R
Req =  +  =  = 150.0 Ω
25 25 25
I
25(150.0 Ω)
R =  = 1875 Ω
2
35. R = 6.0 Ω
The following equations represent the circuits as listed.
(a) Req = 2R = 2(6.0 Ω) = 12.0 Ω
−1
−1
2
(b) Req = 
R
−1
2
= 
6.0 Ω
−1
3
(c) Req = 
R
3
= 
6.0 Ω
−1
= 3.0 Ω
= 2.0 Ω
−1

= 6.0Ω + 
12.0 Ω
1
1
1
= 0.17  + 0.0833  = 0.25 
Ω
Ω Ω
1
1
(d) Req =  + 
R 2R
1
1
−1
Req
−1
2
(e) Req = 
R
36. ∆V = 9.0 V
R1 = 4.5 Ω
R2 = 3.0 Ω
−1
= 4.0 Ω
+ R = 3.0 Ω + 6.0 Ω = 9.0 Ω
−1
−1
−1
= 3.0Ω + 2.0Ω = 0.83 Ω
1
1
a. R23 =  + 
R2 R3
1
1
1
= 1.2 Ω
Req = R1 + R23 = 4.5 Ω + 1.2 Ω = 5.7 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
R3 = 2.0 Ω
∆V 9.0 V
b. I =  =  = 1.6 A
Req 5.7 Ω
c. I1 = I = 1.6 A
∆V23 = IR23 = (1.6 A)(1.2 Ω) = 1.9 V
∆V23 1.9 V
I2 = 
=  = 0.63 A
R2
3.0 Ω
∆V23 1.9 V
I3 = 
=  = 0.95 A
R3
2.0 Ω
d. ∆V1 = I1R1 = (1.6 A)(4.5 Ω) = 7.2 V
∆V2 = ∆V3 = ∆V23 = 1.9 V
Section One—Pupil’s Edition Solutions
I Ch. 20–11
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Givens
Solutions
37. R1 = 18.0 Ω
Req = R1 + R2 = 18.0 Ω + 6.0 Ω = 24.0 Ω
R2 = 6.0 Ω
∆V = 18.0 V
∆V 18.0 V
I1 = I2 = I =  =  = 0.750 A
Req 24.0 Ω
∆V1 = I1 R1 = (0.750 A)(18.0 Ω) = 13.5 V
I
∆V2 = I2 R2 = (0.750 A)(6.0 Ω) = 4.5 V
38. R1 = 30.0 Ω
R2 = 15.0 Ω
R3 = 5.00 Ω
∆V = 30.0 V
−1
−1
 + 
= 
30.0 Ω 15.0 Ω
1
1
1
= 0.0333  + 0.0667  = 0.1000 
Ω
Ω Ω
1
1
b. R12 =  + 
R1 R2
1
1
−1
R12
−1
= 10.00 Ω
Req = R12 + R3 = 10.00 Ω + 5.00 Ω = 15.00 Ω
∆V 30.0 V
c. I3 = I =  =  = 2.00 A
Req 15.00 Ω
∆V12 = IR12 = (2.00 A)(10.00 Ω) = 20.0 V
∆V12 20.0 V
I1 = 
=  = 0.667 A
R1
30.0 Ω
∆V12 20.0 V
I2 = 
=  = 1.33 A
R2
15.0 Ω
d. ∆V1 = ∆V2 = ∆V12 = 20.0 V
∆V3 = I3 R3 = (2.00 A)(5.00 Ω) = 10.0 A
39. R2 = 12 Ω
∆V = 12 V
I1 = 3.0 A
40. R1 = 18.0 Ω
−1
R2 = 6.0 Ω
1
1
Req =  + 
R1 R2
∆V = 18.0 V
Req = 4.3 Ω
∆V1 = ∆V2 = ∆V = 18.0 V
∆V
18.0 V
I1 = 1 =  = 1.00 A
R1
18.0 Ω
∆V
18.0 V
I2 = 2 =  = 3.0 A
R2
6.0 Ω
I Ch. 20–12
Holt Physics Solution Manual
−1
1
1
=  + 
18.0 Ω 6.0 Ω
−1
1
1
= 0.0556  + 0.17 
Ω
Ω
−1
1
= 0.23 
Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆V 12 V
R1 =  =  = 4.0 Ω
I1
3.0 A
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Givens
Solutions
41. R1 = 90.0 Ω
Switch open:
R2 = 10.0 Ω
R12 = R1 + R2 = 90.0 Ω + 10.0 Ω = 100.0 Ω
R3 = 10.0 Ω
R34 = R3 + R4 = 10.0 Ω + 90.0 Ω = 100.0 Ω
R4 = 90.0 Ω
Req = 2Req,S
−1
1
1
R1234 =  + 
R12 R34
−1
1
1
=  + 
100.0 Ω 100.0 Ω
−1
1
= 0.02000 
Ω
= 50.00 Ω
I
Req = R + R1234 = R + 50.00 Ω
Switch closed:
−1
−1
1
1
1
= 0.0111  + 0.100  = 0.111  = 9.01 Ω
Ω Ω Ω
1
1
1
1
=  +  =  +  = 9.01 Ω
R R 10.0 Ω 90.0 Ω
1
1
R13 =  + 
R1 R3
1
1
=  + 
90.0 Ω 10.0 Ω
−1
R13
−1
R24
2
−1
−1
4
Req,S = R + R13 + R24 = R + 9.01 Ω + 9.01 Ω = R + 18.02 Ω
Req = 2Req,S
R + 50.00 Ω = 2(R + 18.02 Ω) = 2R + 36.04 Ω
2R − R = 50.00 Ω − 36.04 Ω
R = 13.96 Ω
42. R = 20.0 Ω
a. Two resistors in series with two parallel resistors:
−1
2
Req = R + R + 
R
−1
2
= 20.0 Ω + 20.0 Ω + 
20.0 Ω
= 50.0 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
b. Four parallel resistors:
−1
4
Req = 
R
R 20.0 Ω
=  =  = 5.00 Ω
4
4
Section One—Pupil’s Edition Solutions
I Ch. 20–13
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Givens
Solutions
43. ∆V = 12.0 V
a. R12 = R1 + R2 = 30.0 Ω + 50.0 Ω = 80.0 Ω
R1 = 30.0 Ω
I
R3 = 90.0 Ω
R123
R4 = 20.0 Ω
−1
−1
1
1
1
= 0.0125  + 0.0111  = 0.0236 
Ω
Ω
Ω
R2 = 50.0 Ω
1
1
R123 =  + 
R12 R3
1
1
=  + 
80.0 Ω 90.0 Ω
−1
−1
= 42.4 Ω
Req = R123 + R4 = 42.4 Ω + 20.0 Ω = 62.4 Ω
∆V 12.0 V
b. I =  =  = 0.192 A
Req 62.4 Ω
c. ∆V123 = IR123 = (0.192 A)(42.4 Ω) = 8.14 V
∆V123 8.14 V
I12 = 
=  = 0.102 A
R12
80.0 Ω
I1 = I12 = 0.102 A
d. ∆V2 = I12 R2 = (0.102 A)(50.0 Ω) = 5.10 V
(∆V2 )2 (5.10 V)2
P2 = 
=  = 0.520 W
R2
50.0 Ω
e. ∆V4 = IR4 = (0.192 A)(20.0 Ω) = 3.84 V
(∆V4 )2 (3.84 V)2
P4 = 
=  = 0.737 W
R4
20.0 Ω
series:
∆VA = 4.0 V (series)
∆V = ∆VA + ∆VB
IB = 2.0 A (parallel)
∆VB = ∆V − ∆VA = 6.0 V − 4.0 V = 2.0 V
2.0 V
∆V
IB = B =  = 0.67 A
3.0 Ω
RB
IA = IB = 0.67 A
4.0 V
∆V
RA = A =  = 6.0 Ω
0.67 A
IA
parallel:
∆VA = ∆VB = 6.0 V
∆V
6.0 V
RB = B =  = 3.0 Ω
IB
2.0 A
I Ch. 20–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
44. ∆V = 6.0 V
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Givens
Solutions
46. R1 = 5.0 Ω
a. R789 = R7 + R8 + R9 = 3.0 Ω + 4.0 Ω + 3.0 Ω = 10.0 Ω
R2 = 10.0 Ω
−1
−1
R3 = 4.0 Ω
R4 = 3.0 Ω
R456789 = R4 + R5789 + R6 = 3.0 Ω + 5.00 Ω + 2.0 Ω = 10.0 Ω
R5 = 10.0 Ω
R6 = 2.0 Ω
R7 = 3.0 Ω
R2456789
1
1
=  + 
10.0 Ω 10.0 Ω
−1
1
1
R5789 =  + 
R5 R789
−1
1
1
=  + 
R2 R456789
1
= 0.200 
Ω
−1
= 5.00 Ω
1
1
=  + 
10.0 Ω 10.0 Ω
−1
1
= 0.200 
Ω
I
= 5.00 Ω
Req = R1 + R2456789 + R3 = 5.0 Ω + 5.00 Ω + 4.0 Ω = 14.0 Ω
R8 = 4.0 Ω
R9 = 3.0 Ω
∆V = 28 V
∆V
28 V
b. I =  =  = 2.0 A
Req 14.0 Ω
I1 = I = 2.0 A
47. P = 4.00 W
R1 = 3.0 Ω
R2 = 10.0 Ω
R3 = 5.0 Ω
R4 = 4.0 Ω
R5 = 3.0 Ω
−1
−1
1
1
1
= 0.100  + 0.20  = 0.30 
Ω Ω Ω
1
1
a. R23 =  + 
R2 R3
1
1
=  + 
10.0 Ω 5.0 Ω
−1
R23
−1
= 3.3 Ω
R234 = R23 + R4 = 3.3 Ω + 4.0 Ω = 7.3 Ω
−1
1
1
=  + 
7.3 Ω 3.0 Ω
−1
R2345
−1
1
1
1
= 0.14  + 0.33  = 0.47 
Ω Ω Ω
1
1
R2345 =  + 
R234 R5
−1
= 2.1 Ω
Req = R1 + R2345 = 3.0 Ω + 2.1 Ω = 5.1 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
b. ∆V = P
R
eq
= (4
.0
0W
)(
5.
1Ω
) = 4.5 V
48. PT = 1200 W
P = I∆V
PC = 1200 W
PT + PC = I∆V
∆V = 120 V
2(1200 W)
I =  = 20 A
120 V
Imax = 15 A
no, because 20 A > 15 A
49. PH = 1300 W
PT = 1100 W
PG = 1500 W
∆V = 120 V
P
1300 W
 =  = 11 A
a. heater: I = H
∆V
120 V
PT 1100 W
toaster: I = 
=  = 9.2 A
∆V
120 V
PG 1500 W
grill: I = 
=  = 12 A
∆V
120 V
b. yes; Itot = 11 A + 9.2 A + 12 A = 32.2 A
Section One—Pupil’s Edition Solutions
I Ch. 20–15
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Magnetism
Chapter 21
I
Practice 21A, p. 775
Givens
Solutions
1. B = 4.20 × 10−2 T
Fmagnetic = 2.40 × 10−14 N
2.40 × 10−14 N
Fmagnetic
 = 
= 3.57 × 106 m/s
v= 
(1.60 × 10−19 C)(4.20 × 10−2 T)
qB
q = 1.60 × 10−19 C
2. B = 2.5 T
−12
Fmagnetic = 3.2 × 10
−19
q = 1.60 × 10
N
3.2 × 10−12 N
Fmagnetic
 = 
v= 
= 8.0 × 106 m/s
(1.60 × 10−19 C)(2.5 T)
qB
C
3. Fmagnetic = 2.0 × 10−14
downward
B = 8.3 × 10−2 T west
2.0 × 10−14 N
Fmagnetic
 = 
= 1.5 × 106 m/s north
v= 
(1.60 × 10−19 C)(8.3 × 10−2 T)
qB
q = 1.60 × 10−19 C
4. B = 1.5 T north
7
v = 2.5 × 10 m/s
downward
Fmagnetic = qvB = (1.60 × 10−19 C)(2.5 × 107 m/s)(1.5 T) = 6.0 × 10−12 m/s N west
q = 1.60 × 10−19 C
5. B = 2.5 T west
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7
v = 1.5 × 10 m/s
Fmagnetic = qvB = (1.60 × 10−19 C)(1.5 × 107 m/s)(2.5 T) = 6.0 × 10−12 N south
q = 1.60 × 10−19 C
6. q = 3.2 × 10−19 C
v = 5.5 × 107 m/s
Fmagnetic
1.5 × 10−14 N
B =  = 
= 8.5 × 10−4 T
qv
(3.2 × 10−19 C)(5.5 × 107 m/s)
Fmagnetic = 1.5 × 10−14 N
Practice 21B, p. 778
1. l = 6.0 m
I = 7.0 A
B=
−6
7.0 × 10 N
 = 
=
(7
.0 A)(6.0 m)
Il
Fmagnetic
1.7 × 10−7 T in the +z direction
Fmagnetic = 7.0 × 10−6 N
2.
l = 1.0 m
Fmagnetic = 0.50 N
Fmagnetic
0.50 N
B =  =  = 0.050 T
Il
(10.0 A)(1.0 m)
I = 10.0 A
Section One—Pupil’s Edition Solutions
I Ch. 21–1
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Givens
Solutions
3. l = 0.15 m
B=
I = 4.5 A
1.0 N
 = 
=
(4.5
A
)(0.15 m)
Il
Fmagnetic
1.5 T
Fmagnetic = 1.0 N
I
4. B = 1.5 T
Fmagnetic = 4.4 N
F
4.4 N
magnetic
 =  =
l= 
IB
(5.0 A)(1.5 T)
0.59 m
I = 5.0 A
Section Review, p. 779
1. q = 0.030 C
Fmagnetic = 1.5 N
Fmagnetic
1.5 N
B =  =  = 0.081 T
qv
(0.030 C)(620 m/s)
v = 620 m/s
3.
l
= 25 cm
I = 5.0 A
Fmagnetic = BI l = (0.60 T)(5.0 A)(0.25 m) = 0.75 N
B = 0.60 T
Chapter Review and Assess, pp. 781–785
30. B = 5.0 × 10−5 T North
Fmagnetic = 3.0 × 10−11 N
upward
3.0 × 10−11 N
Fmagnetic
 = 
= 15 m/s
v=
(4.0 × 10−8 C)(5.0 × 10−5 T)
qB
q = 4.0 × 10−8 C
B = 5.0 × 10−5 T
q = 1.60 × 10−19 C
mg = qvB
−27
2
mg (1.673 × 10 kg)(9.81 m/s )
v =  = 
= 2.1 × 10−3 m/s
qB (1.60 × 10−19 C)(5.0 × 10−5 T)
g = 9.81 m/s2
32. I = 10.0 A
l
= 5.00 m
B=

Il
Fmagnetic
15.0 N
=  = 0.300 T
(10.0 A)(5.00 m)
Fmagnetic = 15.0 N
33.
l
= 1.00 m
m = 50.0 g = 0.0500 kg
I = 0.245 A
mg = BIl
mg (0.0500 kg)(9.81 m/s2)
B =  =  = 2.00 T
(0.245 A)(1.00 m)
Il
g = 9.81 m/s2
I Ch. 21–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
31. m = 1.673 × 10−27 kg
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Givens
Solutions
34. v = 2.50 × 106 m/s
a. mg = qvB
m = 1.673 × 10−27 kg
q = 1.60 × 10−19 C
(1.673 × 10−27 kg)(9.81 m/s2)
mg
B =  = 
= 4.10 × 10−14 T
qv
(1.60 × 10−19 C)(2.50 × 106 m/s)
g = 9.81 m/s2
38. v = 2.0 × 107 m/s
B = 0.10 T
m = 1.673 × 10−27 kg
I
ma = qvB
−19
7
qvB (1.60 × 10 C)(2.0 × 10 m/s)(0.10 T)
a =  = 
= 1.9 × 1014 m/s2
m
1.673 × 10−27 kg
q = 1.60 × 10−19 C
39. q = 1.60 × 10−19 C
a = 2.0 × 1013 m/s2
v = 1.0 × 107 m/s
m = 1.673 × 10−27 kg
40. v = 3.0 × 106 m/s
q = 37°
B = 0.30 T
q = 1.60 × 10−19 C
m = 1.673 × 10−27 kg
41.
l
= 15 cm
I = 5.0 A
m = 0.15 kg
ma = qvB
−27
13
2
ma (1.673 × 10 kg)(2.0 × 10 m/s )
B =  = 
−19
7
qv
(1.60 × 10 C)(1.0 × 10 m/s)
B = 2.1 × 10−2 T, in the negative y direction
a. Fmagnetic = qvB(sin q) = (1.60 × 10−19 C)(3.0 × 106 m/s)(0.30 T)(sin 37°)
Fmagnetic = 8.7 × 10−14 N
F
8.7 × 10−14 N
c. a =  = 
= 5.2 × 1013 m/s2
m 1.673 × 10−27 kg
mg = BIl
mg (0.15 kg)(9.81 m/s2)
B =  =  = 2.0 T, out of the page
Il
(5.0 A)(0.15 m)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
42. I = 15 A
Fmagnetic
 = 0.12 N/m
l
Consider the force on a 1.0 m length of wire.
Fmagnetic
Fmagnetic =  (l ) = (0.12 N/m)(1.0 m) = 0.12 N
l
Fmagnetic = BIl
Fmagnetic
0.12 N
B =  =  = 8.0 × 10−3 T, in the positive z direction
Il
(15 A)(1.0 m)
43. m = 1.673 × 10−27 kg
Use the equation for the force that maintains circular motion from Chapter 7.
q = 1.60 × 10−19 C
Fmagnetic = Fc
∆t = 1.00 × 10−6 s
q vB = mrw 2
∆q = 2p rad
∆q
q rwB = mrw 2 where w = 
∆t
m∆q
(1.673 × 10−27 kg)(2p rad)
B =  = 
= 6.57 × 10−2 T
q∆t
(1.60 × 10−19 C)(1.00 × 10−6 s)
Section One—Pupil’s Edition Solutions
I Ch. 21–3
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I
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Givens
Solutions
44. m = 6.68 × 10−27 kg
Use the equation for the force that maintains circular motion from Chapter 7.
r = 3.00 cm
Fc = Fmagnetic
v = 1.00 × 104 m/s
v2
m = qvB
r
q = 1.60 × 10−19 C
45. r = 1000.0 km
−8
B = 4.00 × 10
Use the equation for the force that maintains circular motion from Chapter 7.
T
−19
q = 1.60 × 10
mv (6.68 × 10−27 kg)(1.00 × 104 m/s)
= 1.39 × 10−2 T, toward the observer
B =  = 
qr
(1.60 × 10−19 C)(0.0300 m)
C
m = 1.673 × 10−27 kg
rE = 6.38 × 106 m
Fc = Fmagnetic
v2
m  = qvB
r + rE
(r + rE)qB (1.0000 × 106 m + 6.38 × 106 m)(1.60 × 10−19 C)(4.00 × 10−8 T)
v = 
= 
m
1.673 × 10−27 kg
v = 2.82 × 107 m/s
46. B = 1.00 × 10−3 T
a. Use the equation for the force that maintains circular motion from Chapter 7.
m = 9.109 × 10−31 kg
Fc = Fmagnetic
q = 1.60 × 10−19 C
mrw 2 = qvB
L
= 4.00 × 10−25 J • s
mw = qB
where v = rw
L
L
where w =  = 2
I mr
L
2 = qB
r
r=
4.00 ×10−25 J • s

= 5.00 × 10−2 m
(1.60 × 10−19 C)(1.00 × 10−3 T)
L
 =
qB
v2
m  = qvB
r
−19
−2
−3
qrB (1.60 × 10 C)(5.00 × 10 m)(1.00 × 10 T)
v =  = 
= 8.78 × 106 m/s
m
9.109 × 10−31 kg
I Ch. 21–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
b. Fc = Fmagnetic
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Induction and Alternating Current
Chapter 22
I
Practice 22A, p. 800
Givens
1. r = 22 cm
Bi = 0.50 T
Solutions
∆B
emf = −NA(cos q)
∆t
Bf = 0.00 T
∆B = Bf − Bi = 0.00 T − 0.50 T = −0.50 T
∆t = 0.25 s
A = pr 2
N =1
∆B
−0.50 T
emf = −Npr 2(cos q) = −(1)(p)(0.22 m)2(cos 0.0°) 
∆t
0.25 s
q = 0.0°
emf = 0.30 V
2. N = 205
R = 23 Ω
A = 0.25 m2
q = 0.0°
∆t = 0.25 s
∆B
emf = −NA(cos q) 
∆t
∆B = Bf − Bi = 0.0 T − 1.6 T = −1.6 T
Bi = 1.6 T
2
−NA(cos q)∆B −(205)(0.25 m )(cos 0.0°)(−1.6 T)
I =  = 
∆tR
(0.25 s)(23 Ω)
Bf = 0.0 T
I = 14 A
3. r = 0.33 m
Bi = 0.35 T
Copyright © Holt, Rinehart and Winston. All rights reserved.
emf
I = 
R
∆B
emf = −NA(cos q) 
∆t
q = 0.0°
∆B = Bf − Bi = −0.25 T − 0.35 T = −0.60 T
Bf = −0.25 T
A = pr 2
∆t = 1.5 s
2
∆B −(1)(p)(0.33 m) (cos 0.0°)(−0.60 T)
emf = −Npr 2(cos q)  = 
∆t
1.5 s
N=1
emf = 0.14 V
4. N = 505
d = 15.5 cm
qi = 0.0°
∆t = 2.77 ms
qf = 90.0°
emf = 0.166 V
∆cos q = cos qf − cos qi = cos 90.0° − cos 0.0° = 0 − 1 = −1
d
A = p 
2
2
(0.166 V)(2.77 × 10−3 s)
(emf )(∆t)
B =  = 
2
−NA(∆cos q)
0.155 m
−(505)(p)  (−1)
2
B = 4.83 × 10−5 T
Section One—Pupil’s Edition Solutions
I Ch. 22–1
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Section Review, p. 802
Givens
Solutions
∆B = Bf − Bi = 0.00 T − 0.25 T = −0.25 T
4. N = 256
2
A = 0.0025 m
I
Bi = 0.25 T
2
∆B −(256)(0.0025 m )(cos 0.0°)(−0.25 T)
emf = −NA(cos q ) = 
∆t
0.75 s
q = 0.0°
emf = 0.21 V
∆t = 0.75 s
Bf = 0.00 T
6. Bmax = 9.0 × 10−4 T
N = 5200
1
1
∆t =  = 
4f
(4)(440 Hz)
A = 5.4 × 10−5 m2
∆B = Bmax − Bmin = (9.0 × 10−4 T) − 0.0 T = 9.0 × 10−4 T
f = 440 Hz
−5 2
−4
∆B −(5200)(5.4 × 10 m )(cos 0.0°)(9.0 × 10 T)
emf = −NA(cos q)  = 
∆t
1

(4)(440 Hz)
q = 0.0°
Bmin = 0.0 T
emf = −0.44 V
Practice 22B, p. 806
1. N = 510
A = (0.082 m)(0.25 m)
maximum emf = NABw = (510)(0.082 m)(0.25 m)(0.65 T)(12.8 rad/s)
maximum emf = 87 V
w = 12.8 rad/s
B = 0.65 T
maximum emf = NABw
N = 17
w = 2pf
B = 1.7 T
A = pr 2
f = 2.0 Hz
maximum emf = N(pr 2)B(2pf ) = 2Nr 2B fp 2
maximum emf = (2)(17)(0.22 m)2(1.7 T)(2.0 Hz)(p 2)
maximum emf = 55 V
3. A = 0.045 m2
N = 120
maximum emf = NABw = (120)(0.045 m2)(2.0 × 10−5 T)(157 rad/s)
maximum emf = 1.7 × 10−2 V
w = 157 rad/s
B = 2.0 × 10−5 T
4. maximum emf = 90.4 V
f = 65 Hz
90.4 V
maximum emf maximum emf
N =  =  = 
2
ABw
AB 2p f
(0.0230 m )(1.2 T)(2p)(65 Hz)
A = 230 cm2
N = 8.0 turns
B = 1.2 T
I Ch. 22–2
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
2. r = 0.22 m
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Practice 22C, p. 810
Givens
1. R = 25 Ω
∆Vrms = 120 V
Solutions
∆Vrms 120 V
Irms = 
=  = 4.8 A
R
25 Ω
I s
4.8 A
Imax = rm
=  = 6.8 A
0.707 0.707
I
∆Vrms 120 V
∆Vmax = 
=  = 170 V
0.707
0.707
2. Irms = 5.5 A
I s
5.5 A
=  = 7.8 A
Imax = rm
0.707 0.707
3. ∆Vrms = 110 V
a. Irms = (0.707)(Imax) = (0.707)(10.5 A) = 7.42 A
Imax = 10.5 A
4. ∆Vrms = 15.0 V
R = 10.4 Ω
∆Vrms 110 V
b. R = 
=  = 14.8 Ω
7.42 A
Irms
∆Vrms 15.0 V
Irms = 
=  = 1.44 A
R
10.4 Ω
I s
1.44 A
Imax = rm
=  = 2.04 A
0.707 0.707
∆Vrms 15.0 V
∆Vmax = 
=  = 21.2 V
0.707
0.707
5. ∆Vmax = 155 V
R = 53 Ω
Copyright © Holt, Rinehart and Winston. All rights reserved.
6. ∆Vmax = 451 V
a. ∆Vrms = (0.707)(∆Vmax) = (0.707)(155 V) = 1.10 × 102 V
∆Vrms 1.10 × 102 V
b. Irms = 
=  = 2.1 A
R
53 Ω
∆Vrms = (0.707)(∆Vmax) = (0.707)(451 V) = 319 V
Section Review, p. 813
1. A = 0.33 m2
w = 281 rad/s
maximum emf = NABw = (37)(0.33 m2)(0.035 T)(281 rad/s)
maximum emf = 1.2 × 102 V
B = 0.035 T
N = 37
2. maximum emf = 2.8 V
N = 25
A = 36 cm2
f = 60.0 Hz
2.8 V
maximum emf maximum emf
B =  =  = 
NAw
NA2p f
(25)(0.0036 m2)(2p)(60.0 Hz)
B = 8.3 × 10−2 T
Section One—Pupil’s Edition Solutions
I Ch. 22–3
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Givens
Solutions
4. Irms = 0.025 mA
R = 4.3 kΩ
I s
0.025 mA
Imax = rm
=  = 0.035 mA
0.707
0.707
∆Vrms = IrmsR = (0.025 × 10−3 A)(4.3 × 103 Ω) = 0.11 V
∆Vmax = ImaxR = (0.035 × 10−3 A)(4.3 × 103 Ω) = 0.15 V
I
Practice 22D, p. 818
1. N1 = 2680 turns
∆V1 = 5850 V
N1∆V2 (2680 turns)(120 V)
 =  = 55.0 turns
N2 = 
5850 V
∆V1
∆V2 = 120 V
2. ∆V1 = 12 V
4
∆V2 = 2.0 × 10 V
N1∆V2 (21 turns)(2.0 × 104 V)
 =  = 3.5 × 104 turns
N2 = 
∆V1
12 V
N1 = 21 turns
3. ∆V1 = 117 V
∆V2 = 119 340 V
N2 ∆V1 (25 500 turns)(117 V)
 =  = 25.0 turns
N1 = 
∆V2
119 340 V
N2 = 25 500 turns
4. ∆V1 = 117 V
∆V2 = 0.750 V
5. N1 = 12 turns
N2 = 2550 turns
N
∆V
117 V
156
1 = 1 =  = 
N2 ∆V2 0.750 V
1
N2 ∆V1 (2550 turns)(120 V)
 =  = 2.6 × 104 V
∆V2 = 
N1
12 turns
6. N1 = 12 500 turns
N2 = 525 turns
N2 ∆V1 (525 turns)(3510 V)
 =  = 147 V
∆V2 = 
N1
12 500 turns
∆V1 = 3510 V
Section Review, p. 819
2. N1 = 50 turns
N2 = 7000 turns
N2 ∆V1 (7000 turns)(120 V)
 =  = 1.7 × 104 V
∆V2 = 
N1
50 turns
∆V1 = 120 V
Chapter Review and Assess, pp. 821–825
10. N = 1
ri = 0.12 m
−N(cos q )B∆A
emf = 
∆t
B = 0.15 T
∆A = Af − Ai = Af − pri2 = (3 × 10−3 m2) − (p)(0.12 m)2
Af = 3 × 10−3 m2
∆A = (3 × 10−3 m2) − 0.045 m2 = −0.042 m2
∆t = 0.20 s
−(1)(cos 0.0°)(0.15 T)(−0.042 m2)
emf =  = 3.2 × 10−2 V
0.20 s
q = 0.0°
I Ch. 22–4
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
∆V1 = 120 V
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Givens
Solutions
11. A = (0.055 m)(0.085 m)
q = 0.0°
∆[AB(cos q)]
∆B
emf = −N  = −NA(cos q) 
∆t
∆t
N = 75
emf = −(75)(0.055 m)(0.085 m)(cos 0.0°)(−3.0 T/s) = 1.05 V
R = 8.7 Ω
emf 1.05 V
I =  =  = 0.12 A
R
8.7 Ω
−∆B
 = 3.0 T/s
∆t
12. N = 52
−3
A = 5.5 × 10
I
−NA(cos q)∆B
emf = 
∆t
2
m
Bi = 0.00 T
∆B = Bf − Bi = 0.55 T − 0.00 T = 0.55 T
Bf = 0.55 T
−(52)(5.5 × 10−3 m2)(cos 0.0°)(0.55 T)
emf =  = −0.63 V
0.25 s
∆t = 0.25 s
q = 0.0°
23. N = 112
maximum emf = NABw = NAB2pf
−2
A = 4.41 × 10
2
m
f = 25.0 Hz
−5
B = 5.00 × 10
maximum emf = (112)(4.41 × 10–2 m2)(5.00 × 10–5 T)(2p)(25.0 Hz)
maximum emf = 3.88 × 10–2 V
T
a. maximum emf = NABw = NAB2pf
24. N = 45
A = 0.12 m
maximum emf = (45)(0.12 m2)(0.118 T)(2p)(60.0 Hz)
B = 0.118 T
maximum emf = 2.4 × 102 V
2
f = 60.0 Hz
Copyright © Holt, Rinehart and Winston. All rights reserved.
R = 35 Ω
25. ∆Vrms = 220 000 V
26. ∆Vmax = 340 V
R = 120 Ω
maximum emf 2.4 × 102 V
b. I =  =  = 6.9 A
R
35 Ω
∆Vrms 220 000 V
=  = 310 000 V
∆Vmax = 
0.707
0.707
a. ∆Vrms = (0.707)(∆Vmax ) = (0.707)(340 V) = 2.40 × 102 V
∆Vrms 2.40 × 102 V
b. Irms = 
=  = 2.00 A
R
120 Ω
Section One—Pupil’s Edition Solutions
I Ch. 22–5
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Givens
Solutions
27. Imax = 0.909 A
a. Irms = (0.707)(Imax ) = (0.707)(0.909 A) = 0.643 A
R = 182 Ω
b. ∆Vrms = IrmsR = (0.643 A)(182 Ω) = 117 V
I
c. P = Irms2R = (0.643 A)2(182 Ω) = 75.2 W
28. P = 996 W
a. Irms = (0.707)(Imax ) = (0.707)(11.8 A) = 8.34 A
Imax = 11.8 A
P
996 W
b. ∆Vrms =  =  = 119 V
Irms 8.34 A
35. ∆V1 = 120 V
∆V N
(9.0 V)(640 turns)
N2 = 21 =  = 48 turns
∆V1
120 V
∆V2 = 9.0 V
N1 = 640 turns
36. ∆V1 = 120 V
∆V N
(6.3 V)(210 turns)
N2 = 21 =  = 11 turns
∆V1
120 V
∆V2 = 6.3 V
N1 = 210 turns
37. N = 1
−2
Bi = 2.5 × 10
T
−3
A = 7.54 × 10
m2
emf = 1.5 V
q = 0.0°
−NA(cos q)∆B
∆t = 
emf
∆B = Bf − Bi = 0.000 T − 2.5 × 10−2 T = −2.5 × 10−2 T
−(1)(7.54 × 10−3 m2)(cos 0.0°)(−2.5 × 10−2 T)
∆t =  = 1.3 × 10−4 s
1.5 V
38. A = 1.886 × 10−3 m2
Bi = 2.5 × 10−2 T
(emf )(∆t )
N = 
−A(cos q )∆B
∆t = 0.25 s
∆B = Bf − Bi = 0.000 T − 2.5 × 10−2 T = −2.5 × 10−2 T
emf = 149 mV
(149 × 10−3 V)(0.25 s)
N = 
= 7.9 × 102 turns
−(1.886 × 10−3 m2)(cos 0.0°)(−2.5 × 10−2 T)
Bf = 0.000 T
q = 0.0°
39. N = 325
A = 19.5 × 10−4 m2
q = 45°
∆t = 1.25 s
(15 × 10−3 V)(1.25 s)
(emf )(∆t )
= −4.2 × 10−2 T
∆B =  = 
−NA(cos q ) −(325)(19.5 × 10−4 m2)(cos 45°)
Bi = Bf − ∆B = 0.0 T − (−4.2 × 10−2 T) = 4.2 × 10−2 T
emf = 15 mV
Bf = 0.0 T
I Ch. 22–6
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
Bf = 0.000 T
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Givens
Solutions
40. N1 = 22 turns
N2 ∆V1 (88 turns)(110 V)
 =  = 4.4 × 102 V
b. ∆V2 = 
N1
22 turns
N2 = 88 turns
∆V1 = 110 V
41. ∆V1 = 20.0 kV
∆V2 = 117 V
42. N = 105
q = 0.0°
r = 0.833 m
Bi = 4.72 × 10−3 T
Bf = 0.00 T
∆t = 10.5 ms
43. P = 5.0 × 103 kW
∆V1 = 4500 V
∆V2 = 510 kV
R = (4.5 × 10−4 Ω/m)
(6.44 × 105 m)
I
N
∆V
20.0 × 103 V
171
1 = 1 =  = 
N2 ∆V2
117 V
1
−NA(cos q)∆B −Npr 2(cos q)∆B
emf =  = 
∆t
∆t
∆B = Bf − Bi = 0.00 T − 4.72 × 10−3 T = −4.72 × 10−3 T
−(105)(p )(0.833 m)2(cos 0.0°)(−4.72 × 10−3 T)
emf = 
10.5 × 10−6 s
emf = 1.03 × 105 V
P
a. I = 
∆V2
2
2
P
5.0 × 106 W
Pdissipated = I 2R =  R = 
(4.5 × 10−4 Ω/m)(6.44 × 105 m)
∆V2
510 × 103 V
Pdissipated = 28 × 103 W = 28 kW
b. If the generator’s output were not stepped up,
P
I = 
∆V1
2
2
P
5.0 × 106 W
Pdissipated = I 2R =  R =  (4.5 × 10−4 Ω/m)(6.44 × 105 m)
∆V1
4500 V
Copyright © Holt, Rinehart and Winston. All rights reserved.
Pdissipated = 3.6 × 108 W = 3.6 × 105 kW
44. emf = (245 V)sin 560t
560
f =  = 89.1 Hz
2p
maximum potential difference = 245 V
45. −M = 1.06 H
∆I = 9.50 A
∆I
9.50 A
emf = −M  = 1.06 H ×  = 300 V
∆t
0.0336 s
∆t = 0.0336 s
Section One—Pupil’s Edition Solutions
I Ch. 22–7
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Atomic Physics
Chapter 23
I
Practice 23A, p. 833
Givens
Solutions
1. E = 8.1 × 10−15 eV
E (8.1 × 10−15 eV)(1.60 × 10−19 J/eV)
= 2.0 Hz
f =  = 
6.63 × 10−34 J • s
h
2. f = 0.56 Hz
E = hf = (6.63 × 10−34 J • s)(0.56 Hz) = 3.7 × 10−34 J
3. E = 5.0 eV
−19
E (1.60 × 10 J/eV)(5.0 eV)
f =  = 
= 1.2 × 1015 Hz
6.63 × 10−34 J•s
h
4. l = 940 mm
8
c 3.00 × 10 m/s
a. f =  = 
= 3.19 × 1011 Hz
−6
l
940 × 10 m
(6.63 × 10−34 J • s)(3.19 × 1011 Hz)
c. E = hf = 
= 1.32 × 10−3 eV
1.60 × 10−19 J/eV
Practice 23B, p. 836
1. E = 5.00 eV
KEmax = E hft
E KEmax [5.00 eV 3.00 eV](1.60 × 10−19 J/eV)
ft = 
= 
6.63 × 10−34 J • s
h
Copyright © Holt, Rinehart and Winston. All rights reserved.
ft = 4.83 × 1014 Hz
2. l = 350 nm
KEmax = 1.3 eV
a. work function = hf − KEmax
c
f = 
l
c
work function = h  − KEmax
l
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
work function = 
− 1.3 eV
(1.60 × 10−19 J/eV)(350 × 10−9 m)
work function = 3.6 eV − 1.3 eV = 2.3 eV
work function (2.3 eV)(1.60 × 10−19 J/eV)
b. ft =  = 
h
6.63 × 10−34 J • s
ft = 5.6 × 1014 Hz
Section One—Pupil’s Edition Solutions
I Ch. 23–1
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Givens
Solutions
3. f = 1.00 × 1015 Hz
KEmax = 2.85 10–19J
KEmax = hf − hft
hft = hf − KEmax
hft = (6.63 × 10−34 J • s)(1.00 × 1015 Hz) − 2.85 × 10−19 J
hft = 6.63 × 10−19 J − 2.85 × 10−19 J
I
hft = 3.78 × 10−19 J
3.78 × 10−19 J
= 2.36 eV
Converting to electron-volts, hft = 
1.60 × 10−19 J
hft,lithium = 2.3 eV
(6.63 × 10−34 J • s)(7.0 × 1014 Hz)
E = hf = 
= 2.9 eV
1.60 × 10−19 J/eV
hft,silver = 4.7 eV
The photoelectric effect will be observed if E > hft , which holds true for lithium
hft,cesium = 2.14 eV
and cesium .
4. f = 7.0 × 1014 Hz
Section Review, p. 839
2. l = 4.5 × 10−7 m
hc (6.63 × 10−34 J • s)(3.00 × 108 m/s)
E = hf =  = 
l
(1.60 × 10−19 J/eV)(4.5 × 10−7 m)
E = 2.8 eV
5. l = 1.00 × 10−7 m
hft = 4.6 eV
c 3.00 × 108 m/s
f =  = 
= 3.00 × 1015 Hz
l 1.00 × 10−7 m
−19
4.6 eV (4.6 eV)(1.60 × 10 J/eV)
ft =  = 
−34
h
6.63 × 10 J • s
ft = 1.1 × 1015 Hz
Because f > ft , electrons are ejected.
(6.63 × 10−34 J • s)(3.00 × 1015 Hz)
KEmax = 
− 4.6 eV
1.60 × 10−19 J/eV
KEmax = 12.4 eV − 4.6 eV = 7.8 eV
Practice 23C, p. 851
1. m = 50.0g = 5.00 × 10−2 kg
l = 3.32 × 10−34 m
2. l = 5.00 × 10−7 m
m = 9.109 × 10−31 kg
3. m = 0.15 kg
l = 5.00 × 10−7 m
I Ch. 23–2
6.63 × 10−34 J • s
h
= 39.9 m/s
v =  = 
l m (5.00 × 10−2 kg)(3.32 × 10−34 m)
6.63 × 10−34 J • s
h
= 1.46 × 103 m/s
v =  = 
lm (5.00 × 10−7 m)(9.109 × 10−31 kg)
6.63 × 10−34 J • s
h
= 8.84 × 10−27 m/s
v =  = 
−7
lm (5.00 × 10 m)(0.15 kg)
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
KEmax = hf − hft
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Givens
4. m = 1375 kg
v = 43 km/h
5. v = 3.5 mm/s
l = 1.9 × 10−13 m
Solutions
h
6.63 × 10−34 J • s
l =  = 
= 4.0 × 10−38 m
mv (1375 kg)(43 × 103 m/h)(1 h/3600 s)
6.63 × 10−34 J • s
h
= 1.0 × 10−15 kg
m =  = 
l v (1.9 × 10−13 m)(3.5 × 10−6 m/s)
I
Section Review, p. 854
3. v = 1.00 × 104 m/s
m = 1.673 × 10−27 kg
6.63 × 10−34 J • s
h
= 3.96 × 10−11 m
l =  = 
mv (1.673 × 10−27 kg)(1.00 × 104 m/s)
Chapter Review and Assess, pp. 856–859
14. E = 2.0 keV
3
−19
E (2.0 × 10 eV)(1.60 × 10 J/eV)
f =  = 
= 4.8 × 1017 Hz
h
6.63 × 10−34 J • s
15. l1 = 5.00 cm
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
hc
a. E1 = hf1 =  = 
= 2.49 × 10−5 eV
l1 (1.60 × 10−19 J/eV)(5.00 × 10−2 m)
l2 = 5.00 × 10−7 m
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
hc
b. E2 = hf2 =  = 
= 2.49 eV
l2 (1.60 × 10−19 J/eV)(5.00 × 10−7 m)
l3 = 5.00 × 10−8 m
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
hc
c. E3 = hf3 =  = 
= 24.9 eV
l3 (1.60 × 10−19 J/eV)(5.00 × 10−8 m)
16. f = 1.5 × 1015 Hz
Copyright © Holt, Rinehart and Winston. All rights reserved.
KEmax = 1.2 eV
hf − KEmax
ft = 
h
(6.63 × 10−34 J • s)(1.5 × 1015 Hz) − (1.2 eV)(1.60 × 10−19 J/eV)
ft = 
6.63 × 10−34 J • s
(9.9 × 10–19 J − 1.9 × 10–19 J)
ft = 
= 1.2 × 1015 Hz
6.63 × 10–34 J• s
17. l = 3.0 × 10−7 m
hft,lithium = 2.3 eV
hft,iron = 3.9 eV
hft,mercury = 4.5 eV
hc (6.63 × 10−34 J • s)(3.00 × 108 m/s)
= 4.1 eV
a. E = hf =  = 
l
(1.60 × 10−19 J/eV)(3.0 × 10−7 m)
The photoelectric effect will be observed if E > hft , which holds true for
lithium and iron .
b. For lithium, KEmax = E − hft,lithium = 4.1 eV − 2.3 eV = 1.8 eV
For iron, KEmax = E − hft,iron = 4.1 eV − 3.9 eV = 0.2 eV
18. ft = 1.14 × 1015 Hz
(6.63 × 10−34 J • s)(1.14 × 1015 Hz)
work function = hft = 
1.60 × 10−19 J/eV
work function = 4.72 eV
Section One—Pupil’s Edition Solutions
I Ch. 23–3
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I
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Givens
Solutions
39. l = 5.2 × 10−11 m
6.63 × 10−34 J • s
h
v =  = 
= 1.4 × 107 m/s
ml (9.109 × 10−31 kg)(5.2 × 10−11 m)
40. m = 0.15 kg
6.63 × 10−34 J • s
h
l =  =  = 9.8 × 10−35 m
mv (0.15 kg)(45 m/s)
v = 45 m/s
41. l = 4.00 × 10−14 m
m = 1.673 × 10−27 kg
42. v = 12 m/s
l = 5.5 × 10−30 m
43. vmax = 4.6 × 105 m/s
l = 625 nm
m = 9.109 × 10−31 kg
6.63 × 10−34 J • s
h
= 9.91 × 106 m/s
v =  = 
l m (4.00 × 10−14 m)(1.673 × 10−27 kg)
6.63 × 10−34 J • s
h
m =  = 
= 1.0 × 10−5 kg
l v (5.5 × 10−30 m)(12 m/s)
hc 1
a. hf t = hf − KEmax =  − 2m(vmax)2
l
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
hf t = 
− (0.5)(9.109 × 10−31 kg)(4.6 × 105 m/s)2
625 × 10−9 m
hf t = (3.18 × 10−19 J) − (9.6 × 10−20 J) = 22.2 × 10−20 J
22.2 × 10−20 J
hf t = 
= 1.39 eV
1.60 × 10−19 J/eV
work function
22.2 × 10−20 J
b. ft =  = 
= 3.35 × 1014 Hz
h
6.63 × 10−34 J • s
m = 1.673 × 10−27 kg
6.63 × 10−34 J • s
h
= 4.0 × 10−14 m
l =  = 
m v (1.673 × 10−27 kg)(1.0 × 107 m/s)
45. E = 2.8 × 10−19 J
E
2.8 × 10−19 J
f =  = 
= 4.2 × 1014 Hz, which is red light
h 6.63 × 10−34 J • s
46. l1 = l
hc
work function = hf − KEmax =  − KEmax
l
hc
2hc
hc
 − KEmax,1 =  − KEmax,2 =  − KEmax,2
1
l
l
l
KEmax,1 = 1.00 eV
1
l2 = 2l
KEmax,2 = 4.00 eV
I Ch. 23–4
2
hc
KEmax,2 − KEmax,1 = 
l
hc
 = 4.00 eV − 1.00 eV = 3.00 eV
l
hc
work function =  − KEmax,1 = 3.00 eV − 1.00 eV = 2.00 eV
l
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
44. v = 1.0 × 107 m/s
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Givens
Solutions
47. m = 0.50 kg
nhc
PE = mgh m = nhf = 
l
h m = 3.0 m
l = 5.0 × 10−7 m
lmghm (5.0 × 10−7 m)(0.50 kg)(9.81 m/s2)(3.0 m)
n = 
= 
hc
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
I
n = 3.7 × 1019 photons
48. l1 = 670.0 nm
l2 = 520.0 nm
KEmax,2 = (1.50)(KEmax,1)
hc
KEmax = hf − hft =  − hft
l
hc
For wavelength l1, KEmax,1 =  − hft
l1
hc
For wavelength l2, KEmax,2 =  − hft
l2
KEmax,2 = (1.50)(KEmax,1)
hc
hc
 − hft = (1.50)  − hft
l2
l1
hc
(1.50)(hc)
 − hft =  − (1.50)(hft )
l2
l1
(1.50)(hc) hc
(1.50)(hft ) − hft =  − 
l1
l2
(1.50)(hc) hc
(0.50)(hft ) =  − 
l1
l2
(1.50)(hc) hc
hft = (2.0)  − 
l1
l2
1.50
1
hft = (2.0)(hc)  − 
l1
l2
1
1.50
hft = (2.0)(6.63 × 10−34 J • s)(3.00 × 108 m/s) 
− 
670.0 × 10−9 m 520.0 × 10−9 m
Copyright © Holt, Rinehart and Winston. All rights reserved.
hft = (2.0)(6.63 × 10−34 J • s)(3.00 × 108 m/s)(2.24 × 106 m−1 − 1.923 × 106 m−1)
(2.0)(6.63 × 10−34 J • s)(3.00 × 108 m/s)(0.32 × 106 m−1)
hft = 
1.60 × 10−19 J/eV
hft = 0.80 eV
49. m = 0.200 kg
∆y = 50.0 m
h
l = 
mv
v = 2g
∆
y
6.63 × 10−34 J • s
h
l= 
= 
= 1.06 × 10−34 m
m/s
m 2g ∆y (0.200 kg) (2
)(
9.
81
2)(5
0.
0m
)
50. ∆t = 1.00 s
P = 100.0 W
l = 589.3 nm
E P∆t P∆tl
n =   =  = 
hf
hf
hc
(100.0 W)(1.00 s)(589.3 × 10−9 m)
n = 
= 2.96 × 1020 photons/s
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
Section One—Pupil’s Edition Solutions
I Ch. 23–5
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Givens
Solutions
51. r = 2.82 × 10−15 m
intensity = 5.00 × 10 W/m
E
E
∆t =  = 
P (intensity)(area)
E = 1.0 eV
area = pr 2
2
2
(1.0 eV)(1.60 × 10−19 J/eV)
E

∆t = 
=
(intensity)(pr 2) (5.00 × 102 W/m2)(p)(2.82 × 10−15 m)2
I
∆t = 1.3 × 107 s
52. work function = 3.44 eV
2
intensity = 0.055 W/m
vmax = 5.2 × 105 m/s
area = 1.0 cm2
∆t = 1.0 s
m = 9.109 × 10−31 kg
hf = work function + KEmax
1
hf = work function + 2m(vmax)2
(intensity)(area)
n
P (intensity)(area)
 =   =  = 
1
∆t E
hf
work function + 2m(vmax)2
(intensity)(area)(∆t)
n = 
1
work function + 2m(vmax)2
(0.055 W/m2)(0.01 m)2(1.0 s)
n = 
−19
(3.44 eV)(1.60 × 10 J/eV) + (0.5)(9.109 × 10−31 kg)(5.2 × 105 m/s)2
(0.055 W/m2)(0.01 m)2(1.0 s)
n = 
(5.50 × 10−19 J) + (1.2 × 10−19 J)
(0.055 W/m2)(0.01 m)2(1.0 s)
n = 
6.7 × 10−19 J
n = 8.2 × 1012 electrons
−31
m = 9.109 × 10
kg
h
a. l = 
mv
h
v = 
lm
2
h
h2
1
1
KE = 2 mv 2 = 2m  = 2
lm
2l m
(6.63 × 10−34 J • s)2
KE = 
(2)(1.0 × 10−11 m)2(9.109 × 10−31 kg)(1.60 × 10−19 J/eV)
KE = 1.5 × 104 eV (15 keV)
hc (6.63 × 10−34 J • s)(3.00 × 108 m/s)
b. E = hf =  = 
l
(1.60 × 10−19 J/eV)(1.0 × 10−11 m)
E = 1.2 × 105 eV (120 keV)
I Ch. 23–6
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
53. l = 1.0 × 10−11 m
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Modern Electronics
Chapter 24
I
Section Review, p. 880
Givens
6. Ib = 2.5 mA
Ic = 0.1 A
b per transistor = 10
Solutions
I
0.1 A
= 4 × 104
a. b = c = 
Ib 2.5 × 10−6 A
b. Number of transistors = x
10x ≥ 4 × 104
xmin = 5 transistors
Section Review, p. 884
4. l1 = 1.0 × 10−10 m
f2 = 100 Mhz
−34
8
hc (6.63 × 10 J • s)(3.00 × 10 m/s)
= 2.0 × 10−15 J
E1 = hf1 =  = 
−10
l1
1.0 × 10 m
E2 = hf2 = (6.63 × 10−34 J • s)(100 × 106 Hz) = 7 × 10−26 J
Chapter Review and Assess, pp. 886–888
25. E = 1.14 eV
−19
E (1.14 eV)(1.60 × 10 J/eV)
a. f =  = 
= 2.75 × 1014 Hz
−34
h
6.63 × 10 J • s
Copyright © Holt, Rinehart and Winston. All rights reserved.
8
c 3.00 × 10 m/s
= 1.09 × 10−6 m
b. l =  = 
f
2.75 × 1014 Hz
26. l = 650 nm
−34
8
hc (6.63 × 10 J • s)(3.00 × 10 m/s)
= 1.9 eV
E = hf =  = 
−9
−19
l (650 × 10 m)(1.60 × 10 J/eV)
27. l = 10−6 m
−34
8
hc (6.63 × 10 J • s)(3.00 × 10 m/s)
= 1 eV
E = hf =  = 
l
(10−6 m)(1.60 × 10−19 J/eV)
Section One—Pupil’s Edition Solutions
I Ch. 24–1
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Subatomic Physics
Chapter 25
I
Practice 25A, p. 902
Givens
1. For 20
10 Ne:
Z = 10
A = 20
atomic mass of Ne-20 =
19.992 435 u
atomic mass of H =
1.007 825 u
mn = 1.008 665 u
For 40
20 Ca:
Z = 20
A = 40
atomic mass of Ca-40 =
39.962 591 u
Solutions
N = A − Z = 20 − 10 = 10
∆m = Z(atomic mass of H) + Nmn − atomic mass of Ne-20
∆m = 10(1.007 825 u) + 10(1.008 665 u) − 19.992 435 u
∆m = 10.078 25 u + 10.086 65 u − 19.992 435 u
∆m = 0.172 46 u
Ebind = (0.172 46 u)(931.50 MeV/u) = 160.65 MeV
N = A − Z = 40 − 20 = 20
∆m = Z(atomic mass of H) + Nmn − atomic mass of Ca-40
∆m = 20(1.007 825 u) + 20(1.008 665 u) − 39.962 591 u
∆m = 20.156 50 u + 20.173 30 u − 39.962 591 u
∆m = 0.367 21 u
Ebind = (0.367 21 u)(931.50 MeV/u) = 342.06 MeV
Copyright © Holt, Rinehart and Winston. All rights reserved.
2. For 31H:
N =A −Z =3−1=2
Z =1
∆m = Z(atomic mass of H) + Nmn − atomic mass of H-3
A =3
∆m = 1(1.007 825 u) + 2(1.008 665 u) − 3.016 049 u
atomic mass of H-3 =
3.016 049 u
atomic mass of H =
1.007 825 u
∆m = 1.007 825 u + 2.017 330 u − 3.016 049 u
∆m = 0.009 106 u
Ebind = (0.009 106 u)(931.50 MeV/u) = 8.482 MeV
mn = 1.008 665 u
For 23He:
N=A−Z=3−2=1
Z=2
∆m = Z(atomic mass of H) + Nmn − atomic mass of He-3
A=3
∆m = 2(1.007 825 u) + 1(1.008 665 u) − 3.016 029 u
atomic mass of He-3 =
3.016 029 u
∆m = 2.015 650 u + 1.008 665 u − 3.016 029 u
∆m = 0.008 286 u
Ebind = (0.008 286 u)(931.50 MeV/u) = 7.718 MeV
The difference in binding energy is 8.482 MeV − 7.718 MeV = 0.7640 MeV .
Section One—Pupil’s Edition Solutions
I Ch. 25–1
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Givens
Solutions
3. atomic mass of Ca-43 =
42.958 767 u
mn = 1.008 665 u
∆m = (atomic mass of Ca-42 + mn) − (atomic mass of Ca-43)
∆m = 41.958 618 u + 1.008 665 u − 42.958 767 u
∆m = 0.008 516 u
Ebind of the last neutron = (0.008 516 u)(931.50 MeV/u) = 7.933 MeV
4. For 238
92U:
N = A − Z = 238 − 92 = 146
Z = 92
∆m = Z(atomic mass of H) + Nmn − atomic mass of U-238
A = 238
∆m = 92(1.007 825 u) + 146(1.008 665 u) − 238.050 784 u
atomic mass of U-238 =
238.050 784 u
atomic mass of H =
1.007 825 u
mn = 1.008 665 u
23
5. For 11
Na:
∆m = 92.719 90 u + 147.2651 u − 238.050 784 u
∆m = 1.9342 u
(1.9342 u)(931.50 MeV/u)
E ind
E
b
 =  = 7.5702 MeV/nucleon
= bind
nucleon
A
238
N = A − Z = 23 − 11 = 12
Z = 11
∆m = Z(atomic mass of H) + Nmn − atomic mass of Na-23
A = 23
∆m = 11(1.007 825 u) + 12(1.008 665 u) − 22.989 767 u
atomic mass of Na-23 =
22.989 767 u
atomic mass of H =
1.007 825 u
mn = 1.008 665 u
23
For 12
Mg:
∆m = 11.086 08 u + 12.103 98 u − 22.989 767 u
∆m = 0.200 29 u
(0.200 29 u)(931.50 MeV/u)
E nd
E
bi
 =  = 8.1117 MeV/nucleon
= bind
nucleon
A
23
N = A − Z = 23 − 12 = 11
Z = 12
∆m = Z(atomic mass of H) + Nmn − atomic mass of Mg-23
A = 23
∆m = 12(1.007 825 u) + 11(1.008 665 u) − 22.994 124 u
atomic mass of Mg-23 =
22.994 124 u
∆m = 12.093 90 u + 11.095 32 u − 22.994 124 u
∆m = 0.195 10 u
E nd
E
(0.195 10 u)(931.50 MeV/u)
bi
 =  = 7.9016 MeV/nucleon
= bind
nucleon
A
23
The difference in binding energy per nucleon is
8.1117 MeV − 7.9016 MeV = 0.2101 MeV .
I Ch. 25–2
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
I
atomic mass of Ca-42 =
41.958 618 u
∆m = munbound − mbound
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Section Review, p. 902
Givens
Solutions
6. atomic mass of H =
1.007 825 u
a. N = A − Z = 93 − 41 = 52
∆m = Z(atomic mass of H) + Nmn − atomic mass of Nb-93
mn = 1.008 665 u
∆m = 41(1.007 825 u) + 52(1.008 665 u) − 92.906 376 u
For 93
41Nb:
I
∆m = 41.320 82 u + 52.450 58 u − 92.906 376 u
Z = 41
∆m = 0.865 02 u
A = 93
Ebind = (0.865 02 u)(931.50 MeV/u) = 805.77 MeV
atomic mass of Nb-93 =
92.906 376 u
For 197
79 Au:
b. N = A − Z = 197 − 79 = 118
Z = 79
∆m = Z(atomic mass of H) + Nmn − atomic mass of Au-197
A = 197
∆m = 79(1.007 825 u) + 118(1.008 665 u) − 196.966 543 u
atomic mass of Au-197 =
196.996 543 u
∆m = 79.618 18 u + 119.0225 u − 196.966 543 u
∆m = 1.6741 u
Ebind = (1.6741 u)(931.50 MeV/u) = 1559.4 MeV
For 27
13 Al:
c. N = A − Z = 27 − 13 = 14
Z = 13
∆m = Z(atomic mass of H) + Nmn − atomic mass of Al-27
A = 27
∆m = 13(1.007 825 u) + 14(1.008 665 u) − 26.981 534 u
atomic mass of Al-27 =
26.981 534 u
∆m = 13.101 72 u + 14.121 31 u − 26.981 534 u
∆m = 0.241 50 u
Ebind = (0.241 50 u)(931.50 MeV/u) = 224.96 MeV
Practice 25B, p. 908
Copyright © Holt, Rinehart and Winston. All rights reserved.
1.
0
12
5 B → ? + −1e
+ v
A = 12 − 0 = 12
Z = 5 − (−1) = 6, which is carbon, C
?=
2.
212
4
83Bi → ? + 2He
12
6C
A = 212 − 4 = 208
Z = 83 − 2 = 81, which is thallium, Tl
?=
3. ? → 147 N + −01e + v
208
81 Tl
A = 14 + 0 = 14
Z = 7 + (−1) = 6, which is carbon, C
?=
4.
225
221
89 Ac → 87 Fr + ?
14
6C
A = 225 − 221 = 4
Z = 89 − 87 = 2, which is helium, He
?=
4
2He
Section One—Pupil’s Edition Solutions
I Ch. 25–3
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Givens
Solutions
5. Nickel-63 decays by b − to
copper-63.
b − decay involves an electron and an antineutrino.
63
63
0
28Ni → 29 Cu + −1e
I
6.
56
56
26 Fe → 27 Co + ?
+ v
a. A = 56 − 56 = 0
Z = 26 − 27 = −1
? = −01e, so the decay is b −
b. b − decay involves an electron and an antineutrino.
56
56
0
26 Fe → 27Co + − 1e
+ v
Practice 25C, p. 911
1. T1/2 = 164 × 10–6s
6
N = 2.0 × 10
0.693
0.693
l =  = 
= 4.23 × 103 s–1
T1/2
164 × 10–6s
(4.23 × 103 s–1)(2.0 × 106)
activity = lN = 
= 0.23 Ci
3.7 × 1010 s–1/Ci
2. T1/2 = 19.7 min
9
N = 2.0 × 10
0.693
0.693
l =  =  = 5.86 × 10–4 s–1
(19.7min)(60 s/min)
T1/2
(5.86 × 10–4 s–1)(2.0 × 109)
activity = lN = 
= 3.2 × 10–5 Ci
3.7 × 1010 s–1/Ci
3. T1/2 = 8.07 days
0.693
0.693
l =  = 
T1/2
(8.07 days)(24 h/day)(3600 s/h)
N = 2.5 × 1010
(9.94 × 10−7 s−1)(2.5 × 1010)
activity = lN = 
= 6.7 × 10−7 Ci
3.7 × 1010 s−1/Ci
t = 2.0 h
mf
0.25 × 10−3 g 1
 = 
= 
mi
1.00 × 10−3 g 4
mf = 0.25 × 10−3 g
2×
4. mi = 1.00 × 10−3 g
1

4
1
= 2, so we know that 2 half-lives have passed in 2.0 h.
2.0 h
T1/2 =  =
2
5. T1/2 = 3.82 days
8
N = 4.0 × 10
1.0 h
12 days
a.  ≈ 3 half-lives
3.82 days
1
1 1 1
 ×  ×  = 
2
2 2 8
1

8
(4.0 × 108) = 5.0 × 107 atoms
b. Ndecayed = N − Nremaining
Ndecayed = 4.0 × 108 atoms − 5.0 × 107 atoms
Ndecayed = 3.5 × 108 atoms
I Ch. 25–4
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
l = 9.94 × 10−7 s−1
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Section Review, p. 912
Givens
2.
Solutions
232
4
90 Th → ? + 2He
a. A = 232 − 4 = 228
Z = 90 − 2 = 88, which is radium, Ra
?=
0
12
5B → ? + −1e
+ v
activity = 4.2 h−1
12
6C
c. A = 4 + 145 = 149
Z = 2 + 60 = 62, which is samarium, Sm
?=
3. N = 5.3 × 105 nuclei
I
b. A = 12 − 0 = 12
Z = 5 − (−1) = 6, which is carbon, C
?=
? → 42He + 145
60 Nd
228
88 Ra
149
62 Sm
activity (4.2 h−1)(1 h/3600 s)
a. l =  = 
N
5.3 × 105
l = 1.2 × 10−10 s−1
0.693
0.693
b. T1/2 =  = 
= 5.8 × 109 s
l
1.2 × 10−10 s−1
or (5.8 × 109 s)(1 h/3600 s)(1 day/24 h)(1 year/365.25 days) = 180 years
4. T1/2 = 5730 years
14
C = (0.125)(original 14C)
13
The 14C has been reduced by 0.125 = 8 = 2 = 3 half-lives. Thus, the age of the site is
1
3T1/2 = (3)(5730 years) = 17 200 years .
Copyright © Holt, Rinehart and Winston. All rights reserved.
Chapter Review and Assess, pp. 927–931
7. For 126C:
Z =6
A = 12
atomic mass of C-12 =
12.000 000 u
atomic mass of H =
1.007 825 u
N = A − Z = 12 − 6 = 6
∆m = Z(atomic mass of H) + Nmn − atomic mass of C-12
∆m = 6(1.007 825 u) + 6(1.008 665 u) − 12.000 000 u
∆m = 6.046 950 u + 6.051 990 u − 12.000 000 u
∆m = 0.098 940 u
Ebind = (0.098 940 u)(931.50 MeV/u) = 92.163 MeV
mn = 1.008 665 u
8. For 31H:
N =A −Z =3−1=2
Z =1
∆m = Z(atomic mass of H) + Nmn − atomic mass of H-3
A =3
∆m = 1(1.007 825 u) + 2(1.008 665 u) − 3.016 049 u
atomic mass of H-3 =
3.016 049 u
atomic mass of H =
1.007 825 u
∆m = 1.007 825 u + 2.017 330 u − 3.016 049 u
∆m = 0.009 106 u
Ebind = (0.009 106 u)(931.50 MeV/u) = 8.482 MeV
mn = 1.008 625 u
Section One—Pupil’s Edition Solutions
I Ch. 25–5
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Givens
Solutions
For 32He:
I
N =A −Z =3−2=1
Z =2
∆m = Z(atomic mass of H) + Nmn − atomic mass of He-3
A =3
∆m = 2(1.007 825 u) + 1(1.008 665 u) − 3.016 029 u
atomic mass of He-3 = 3.016
029 u
∆m = 2.015 650 u + 1.008 665 u − 3.016 029 u
∆m = 0.008 286 u
Ebind = (0.008 286 u)(931.50 MeV/u) = 7.718 MeV
9. For 24
12Mg:
Z = 12
A = 24
atomic mass of Mg-24 =
23.985 042 u
atomic mass of H =
1.007 825 u
N = A − Z = 24 − 12 = 12
∆m = Z(atomic mass of H) + Nmn − atomic mass of Mg-24
∆m = 12(1.007 825 u) + 12(1.008 665 u) − 23.985 042 u
∆m = 12.093 90 u + 12.103 98 u − 23.985 042 u
∆m = 0.212 84 u
(0.212 84 u)(931.50 MeV/u)
E nd
E
bi
 =  = 8.2609 MeV/nucleon
= bind
nucleon
A
24
mn = 1.008 665 u
For 85
37 Rb:
Z = 37
A = 85
atomic mass of Rb-85 =
84.911 793 u
N = A − Z = 85 − 37 = 48
∆m = Z(atomic mass of H) + Nmn − atomic mass of Rb-85
∆m = 37(1.007 825 u) + 48(1.008 665 u) − 84.911 793 u
∆m = 37.289 52 u + 48.415 92 u − 84.911 793 u
∆m = 0.793 65 u
(0.793 65 u)(931.50 MeV/u)
E nd
E
bi
 =  = 8.6975 MeV/nucleon
= bind
nucleon
A
85
23. 73 Li + 42 He → ? + 10n
A = 7 + 4 − 1 = 10
Z = 3 + 2 = 5, which is Boron, B
24. 42 He + 94 Be → 126 C + X
A = 4 + 9 − 12 = 1
Z =2+4−6=0
X=
25. ? + 147 N → 11H + 178O
1
0n
a. A = 1 + 17 − 14 = 4
Z = 1 + 8 − 7 = 2, which is helium, He
?=
7
1
4
3 Li + 1H → 2He + ?
4
2He
b. A = 7 + 1 − 4 = 4
Z = 3 + 1 − 2 = 2, which is helium, He
?=
26. T1/2 = 2.42 min
N = 1.67 × 1011
I Ch. 25–6
4
2He
0.693
0.693
l =  =  = 4.77 × 10−3 s−1
(2.42
min)(60.0
s/min)
T1/2
(4.77 × 10−3 s−1)(1.67 × 1011)
activity = lN = 
= 2.2 × 10−2 Ci
3.7 × 1010 s−1/Ci
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
10
5B
?=
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Givens
Solutions
27. T1/2 = 140 days
1
Nf = Ni
16
1
14
 =  , so 4 half-lives have passed.
16
2
28.
4T1/2 = (4)(140 days) = 560 days
14
14
6C = (0.0625)(original 6C)
T1/2 = 5730 years
I
x
= 0.0625
1

2
x = 4, so 4 half-lives have passed.
4T1/2 = (4)(5730 years) = 22 900 years
29. age of the material =
23 000 years
T1/2 = 5730 years
23 000 years
 ≈ 4.0 half-lives
5730 years
14
 = 0.062
2
Approximately 6.2% of the material remains, hence about 100.0% − 6.2% = 93.8%
of the material has decayed.
38. rH = 0.53 × 10−10 m
rnuclear = 2.3 × 1017 kg/m3
mH ≈ mp = 1.673 × 10−27 kg
39. m = 1.99 × 1030 kg
17
ratomic
(3)(1.673 × 10−27 kg)

 = 
= 1.2 × 10−14
rnuclear (4p)(0.53 × 10−10 m)3(2.3 × 1017 kg/m3)
4
3
r = 2.3 × 10 kg/m
40. For 31H:
Copyright © Holt, Rinehart and Winston. All rights reserved.
mH
mH


4 
p
rH3
V
ratomic
3mH
H
3

=  =  =

rnuclear rnuclear rnuclear 4prH3rnuclear
m = Vr = 3pr 3r
1/3
(3)(1.99 × 1030 kg)
= 
(4)(p)(2.3 × 1017 kg/m3)
3m
r = 
4pr
1/3
= 1.3 × 104 m
∆m = Z(atomic mass of H) + Nmn − atomic mass of H-3
Z=1
∆m = 1(1.007825 u) + 2(1.008665 u) − 3.016049 u
N=2
∆m = 1.007825 u + 2.017330 u − 3.016049 u
atomic mass of H = 1.007825 u
∆m = 9.01600 × 10–3 u
mn = 1.008665 u
E bind = (9.01600 × 10–3 u)(931.50 MeV/u) = 8.3984 MeV
For 23H:
∆m = Z(atomic mass of H) + Nmn − atomic mass of He-3
Z=2
∆m = 2(1.007825 u) + 1(1.008665 u) − 3.016029 u
N=1
∆m = 2.015650 u + 1.008665 u − 3.016029 u
atomic mass of He = 3.016029 u
∆m = 8.28600 × 10–3 u
E bind = (8.28600 × 10–3 u)(931.50 MeV/u) = 7.7184 MeV
The energy released is 8.3984 MeV – 7.7184 MeV = 0.6800 MeV
Section One—Pupil’s Edition Solutions
I Ch. 25–7
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Givens
Solutions
41. m of U-238 = 238.050 784 u
∆m = munbound − mbound
m of Th-234 = 234.043 593 u
∆m = (m of U-238) − (m of Th-234 + m of He-4)
m of He-4 = 4.002 602 u
∆m = 238.050 784 u − (234.043 593 u + 4.002 602 u)
∆m = 0.004 589 u
I
E = (0.004 589 u)(931.50 MeV/u) = 4.275 MeV
42. mp = 1.007 276 u
mn = 1.008 665 u
proton = uud
neutron = udd
2mu + md = mp
mu + 2md = mn
mu = mn − 2md
2(mn − 2md) + md = mp
2mn − 4md + md = mp
2mn − mp = 3md
2mn − mp 2(1.008 665 u) − 1.007 276 u
md =  = 
3
3
2.017 330 u − 1.007 276 u 1.010 054 u
md =  =   = 0.336 6847 u
3
3
mu = mn − 2md = 1.008 665 u − 2(0.336 6847 u)
mu = 1.008 665 u − 0.673 3694 u = 0.335 296 u
43. atomic mass of H =
1.007 825 u
mn = 1.008 665 u
atomic mass of O-15 =
15.003 065 u
For 158O:
N = A − Z = 15 − 8 = 7
∆m = Z(atomic mass of H) + Nmn − atomic mass of O-15
∆m = 8(1.007 825 u) + 7(1.008 665 u) − 15.003 065 u
∆m = 8.062 600 u + 7.060 655 u − 15.003 065 u
∆m = 0.120 190 u
Ebind = (0.120 190 u)(931.50 MeV/u) = 111.96 MeV
A = 15
atomic mass of N-15 =
15.000 108 u
For 157 N:
Z =7
A = 15
N = A − Z = 15 − 7 = 8
∆m = Z(atomic mass of H) + Nmn − atomic mass of N-15
∆m = 7(1.007 825 u) + 8(1.008 665 u) − 15.000 108 u
∆m = 7.054 775 u + 8.069 320 u − 15.000 108 u
∆m = 0.123 987 u
Ebind = (0.123 987 u)(931.50 MeV/u) = 115.49 MeV
The difference in binding energy is 115.49 MeV − 111.96 MeV = 3.53 MeV .
44. 10 n + 42He → 21H + 31H
KEmin = Eafter − Ebefore
mn = 1.008 665 u
KEmin = ∆m(931.50 MeV/u)
m of He-4 = 4.002 602 u
KEmin = [(m of H-2 + m of H-3) − (mn + m of He-4)](931.50 MeV/u)
m of H-2 = 2.014 102 u
KEmin = [(2.014 102 u + 3.016 049 u) − (1.008 665 u + 4.002 602 u)](931.50 MeV/u)
m of H-3 = 3.016 049 u
KEmin = (0.018 884 u)(931.50 MeV/u) = 17.590 MeV
I Ch. 25–8
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
Z =8
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Givens
Solutions
45. Ephoton = 2.09 GeV
KEantiproton = 2.09 GeV − 95 MeV − (1.007 276 u)(931.50 MeV/u)(2)
mproton = 1.007 276 u
mantiproton = mproton
Eproton = 95 MeV
0
46. 10 n + 197
79 Au → ? + − 1e + v
KEantiproton = (2.09 × 103 MeV) − 95 MeV − 1876.6 MeV
KEantiproton = 120 MeV
I
a. A = 197 + 1 = 198
Z = 79 − (−1) = 80, which is mercury, Hg
? = 198
80 Hg
0
1
197
198
0 n + 79 Au → 80 Hg + −1e
atomic mass of Au-197 =
196.966 543 u
b. ∆m = munbound − mbound
∆m = (atomic mass of Au-197 + mn) − (atomic mass of Hg-198)
atomic mass of Hg-198 =
197.966 743 u
∆m = 196.966 543 u + 1.008 665 u − 197.966 743 u
∆m = 0.008 465 u
mn = 1.008 665 u
47.
140
Xe and 94Sr are released
as fission fragments.
+ v
E = (0.008 465 u)(931.50 MeV/u) = 7.885 MeV
a.
235
U + 10 n → 140Xe + 94Sr + ?
A = 235 + 1 − (140 + 94) = 2
? = 2 10 n
132
Sn and 101Mo are released as fission fragments.
b.
235
U + 10 n → 132Sn + 101Mo + ?
A = 235 + 1 − (132 + 101) = 3
? = 3 10 n
48. 63 Li + 11p Æ 42He + ?
A =6+1−4=3
Copyright © Holt, Rinehart and Winston. All rights reserved.
Z = 3 + 1 − 2 = 2, which is helium, He
?=
49.
10
4
1
5B + 2He → 1p
+?
3
2He
A = 10 + 4 − 1 = 13
Z = 5 + 2 − 1 = 6, which is carbon, C
?=
50.
18
1
8O + 1p
→ 189F + ?
13
6C
A = 18 + 1 − 18 = 1
Z =8+1−9=0
?=
51.
27
4
30
13 Al + 2He → ? + 15P
1
0n
a. A = 27 + 4 − 30 = 1
Z = 13 + 2 − 15 = 0
?=
1
0n
+ ? → 24He + 37Li
1
0n
b. A = 4 + 7 − 1 = 10
Z = 2 + 3 − 0 = 5, which is boron, B
?=
10
5B
Section One—Pupil’s Edition Solutions
I Ch. 25–9
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Givens
Solutions
52. U-235 absorbs a neutron;
barium-141 and krypton-92
are formed.
a. (1 + 235) − (141 + 92) = 3
53. 2 42He → nucleus A + g
a. A = 2(4) = 8
1
235
U → 141Ba + 92Kr + 3 10 n
0n +
Z = 2(2) = 4, which is beryllium, Be
nucleus A =
8
4
4Be + 2He → nucleus B + g
8
4Be
b. A = 8 + 4 = 12
Z = 4 + 2 = 6, which is carbon, C
nucleus B =
54. N = 7.96 × 1010 atoms
T1/2 = 5730 years
12
6C
0.693
0.693
a. l =  = 
T1/2
(5730 years)(365.25 days/year)(24 h/day)(60 min/h)
l = 2.30 × 10−10 min−1
lN = (2.30 × 10−10 min−1)(7.96 × 1010) = 18.3 decays/min
55. T1/2 = 5.76 years
9
N = 2.0 × 10
0.693
0.693
l =  = 
= 3.81 × 10–9 s–1
(5.76 years)(3.156 × 107 s/year)
T1/2
(3.81 × 10−9 s−1)(2.0 × 109)
activity = lN = 
= 2.1 × 10–10 Ci
3.7 × 1010 s–1/Ci
56. lN = 240.0 mCi
T1/2 = 14 days
0.693
0.693
l =  = 
T1/2
(14 days)(24 h/day)(3600 s/h)
−3
10 −1
lN (240.0 × 10 Ci)(3.7 × 10 s /Ci)
N =  = 
l
5.7 × 10−7 s−1
N = 1.6 × 1016 nuclei
57. activity = 5.0 MCi
9
N = 1.0 × 10
58. T1/2 = 432 years
0.693
T1/2 = 
l
activity (5.0 × 10−6 Ci)(3.7 × 1010 s–1/Ci)
l =  = 
= 1.8 × 10–4 s–1
1.0 × 109
N
0.693
T1/2 = 
= 3.8 × 103 s
1.8 × 10–4 s–1
It takes 10 half-lives to reach 0.1% of its original activity.
Total length of time to reach 0.1% is (432 years)(10) = 4320 years
I Ch. 25–10
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
l = 5.7 × 10−7 s−1
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Givens
Solutions
59. mtot =
(9.1 × 1011 kg)(0.0070)
m t

N = to
atomic mass
atomic mass of U-235 =
3.9 × 10−25 kg
P = 7.0 × 1012 J/s
E = 208 MeV/fission event
E = N(208 MeV)
E N(208 MeV)
(mtot )(208 MeV)

∆t =  =  = 
P
P
(atomic mass)(P )
I
(9.1 × 1011 kg)(0.0070)(208 × 106 eV)(1.60 × 10−19 J/eV)
∆t = 
(3.9 × 10−25 kg)(7.0 × 1012 J/s)
∆t = 7.8 × 1010 s
or (7.8 × 1010 s)(1 h/3600 s)(1 d/24 h)(1 year/365.25 d) = 2500 years
60. E = 208 MeV/fission event
P = 100.0 W
∆t = 1.0 h
Etot = P∆t = N(208 MeV)
P∆t
(100.0 W)(1.0 h)(3600 s/h)
N =  = 
208 MeV (208 × 106 eV)(1.60 × 10−19 J/eV)
N = 1.1 × 1016 fission events
61. P = 1.0 × 103 MW
E = 208 MeV/fission event
conversion efficiency =
30.0%
∆t = 24 h
62. P = 2.0 × 103 kW • h/month
conversion efficiency =
100.0%
Copyright © Holt, Rinehart and Winston. All rights reserved.
E = 208 MeV/fission event
∆t = 1 year
E = P∆t = (N)(208 MeV)(0.300)
(1.0 × 109 W)(24 h)(3600 s/h)
P∆t
N =  = 
(208 MeV)(0.300) (208 × 106 eV)(1.60 × 10−19 J/eV)(0.300)
N = 8.7 × 1024 atoms
E = P∆t = (N)(208 MeV)
P∆t
(2.0 × 106 W• h/month)(12 months)(3600 s/h)
N =  = 
208 MeV
(208 × 106 eV)(1.60 × 10−19 J/eV)
N = 2.6 × 1021 atoms
Section One—Pupil’s Edition Solutions
I Ch. 25–11
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Section
Problem Workbook
Solutions
II
Holt Physics
II
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The Science of Physics
Chapter
1
Additional Practice 1A
Givens
1. distance = 4.35 light years
Solutions
9.461 × 1015 m
distance = 4.35 light years ×  = 4.12 × 1016 m
1 light year
1 Mm
a. distance = 4.12 × 1016 m × 6 = 4.12 × 1010 Mm
10 m
1 pm
 = 4.12 × 1028 pm
b. distance = 4.12 × 1016 m × 
10−12 m
2. energy = 1.2 × 1044 J
1 kJ
a. energy = 1.2 × 1044 J × 
= 1.2 × 1041 kJ
103 J
II
1 nJ
b. energy = 1.2 × 1044 J × −
= 1.2 × 1053 nJ
10 9 J
3. m = 1.0 × 10−16 g
1 Pg
a. m = 1.0 × 10−16 g × 1
= 1.0 × 10−31 Pg
10 5 g
1 fg
b. m = 1.0 × 10−16 g × −1
= 0.10 fg
10 5 g
1 ag
c. m = 1.0 × 10−16 g × −1
= 1.0 × 102 ag
10 8 g
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. distance = 152 100 000 km
1000 m
1 ym
 = 1.521 × 1035 ym
a. distance = 152 100 000 km ×  × 
1 km
10−24 m
1000 m
1 Ym
b. distance = 152 100 000 km ×  × 24 = 1.521 × 10−13 Ym
1 km
10 m
5. energy = 2.1 × 1015 W • h
1 J/s 3600 s
a. energy = 2.1 × 1015 W • h ×  ×  = 7.6 × 1018 J
1W
1h
1 GJ
b. energy = 7.6 × 1018 J × 
= 7.6 × 109 GJ
109 J
6. m = 1.90 × 105 kg
1 eV
m = 1.90 × 105 kg × 
= 1.07 × 1041 eV
1.78 × 10−36 kg
1 MeV
a. m = 1.07 × 1041 eV × 6 = 1.07 × 1035 MeV
10 eV
1 TeV
 = 1.07 × 1029 TeV
b. m = 1.07 × 1041 eV × 
1012 eV
Section Two — Problem Workbook Solutions
II Ch. 1–1
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Givens
Solutions
7. m = (200)(2 × 1030 kg) =
4 × 1032 kg
103 g 103mg
a. m = 4 × 1032 kg × × = 4 × 1038 mg
1 kg
1g
1 Eg
103 g
b. m = 4 × 1032 kg ×  × 1
= 4 × 1017 Eg
10 8 g
1 kg
8. area = 166 241 700 km2
depth = 3940 m
V = volume = area × depth
17
2
1000 m
V = (166 241 700 km2)(3940 m) × 
1 km
3
V = 6.55 × 10 m
106 cm3
a. V = 6.55 × 1017 m3 × 
= 6.55 × 1023 cm3
1 m3
109 mm3
b. V = 6.55 × 1017 m3 × 
= 6.55 × 1026 mm3
1 m3
Copyright © by Holt, Rinehart and Winston. All rights reserved.
II
II Ch. 1–2
Holt Physics Solution Manual
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Motion In One Dimension
Chapter
2
Additional Practice 2A
Givens
1. ∆x = 443 m
vavg = 0.60 m/s
2. vavg = 72 km/h
∆x = 1.5 km
3. ∆x = 5.50 × 102 m
Solutions
∆x
443 m
∆t =  =  = 740 s = 12 min, 20 s
vavg 0.60 m/s
1.5 km
∆x
∆t =  =  = 75 s
1h
km
vavg
72  
h 3600 s
vavg = 1.00 × 102 km/h
5.50 × 102 m
∆x
a. ∆t =  =  = 19.8 s
vavg
1h
km
1000 m
1.00 × 102   
h 3600 s
1 km
vavg = 85.0 km/h
b. ∆x = ∆v avg ∆t
1h
∆x = (85.0 km/h) 
3600 s
4. ∆x1 = 1.5 km
v1 = 85 km/h
∆x1 = 0.80 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v2 = 67 km/h
II
103 m
 (19.8 s) = 468 m
1 km
∆x ∆x
a. ∆ttot = ∆t1 + ∆t2 = 1 + 2
v1
v2
1.5 km
0.80 km
∆ttot =  +  = 64 s + 43 s = 107 s
km
km
1h
1h
85  
67  
h 3600 s
h 3600 s
1.5 km + 0.80 km
2.3 km
∆x1 + ∆x2
b. v avg = 
=  =  = 77 km/h
1h
1h
∆t1 + ∆t2
(64 s + 43 s) 
(107 s) 
3600
3600 s
5. r = 7.1 × 104 km
∆t = 9 h, 50 min
∆x = 2πr
∆x
2π (7.1 × 107 m)
4.5 × 108 m
vavg =  =  = 
∆t
60 s
60 s
60 min
(9 h)  + 50 min 
(540 min + 50 min) 
1 min
1 min
1h
4.5 × 108 m
vavg = 
60 s
(590 min) 
1 min
vavg = 1.3 × 104 m/s
Thus the average speed = 1.3 × 104 m/s.
On the other hand, the average velocity for this point is zero, because the point’s displacement is zero.
Section Two — Problem Workbook Solutions
II Ch. 2–1
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Givens
Solutions
6. ∆x = –1.73 km
∆t = 25 s
7. vavg,1 = 18.0 km/h
∆t1 = 2.50 s
∆t2 = 12.0 s
∆x –1.73 × 103 m
vavg =  =  = –69 m/s = –250 km/h
∆t
25 s
1h
103 m
a. ∆x1 = vavg,1∆t1 = (18.0 km/h)   (2.50s) = 12.5 m
3600 s 1 km
∆x2 = –∆x1 = –12.5 m
∆x
–12.5 m
vavg,2 = 2 =  = –1.04 m/s
∆t2
12.0 s
∆x1 + ∆x2 12.5 m + (−12.5 m) 0.0 m
 =  =  = 0.0 m/s
b. vavg,tot = 
14.5 s
∆t1 + ∆t2
2.50 s + 12.0 s
c. total distance traveled = ∆x1 – ∆x2 = 12.5 m – (–12.5 m) = 25.0 m
total time of travel = ∆t1 + ∆t2 = 2.50 s + 12.0 s = 14.5 s
total distance
25.0 m
average speed =  =  = 1.72 m/s
total time
14.5 s
II
8. ∆x = 2.00 × 102 km
∆t = 5 h, 40 min, 37 s
2.00 × 105 m
∆x
2.00 × 105m
a. vavg =  =  = 
∆t
20 437 s
3600 s
60 s
5 h  + 40 min  + 37 s
h
min
vavg = 9.79 m/s = 35.2 km/h
1
∆x⬘ = 2∆x
1h
∆t = (9.73 × 103 s)  = 2.70 h
3600 s
60 min
(0.70 h)  = 42 min
1h
∆t = 2 h, 42 min
II Ch. 2–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vavg ⬘ = (1.05)vavg
2.00 × 105 m

2
∆x⬘
b. ∆t =  =  = 9.73 × 103 s
vavg⬘
m
(1.05) 9.79 
s
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Additional Practice 2B
Givens
Solutions
1. vi = 0 km/h = 0 m/s
60 s
vf = aavg ∆t + vi = (1.80 m/s2)(1.00 min)  + 0 m/s = 108 m/s
1 min
aavg = 1.8 m/s2
3600 s 1 km
vf = 108 m/s = (108 m/s)  3 = 389 km/h
1 h 10 m
∆t = 1.00 min
2. ∆t = 2.0 min
60 s
vf = aavg ∆t + vi = (0.19 m/s2) (2.0 min)  + 0 m/s = 23 m/s
1 min
2
aavg = 0.19 m/s
vi = 0 m/s
3. ∆t = 45.0 s
vf = aavg ∆t + vi = (2.29 m/s2)(45.0 s) + 0 m/s = 103 m/s
aavg = 2.29 m/s2
vi = 0 m/s
4. ∆x = 29 752 m
∆t = 2.00 h
∆x
29 752 m
a. vavg =  =  = 4.13 m/s
∆t
3600 s
(2.00 h) 
1h
vi = 3.00 m/s
∆v 4.13 m/s − 3.00 m/s
1.13 m/s
b. aavg =  =  =  = 3.77 × 10−2 m/s2
∆t
30.0 s
30.0 s
vf = 4.13 m/s
II
∆t = 30.0 s
10.0 m
5. ∆x = (15 hops) 
1 hop
= 1.50 × 102 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 60.0 s
∆x 1.50 × 102 m
a. vavg =  =  = +2.50 m/s
∆t
60.0 s
vf − vi 0 m/s − 2.50 m/s
−2.50 m/s
 =  =  = −1.0 × 101 m/s2
b. aavg = 
∆tstop
0.25 s
0.25 m/s
∆t stop = 0.25 s
vf = 0 m/s
vi = vavg = +2.50 m/s
6. ∆x = 1.00 × 102 m, backward
= −1.00 × 102 m
∆t = 13.6 s
∆t ⬘ = 2.00 s
∆x −1.00 × 1 02 m
vavg =  =   = −7.35 m/s
∆t
13.6 s
−7.35
m/s − 0 m/s
vf − vi
 =  = 3.68 m/s2
aavg = 
2.00 s
∆t⬘
vi = 0 m/s
vf = vavg
7. ∆x = 150 m
vi = 0 m/s
vf = 6.0 m/s
vavg = 3.0 m/s
∆x
150 m
a. ∆t =  =  = 5.0 × 101 s
vavg 3.0 m/s
vf − vi 6.0 m/s − 0 m/s
b. aavg =  = 
= 0.12 m/s2
∆t
5.0 × 101
Section Two — Problem Workbook Solutions
II Ch. 2–3
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Givens
Solutions
8. vi = +245 km/h
103 m
 = +68.1 m/s
1 km
aavg = −3.0 m/s2
km 1 h
vi = 245  
h 3600 s
vf = vi −(0.200) vi
vf = (1.000 − 0.200) vi = (0.800)(68.1 m/s) = +54.5 m/s
vf − vi 54.5 m/s − 68.1 m/s
−13.6 m/s
∆t =  = 
= 2 = 4.5 s
2
− 3.0 m/s
aavg
−3.0 m/s
9. ∆x = 3.00 km
∆t = 217.347 s
aavg = −1.72 m/s2
vf = 0 m/s
10. ∆x = +5.00 × 102 m
∆t = 35.76 s
vi = 0 m/s
II
∆t ⬘ = 4.00 s
3
∆x 3.00 × 10 m
vi = vavg =  =  = 13.8 m/s
217.347 s
∆t
vf − vi 0 m/s − 13.8 m/s
−13.8 m/s
tstop =  = 
= 2 = 8.02 s
−1.72 m/s2
−1.72 m/s
aavg
∆x
5.00 × 102 m
vf = vmax = (1.100)vavg = (1.100)  = (1.100)  = +15.4 m/s
∆t
35.76 s
15.4 m/s − 0 m/s
∆v vf − vi
aavg =  =  =  = + 3.85 m/s2
4.00 s
∆t ⬘
∆t ⬘
vmax = vavg + (0.100) vavg
Additional Practice 2C
1. ∆x = 115 m
vi = 4.20 m/s
(2)(115 m)
2∆x
(2)(115 m)
∆t =  =  =  = 25.0 s
vi + vf
4.20 m/s + 5.00 m/s
9.20 m/s
vf = 5.00 m/s
2. ∆x = 180.0 km
vi = 3.00 km/s
2∆x
360.0 km
(2)(180.0 km)
∆t =  =  =  = 1.2 × 102 s
vi + vf
3.00 km/s
3.00 km/s + 0 km/s
3. vi = 0 km/h
vf = 1.09 × 103 km/h
∆x = 20.0 km
(2)(20.0 × 103 m)
2∆x
a. ∆t =  = 
vi + vf
1 h 1000 m
(1.09 × 103 km/h + 0 km/h)  
3600 s 1 km
40.0 × 103 m
∆t =  = 132 s
1h
1000 m
(1.09 × 103 km/h)  
3600 s 1 km
∆x = 5.00 km
vi = 1.09 × 103 km/h
vf = 0 km/h
(2)(5.00 × 103 m)
2∆x
b. ∆t =  = 
vi + vf
1 h 1000 m
(1.09 × 103 km/h + 0 km/h)  
3600 s 1 km
10.0 × 103 m
∆t =  = 33.0 s
1h
1000 m
(1.09 × 103 km/h)  
3600 s 1 km
II Ch. 2–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vf = 0 km/s
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Givens
Solutions
4. vi = vavg = 518 km/h
vf = (0.600) vavg
1h
km
103m
vavg = 518    = 8.63 × 103 m/min
h 60 min 1 km
∆t = 2.00 min
∆x = 2(vi + vf )∆t = 2[vavg + (0.600) vavg ]∆t = 2(1.600)(8.63 × 103 m/min)(2.00 min)
1
1
1
∆x = 13.8 × 103 m = 13.8 km
5. ∆t = 30.0 s
vi = 30.0 km/h
vf = 42.0 km/h
1h
1
1
∆x = 2(vi + vf )∆t = 2(30.0 km/h + 42.0 km/h)  (30.0 s)
3600 s
1h
km
1
∆x = 2 72.0   (30.0 s)
h 3600 s
∆x = 3.00 × 10−1 km = 3.00 × 102 m
6. vf = 96 km/h
vi = 0 km/h
∆t = 3.07 s
1h
103 m
1
1
∆x = 2(vi + vf )∆t = 2(0 km/h + 96 km/h)   (3.07 s)
3600 s 1 km
∆x =
7. ∆x = 290.0 m
1

2
96 × 10 h(8.53 + × 10
3m
−4
h) = 41 m
∆t = 10.0 s
2∆x
(2)(290.0 m)
vi =  − vf =  − 0 m/s = 58.0 m/s = 209 km/h
∆t
10.0 s
vf = 0 km/h = 0 m/s
(Speed was in excess of 209 km/h.)
8. ∆x = 5.7 × 103 km
∆t = 86 h
vf = vi + (0.10) vi
II
2∆x
vf + vi = 
∆t
2∆x
vi (1.00 + 0.10) + vi = 
∆t
(2)(5.7 × 103 km)
vi =  = 63 km/h
(2.10)(86 h)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. vi = 2.60 m/s
vf = 2.20 m/s
60 s
1
1
1
∆x = 2(vi + vf )∆t = 2(2.60 m/s + 2.20 m/s)(9.00 min)  = 2(4.80 m/s)(5.40 × 102 s)
min
∆t = 9.00 min
∆x = 1.30 × 103 m = 1.30 km
Additional Practice 2D
1. vi = 186 km/h
vf = 0 km/h = 0 m/s
a = −1.5 m/s2
2. vi = −15.0 m/s
vf = 0 m/s
2
a = +2.5 m/s
vi = 0 m/s
vf = +15.0 m/s
a = +2.5 m/s
1h
103 m
0 m/s − (186 km/h)  
vf − vi
3600 s 1 km
−51.7 m/s
∆t =  = 
= 2 = 34 s
a
−1.5 m/s
−1.5 m/s2
For stopping:
vf − vi 0 m/s − (−15.0 m/s) 15.0 m/s
∆t1 =  = 
= 2 = 6.0 s
a
2.5 m/s2
2.5 m/s
For moving forward:
vf − vi 15.0 m/s − 0.0 m/s 15.0 m/s
∆t 2 =  = 
= 2 = 6.0 s
2.5 m/s2
a
2.5 m/s
∆t tot = ∆t 1 + ∆t 2 = 6.0 s + 6.0 s = 12.0 s
Section Two — Problem Workbook Solutions
II Ch. 2–5
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Givens
Solutions
vf − vi
∆t = 
a
3. vi = 24.0 km/h
vf = 8.0 km/h
1h
103 m
(8.0 km/h − 24.0 km/h)  
3600 s 1 km
∆t = 
−0.20 m/s2
a = −0.20 m/s2
1h
km
103 m
−16.0   
h 3600 s 1 km
∆t = 
= 22 s
−0.20 m/s2
4. v1 = 65.0 km/h
For cage 1:
∆x = v1∆t1
vi,2 = 0 km/h
−2
a2 = 4.00 × 10
2
m/s
∆x = 2072 m
∆x
2072 m
∆t1 =  =  = 115 s
v1
103 m
1h
(65.0 km/h)  
3600 s 1 km
For cage 2:
1
∆x = vi,2 ∆t2 + 2a2 ∆t2 2
II
Because vi,2 = 0 km/h,
∆t2 =
72 m)
0(20)×(2100
m/s = 322 s
2a∆x = 
4.
−2
2
2
Cage 1 reaches the bottom of the shaft in nearly a third of the time required for cage 2.
v = 105.4 km/h
vi,car = 0 m/s
∆x
2.00 × 102 m
a. ∆t =  =  = 6.83 s
v
km
1h
103 m
105.4   
h 3600 s 1 km
1
b. ∆x = vi,car ∆t + 2acar ∆t2
2∆x (2)(2.00 × 102 m)
acar = 
= 
= 8.57 m/s2
∆t2
(6.83 s)2
6. vi = 6.0 m/s
2
a = 1.4 m/s
1
1
∆x = vi∆t + 2a∆t 2 = (6.0 m/s)(3.0 s) + 2(1.4 m/s2)(3.0 s)2 = 18 m + 6.3 m = 24 m
∆t = 3.0 s
7. vi = 3.17 × 102 km/h
vf = 2.00 × 102 km/h
∆t = 8.0 s
1h
103 m
(2.00 × 102 km/h − 3.17 × 102 km/h)  
3600 s 1 km
vf − v
a = i = 
8.0 s
∆t
1h
103 m
(−117 km/h)  
3600 s 1 km
a =  = −4.1 m/s2
8.0 s
1h
103 m
1
1
∆x = vi∆t + 2a∆t2 = (3.17 × 102 km/h)   (8.0 s) + 2(−4.1 m/s2)(8.0 s)2
3600 s 1 km
∆x = (7.0 × 102 m) + (−130 m) = +570 m
II Ch. 2–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. ∆x = 2.00 × 102 m
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Givens
Solutions
8. vi = 0 m/s
vf − vi 3.06 m/s − 0 m/s
∆t1 =  = 
= 3.82
a
0.800 m/s2
vf = 3.06 m/s
a = 0.800 m/s2
1
1
∆x1 = vi∆t1 + 2a∆t1 2 = (0 m/s) (3.82 s) + 2(0.800 m/s2) (3.82 s)2 = 5.84 m
∆t2 = 5.00 s
∆x2 = vf ∆t2 = (3.06 m/s)(5.00 s) = 15.3 m
∆xtot = ∆x1 + ∆x2 = 5.84 m + 15.3 m = 21.1 m
9. vf = 3.50 × 102 km/h
vi = 0 km/h = 0 m/s
a = 4.00 m/s2
1h
103 m
(3.50 × 102 km/h − 0 km/h)  
(vf − vi)
3600 s 1 km
∆t =  = 
= 24.3 s
(4.00 m/s2)
a
1
1
∆x = vi∆t + 2a∆t 2 = (0 m/s)(24.3 s) + 2(4.00 m/s2)(24.3 s)2
∆x = 1.18 × 103 m = 1.18 km
10. vi = 24.0 m/s
2
a = −0.850 m/s
vf = vi + a∆t = 24.0 m/s + (− 0.850 m/s2)(28.0 s) = 24.0 m/s − 23.8 m/s = +0.2 m/s
II
∆t = 28.0 s
11. a = +2.67 m/s2
∆t = 15.0 s
∆x = +6.00 × 102m
∆x 1
6.00 × 102 m 1
vi =  − 2a∆t =  − 2(2.67 m/s2)(15.0 s) = 40.0 m/s − 20.0 m/s = +20.0 m/s
∆t
15.0 s
vi = vf − a∆t
12. a = 7.20 m/s2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
vi ∆t = ∆x − 2a∆t 2
∆t = 25.0 s
vi = (3.00 × 102 m/s) − (7.20 m/s2)(25.0 s) = (3.00 × 102 m/s) − (1.80 × 102 m/s)
vf = 3.00 × 102 ms
vi = 1.20 × 102 m/s
13. vi = 0 m/s
1
2
∆x = 1.00 × 10 m
∆t = 12.11 s
14. vi = 3.00 m/s
∆x = 1.00 × 102 m
∆t = 12.11 s
∆x = vi ∆t + 2a∆t2
Because vi = 0 m/s,
2∆x (2)(1.00 × 102 m)
a = 2 = 
= 1.36 m/s2
∆t
(12.11 s)2
2(∆x − vi∆t) (2)[1.00 × 102 m − (3.00 m/s)(12.11 s)]
a = 
= 
∆t 2
(12.11 s)2
(2)(1.00 × 102 m − 36.3 m)
a = 
(12.11 s)2
(2)(64 m)
a = 2 = 0.87 m/s2
(12.11 s)
15. vf = 30.0 m/s
vi = 18.0 m/s
vf − vi 30.0 m/s − 18.0 m/s 12.0 m/s
a =  =  =  = 1.5 m/s2
∆t
8.0 s
8.0 s
∆t = 8.0 s
Section Two — Problem Workbook Solutions
II Ch. 2–7
Additional Practice 2E
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Givens
Solutions
1 h 2 103 m 2
(965 km/h) 2 − (0 km/h)2  
3600 s 1 km
vf − vi
∆x =  = 
2
(2)(4.0 m/s )
2a
1. vi = 0 km/h
2
vf = 965 km/h
a = 4.0 m/s2
2
7.19 × 104m2/s2
∆x = 
= 9.0 × 103 m = 9.0 km
8.0 m/s2
2. vi = (0.20) vmax
3
vmax = 2.30 × 10 km/h
vf = 0 km/h
a = −5.80 m/s2
3. vf = 9.70 × 102 km/h
II
vi = (0.500)vf
a = 4.8 m/s2
1 h 2 103 m 2
(0 km/h)2 − (0.20)2 (2.30 × 103 km/h)2  
3600 s 1 km
∆x =  = 
2
2a
(2)(−5.80 m/s )
vf2 − vi2
−1.63 × 104m2/s2
∆x = 
= 1.41 × 103 m = 1.41 km
−11.6 m/s2
1 h 2 103 m 2
(9.70 × 102 km/h)2 −(0.50)2 (9.70 × 102 km/h)2  
vf − vi
3600 s 1 km
∆x =  = 
2a
(2)(4.8 m/s2)
2
2
1 h 2 103 m 2
(9.41 × 105 km2/h)2 −2.35 × 105 km2/h2)  
3600 s 1 km
∆x = 
(2)(4.8 m/s2)
1 h 2 103 m 2
(7.06 × 105 km2/h2)  
3600 s 1 km
∆x = 
(2)(4.8 m/s2)
5.45 × 104m2/s2
∆x = 
= 5.7 × 103 m = 5.7 km
9.6 m/s2
∆x = 40.0 m
a = 2.00 m/s2
5. ∆x = +9.60 km
a = −2.0 m/s2
vf = 0 m/s
6. a = +0.35 m/s2
vi = 0 m/s
∆x = 64 m
7. ∆x = 44.8 km
∆t = 60.0 min
a = −2.0 m/s2
∆x = 20.0 m
vi = 12.4 m/s
II Ch. 2–8
2
vf = 2a
)2 = 1.
02m
s2
+64m
s2
∆
x+v
)(
2.
0m
/s
2)(4
0.
m
)+(8.
0m
/s
60
×1
2/
2/
i = (2
vf = 22
s2 = ± 15 m/s = 15 m/s
4m
2/
vi =
vf2
−2a∆
x =
(0
m/s
)2
−(2)
(−
2.
0m
/s
2)(9
.6
0×1
03
m)
vi = 3.
04m
s2 = ±196 m/s = +196 m/s
84
×1
2/
2
vf = 2a
m/s
)2
∆
x+v
)(
0.
35
2)(6
4m
)+(0m
/s
i = (2
vf = 45
m2s2 = ±6.7 m/s = +6.7 m/s
∆x
44.8 × 103 m
a. vavg =  =  = 12.4 m/s
∆t (60.0 min)(60 s/min)
2
b. vf = 2a
m/s
)2
∆
x+v
)(
−2.
0m
/s
2)(2
0.
0m
)+(12
.4
i = (2
vf = (−
m2/
s2)+
m2/
s2 = 74
m2/
s2 = ±8.6 m/s = 8.6 m/s
80
.0
154
Holt Physics Solution Manual
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4. vi = 8.0 m/s
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Givens
8. ∆x = 2.00 × 102 m
a = 1.20 m/s2
vf = 25.0 m/s
9. ∆x = 4.0 × 102 m
∆t = 11.55
vi = 0 km/h
vf = 2.50 × 102 km/h
10. vi = 25.0 km/h
vf = 0 km/h
∆x = 16.0 m
Solutions
vf2−
)2
−(2)
02
m)
2a∆
x = (2
5.
0m
/s
(1
.2
0m
/s
2)(2
.0
0×1
vi = 62
5m
2/s2−4.8
0×102m
2/s2
vi = 14
5m
2/s2 = ±12.0 m/s = 12.0 m/s
vi =
2
1 h 2 103 m
2
2
2

 
(2.50
×
10
km/h)
−
(0
km/h)
3600 s 1 km
vf2 − vi2

a =  =
(2)(4.0 × 102 m)
2∆x
4.82 × 103m2/s2
= 6.0 m/s2
a = 
8.0 × 102 m
2
1 h 2 103 m
(0 km/h)2 − (25.0 km/h)2  
3600 s 1 km
a =  = 
2∆x
(2)(16.0 m)
vf2 − vi2
−4.82 m2/s2
a =  = −1.51 m/s2
32.0 m
II
Additional Practice 2F
1. ∆y = −343 m
a = −9.81 m/s2
vi = 0 m/s
2. ∆y = +4.88 m
vi = +9.98 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
a = −9.81 m/s2
3. ∆y = −443 m + 221 m
= −222 m
a = −9.81 m/s2
2
vf = 2a
m/s
m2/s2
∆
y+v
)(
−9.
81
2)(−
34
3m
)+(0m
/s
)2 = 67
30
i = (2
vf = ±82.0 m/s = −82.0 m/s
2
2a
m/s
m/s
m2/s2
+99.
∆
y+v
)(
−9.
81
2)(4
.8
8m
)+(9.
98
)2 = −95
.7
6m
2/s2
i = (2
vf = 3.
m2/s2 = ±1.97 m/s = ±1.97 m/s
90
vf =
m/s
)2 = 43
m2/
s2
vf = 2a
∆
y−vi2 = (2
)(
−9.
81
2)(−
22
2m
)−(0m
/s
60
vf = ±66.0 m/s = −66.0 m/s
vi = 0 m/s
4. ∆y = +64 m
a = −9.81 m/s2
∆t = 3.0 s
1
∆y = vi ∆t + 2a∆t2
1
1
∆y − 2a∆t2
64 m − 2(−9.81 m/s2)(3.0 s)2
64 m + 44 m
vi =  =  = 
∆t
3.0 s
3.0 s
108 m
vi =  = 36 m/s
3.0 s
5. ∆y = −111 m
∆t = 3.80 s
a = −9.81 m/s2
initial speed of arrow = 36 m/s
1
∆y = vi ∆t + 2a∆t 2
1
1
∆y − 2a∆t2
−111 m − 2(−9.81 m/s2)(3.80 s)2
−111 m + 70.8 m
vi = 
= 
= 
∆t
3.80 s
3.80 s
−40.2 m
vi =  = −10.6 m/s
3.80 s
Section Two — Problem Workbook Solutions
II Ch. 2–9
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Givens
Solutions
When vi = 0 m/s,
6. ∆y = −228 m
a = −9.81 m/s2
∆t =
vi = 0 m/s
2∆y
 = 6.82 s
a = −9.81 m/s
(2)(−228 m)
2
In the presence of air resistance, the sandwich would require more time to fall because the downward acceleration would be reduced.
7. vi = 12.0 m/s, upward =
+12.0 m/s
vf = 3.0 m/s, upward =
+3.0 m/s
a = −9.81 m/s 2
yi = 1.50 m
−135 m2/s2
∆y = 2 = 6.88 m
−19.6 m/s
height of nest from ground = h
∆y = h − yi
8. ∆y = +43 m
2
a = −9.81 m/s
II
vf 2 − vi2 (3.0 m/s)2 − (12.0 m/s)2 9.0 m3/s2 − 144 m2/s2
= 
∆y =  = 
(2 )(−9.81 m/s2)
(2)(−9.81 m/s2)
2a
vf = 0 m/s
h = ∆y + yi = 6.88 m + 1.50 m = 8.38 m
Because it takes as long for the ice cream to fall from the top of the flagpole to the
ground as it does for the ice cream to travel up to the top of the flagpole, the free-fall
case will be calculated.
1
Thus, vi = 0 m/s, ∆y = −43 m, and ∆y = 2a∆t2.
∆t =
9. ∆ymax = +21 cm
a = −9.81 m/s2
vf = 0 m/s
∆y = +7.0 cm
vi =
 = 3.0 s
2a∆y = −9.81 m/s
(2)(−43 m)
2
2 )(2 1 × 10−1 m) = 4.1 m2/s2
vf 2−
ymax
m/s
)2
−
(2)(
2a∆
= (0
−9
.8
1m
/s
. vi = +2.0 m/s
For the flea to jump +7.0 cm = +7.0 × 10−2 m = ∆y,
1
∆y = vi ∆t + 2a∆t 2
1
a∆t 2
2
or
+ vi ∆t − ∆y = 0
a
(vi )2 − 4  (−∆y )
2
∆t = 
a
2 
2
−vi ±
(2
m/s
)2
−(2)
m/s
.0
(−
9.
81
2)(−
7.
0×10−2m
)
∆t = 
2
−2.0 m/s ±
−9.81 m/s
−2.0 m/s ± 4.
s2
−1.4
m2/
s2
0m
2/
∆t = 
=
−9.81 m/s2
2.0 m/s ± 2.
6m
2/s2

2
9.81 m/s
2.0 m/s ± 1.6 m/s
∆t = 
= 0.37 s or 0.04 s
9.81 m/s2
To choose the correct value for ∆t, insert ∆t, a, and vi into the equation for vf .
vf = a∆t + vi = (−9.81 m/s2)(0.37 s) + 2.0 m/s
vf = (−3.6 m/s) + 2.0 m/s = −1.6 m/s
vf = a∆t + vi = (−9.81 m/s2)(0.04 s) + 2.0 m/s
vf = (−0.4 m/s) + 2.0 m/s = +1.6 m/s
Because vf is still directed upward, the shorter time interval is correct. Therefore,
∆t = 0.04 s
II Ch. 2–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Solving for ∆t by means of the quadratic equation,
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Two-Dimensional Motion and Vectors
Chapter
3
Additional Practice 3A
Givens
1. ∆tx = 7.95 s
Solutions
d 2 = ∆x 2 + ∆y 2
∆y = 161 m
∆x = d 2−
∆y2 = (2
m
)2
−(16
)2 = 5.
04
m2−
04
m2
26
1m
11
×1
2.5
9×1
d = 226 m
∆x = 2.
04
m2 = 159 m
52
×1
∆x = 159 m
∆x 159 m
v =  =  = 20.0 m/s
∆tx 7.95 s
2. d1 = 5.0 km
θ1 = 11.5°
2
d = 1.0 km
q2 = −90.0°
∆xtot = d1(cos q1) + d2(cos q2) = (5.0 km)(cos 11.5°) + (1.0 km)[cos(−90.0°)]
II
∆xtot = 4.9 km
∆ytot = d1(sin q1) + d2(sin q2) = (5.0 km)(sin 11.5°) + (1.0 km)[sin(−90.0°)]
= 1.0 km − 1.0 km
∆ytot = 0.0 km
d = (∆
xt
)2
+(∆
ytot
)2 = (4
)2
+(0.
)2
.9
km
0km
ot
d = 4.9 km
∆ytot
0.0 km
= tan−1  = 0.0°, or due east
q = tan−1 
∆xtot
4.9 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. ∆x = 5 jumps
1 jump = 8.0 m
d = 68 m
d 2 = ∆x 2 + ∆y 2
∆y = d 2−
∆x2 = (6
)2
−[(5
03
m2−
03
m2
8m
)(
8.
0m
)]
2 = 4.
6×1
1.6
×1
∆y = 3.
03
m = 55 m
0×1
55 m
number of jumps northward =  = 6.9 jumps = 7 jumps
8.0 m/jump
∆x
(5)(8.0 m)
q = tan−1  = tan−1  = 36° west of north
∆y
55 m
4. ∆x = 25.2 km
∆y = 21.3 km
d= ∆
∆y2 = (2
)2
+(21
)2
x2+
5.
2km
.3
km
d = 63
5km
2+454
km
2 = 10
89
km
2
d = 33.00 km
∆y
21.3 km
q = tan−1  = tan−1 
∆x
25.2 km
q = 42.6° south of east
Section Two — Problem Workbook Solutions
II Ch. 3–1
Givens
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5. ∆y = −483 m
∆x = 225 m
Solutions
∆y
−483
q = tan−1  = tan−1  = −65.0° = 65.0° below the waters surface
∆x
225 m
2
d= ∆
∆y2 = (2
m
)2
+(−
)2
x+
25
48
3m
d = 5.
m2+
m2 = 2.
m2
06
×104
2.3
3×105
84
×105
d = 533 m
6. v = 15.0 m/s
∆tx = 8.0 s
d = 180.0 m
d 2 = ∆x 2 + ∆y 2 = (v∆tx)2 + (v∆ty)2
d 2 = v 2(∆tx2 + ∆ty2)
∆ty =
180.0 m 2
 − (8.0 s)2 = 14
s2
−64
s2 = 8.
01s2
4
0×1
15.0 m/s
d

v
2
− ∆tx2 =
∆ty = 8.9 s
7. v = 8.00 km/h
∆tx = 15.0 min
II
∆ty = 22.0 min
d= ∆
∆y2 = (v
tx
)2
+(v∆
ty
)2
x2+
∆
=v ∆
tx2+
∆
ty2
(15.0min) +(22.0min)
1h d = (8.00 km/h)  22
60 min 5min+484min
8.00 km d =  70
60 min 9min = 3.55 km
v∆t
∆t
∆y
22.0 min
q = tan  = tan  = tan  = tan 
∆x v∆t ∆t 15.0 min
1h
d = (8.00 km/h) 
60 min
2
2
2
2
2
−1
−1
−1
y
x
−1
y
x
q = 55.7° north of east
1. d = (5)(33.0 cm)
∆y = 88.0 cm
88.0 cm
∆y
q = sin−1  = sin−1  = 32.2° north of west
(5)(33.0 cm)
d
∆x = d(cos q) = (5)(33.0 cm)(cos 32.2°) = 1.40 × 102 cm to the west
2. q = 60.0°
d = 10.0 m
3. d = 10.3 m
∆y = −6.10 m
∆x = d(cos q) = (10.0 m)(cos 60.0°) = 5.00 m
∆y = d(sin q) = (10.0 m)(sin 60.0°) = 8.66 m
Finding the angle between d and the x-axis yields,
∆y
−6.10 m
q1 = sin−1  = sin−1  = −36.3°
d
10.3 m
The angle between d and the negative y-axis is therefore,
q = −90.0 − (−36.3°) = −53.7°
q = 53.7° on either side of the negative y-axis
d 2 + ∆x 2 + ∆y2
∆x = d 2−
∆
y2 = (1
)2
−(−
m)2 = 10
m2
0.
3m
6.
10
6m
2−37.
2m
2 = 69
∆x = ±8.3 m
II Ch. 3–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Additional Practice 3B
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Givens
4. d = (8)(4.5 m)
q = 35°
Solutions
∆x = d(cos q) = (8)(4.5 m)(cos 35°) = 29 m
∆y = d(sin q) = (8)(4.5 m)(sin 35°) = 21 m
5. v = 347 km/h
vx = v(cos q) = (347 km/h)(cos 15.0°) = 335 km/h
q = 15.0°
vy = v(sin q) = (347 km/h)(sin 15.0°) = 89.8 km/h
6. v = 372 km/h
∆t = 8.7 s
q = 60.0°
1h
d = v∆t = (372 km/h)  (103 m/km)(8.7 s) = 9.0 × 102 m
3600 s
∆x = d(cos q) = (9.0 × 102 m)(cos 60.0°) = 450 m east
∆y = d(sin q ) = (9.0 × 102 m)(sin 60.0°) = 780 m north
7. d = 14 890 km
q = 25.0°
d 1.489 × 104 km
vavg =  =  = 805 km/h
∆t
18.45 h
∆t = 18.5 h
vx = vavg (cos q) = (805 km/h)(cos 25.0°) = 730 km/h east
vy = vavg (sin q) = (805 km/h)(sin 25.0°) = 340 km/h south
8. vi = 6.0 × 102 km/h
vf = 2.3 × 103 km/h
∆t = 120 s
q = 35° with respect to
horizontal
II
∆v vf − vi
a =  =  
∆t
∆t
1h
(2.3 × 103 km/h − 6.0 × 102 km/h)  (103 m/km)
3600 s
a = 
1.2 × 102 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1h
(1.7 × 103 km/h)  (103 m/km)
3600 s
a = 
1.2 × 102 s
2
a = 3.9 m/s
ax = a(cos q) = (3.9 m/s2)(cos 35°) = 3.2 m/s2 horizontally
ay = a(sin q) = (3.9 m/s2)(sin 35°) = 2.2 m/s2 vertically
Additional Practice 3C, p. 21
1. ∆x1 = 250.0 m
∆x2 = d2 (cos q 2 ) = (125.0 m)(cos 120.0°) = −62.50 m
d2 = 125.0 m
∆y2 = d2 (sin q 2 ) = (125.0 m)(sin 120.0°) = 108.3 m
q 2 = 120.0°
∆xtot = ∆x1 + ∆x2 = 250.0 m − 62.50 m = 187.5 m
∆ytot = ∆y 1 + ∆y2 = 0 m + 108.3 m = 108.3 m
d = (∆
xtot
)2
+(∆
ytot
)2 = (1
m
)2
+(10
)2
87
.5
8.
3m
d = 3.
04
m2+
04
m2 =
51
6×1
1.1
73
×1
4.
m2
68
9×104
d = 216.5 m
∆ytot
108.3 m
= tan−1  = 30.01° north of east
q = tan−1 
∆xtot
187.5 m
Section Two — Problem Workbook Solutions
II Ch. 3–3
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Givens
Solutions
2. v = 3.53 × 103 km/h
∆t1 = 20.0 s
∆t2 = 10.0 s
q1 = 15.0°
q2 = 35.0°
∆x1 = v∆t1(cos q 1 )
1h
∆x1 = (3.53 × 103 km/h)  (103 m/km)(20.0 s)(cos 15.0°) = 1.89 × 104 m
3600 s
∆y1 = v∆t1(sin q 1 )
1h
∆y1 = (3.53 × 103 km/h)  (103 m/km)(20.0 s)(sin 15.0°) = 5.08 × 103 m
3600 s
∆x2 = v∆t2 (cos q 2 )
1h
∆x2 = (3.53 × 103 km/h)  (103 m/km)(10.0 s)(cos 35.0°) = 8.03 × 103 m
3600 s
∆y2 = v∆t2 (sin q 2 )
1h
∆y2 = (3.53 × 103 km/h)  (103 m/km)(10.0 s)(sin 35.0°) = 5.62 × 103 m
3600 s
∆ytot = ∆y1 + ∆y2 = 5.08 × 103 m + 5.62 × 103 m = 1.07 × 104 m
∆xtot = ∆x1 + ∆x2 = 1.89 × 104 m + 8.03 × 103 m = 2.69 × 104 m
d = (∆
xtot
)2
+(∆
ytot
)2 = (2
04
m
)2
+(1.
04
m
)2
.6
9×1
07
×1
II
d = 7.
08
m2+
1×
08
m2 = 8.
08
m
24
×1
1.1
1
35
×1
d = 2.89 × 104 m
∆ytot
1.07 × 104 m
q = tan−1  = tan−1 
∆xtot
2.69 × 104 m
q = 21.7° above the horizontal
∆y1 + ∆y2 = 0
q1 = 30.0°
q2 = −45.0°
v = 11.6 km/h
∆y1 = d1(sin q1) = −∆y2 = −d2(sin q2)
sin q
sin(− 45.0°)
d1 = −d2 2 = −d2  = 1.41d2
sin q1
sin 30.0°
∆x1 = d1(cos q1) = (1.41d2)(cos 30.0°) = 1.22d2
∆x2 = d2(cos q2) = d2[cos(−45.0°)] = 0.707d2
∆x1 + ∆x2 = d2(1.22 + 0.707) = 1.93d2 = 2.00 × 102 m
d2 = 104 m
d1 = (1.41)d2 = (1.41)(104 m) = 147 m
1h
v = 11.6 km/h = (11.6 km/h)  (103 m/km) = 3.22 m/s
3600 s
104 m
d
∆t =  =  = 32.3 s
3.22 m/s
v
147 m
d
∆t1 = 1 =  = 45.7 s
3.22 m/s
v
2
2
∆ttot = ∆t1 + ∆t2 = 45.7 s + 32.3 s = 78.0 s
II Ch. 3–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. ∆x1 + ∆x2 = 2.00 × 102 m
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Givens
Solutions
4. v = 925 km/h
d1 = v∆t1 = (925 km/h)(103 m/km)(1.50 h) = 1.39 × 106 m
∆t1 = 1.50 h
d2 = v∆t2 = (925 km/h)(103 m/km)(2.00 h) = 1.85 × 106 m
∆t2 = 2.00 h
∆x1 = d1 = 1.39 × 106 m
q 2 = 135°
∆y1 = 0 m
∆x2 = d2 (cos q 2 ) = (1.85 × 106 m)(cos 135°) = −1.31 × 106 m
∆y2 = d2 (sin q 2 ) = (1.85 × 106 m)(sin 135°) = 1.31 × 106 m
∆xtot = ∆x1 + ∆x2 = 1.39 × 106 m + (− 1.31 × 106 m) = 0.08 × 106 m
∆ytot = ∆y1 + ∆y2 = 0 m + 1.31 × 106 m = 1.31 × 106 m
d = (∆
xtot
)2
+(∆
ytot
)2 = (0
06
m
)2
+(1.
06
m
)2
.0
8×1
31
×1
2 2
d = 6×
09
m2+
01
m
012m
1
1.7
2×1
= 1.
73
×1
2
d = 1.32 × 106 m = 1.32 × 103 km
∆ytot
1.31 × 106 m
q = tan−1 
= tan−1 
= 86.5° = 90.0° − 3.5°
∆xtot
0.08 × 106 m
II
q = 3.5° east of north
5. v = 57.2 km/h
d1 = v∆t1 = (57.2 km/h)(2.50 h) = 143 km
∆t1 = 2.50 h
d2 = v∆t2 = (57.2 km/h)(1.50 h) = 85.8 km
∆t2 = 1.50 h
∆tot = d1 + d2(cos q2) = 143 km + (85.8 km)(cos 30.0°) = 143 km + 74.3 km = 217 km
θ2 = 30.0°
∆ytot = d2(sin q2) = (85.8 km)(sin 30.0°) = 42.9 km
xt
)2
+(∆
yt
)2 = (2
)2
+(42
d = (∆
17
km
.9
km
)
ot
ot
d = 4.
04km
03km
04km
71
×1
2+1.8
4×1
2 = 4.
89
×1
2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 221 km
∆ytot
42.9 km
q = tan−1  = tan−1  = 11.2° north of east
217 km
∆xtot
Additional Practice 3D
1. vx = 9.37 m/s
∆y = −2.00 m
∆t =
2∆y ∆x
−g = v
x
2
g = 9.81 m/s
∆x = vx
(2)(−2.00 m)
2∆y
−g = (9.37 m/s) 
−9.81
m/s = 5.98 m
2
The river is 5.98 m wide.
2. ∆x = 7.32 km
∆y = −8848 m
∆t =
x
2
g = 9.81 m/s
2∆y ∆x
−g = v
vx =
−g
 ∆x =
2∆y
−9.81 m/s2
 (7.32 × 103 m) = 172 m/s
(2)(−8848 m)
No. The arrow must have a horizontal speed of 172 m/s, which is much greater than
100 m/s.
Section Two — Problem Workbook Solutions
II Ch. 3–5
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Givens
Solutions
3. ∆x = 471 m
vi = 80.0 m/s
g = 9.81 m/s2
∆x
∆t = 
vx
1
−g ∆x2 −(9.81 m/s2)(471 m)2
∆y = −  g ∆t 2 = 
= 
= −1.70 × 102 m
2
2vx2
(2)(80.0 m/s)2
The cliff is 1.70 × 102 m high.
4. vx = 372 km/h
∆x = 40.0 m
g = 9.81 m/s2
∆x
∆t = 
vx
1
−g∆x 2
−(9.81 m/s2)(40.0 m)2
∆y = − g ∆t 2 = 
2 = 
2
2
2vx
1 h 103 m
(2) (372 km/h)  
3600 s 1 km
∆y = −0.735 m
The ramp is 0.735 m above the ground.
II
5. ∆x = 25 m
vx = 15 m/s
g = 9.81 m/s2
h = 25 m
∆x
∆t = 
vx
1
−g∆x 2 −(9.81 m/s2)(25 m)2
∆y = − g∆t 2 = 
= 
2
2vx2
(2)(15 m/s)2
∆y = h − h′ = −14 m
h′ = h − ∆y = 25 m − (−14 m)
= 39 m
l = 420 m
−l
∆y = 
2
∆t =
∆x
x
vx =
∆x = l
2∆y
−g = v
m/s
(−29)(.8−121
0m) (420 m) = 64 m/s
2−∆gy ∆x = 
2
g = 9.81 m/s2
7. ∆y = −2.45 m
v = 12.0 m/s
g = 9.81 m/s2
vy 2 = −2g∆y
v 2 = vx2 + vy2 = vx2 − 2g∆y
vx = v 2+
m/s
m)
2g∆
y = 12
.0
2+(2)
(9
.8
1m
/s
2)(−
2.
45
vx = 14
s2
−48.
s2
4m
2/
1m
2/
= 96
m2/
s2
vx = 9.8 m/s
II Ch. 3–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6.
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Givens
Solutions
8. ∆y = −1.95 m
vy 2 = −2g ∆y
vx = 3.0 m/s
g = 9.81 m/s2
v = vx 2+
vy 2 = vx2−2g∆
y
v = (3
m/s
)2
−(2)
m)
.0
(9
.8
1m
/s
2)(−
1.
95
v = 9.
s2
+38.
s2 = 47
m2/
s2 = 6.88 m/s
0m
2/
3m
2/
.3
−1
q = tan
2g
∆y
−
−
m/s
vy
(2
)(
9.
81
2)(–1
.9
5m
)
 = tan−1  = tan−1 
v
3.0 m/s
vx
x
q = 64° below the horizontal
Additional Practice 3E
1. ∆x = 201.24 m
q = 35.0°
1
1
∆y = vi (sin q) ∆t − g∆t 2 = vi (sin q) − g∆t = 0
2
2
g = 9.81 m/s2
∆x = vi (cos q)∆t
∆x
∆t = 
vi(cos q)
II
(9.81 m/s )(201.24 m)

v = 
2(singq∆)
(xcosq) = (2)(sin 35.0°)(cos 35.0°)
∆x
1
vi(sin q) = g  
2 vi (cos q)
2
i
vi = 45.8 m/s
2. ∆x = 9.50 × 102 m
q = 45.0°
Using the derivation shown in problem 1,
vi =
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
g ∆x
 =
2(sin q )(cos q )
(9.81 m/s2)(9.50 × 102 m)

(2)(sin 45.0°)(cos 45.0°)
vi = 96.5 m/s
At the top of the arrow’s flight:
v = vx = vi (cos q ) = (96.5 m/s)(cos 45.0°) = 68.2 m/s
3. ∆x = 27.5 m
Using the derivation shown in problem 1,
q = 50.0°
2
g = 9.81 m/s
vi =
g ∆x
 =
2(sin q )(cos q )
(9.81 m/s2)(27.5 m)

2(sin 50.0°)(cos 50.0°)
vi = 16.6 m/s
4. ∆x = 44.0 m
Using the derivation shown in problem 1,
q = 45.0°
2
g = 9.81 m/s
a. vi =
x
(9.81 m/s )(44.0 m)


2(sinqg∆)(
cosq ) = (2)(sin 45.0°)(cos 45.0°)
2
vi = 20.8 m/s
Section Two — Problem Workbook Solutions
II Ch. 3–7
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Givens
Solutions
b. At maximum height, vy, f = 0 m/s
vy, f2 = vy, i2 − 2g∆y = 0
vy, i2 vi 2(sin q)2
(20.8 m/s)2(sin 45.0°)2
 = 
∆ymax =  = 
= 11.0 m
2g
(2)(9.81 m/s2)
2g
The brick’s maximum height is 11.0 m.
(20.8 m/s)2
v2
c. ∆ymax = i = 2 = 22.1 m
2g (2)(9.81 m/s
The brick’s maximum height is 22.1 m.
5. ∆x = 76.5 m
At maximum height, vy, f = 0 m/s.
q = 12.0°
vy, f 2 = vy, i 2 − 2g∆y = 0
g = 9.81 m/s2
vy, i2 vi 2(sin q)2

∆ymax =  = 
2g
2g
Using the derivation for vi 2 from problem 1,
g∆x
(sin q)2 ∆x (sin q) ∆x (tan q)
∆ymax =   =  = 
2(sin q)(cos q)
2g
4(cos q)
4
II
(76.5 m)(tan 12.0°)
∆ymax =  = 4.07 m
4
6. vrunner = 5.82 m/s
vi,ball = 2vrunner
In x-direction,
vi,ball (cos q ) = 2vrunner (cos q) = vrunner
2(cos q ) = 1
q = cos−12 = 60°
7. vi = 8.42 m/s
q = 55.2°
∆t = 1.40 s
g = 9.81 m/s2
For first half of jump,
1.40 s
∆t1 =  = 0.700 s
2
1
1
∆y = vi (sin q)∆t1 − 2g∆t12 = (8.42 m/s)(sin 55.2°)(0.700 s) − 2(9.81 m/s2)(0.700 s)2
∆y = 4.84 m − 2.40 m = 2.44 m
The fence is 2.44 m high.
∆x = vi (cos q)∆t
∆x = (8.42 m/s) (cos 55.2°)(1.40 s) = 6.73 m
8. vi = 2.2 m/s
q = 21°
∆t = 0.16 s
g = 9.81 m/s2
∆x = vi (cos q)∆t = (2.2 m/s) (cos 21°)(0.16 s) = 0.33 m
∆t
Maximum height is reached in a time interval of 
2
2
∆t 1 ∆t


∆ymax = vi (sin q)  − 2g 
2
2
∆ymax
0.16 s
= (2.2 m/s)(sin 21°)  −
2
−2
−2
∆ymax = 6.3 × 10 m − 3.1 × 10
1

2
m = 3.2 × 10 m = 3.2 cm
The flea’s maximum height is 3.2 cm.
II Ch. 3–8
Holt Physics Solution Manual
−2
2
0.16 s
(9.81 m/s2) 
2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
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Additional Practice 3F
Givens
1. vse = 126 km/h north
vgs = 40.0 km/h east
Solutions
vge =
2
2
vg
+v
s se =
0.
(4
0km
/h
)2
+(12
6km
/h
)2
vge =
4
60
1.
×1
03km
2/h2+1.5
9×1
0
k
m2/
h2
vge =
75
1.
×1
04km
/h
= 132 km/h
v
126 km/h
q = tan−1 se = tan−1  = 72.4° north of east
vgs
40.0 km/h
2. vwe = −3.00 × 102 km/h
vpw = 4.50 × 102 km/h
∆x = 250 km
3. vtw = 9.0 m/s north
vwb = 3.0 m/s east
∆t = 1.0 min
vpe = vpw + vwe = 4.50 × 102 km/h − 3.00 × 102 km/h = 1.50 × 102 km/h
∆x
250 km
∆t =  = 
= 1.7 h
vpe 1.50 × 102 km/h
vtb = vtw + vwb
2
vtb = vtw
m/s
)2
+(3.
)2 = 81
m2/
s2
+9.0
m2/
s2
2+
vw
.0
0m
/s
b = (9
II
vtb = 9.
01
m2/
s2
0×1
vtb = 9.5 m/s
60 s
∆x = vtb∆t = (9.5 m/s)(1.0 min)  = 570 m
1 min
v
3.0 m/s
q = tan−1 wb = tan−1  = 18° east of north
vtw
9.0 m/s
4. vsw = 40.0 km/h forward
vfw = 16.0 km/h forward
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆x = 60.0 m
5. v1E = 90.0 km/h
v2E = −90.0 km/h
∆t = 40.0 s
vsf = vsw − vfw = 40.0 km/h − 16.0 km/h = 24.0 km/h toward fish
∆x
60.0 m
∆t =  =  = 9.00 s
vs f
1 h 103 m
(24.0 km/h)  
3600 s 1 km
v12 = v1E − v2E
v12 = 90.0 km/h − (−90.0 km/h) = 1.80 × 102 km/h
1h
103 m
∆x = v12 ∆t = (1.80 × 102 km/h)   (40.0 s) = 2.00 × 103 m = 2.00 km
3600 s 1 km
The two geese are initially 2.00 km apart
6. vme = 18.0 km/h forward
Vre = 0.333 Vme
= 6.00 km/h forward
∆x = 12.0 m
vmr = vme − vre
vmr = 18.0 km/h − 6.0 km/h = 12.0 km/h
∆x
12.0 m
3600 s 1 km
∆t =  =   3
vmr (12.0 km/h)
1h
10 m
∆t = 3.60 s
Section Two — Problem Workbook Solutions
II Ch. 3–9
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Two-Dimensional Motion and Vectors
Chapter
4
Additional Practice 4A
Givens
Solutions
1. mw = 75 kg
The normal force exerted by the platform on the weight lifter’s feet is equal to and
opposite of the combined weight of the weightlifter and the pumpkin.
mp = 275 kg
g = 9.81 m/s2
Fnet = Fn − mwg − mpg = 0
Fn = (mw + mp)g = (75 kg + 275 kg) (9.81 m/s2)
Fn = (3.50 × 102 kg)(9.81 m/s2) = 3.43 × 103 N
Fn = 3.43 × 103 N upward against feet
Fnet = Fn,1 + Fn,2 − mbg − mwg = 0
2. mb = 253 kg
mw = 133 kg
g = 9.81 m/s2
The weight of the weightlifter and barbell is distributed equally on both feet, so the
normal force on the first foot (Fn,1) equals the normal force on the second foot (Fn,2).
2Fn,1 = (mb + mw)g = 2Fn,2
II
m
(253 kg + 33 kg) 9.81 2
s
(mb + mb)g
 = 
Fn,1 = Fn,2 = 
2
2
(386 kg)(9.81 m/s2)
Fn,1 = Fn,2 =  = 1.89 × 103 N
2
Fn,1 = Fn,2 = 1.89 × 103 N upward on each foot
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. Fdown = 1.70 N
Fnet = 4.90 N
Fnet2 = Fforward2 + Fdown2
2
2
Fforward = F
−
Fdo
)2
−(1.
N
)2
w
.9
0N
70
net
n = (4
Fforward = 21
N2 = 4.59 N
.1
4. m = 3.10 × 102 kg
Fx,net = ΣFx = FT,1(sin q1) + FT,2(sin q2) = 0
g = 9.81 m/s
Fy,net = ΣFy = FT,1(cos q1) + FT,2(cos q2) + Fg = 0
q1 = 30.0°
FT,1(sin 30.0°) = −FT,2[sin (−30.0°)]
q2 = − 30.0°
FT,1 = FT,2
2
FT,1(cos q1) + FT,1(cos q2) = −Fg = mg
FT,1(cos 30.0°) + FT,1[cos (−30.0°)] = (3.10 × 102 kg)(9.81 m/s2)
(3.10 × 102 kg)(9.81 m/s2)
FT,1 = 
(2)(cos 30.0°)[cos(−30.0°)]
FT,1 = FT,2 = 1.76 × 103 N
As the angles q1 and q2 become larger, cos q1 and cos q2 become smaller. Therefore,
FT,1 and FT,2 must become larger in magnitude.
Section Two — Problem Workbook Solutions
II Ch. 4–1
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Givens
Solutions
5. m = 155 kg
FT,1 = 2FT,2
g = 9.81 m/s2
q1 = 90° − q2
Fx,net = FT,1(cos q1) − FT,2(cos q2) = 0
Fy,net = FT,1(sin q1) + FT,2(sin q2) − mg = 0
1
FT,1[(cos q1) − (cos q2)] = 0
2
2 (cos q1) = cos q2 = cos(90° − q1) = sin q1
2 = tan q1
q1 = tan−1(2) = 63°
q2 = 90° − 63° = 27°
F ,1
FT,1(sin q1) + T(sin
q2) = mg
2
mg
FT,1 = 
1
(sin θ1) +  (sin θ2)
2
(155 kg)(9.81 m/s2)
(155 kg)(9.81 m/s2) (155 kg)(9.81 m)
 = 
FT,1 = 
(sin 27°) = 
0.89 + 0.23
1.12
(sin 63°) + 
2
FT,1 = 1.36 × 1.36 × 103 N
II
FT,2 = 6.80 × 102 N
Additional Practice 4B
vf = 0 km/h
[(0 km/h)2 − (173 km/h)2](103 m/km)2(1 h/3600 s)2
vf 2 − vi2
a = 
= 
(2)(0.660 m)
2∆x
∆x = 0.660 m
a = −1.75 × 103 m/s2
m = 70.0 kg
F = ma = (70.0 kg)(−1.75 × 103 m/s2) = −1.22° × 105 N
1. vi = 173 km/h
g = 9.81 m/s2
Fg = mg = (70.0 kg)(9.81 m/s2) = 6.87 × 102 N
2. m = 2.232 × 106 kg
a. Fnet = manet = Fup − mg
g = 9.81 m/s2
Fup = manet + mg = m(anet + g) = (2.232 × 106 kg)(0 m/s2 + 9.81 m/s2)
anet = 0 m/s2
Fup = 2.19 × 107 N = mg
b. Fdown = mg(sin q)
F
Fup − Fdown mg − mg(sin q)
 = 
 = 
anet = net
m
m
m
9.81 m/s2
anet = g(1 − sin q) = (9.81 m/s2)[1.00 − (sin 30.0°)] =  = 4.90 m/s2
2
anet = 4.90 m/s2 up the incline
3. m = 40.00 mg
= 4.00 × 10−5 kg
g = 9.807 m/s2
anet = (400.0)g
Fnet = Fbeetle − Fg = manet = m(400.0) g
Fbeetle = Fnet + Fg = m(400.0 + 1)g = m(401)g
Fbeetle = (4.000 × 10−5 kg)(9.807 m/s2)(401) = 1.573 × 10−1 N
Fnet = Fbeetle − Fg = m(400.0) g = (4.000 × 10−5 kg)(9.807 m/s2)(400.0)
Fnet = 1.569 × 10−1 N
The effect of gravity is negligible.
II Ch. 4–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
The force of deceleration is nearly 178 times as large as David Purley’s weight.
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Givens
Solutions
4. ma = 54.0 kg
The net forces on the lifted weight is
Fw,net = mwanet = F ′ − mwg
mw = 157.5 kg
2
anet = 1.00 m/s
where F′ is the force exerted by the athlete on the weight.
g = 9.81 m/s2
The net force on the athlete is
Fa,net = Fn,1 + Fn,2 − F ′ − mag = 0
where Fn,1 and Fn,2 are the normal forces exerted by the ground on each of the athlete’s feet, and −F′ is the force exerted by the lifted weight on the athlete.
The normal force on each foot is the same, so
Fn,1 = Fn,2 = Fn
and
F ′ = 2Fn − mag
Using the expression for F ′ in the equation for Fw,net yields the following:
mwanet = (2Fn − mag) − ma g
2Fn = mw(anet + g) + mag
2
2
mw(anet + g) + ma g (157.5 kg)(1.00 m/s + 9.81 m/s ) + (54.0 kg)
 = 
Fn = 
2
2
(157.5 kg)(10.81 m/s2) + (54.0 kg)(9.81 m/s2)
Fn = 
2
II
1702 N + 5.30 × 102 N 2232 N
Fn =  =  = 1116 N
2
2
Fn,1 − Fn,2 = Fn = 1116 N upward
5. m = 2.20 × 102 kg
Fnet = manet = Favg − mg
anet = 75.0 m/s
Favg = m(anet + g) = (2.20 × 102 kg)(75.0 m/s2 + 9.81 m/s2)
g = 9.81 m/s2
Favg = (2.20 × 102 kg)(84.8 m/s2) = 1.87 × 104 N
2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Favg = 1.87 × 104 N upward
6. m = 2.00 × 104 kg
∆t = 2.5
vf − v
(1.0 m/s − 0.0 m/s)
anet = i =  = 0.40 m/s2
∆t
2.5 s
Fnet = manet = FT − mg
vi = 0 m/s
FT = manet + mg = m(anet + g)
vf = 1.0 m/s
FT = (2.00 × 104 kg)(0.40 m/s2 + 9.81 m/s2)
g = 9.81 m/s2
FT = (2.00 × 104 kg)(10.21 m/s2) = 2.04 × 105 N
FT = 2.04 × 105 N
7. m = 2.65 kg
Fx,net = FT,1(cos q1) − FT,2(cos q2) = 0
q1 = q2 = 45.0°
2
anet = 2.55 m/s
2
g = 9.81 m/s
FT,1(cos 45.0°) = FT,2(cos 45.0°)
FT,1 = FT,2
Fy,net = manet = FT,1(sin q1) + FT,2(sin q2) − mg
FT = FT,1 = FT,2
q = q1 = q2
FT(sin q) + FT(sin q) = m(anet + g)
2FT(sin q) = m(anet + g)
Section Two — Problem Workbook Solutions
II Ch. 4–3
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Givens
Solutions
m(anet + g) (2.65 kg)(2.55 m/s2 = 9.81 m/s2)
 = 
FT = 
(2)(sin 45.0°)
2 (sin q )
(2.65 kg)(12.36 m/s2)
FT =  = 23.2 N
(2)(sin 45.0°)
FT,1 = 23.2 N
FT,2 = 23.2 N
8. m = 20.0 kg
∆x = 1.55 m
vf 2 − vi2
(0.550 m/s)2 − (0.00 m/s)2
anet = 
=  = 9.76 × 10−2 m/s2
2∆x
(2)(1.55 m)
vi = 0 m/s
Fnet = manet = (20.0 kg)(9.76 × 10−2 m/s2) = 1.95 N
vf = 0.550 m/s
9. mmax = 70.0 kg
II
Fmax = mmaxg = FT
m = 45.0 kg
Fmax = (70.0 kg)(9.81 m/s2) = 687 N
g = 9.81 m/s2
Fnet = manet = FT − mg = Fmax − mg
687 N
F ax
anet = m
− g =  − 9.81 m/s2 = 15.3 m/s2 − 9.81 m/s2 = 5.5 m/s2
m
45.0 kg
anet = 5.5 m/s2 upward
10. m = 3.18 × 105 kg
Fnet = Fapplied − Ffriction = (81.0 × 103 − 62.0 × 103 N)
Fapplied = 81.0 × 103 N
Fnet = 19.0 × 103 N
Ffriction = 62.0 × 103 N
19.0 × 103 N
F
 = 
anet = net
= 5.97 × 10−2 m/s2
3.18 × 105 kg
m
3
Fapplied = 4.00 × 10 N
q = 20.0°
Fopposing = (0.120) mg
g = 9.81 m/s2
Fapplied(cos q) − (0.120) mg
anet = 
m
(4.00 × 103 N)(cos 20.0°) − (0.120)(3.00 × 103 kg)(9.81 m/s2)
anet = 
3.00 × 103 kg
3.76 × 103 N − 3.53 × 103 N
2.3 × 102 N
= 
anet = 
3
3.00 × 10 kg
3.00 × 103 kg
anet = 7.7 × 10−2 m/s2
12. mc = 1.600 × 103 kg
For the counterweight: The tension in the cable is FT.
mw = 1.200 × 103 kg
Fnet = FT − mwg = mwanet
vi = 0 m/s
For the car:
g = 9.81 m/s2
Fnet = mcg − FT = mcanet
∆y = 25.0 m
Adding the two equations yields the following:
mcg − mwg = (mw + mc)anet
3
3
2
(mc − mw)g (1.600 × 10 kg − 1.200 × 10 kg)(9.81 m/s )
 = 
anet = 
1.600 × 103 kg + 1.200 × 103 kg
mc + mw
(4.00 × 102 kg)(9.81 m/s2)
anet = 
= 1.40 m/s2
2.800 × 103 kg
II Ch. 4–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fnet = manet = Fapplied(cos q) − Fopposing
11. m = 3.00 × 103 kg
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Givens
Solutions
vf = 2a
m/s
)2
ne
+vi2 = (2
)(
1.
40
2)(2
5.
0m
)+(0m
/s
t∆y
vf = 8.37 m/s
13. m = 409 kg
a. Fnet = Fapplied − mg(sin q) = 2080 N − (409 kg)(9.81 m/s2)(sin 30.0°)
d = 6.00 m
Fnet = 2080 N − 2010 N = 70 N
q = 30.0°
Fnet = 70 N at 30.0° above the horizontal
2
g = 9.81 m/s
Fapplied = 2080 N
vi = 0 m/s
70 N
F
 =  = 0.2 m/s2
b. anet = net
409 kg
m
anet = 0.2 m/s2 at 30.0° above the horizontal
1
1
c. d = vi∆t + anet ∆t 2 = (0 m/s)∆t + (0.2 m/s2)∆t2
2
2
∆t =
14. amax = 0.25 m/s2
Fmax = 57 N
Fapp = 24 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
15. m = 2.55 × 103 kg
0 m)
(
(20).(26.m0
/s) = 8 s
2
F ax
57 N
a. m = m
= 2 = 2.3 × 102 kg
amax 0.25 m/s
II
b. Fnet = Fmax − Fapp = 57 N − 24 N = 33 N
F
33 N
 = 
anet = net
= 0.14 m/s2
m
2.3 × 102 kg
a. Fx,net = ΣFx = max,net = FT(cos qT) + Fwind
FT = 7.56 × 103 N
Fx,net = (7.56 × 103 N)[cos(−72.3°)] − 920 N = 2.30 × 103 N − 920 N = 1.38 × 103 N
qT = −72.3°
Fy,net = ΣFy = may,net = FT(sin qT) + Fbuoyant + Fg = FT(sin qT) + Fbuoyant − mg
Fbuoyant = 3.10 × 104 N
Fy,net = (7.56 × 103 N)[sin(−72.3°)] = 3.10 × 104 N − (2.55 × 103 kg)(9.81 m/s2)
Fwind = −920 N
Fy,net = −7.20 × 103 N + 3.10 × 104 N − 2.50 × 104 = −1.2 × 103 N
g = 9.81 m/s2
2
Fnet = (F
)2 = (1
03
N
)2
+(−
03
N
)2
x,
(Fy,n
et
.3
8×1
1.
2×1
ne
t)+
Fnet = 1.
06
N2+
06
N2
90
×1
1.4
×1
Fnet = 3.
06
N2 = 1.8 × 103 N
3×1
Fy, net
−1.2 × 103 N
 = tan−1 
q = tan−1 
Fx, net
1.38 × 103 N
q = −41°
Fnet = 1.8 × 103 N at 41° below the horizontal
1.8 × 103 N
F
 = 
b. anet = net
2.55 × 103 kg
m
∆y = −45.0 m
vi = 0 m/s
anet = 0.71 m/s2
c. Because vi = 0
1
∆y =  ay,net ∆t2
2
1
∆x =  ax,net ∆t2
2
ax,net

∆x = 
ay,net
anet(cos q)
∆y = 
anet(sin q)
∆y
∆y = 
tan q
−45.0 m
∆x =  = 52 m
tan(−41°)
Section Two — Problem Workbook Solutions
II Ch. 4–5
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Additional Practice 4C
Givens
Solutions
1. m = 11.0 kg
Fk = mkFn = mkmg
mk = 0.39
Fk = (0.39) (11.0 kg)(9.81 m/s2) = 42.1 N
g = 9.81 m/s2
2. m = 2.20 × 105 kg
ms = 0.220
g = 9.81 m/s2
3. m = 25.0 kg
Fs,max = msFn = msmg
Fs,max = (0.220)(2.20 × 105 kg)(9.91 m/s2) = 4.75 × 105 N
Fs,max = msFm
Fapplied = 59.0 N
Fn = mg(cos q) + Fapplied
q = 38.0°
Fs,max = ms[mg(cos q) = Fapplied] = (0.599)[(25.0 kg)(9.81 m/s2)(cos 38.0° + 59.0 N]
ms = 0.599
g = 9.81 m/s2
II
Fs,max = (0.599)(193 N + 59 N) = (0.599)(252 N) = 151 N
Alternatively,
Fnet = mg(sin q) − Fs,max = 0
Fs,max = mg(sin q) = (25.0 kg)(9.81 m/s2)(sin 38.0°) = 151 N
4. q = 38.0°
Fnet = mg(sin q) − Fk = 0
2
g = 9.81 m/s
Fk = mkFn = mkmg(cos q)
mkmg(cos q) = mg(sin q)
sin q
mk =  = tan q = tan 38.0°
cos q
mk = 0.781
Fnet = mg(sin q) − Fk = 0
2
g = 9.81 m/s
Fk = mkFn = mkmg(cos q)
mkmg(cos q) = mg(sin q)
sin q
mk =  = tan q = tan 5.2°
cos q
mk = 0.091
II Ch. 4–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. q = 5.2°
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Givens
6. m = 281.5 kg
q = 30.0°
Solutions
F net = 3mg(sin q) − ms(3mg)(cos q) − Fapplied = 0
Fapplied = mg
(3)(sin 30.0°) − 1.00
3mg(sin q) − mg
3(sin q) − 1.00
ms =  =  = 
(3)(cos 30.0°)
3mg(cos q)
3(cos q )
1.50 − 1.00
0.50
ms =  = 
(3)(cos 30.0°) (3)(cos 30.0°)
ms = 0.19
7. m = 1.90 × 105 kg
Fnet = Fapplied − Fk = 0
ms = 0.460
Fk = mkFn = mkmg
g = 9.81 m/s2
Fapplied = mkmg = (0.460)(1.90 × 105 kg)(9.81 m/s2)
Fapplied = 8.57 × 105 N
8. Fapplied = 6.0 × 103 N
Fnet = Fapplied − Fk = 0
mk = 0.77
Fk = mkFn
g = 9.81 m/s2
Fapplied 6.0 × 103 N
 =  = 7.8 × 103 N
Fn = 
mk
0.77
II
Fn = mg
F
7.8 × 103 N
m = n = 
= 8.0 × 102 kg
g
9.81 m/s2
9. Fapplied = 1.13 × 108 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ms = 0.741
Fnet = Fapplied − Fs,max = 0
Fs,max = msFn = msmg
Fapplied
1.13 × 108 N
 = 2 = 1.55 × 102 kg
m= 
msg
(0.741)(9.81 m/s
10. m = 3.00 × 103 kg
Fnet = mg(sin q) − Fk = 0
q = 31.0°
Fk = mkFn = mkmg(cos q)
g = 9.81 m/s2
mkmg(cos q) = mg(sin q)
sin q
mk =  = tan q = tan 31.0°
cos q
mk = 0.601
Fk = mkmg(cos q) = (0.601)(3.00 × 103 kg)(9.81 m/s2)(cos 31.0°)
Fk = 1.52 × 104 N
Alternatively,
Fk = mg(sin q) = (3.00 × 103 kg)(9.81 m/s2)(sin 31.0°) = 1.52 × 104 N
Section Two — Problem Workbook Solutions
II Ch. 4–7
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Additional Practice 4D
Givens
Solutions
1. Fapplied = 130 N
2
anet = 1.00 m/s
mk = 0.158
Fnet = manet = Fapplied − Fk
Fk = mkFn = mkmg
manet + mkmg = Fapplied
g = 9.81 m/s2
m(anet + mkg) = Fapplied
130 N
Fapplied
 = 
m= 
2
1.00
m/s
+
(0.158)(9.81
m/s2)
anet + mkg
130 N
130 N
m = 
= 51 kg
2
2 = 
1.00 m/s + 1.55 m/s
2.55 m/s2
2. Fnet = −2.00 × 104 N
II
Fnet = manet = mg(sin q) − Fk
q = 10.0°
Fk = mkFn = mkmg(cos q)
mk = 0.797
m[g(sin q) − mkg(cos q)] = Fnet
g = 9.81 m/s2
−2.00 × 104 N
F et
m = n
= 
2
(9.81 m/s )[(sin 10.0°) − (0.797)(cos 10.0°)]
g[sin q − mk(cos q)]
−2.00 × 104 N
−2.00 × 104 N
m = 
= 2
2
(9.81 m/s )(0.174 − 0.785) (9.81 m/s )(−0.611)
m = 3.34 × 103 kg
Fn = mg(cos q) = (3.34 × 103 kg)(9.81 m/s2)(cos 10.0°) = 3.23 × 104 N
Fnet = manet = mg(sin q) − Fk
q = 45.0°
Fk = mkFn = mkmg(cos q)
mk = 0.597
m[g(sin q) − mkg(cos q)] = Fnet
6.99 × 103 N
F et
m = n
= 
2
(9.81 m/s )[(sin 45.0°) − (0.597)(cos 45.0°)]
g[sin q − mk(cos q)]
6.99 × 103 N
6.99 × 103 N
m = 
= 
2
(9.81 m/s )(0.707 − 0.422) (9.81 m/s2)(0.285)
m = 2.50 × 103 kg
Fn = mg(cos q) = (2.50 × 103 kg)(9.81 m/s2)(cos 45.0°) = 1.73 × 104 N
4. m = 9.50 kg
Fnet = manet = Fapplied − Fk − mg(sin q)
q = 30.0 °
Fk = mkFn = mkmg(cos q)
Fapplied = 80.0 N
2
anet = 1.64 m/s
g = 9.81 m/s2
II Ch. 4–8
mkmg(cos q) = Fapplied − manet − mg(sin q)
Fapplied − m[anet + g (sin q)]
mk = 
mg(cos q)
80.0 N − (9.50 kg)[1.64 m/s2 + (9.81 m/s2)(sin 30.0°)]
mk = 
(9.50 kg)(9.81 m/s2)(cos 30.0°)
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. Fnet = 6.99 × 103 N
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Givens
Solutions
80.0 N − (9.50 kg)(6.54 m/s2)
80.0 N − (9.50 kg)[1.64 m/s2 + 4.90 m/s2)
mk = 
= 
2
(9.50 kg)(9.81 m/s )(cos 30.0°)
(9.50 kg)(9.81 m/s2)(cos 30.0°)
80.0 N − 62.1 N
17.9 N
= 
mk = 
(9.50 kg)(9.81 m/s2)(cos 30.0°)
(9.50 kg)(9.81 m/s2)(cos 30.0°)
mk = 0.222
5. m = 1.89 × 105 kg
Fnet = manet = Fapplied − Fk
Fapplied = 7.6 × 10 N
Fk = Fapplied − manet = 7.6 × 105 N − (1.89 × 105)(0.11 m/s2) = 7.6 × 105 N − 2.1 × 104 N
anet = 0.11 m/s2
Fk = 7.4 × 105 N
5
6. q = 38.0°
Fnet = manet = mg(sin q) − Fk
mk = 0.100
Fk = mkFn = mkmg(cos q)
g = 9.81 m/s2
manet = mg[sin q − mk(cos q)]
anet = g[sin q − mk(cos q)] = (9.81 m/s2)[(sin 38.0°) − (0.100)(cos 38.0°)]
anet = (9.81 m/s2)(0.616 − 7.88 × 10−2) = (9.81 m/s2)(0.537)
anet = 5.27 m/s2
II
Acceleration is independent of the rider’s and sled’s masses. (Masses cancel.)
Fnet = manet = mg(sin q) − Fk
7. ∆t = 6.60 s
Fk = mkFn = mkmg(cos q)
q = 34.0°
manet = mg[sin q − mk(cos q)]
mk = 0.198
g = 9.81 m/s
anet = g[sin q − mk(cos q)] = (9.81 m/s2)[(sin 34.0°) − (0.198)(cos 34.0°)]
vi = 0 m/s
anet = (9.81 m/s2)(0.559 − 0.164) = (9.81 m/s2)(0.395)
2
anet = 3.87 m/s2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vf = vi + anet∆t = 0 m/s + (3.87 m/s2)(6.60 s)
vf = 25.5 m/s2 = 92.0 km/h
Section Two — Problem Workbook Solutions
II Ch. 4–9
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Work and Energy
Chapter
5
Additional Practice 5A
Givens
Solutions
1. W = 1.15 × 103 J
W = Fd(cos q) = mgd(cos q)
1.15 × 105 J
W
d =  = 
mg(cos q) (60.0 kg)(9.81 m/s2)(cos 0°)
m = 60.0 kg
g = 9.81 m/s2
d=
q = 0°
2. m = 1.45 × 106 kg
2
g = 9.81 m/s
195 m
W = Fd(cos q)
q = 0°
1.00 × 108 J
W
d =  = 
−2
F(cos q) (2.00 × 10 )(1.45 × 106 kg)(9.81 m/s2)(cos 0.00°)
W = 1.00 × 102 MJ
d=
352 m
II
F = (2.00 × 10−2) mg
Fnet = manet = F − mg
3. m = 1.7 g
F = manet + mg
W = 0.15 J
anet = 1.2 m/s
W = Fd(cos q) = m(anet + g)d(cos q)
q = 0°
0.15 J
W
d =  = 
−3
2
(1.7
×
10
kg)(1.2
m/s
+ 9.81 m/s2)(cos 0°)
m(anet + g)(cos q)
2
2
g = 9.81 m/s
0.15 J
d = 
(1.7 × 10−3 kg)(11.0 m/s2)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 8.0 m
4. m = 5.40 × 102 kg
W = Fd(cos q′) = Fd
W = 5.30 × 104 J
F = mg(sin q)
g = 9.81 m/s2
W = mg(sin q)d
q = 30.0°
5.30 × 104 J
W
d =   = 
2
mg(sin q) (5.40 × 10 kg)(9.81 m/s3)(sin 30.0°)
q ′ = 0°
d = 20.0 m
Section Two — Problem Workbook Solutions
II Ch. 5–1
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Givens
Solutions
Fnet = Flift − Fg = 0
5. d = 5.45 m
W = 4.60 × 10 J
F = Flift = Fg
q = 0°
W = Fd(cos q) = Fgd(cos q)
4
W
4.60 × 104 J
Fg =  =  = 8.44 ⫻ 103 N
d(cos q) (5.45 m)(cos 0°)
6. d = 52.0 m
m = 40.0 kg
W
2.04 × 104 J
F =  =   = 392 N
d(cos q) (52.0 m)(cos 0°)
W = 2.04 × 104 J
q = 0°
7. d = 646 m
W = 2.15 × 105 J
W
2.15 × 105 J
F =  =  = 333 N
d(cos q) (646 m)(cos 0°)
q = 0°
8. m = 1.02 × 103 kg
Fnet = Fg − Fk = mg (sin q) − mkmg(cos q)
d = 18.0 m
Wnet = Fnetd(cos q′) = mgd(cos q′)[(sin q) − mk(cos q)]
angle of incline = q = 10.0°
Wnet = (1.02 × 103 kg)(9.81 m/s2)(18.0 m)(cos 0°)[(sin 10.0°) − (0.13)(cos 10.0°)]
q′ = 0°
Wnet = (1.02 × 103 kg)(9.81 m/s2)(18.0 m)(0.174 − 0.128)
g = 9.81 m/s2
Wnet = (1.02 × 103 kg)(9.81 m/s2)(18.0 m)(0.046)
mk = 0.13
Wnet = 8.3 × 103 J
9. d = 881.0 m
Wnet = Fnetd(cos q′)
Fapplied = 40.00 N
Fnet = Fapplied (cos q) − Fk
q = 45.00°
Wnet = [Fapplied (cos q) − Fk]d(cos q′)
Fk = 28.00 N
Wnet = [40.00 N(cos 45.00°) − 28.00° N](881.0 m)(cos q)
q′ = 0°
Wnet = (28.28 N − 28.00 N)(881.0 m) = (0.28 N)(881.0 m)
Wnet = 246.7 J
10. m = 9.7 × 103 kg
Wnet = Fnetd(cos q) = (F1 + F2)d(cos q) = 2Fd(cos q)
Wnet = (2)(1.2 × 103 N)(12 m)(cos 45°) = 2.0 × 104 J
q = 45°
F = F1 = F2 = 1.2 × 103 N
d = 12 m
11. m = 1.24 × 103 kg
Only F2 contributes to the work done in moving the flag south.
F1 = 8.00 × 103 N east
q = 90.0° − 30.0° = 60.0°
F2 = 5.00 × 103 N 30.0°
Wnet = Fnetd(cos q) = F2d(cos q) = (5.00 × 103 N)(20.0 m)(cos 60.0°)
south of east
Wnet = 5.00 × 104 J
d = 20.0 m south
II Ch. 5–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
II
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Additional Practice 5B
Givens
Solutions
1. ∆x = 1.00 × 102 m
∆t = 9.85 s
KE = 3.40 × 103 J
∆x
v = 
∆t
2
∆x
1
1
KE = 2mv 2 = 2m 
∆t
2KE∆t2 (2)(3.40 × 103 J)(9.85 s)2
m = 
= 
= 66.0 kg
(1.00 × 102 m)2
∆ x2
2. v = 4.00 × 102 km/h
KE = 2.10 × 107 J
3. v = 50.3 km/h
3
KE = 6.54 × 10 J
4. v = 318 km/h
(2)(6.54 × 103 J)
2KE
= 67.0 kg
m = 
2 = 
(50.3 km/h)2(103 m/km)2 (1 h/3600 s)
v
(2)(3.80 × 106 J)
2KE
m = 
= 974 kg
2 = 
(318 km/h)2 (103 m/km)2 (1 h/3600 s)
v
KE = 3.80 MJ
5. m = 51.0 kg
4
KE = 9.96 × 10 J
v=
(2)(9.96 × 104 J)
 = 62.5 m/s = 225 km/h
51.0 kg
∆t = 24.00 h
m = 55 kg
b. KE = 2mv 2 = 2(55 kg)(1.084 m/s)2 =
7. m = 3.38 × 1031 kg
KE = 1.10 × 1042 J
8. m = 680 kg
v = 56.0 km/h
KELB = 3.40 × 103 J
9. v = 11.2 km/s
1
v=
II
2 KE
 =
m
∆x
9.3625 × 104 m
a. vavg =  =  = 1.084 m/s
∆t
(24.00 h)(3600 s/h)
6. ∆x = 93.625 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(2)(2.10 × 107 J)
2KE

=
= 3.40 × 103 kg
m = 
(4.00 × 102 km/h)2 (103 m km)2 (1 h/3600 s)2
v2
1
32 J
2 KE
 =
m
1
(2)(1.10 × 1042 J)

= 2.55 × 105 m/s = 255 km/s
3.38 × 1031 kg
1
a. KE = 2mv 2 = 2(680 kg)[(56.0 km/h)(103 m/km)(1 h/3600 s)]2 = 8.23 × 104 J
24
KEp
8.2 × 104 J
b. b = 
= 
1
KELB 3.40 × 103 J
1
1
KE = 2mv 2 = 2(2.3 × 105 kg)(11.2 × 103 m/s)2 = 1.4 × 1013 J
m = 2.3 × 105 kg
Section Two — Problem Workbook Solutions
II Ch. 5–3
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Additional Practice 5C
Givens
Solutions
Wnet = ∆KE = KEf − KEi = KEf
1. d = 227 m
Wnet = Fnetd(cos q)
m = 655 g
g = 9.81 m/s
Fnet = Fg − Fresistance = mg − (0.0220)mg = mg(1 − 0.0220)
Fresistance = (0.0220)mg
KEf = mg(1 − 0.0220)d(cos q) = (655 × 10−3 kg)(9.81 m/s2)(1 − 0.0220)(227 m)(cos 0°)
q = 0°
KEf = (0.655 kg)(9.81 m/s2)(0.9780)(227 m)
KEi = 0 J
KEf = 1.43 × 103 J
2
2. vi = 12.92 m/s
Wnet is the work done by friction.
1
Wnet = −2830 J
Wnet = ∆KE = KEf − KEi = KEf − 2mvi2
m = 55.0 kg
KEf = Wnet + 2mvi2 = − 2830 J + 2(55.0 kg)(12.92 m/s)2 = −2830 J + 4590 J
1
1
KEf = 1.76 × 103 J
1
II
1
Wnet = ∆KE = KEf − KEi = 2 mvf 2 − 2mvi2
3. m = 25.0 g
hi = 553 m
Wnet = Fnetd(cos q)
hf = 353 m
Fnet = Fg − Fr = mg − Fr
vi = 0 m/s
d = hi − hf
vf = 30.0 m/s
2
g = 9.81 m/s
q = 0°
Wnet = (mg − Fr)(hi − hf)(cos q)
mg − Fr =
1

2
m (vf 2 − vi 2)
(hi – hf )(cos q)
(30.0 m/s)2 − (0 m/s)2
(vf 2 − vi2)
 = (25.0 × 10−3 kg) 9.81 m/s2 − 
Fr = m g − 
(2)(553 m − 353 m)(cos 0°)
2(hi − hf)(cos q)
2
2 2
9.00 × 10 m /s
Fr = (25.0 × 10−3 kg) 9.81 m/s2 − 
(2)(2.00 × 102 m)
Fr = (25.0 × 10−3 kg)(9.81 m/s2 − 2.25 m/s2) = (25.0 × 10−3 kg)(7.56 m/s2)
4. vi = 404 km/h
Wnet = −3.00 MJ
m = 1.00 × 103 kg
1
1
Wnet = ∆KE = KEf − KEi = 2mvf 2 − 2mvi2
1
mv 2
f
2
vf =
1
= 2mvi2 + Wnet
2Wnet
vi2 + 
=
m
(2)(−3.00 × 106 J)
[(404 km/h)(103 m/km)(1 h/3600 s)]2 + 
1.00 × 103 kg
vf = 1.
04
m2/
s2
−6.0
03
m2/
s2 = 6.
03
m2/
s2
26
×1
0×1
6×1
vf = 81 m/s = 290 km/h
5. m = 45.0 g
hi = 8848.0 m
hf = 8806.0 m
vi = 0 m/s
vf = 27.0 m/s
g = 9.81 m/s2
q = 0°
II Ch. 5–4
1
1
Wnet = ∆KE = KEf − KEi = 2mvf 2 − 2mvi2
Wnet = Fnetd(cos q)
Fnet = Fg − Fr = mg − Fr
d = hi − hf
Wnet = mg(hi − hf )(cos q) − Fr(hi − hf )(cos q)
− Fr(hi − hf )(cos q) = Fr(hi − hf )(cos 180° + q) = Wr
1
m(v 2
f
2
− vi2) = mg(hi − hf )(cos q) + Wr
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fr = 0.189 N
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Givens
Solutions
Wr = m[2(vf 2 − vi2) − g(hi − hf)(cos q)] = (45.0 × 10−3 kg)2(27.0 m/s)2 − 2(0 m/s)2
1
1
1
−(9.81 m/s2)(8848.0 m − 8806.0 m)(cos 0°)
Wr = (45.0 × 10−3 kg)[364 m2/s2 − (9.81 m/s2)(42.0 m)]
Wr = (45.0 × 10−3 kg)(364 m2/s2 − 412 m2/s2)
Wr = (45.0 × 10−3 kg)(−48 m2/s2) = −2.16 J
6. vf = 35.0 m/s
vi = 25.0 m/s
Wnet = 21 kJ
1
1
Wnet = ∆KE = KEf − KEi = 2 mvf 2 − 2 mvi2
2Wnet
(2)(21 × 103 J)
42 × 103 J
2 = 
m= 
= 
2
2
2
vf − vi
(35.0 m/s) − (25.0 m/s)
1220 m2/s2 − 625 m2/s2
42 × 103 J
= 7.0 × 101 kg
m = 
6.0 × 102 m2/s2
7. vi = 104.5 km/h
vf =
1
v
2 i
1
1
Wnet = ∆KE = KEf − KEi = 2mvf2 − 2mvi2
Wnet = Wkd(cos q) = Fkd(cos q) = mkmgd(cos q)
mk = 0.120
2
g = 9.81 m/s
q = 180°
1
m(v 2
f
2
II
− vi2) = mkmgd(cos q)
2
[(104.5 km/h)(103 m/km)(1 h/3600
− (1)2]
vf 2 − vi2
= 
d = 
(2)(0.120)(9.81 m/s2)(cos 180°)
2mkg(cos q)
1
s)]2[ 2
2
2
104.5
104.5
1
 m/s 4 − 1
−(3)  m/s
3.600
3.600
d = 
2 = 
−(2)(0.120)(9.81 m/s )
−(8)(0.120)(9.81 m/s2)
d = 268 m
Additional Practice 5D
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. h = 6.13/2 m = 3.07 m
PEg = 4.80 kJ
PEg
4.80 × 103 J
m =  = 2 = 1.59 × 102 kg
gh
(9.81 m/s )(3.07 m)
g = 9.81 m/s2
2. h = 1.70 m
PEg = 3.04 × 103 J
3.04 × 103 J
PEg
m =  = 2 = 182 kg
(9.81 m/s )(1.70 m)
gh
g = 9.81 m/s2
3. PEg = 1.48 × 107 J
h = (0.100)(180 km)
1.48 × 107 J
PEg
m =  = 
= 83.8 kg
2
(9.81 m/s )(0.100)(180 × 103 m)
gh
g = 9.81 m/s2
4. m = 3.6 × 104 kg
PEg = 8.88 × 108 J
8.88 × 108 J
PEg
h =  = 
= 2.5 × 103 m = 2.5 km
(3.6 × 104 kg)(9.81 m/s2)
mg
g = 9.81 m/s2
Section Two — Problem Workbook Solutions
II Ch. 5–5
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Givens
Solutions
PEg
5.  = 20.482 m2/s2
m
g = 9.81 m/s2
20.482 m2/s2 20.482 m2/s2
h =  = 
= 2.09 m
g
9.81 m/s2
6. k = 3.0 × 104 N/m
2
PEelastic = 1.4 × 10 J
7. m = 51 kg
x=±
2P
Ek
= ± (
32.)0(1×.410×
1N0/mJ) = +9.7 × 10
2
elastic
4
−2
m = 9.7 cm
PEtot = PEg + PEelastic
2
g = 9.81 m/s
Set PEg = 0 J at the river level.
h = 321 m − 179 m = 142 m
PEg = mgh = (51 kg)(9.81 m/s2)(142 m) = 7.1 × 104 J
k = 32 N/m
x = 179 m − 104 m = 75 m
II
PEg
 = gh = 20.482 m2/s2
m
8. h2 = 4080 m
h1 = 1860 m
m = 905 kg
1
1
PEelastic = 2 kx 2 = 2 (32 N/m)(75 m)2 = 9.0 × 104 J
PEtot = (7.1 × 104 J) + (9.0 × 104 J) = 1.6 × 105 J
∆PEg = PEg,2 − PEg,1 = mg(h2 − h1) = (905 kg)(9.81 m/s2)(4080 m − 1860 m)
∆PEg = (905 kg)(9.81 m/s2)(2220 m) = 1.97 × 107 J
g = 9.81 m/s2
k = 9.50 × 103 N/m
g = 9.81 m/s2
x = 59.0 cm
h1 = 1.70 m
h2 = h1 − x
1
1
a. PEelastic = 2 kx 2 = 2 (9.50 × 103 N/m)(0.590 m)2 = 1.65 × 103 J
b. PEg,1 = mgh1 = (286 kg)(9.81 m/s2)(1.70 m) = 4.77 × 103 J
c. h2 = 1.70 m − 0.590 m = 1.11 m
PEg,2 = mgh2 = (286 kg)(9.81 m/s2)(1.11 m) = 3.11 × 103 J
d. ∆PEg = PEg,2 − PEg,1 = (3.11 × 103 J) − (4.77 × 103 J) = −1.66 × 103 J
The answer in part (d) is approximately equal in magnitude to that in (a); the
slight difference arises from rounding. The increase in elastic potential energy
corresponds to a decrease in gravitational potential energy; hence the difference
in signs for the two answers.
II Ch. 5–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. m = 286 kg
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Givens
Solutions
10. ∆x = 9.50 × 102 m
∆x
a. vx = vi (cos q) = 
∆t
∆x
∆t = 
vi(cos q)
q = 45.0°
m = 65.0 g
g = 9.81 m/s2
∆t
vertical speed of the arrow for the first half of the flight = vi (sin q) = g 
2
g∆x
vi (sin q) = 
2vi(cos q)
vi =
g∆x
 =
2(sin q )(cos q )
(9.81 m/s2)(9.50 × 102 m)
 = 96.5 m/s
(2)(sin 45.0°)(cos 45.0°)
KEi = 2 mvi2 = 2 (65.0 × 10−3 kg)(96.5 m/s)2 = 303 J
1
x = 55.0 cm
1
b. From the conservation of energy,
PEelastic = KEi
1
 kx 2
2
= KEi
2KEi
(2)(303 J)
k = 
= 
= 2.00 × 103 N/m
2
x
(55.0 × 10−2 m)2
II
c. KE i = PEg,max + KE f
KE f = 2 mvx 2 = 2 m[(vi (cos q)]2 = 2 (65.0 × 10−3 kg)(96.5 m/s)2(cos 45.0°)2 = 151 J
1
1
1
PEg,max = KE i − KE f = 303 J − 151 J = 152 J
PEg ,max
152 J
hmax = 
=  
mg
(65.0 × 10−3 kg)(9.81 m/s2)
h = 238 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Additional Practice 5E
PEi + KEi = PEf + KEf
1. m = 118 kg
1
hi = 5.00 m
mghi + 2 mvi2 = mghf + KEf
g = 9.81 m/s2
mghf = mghi + 2 mvi2 − KEf
vi = 0 m/s
KEf = 4.61 kJ
1
(0 m/s)2
4.61 × 103 J
v 2 KEf
hf = hi + i −  = 5.00 m + 
− 
2
(2)(9.81 m/s )
(118 kg)(9.81 m/s2)
2g mg
hf = 5.00 m − 3.98 m = 1.02 m above the ground
2. vf = 42.7 m/s
PEi + KEi = PEf + KEf
1
1
mghi + 2mvi 2 = mghf + 2mvf 2
hf = 50.0 m
(42.7 m/s)2 − (0 m/s)2
vf2 − vi2
hi = hf + 
= 50.0 m + 
= 50.0 m + 92.9 m
(2)(9.81 m/s2)
2g
vi = 0 m
hi = 143 m
2
g = 9.81 m/s
The mass of the nut is not needed for the calculation.
Section Two — Problem Workbook Solutions
II Ch. 5–7
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Givens
Solutions
PEi + KEi = PEf + KEf
3. hi = 3150 m
1
mghi = mghf + 2mvf 2
(60.0 m/s)2
vf 2
hf = hi −  = 3150 m − 
= 3150 m − 183 m
(2)(9.81 m/s2)
2g
vf = 60.0 m/s
KEi = 0 J
g = 9.81 m/s2
hf = 2970 m
PEi + KEi = PEf + KEf
4. hi = 1.20 × 102 m
PEi − PEf = KEf
hf = 30.0 m
KEf = ∆PE = mg(hi − hf )
m = 72.0 kg
g = 9.81 m/s
KEf = (72.0 kg)(9.81 m/s2)(1.20 × 102 m − 30.0 m) = (72.0 kg)(9.81 m/s2)(9.0 × 101 m)
KEi = 0 J
KEf = 6.4 × 104 J
2
vf =
2KEf
 =
m
(2)(6.4 × 104 J)

72.0 kg
vf = 42 m/s
II
5. hf = 250.0 m
1
5
∆ME = −2.55 × 10 J
m = 250.0 kg
2
g = 9.81 m/s
∆ME = PEf − KEi = mghf − 2mvi2
vi =
2∆ ME
2ghf −   =
m
(2)(−2.55 × 105 J)
(2)(9.81 m/s2)(250.0 m) −  
250.0 kg
vi = 4.
03
m2/
s2
+2.0
03
m2/
s2 = 6.
03
m2/
s2
90
×1
4×1
94
×1
vi = 83.3 m/s = 3.00 × 102 km/h
PEi + KEi = PEf + KEf
m = 120.0 g
PEi − PEf = KEf
g = 9.81 m/s2
KEf = PEi − PEf = mghi − mghf = mg∆h
KEi = 0 J
KEf = mg∆h = (0.1200 kg)(9.81 m/s2)(3.2 × 103 m) = 3.8 × 103 J
∆h = hi − hf = 3.2 km
PEf = mghf = mg(hi − ∆h) = (0.1200 kg)(9.81 m/s2)(12.3 × 103 m − 3.2 × 103 m)
PEf = (0.1200 kg)(9.81 m/s2)(9.1 × 103 m) = 1.1 × 104 J
Alternatively,
PEf = PEi − KEf = mghi − KEf
PEf = (0.1200 kg)(9.81 m/s2)(12.3 × 103 m) − 3.8 × 103 J = 1.45 × 104 J − 3.8 × 103 J
PEf = 1.07 × 104 J
7. h = 68.6 m
ME i = PE i = mgh
1
v = 35.6 m/s
MEf = KEf = 2 mv 2
g = 9.81 m/s2
mgh − 2 mv 2)
(MEi − MEf )(100)
percent of energy dissipated =  = 
(100)
MEi
mgh
PEf = 0 J
KEi = 0 J
1
1
gh − 2 v 2
percent of energy dissipated = 
(100)
gh
1
(9.81 m/s2)(68.6 m) − 2 (35.6 m/s)2
(100)
percent of energy dissipated = 
(9.81 m/s2)(68.6 m)
(673 J − 634 J)(100)
(39 J) (100)
percent of energy dissipated =  =  = 5.8 percent
673 J
673 J
II Ch. 5–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. hi = 12.3 km
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Additional Practice 5F
Givens
1. P = 56 MW
∆t = 1.0 h
2. ∆t = 62.25 min
Solutions
W = P∆t = (56 × 106 W)(1.0 h)(3600 s/h) = 2.0 × 1011 J
W = P∆t = (585.0 W)(62.25 min)(60 s/min) = 2.185 × 106 J
P = 585.0 W
3. h = 106 m
m = 14.0 kg
g = 9.81 m/s2
W = Fgd(cos q) = Fgd = mgh
W mgh (14.0 kg)(9.81 m/s2)(106 m)
∆t =  =   = 
= 48.5 s
3.00 × 102 W
P
P
P = 3.00 × 102 W
q = 0°
4. P = 2984 W
W = 3.60 × 104 J
5. ∆t = 3.0 min
W = 54 kJ
6. ∆t = 16.7 s
h = 18.4 m
m = 72.0 kg
g = 9.81 m/s2
II
W
54 × 103 J
P =  =  = 3.0 × 102 W
∆t (3.0 min)(60 s/min)
W = Fgd (cos q) = mgh
W mgh (72.0 kg)(9.81 m/s2)(18.4 m)
P =  =  = 
∆t
∆t
16.7 s
P = 778 W
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 0°
W 3.60 × 104 J
∆t =  =  = 12.1 s
P
2984 W
Section Two — Problem Workbook Solutions
II Ch. 5–9
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Momentum and Collisions
Chapter
6
Additional Practice 6A
Givens
Solutions
1. v = 40.3 km/h
p = 6.60 × 102 kg • m/s
6.60 × 102 kg • m/s
p
m =  = 
= 59.0 kg
v (40.3 × 103 m/h)(1 h/3600 s)
ptot = mh v + mp v
2. mh = 53 kg
v = 60.0 m/s to the east
ptot = 7.20 × 103 kg • m/s to
the east
7.20 × 103 kg • m/s − (53 kg)(60.0 m/s)
ptot − mhv
 = 
mp = 
v
60.0 m/s
7.20 × 103 kg • m/s − 3.2 × 103 kg • m/s 4.0 × 103 kg • m/s
mp =  =  = 67 kg
60.0 m/s
60.0 m/s
II
2
3. m1 = 1.80 × 10 kg
m2 = 7.0 × 101 kg
ptot = 2.08 × 104 kg•m/s to
the west
−2.08 × 104 kg•m/s
−2.08 × 104 kg•m/s
p
 = 
v = tot
= 
2
1
2.50 × 102 kg
m1 + m2 1.80 × 10 kg + 7.0 × 10 kg
v = −83.2 m/s = 83.2 m/s to the west
= −2.08 × 104 kg•m/s
4. m = 83.6 kg
HRW material copyrighted under notice appearing earlier in this book.
p = 6.63 × 105 kg•m/s
5. m = 6.9 × 107 kg
p 6.63 × 105 kg•m/s
v =  =  = 7.93 × 103 m/s = 7.93 km/s
m
83.6 kg
p = mv = (6.9 × 107 kg)(33 × 103 m/h)(1 h/3600 s) = 6.3 × 108 kg • m/s
v = 33 km/h
6. h = 22.13 m
m = 2.00 g
g = 9.81 m/s2
1
mgh = 2mvf 2
vf = 2g
h
p = mvf = m 2g
m/s
m)
h = (2.00 × 10−3 kg) (2
)(
9.
81
2)(2
2.
13
p = 4.17 × 10−2 kg • m/s downward
Additional Practice 6B
1. m = 9.0 × 104 kg
vi = 0 m/s
vf = 12 cm/s upward
F = 6.0 × 103 N
mvf − mvi (9.0 × 104 kg)(0.12 m/s) − (9.0 × 104 kg)(0 m/s)
∆t =  = 
F
6.0 × 103 N
(9.0 × 104 kg)(0.12 m/s)
∆t = 
= 1.8 s
6.0 × 103 N
Section Two—Problem Workbook Solutions
II Ch. 6–1
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Givens
Solutions
2. m = 1.00 × 106 kg
vi = 0 m/s
∆p = mvf − mvi = (1.00 × 106 kg)(0.20 m/s) − (1.00 × 106 kg)(0 m/s)
∆p = 2.0 × 105 kg • m/s
vf = 0.20 m/s
5
∆p 2.0 × 10 kg • m/s
= 16 s
∆t =  = 
F
12.5 × 103 N
F = 12.5 kN
3. h = 12.0 cm
F = 330 N, upward
m = 65 kg
The speed of the pogo stick before and after it presses against the ground can be determined from the conservation of energy.
PEg = KE
1
mgh = 2mv 2
g = 9.81 m/s2
v = ± 2g
h
For the pogo stick’s downward motion,
vi = − 2g
h
For the pogo stick’s upward motion,
vf = + 2g
h
∆p = mvf − mvi = m 2g
h − m − 2g
h
II
∆p = 2m 2g
h
m/s
m)
h (2)(65 kg) (2
∆p 2m 2g
)(
9.
81
2)(0
.1
20
∆t =  =  = 
330 N
F
F
∆t = 0.60 s
F = 8.0 kN to the east
F∆t + mvi (8.0 × 103 N)(8.0 s) + (6.0 × 103 kg)(0 m/s)
vf =  = 
m
6.0 × 103 kg
∆t = 8.0 s
vf = 11 m/s, east
4. m = 6.0 × 103 kg
5. vi = 125.5 km/h
2
m = 2.00 × 10 kg
F = −3.60 × 102 N
∆t = 10.0 s
F∆t + mvi
vf = 
m
(−3.60 × 102 N)(10.0 s) + (2.00 × 102 kg)(125.5 × 103 m/h)(1 h/3600 s)
vf = 
2.00 × 102 kg
3
−3.60 × 103 N • s + 6.97 × 103 kg • m/s 3.37 ⫻ 10 kg • m/s

vf = 
=
= 16.8 m/s
2.00 ⫻ 102 kg
2.00 × 102 kg
or vf = (16.8 × 10−3 km/s)(3600 s/h) = 60.5 km/h
6. m = 45 kg
3
F = 1.6 × 10 N
∆t = 0.68 s
vi = 0 m/s
II Ch. 6–2
F∆t + mv (1.6 × 103 N)(0.68 s) + (45 kg)(0 m/s)
vf = i = 
45 kg
m
(1.6 × 103 N)(0.68 s)
vf =  = 24 m/s
45 kg
Holt Physics Solution Manual
HRW material copyrighted under notice appearing earlier in this book.
vi = 0 m/s
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Givens
Solutions
7. m = 4.85 × 105 kg
vi = 20.0 m/s northwest
vf = 25.0 m/s northwest
∆t = 5.00 s
mvf − mvi (4.85 × 105 kg)(25.0 m/s) − (4.85 × 105 kg)(20.0 m/s)
F = 
= 
∆t
5.00 s
6
1.21 × 107 kg•m/s − 9.70 × 106 kg•m/s 2.4 × 10 kg•m/s
F =  = 
5.00 s
5.00 s
F = 4.8 × 105 N northwest
8. vf = 12.5 m/s upward
m = 70.0 kg
∆t = 4.00 s
mvf − mvi (70.0 kg)(12.5 m/s) − (70.0 kg)(0 m/s)
F =  =  = 219 N
∆t
4.00 s
F = 219 N upward
vi = 0 m/s
9. m = 12.0 kg
From conservation of energy, vi = − 2g
h
h = 40.0 m
∆p = mvf − mvi = mvf − m – 2g
h
∆t = 0.250 s
∆p = (12.0 kg)(0 m/s) + (12.0 kg) (2
m/s
)(
9.
81
2)(4
0.
0m
) = 336 kg • m/s
vf = 0 m/s
2
g = 9.81 m/s
II
336 kg • m/s
∆p
F =  =  = 1340 N = 1340 N upward
∆t
0.250 s
Additional Practice 6C
1. F = 2.85 × 106 N backward
= −2.85 × 106 N
∆p = F∆t = (−2.85 × 106 N)(21 s)
∆p = −6.0 × 107 kg•m/s forward or 6.0 × 107 kg•m/s backward
7
m = 2.0 × 10 kg
vi = 3.0 m/s forward
= +3.0 m/s
1
1
∆x = 2(vi + vf)∆t = 2(3.0 m/s + 0 m/s)(21 s) = 32 m forward
vf = 0 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 21 s
2. m = 6.5 × 104 kg
F = −1.7 × 106 N
vi = 1.0 km/s
∆t = 30.0 s
∆p = F∆t = (−1.7 × 106 N)(30.0 s) = −5.1 × 107 kg•m/s
∆p + mv −5.1 × 107 kg•m/s + (6.5 × 104 kg)(1.0 × 103 m/s)
vf = i = 
6.5 × 104 kg
m
−5.1 × 107 kg•m/s + 6.5 × 107 kg•m/s 1.4 × 107 kg•m/s
= 
= 220 m/s
vf = 
6.5 × 104 kg
6.5 × 104 kg
1
1
1
∆x = 2(vi + vf )∆t = 2(1.0 × 103 m/s + 220 m/s)(30.0 s) = 2(1.2 × 103 m/s)(30.0 s)
∆x = 1.8 × 104 m = 18 km
Section Two—Problem Workbook Solutions
II Ch. 6–3
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Givens
Solutions
3. m = 2.03 × 104 kg
vi = 5.00 m/s to the east
= 5.00 m/s
∆t = 20.3 s
F = 1.20 × 103 N to the west
∆p = F∆t = (−1.20 × 103 N)(20.3 s) = 2.44 × 104 kg•m/s to the west
∆p + mv −2.44 × 104 kg•m/s + (2.03 × 104 kg)(5.00 m/s)
vf = i = 
2.3 × 104 kg
m
−2.44 × 104 kg•m/s + 1.02 × 105 kg•m/s 7.58 × 104 kg•m/s
vf = 
= 
2.03 × 104 kg
2.03 × 104 kg
vf = 3.73 m/s
1
1
1
∆x = 2(vi + vf )∆t = 2[5.00 m/s + (3.73 m/s)](20.3 s) = 2(8.73 m/s)(20.3 s)
∆x = 88.6 m = 88.6 m to the east
4. m = 113 g
vi = 2.00 m/s to the right
mvf − mvi (0.113 kg)(0 m/s) − (0.113 kg)(2.00 m/s) −(0.113 kg)(2.00 m/s)
 =  = 
F= 
0.80 s
0.80 s
∆t
vf = 0 m/s
F = −0.28 N = 0.28 N to the left
∆t = 0.80 s
∆x = 2(vi + vf )∆t = 2(2.00 m/s + 0 m/s)(0.80 s)
1
1
∆x = 0.80 m to the right
II
5. m = 4.90 × 106 kg
vi = 0.200 m/s
6
6
∆p mvf − mvi (4.90 × 10 kg)(0 m/s) − (4.90 × 10 kg)(0.200 m/s)
F =  =  = 
∆t
∆t
10.0 s
–(4.90 ⫻ 106 kg)(0.200 m/s)
F =  ⫽ –9.80 ⫻ 104 N
10.0 s
vf = 0 m/s
∆t = 10.0 s
F = 9.80 × 104 N opposite the palace’s direction of motion
1
1
∆x = 2(vi + vf)∆t = 2(0.200 m/s + 0 m/s)(10.0 s)
∆x = 1.00 m
6. h = 68.6 m
From conservation of energy,
vi = 2g
h
F = −2.24 × 104 N
∆p = mvf − mvi = mvf − m 2g
h
g = 9.81 m/s2
h (1.00 × 103 kg)(0 m/s) − (1.00 × 103 kg) (2
m/s
)(
9.
81
.
2)(6
8.
6m
)
∆p mvf − m 2g
∆t =  =  = 
4
F
−2.24 × 10 N
F
vf = 0 m/s
−(1.00 × 103 kg) (2
m/s
)(
9.
81
2)(6
8.
6m
)
= 1.64 s

−2.24 × 104 N
1
1
∆x = 2(vi + vf )∆t = 2( 2g
h + vf )∆t
1
∆x = 2 (2
m/s
)(
9.
81
2)(6
8.
6m
) + 0 m/s(1.64 s) = 30.1 m
∆t =
II Ch. 6–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m = 1.00 × 10 kg
3
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Givens
Solutions
7. m = 100.0 kg
2
vi = 4.5 × 10 m/s
vf = 0 m/s
F = −188 N
mvf − mv (100.0 kg)(0 m/s) − (100.0 kg)(4.5 × 102 m/s)
∆t = i = 
−188 N
F
−(100.0 kg)(4.5 × 102 m/s)
∆t =  = 240 s
−188 N
1
1
∆x = 2(vi + vf)∆t = 2(4.5 × 102 m/s + 0 m/s)(240 s) = 5.4 × 104 m = 54 km
The tunnel is 54 km long.
Additional Practice 6D
1. m1 = 3.3 × 103 kg
v1,i = 0 m/s
v2,i = 0 m/s
Because the initial momentum is zero, the final momentum is also zero, and so
3
−m v1,f −(3.3 × 10 kg)(−0.050 m/s)
m2 = 1
=  = 66 kg
v2,f
2.5 m/s
v2,f = 2.5 m/s to the right
= ⫹2.5 m/s
II
v1,f = 0.050 m/s to the left
= –0.050 m/s
2. m1 = 1.25 × 103 kg
v1,i = 0 m/s
v2,i = 0 m/s
v2,f = 1.40 m/s backward
= –1.40 m/s
Because the initial momentum is zero, the final momentum is also zero, and so
∆x
0.24 m
v1,f = 1 =  = 0.060 m/s forward
∆t1
4.0 s
3
−m v1,f −(1.25 × 10 kg)(0.060 m/s)
m2 = 1
=  = 54 kg
v2,f
−1.40 m/s
∆t1 = 4.0 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆x1 = 24 cm forward
= ⫹24 cm
3. m1 = 114 kg
v2, f = 5.32 m/s backward
= −5.32 m/s
v1, f = 3.41 m/s forward
= +3.41 m/s
m2 = 60.0 kg
4. m1 = 5.4 kg
v1, f = 7.4 m/s forward
= +7.4 m/s
v2, f = 1.4 m/s backward
= −1.4 m/s
m2 = 50.0 kg
m1vi + m2vi = m1v1, f + m2v2, f
m1v1, f + m2v2, f (114 kg)(3.41 m/s) + (60.0 kg)(−5.32 m/s)
 = 
vi = 
114 kg + 60.0 kg
m1 + m2
389 kg•m/s − 319 kg•m/s 7.0 × 101 kg•m/s
vi =  =  = 0.40 m/s
174 kg
174 kg
vi = 0.40 m/s forward
m1vi + m2vi = m1v1, f + m2v2, f
m1v1, f + m2v2, f (5.4 kg)(7.4 m/s) + (50.0 kg)(−1.4 m/s)
vi =  = 
5.4 kg + 50.0 kg
m1 + m2
4.0 × 101 kg•m/s − 7.0 × 101 kg•m/s −3.0 × 101 kg•m/s
vi =  =  = –0.54 m/s
55.4 kg
55.4 kg
vi = 0.54 m/s backward
Section Two—Problem Workbook Solutions
II Ch. 6–5
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Givens
Solutions
5. m1 = 3.4 × 102 kg
m1vi + m2vi = m1v1, f + m2v2, f
v2, f = 9.0 km/h northwest
= −9.0 km/h
m1v1, f + m2v2, f (3.4 × 102 kg)(28 km/h) + (2.6 × 102 kg)(−9.0 km/h)
vi =  = 
m1 + m2
3.4 × 102 kg + 2.6 × 102 kg
v1, f = 28 km/h southeast
= +28 km/h
9.5 × 103 kg•km/h − 2.3 × 103 kg•km/h 7.2 × 103 kg•km/h
vi = 
= 
6.0 × 102 kg
6.0 × 102 kg
m2 = 2.6 × 102 kg
6. mi = 3.6 kg
vi = 12 km/h to the southeast
Because the initial momentum is zero, the final momentum must also equal zero.
m2 = 3.0 kg
mi v1,f = −m2 v2,f
v1,i = 0 m/s
−m v2,f −(3.0 kg)(−2.0 m/s)
v1,f = 2
=  = 1.7 m/s = 1.7 m/s to the right
m1
3.6 kg
v2,i = 0 m/s
v2,f = 2.0 m/s to the left
= –2.0 m/s
II
7. m1 = 449 kg
v1,i = 0 m/s
v2,i = 0 m/s
v2,f = 4.0 m/s backward
= –4.0 m/s
Because the initial momentum is zero, the final momentum must also equal zero.
−m v2,f −(60.0 kg)(−4.0 m/s)
=  = 0.53 m/s = 0.53 m/s forward
v1,f = 2
m1
449 kg
∆x = v1,f ∆t = (0.53 m/s)(3.0 s) = 1.6 m forward
m2 = 60.0 kg
∆t = 3.0 s
Additional Practice 6E
v1,i = 6.0 m/s forward
v2,i = 0 m/s
vf = 2.2 m/s forward
m1v1,i − m1vf (155 kg)(6.0 m/s) − (155 kg)(2.2 m/s)
m2 =  = 
vf − v2,i
2.2 m/s − 0 m/s
930 kg•m/s –340 kg•m/s 590 kg•m/s
m2 =  = 
2.2 m/s
2.2 m/s
m2 = 270 kg
2. v1,i = 10.8 m/s
v2,i = 0 m/s
vf = 10.1 m/s
m1 = 63.0 kg
3. v1, i = 4.48 m/s to the right
m1v1,i − m1vf (63.0 kg)(10.8 m/s) − (63.0 kg)(10.1 m/s)
m2 =  = 
10.1 m/s − 0 m/s
vf − v2,i
6.80 × 102 kg • m/s − 6.36 × 102 kg • m/s 44 kg • m/s
m2 =  =  = 4.4 kg
10.1 m/s
10.1 m/s
v2, i = 0 m/s
(54 kg)(4.00 m/s) − (54 kg)(0 m/s) (54 kg)(4.00 m/s)
m1 =  = 
4.48 m/s − 4.00 m/s
0.48 m/s
vf = 4.00 m/s to the right
m1 = 450 kg
m 2 = 54 kg
II Ch. 6–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. m1 = 155 kg
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Givens
Solutions
4. m1 = 28 × 103 kg
m2 = 12 × 103 kg
v1,i = 0 m/s
vf = 3.0 m/s forward
(m1 + m2)vf − m1v1,i
v2,i = 
m2
(28 × 103 kg + 12 × 103 kg)(3.0 m/s) − (28 × 103 kg)(0 m/s)
v2,i = 
12 × 103 kg
(4.0 ⫻ 104 kg)(3.0 m/s)
v2,i = 
12 ⫻ 103 kg
v2,i = 1.0 × 101 m/s forward
m2 = 267 kg
(m1 + m2)vf − m1v1,i
v2,i = 
m2
v1,i = 4.00 m/s to the left
= – 4.00 m/s
(227 kg + 267 kg)(0 m/s) − (227 kg)(– 4.00 m/s)
v2,i =  = 3.40 m/s
267 kg
5. m1 = 227 kg
vf = 0 m/s
v2,i = 3.40 m/s to the right
6. m1 = 9.50 kg
v1, i = 24.0 km/h to the north
m2 = 32.0 kg
vf = 11.0 km/h to the north
II
(m1 + m2)vf − m1v,1
v2, i = 
m2
(9.5 kg + 32.0 kg)(11.0 km/h) − (9.50 kg)(24.0 km/h)
v2, i = 
32.0 kg
(41.5 kg)(11.0 km/h) − 228 kg•km/h 456 kg•km/h − 228 kg•km/h
v2, i =  = 
32.0 kg
32.0 kg
228 kg•km/h
= 
32.0 kg
v2, i = 7.12 km/h to the north
v1,i = 89 km/h
v1,i + v2,i 89 km/h + 69 km/h 158 km/h
 =  =  = 79 km/h
Because m1 = m2, vf = 
2
2
2
v2,i = 69 km/h
vf = 79 km/h
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. m1 = m2
8. m1 = 3.0 × 103 kg
m2 = 2.5 × 102 kg
m1v1,i + m2v2,i (3.0 × 103 kg)(1.0 m/s) + (2.5 × 102 kg)(−3.0 m/s)
vf =  = 
(3.0 × 103 kg) + (2.5 × 102 kg)
m1 + m2
v2,i = 3.0 m/s down
= –3.0 m/s
3.0 ⫻103 kg•m/s –7.5 ⫻ 102 kg•m/s 2.2 ⫻ 103 kg•m/s
vf = 
= 
3.2 ⫻ 103 kg
3.2 ⫻ 103 kg
v1,i = 1.0 m/s up = +1.0 m/s
vf = 0.69 m/s = 0.69 m/s upward
9. m1 = (2.267 × 103 kg) +
(5.00 × 102 kg) = 2.767 ×
103 kg
3
m2 = (1.800 × 10 kg) +
(5.00 × 102 kg) = 2.300 ×
103 kg
v1,i = 2.00 m/s to the left
= –2.00 m/s
m1v1,i + m2v2,i (2.767 × 103 kg)(−2.00 m/s) + (2.300 × 103 kg)(1.40 m/s)
vf =  = 
2.767 × 103 kg + 2.300 × 103 kg
m1 + m2
–5.53 ⫻103 kg • m/s + 3220 kg • m/s –2310 kg•m/s
=  = –0.456 m/s
vf = 
5.067 × 103 kg
5067 kg
vf = 0.456 m/s to the left
v2,i = 1.40 m/s to the right
= +1.40 m/s
Section Two—Problem Workbook Solutions
II Ch. 6–7
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Additional Practice 6F
Givens
Solutions
1. m1 = 2.0 g
v1,i = 2.0 m/s forward
= +2.0 m/s
m2 = 0.20 g
v2,i = 8.0 m/s backward
= −8.0 m/s forward
m1v1,i + m2v2,i
vf = 
m1 + m2
(2.0 ⫻ 10–3 kg)(2.0 m/s) + (0.20 ⫻ 10–3 kg)(−8.0 m/s)
vf = 
2.0 ⫻ 10–3 kg + 0.20 ⫻ 10–3 kg
4.0 ⫻ 10–3 kg•m/s − 1.6 ⫻ 10–3 kg•m/s
vf = 
2.2 ⫻ 10–3 kg
–
2.4 ⫻ 10 3 kg•m/s
= 1.1 m/s forward
vf = 
2.2 ⫻ 10–3 kg
1
1
KEi = 2m1v1,i 2 + 2m2 v2,i 2
KEi = 2(2.0 × 10−3 kg)(2.0 m/s)2 + 2(0.20 × 10−3 kg)(−8.0 m/s)2
1
1
KEi = 4.0 × 10−3 J + 6.4 × 10−3 J = 1.04 × 10−3 J
KEf = 2(m1 + m2 )vf 2 = 2(2.0 × 10−3 kg + 0.20 × 10−3 kg)(1.1 m/s)2
1
1
KEf = 2(2.2 × 10−3 kg)(1.1 m/s)2
1
∆KE = KEf − KEi = 1.3 × 10−3 J − 1.04 × 10−2 J = −9.1 × 10−3 J
II
−3
∆KE 9.1 × 10 J
fraction of total KE dissipated =  = 
= 0.88
KEi 1.04 × 10−2 J
2. m1 = 313 kg
v1,i = 6.00 m/s away from
shore
v2,i = 0 m/s
vf = 2.50 m/s away from
shore
m1v1,i − m1vf
(313 kg)(6.00 m/s) − (313 kg)(2.50 m/s)
m2 =  = 
vf − v2,i
2.50 m/s − 0 m/s
1880 kg • m/s − 782 kg • m/s 1.10 × 103 kg • m/s
m2 =  =  = 4.4 × 102 kg
2.50 m/s
2.50 m/s
1
1
KEi = 2m1v1,i 2 + 2m2v2,i 2
1
1
KEi = 2 (313 kg)(6.00 m/s)2 + 2 (4.40 × 102 kg)(0 m/s)2 = 5630 J
1
KEf = 2 (m1 + m2)vf 2
1
∆KE = KEf − KEi = 2350 J − 5630 J = −3280 J
3. m1 = m2 = 111 kg
v1, i = 9.00 m/s to the right
= +9.00 m/s
v2, i = 5.00 m/s to the left
= −5.00 m/s
m1v1, i + m2v2, i (111 kg)(9.00 m/s) + (111 kg)(−5.00 m/s)
 = 
vf = 
111 kg + 111 kg
m1 + m2
999 kg•m/s − 555 kg•m/s 444 kg•m/s
vf =  =  = 2.00 m/s to the right
222 kg
222 kg
1
1
1
1
KEi = 2 m1 v1,i 2 + 2 m2 v2,i 2 = 2 (111 kg)(9.00 m/s)2 + 2 (111 kg)(–5.00 m/s)2
KEi = 4.50 × 103 J + 1.39 × 103 J = 5.89 × 103 J
1
1
1
KEf = 2(m1 + m2 )vf 2 = 2 (111 kg ⫹ 111 kg)(2.00 m/s)2 = 2 (222 kg)(2.00 m/s)
KEf = 444 J
∆KE = KEf − KEi = 444 J − 5.89 × 103 J = −5450 J
II Ch. 6–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
KEf = 2 (313 kg + 4.40 × 102 kg)(2.50 m/s)2 = 2 (753 kg)(2.50 m/s)2 = 2350 J
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Givens
4. m1 = m2 = 60.0 kg + 50.0 kg
= 110.0 kg
v1, i = 106.0 km/h to the east
= +106.0 km/h
v2, i = 75.0 km/h to the west
= −75.0 km/h
Solutions
m1v1,i + m2v2,i (110.0 kg)(106.0 km/h) + (110.0 kg)(−75.0 km/h)
vf =  = 
m1 + m2
110.0 kg + 110.0 kg
1.166 × 104 kg•km/h − 8.25 × 103 kg•km/h 3.41 × 103 kg•km/h
vf =  = 
220.0 kg
220.0 kg
vf = 15.5 km/h to the east
1
1
KEi = 2m1v1, i2 + 2m2v2, i2
1
1
KEi = 2(110.0 kg)(106.0 × 103 m/h)2(1 h/3600 s)2 + 2(110.0 kg)(−75.0 × 103 m/h)2
(1 h/3600 s)2
KEi = 4.768 × 104 J + 2.39 × 104 J = 7.16 × 104 J
1
KEf = 2(m1 + m2)vf 2
1
KEf = 2(110.0 kg + 110.0 kg)(15.5 × 103 m/h)2(1 h/3600 s)2
1
= 2(220.0 kg)(15.5 × 103 m/h)2(1 h/3600 s)2
KEf = 2.04 × 103 J
∆KE = KEf − KEi = 2.04 × 103 J − 7.16 × 104 J = −6.96 × 104 J
5. m1 = 4.00 × 105 kg
v1, i = 32.0 km/h
m2 = 1.60 × 105 kg
v2, i = 45.0 km/h
m1v1, i + m2v2,i (4.00 × 105 kg)(32.0 km/h) + (1.60 × 105 kg)(45.0 km/h)
 = 
vf = 
4.00 × 105 kg + 1.60 × 105 kg
m1 + m2
II
1.28 × 107 kg•km/h + 7.20 × 106 kg•km/h 2.00 × 107 kg•km/h
vf = 
= 
5.60 × 105 kg
5.60 × 105 kg
vf = 35.7 km/h
1
1
KEi = 2m1v1,i2 + 2m2v2,i2
1
1
KEi = 2(4.00 × 105 kg)(32.0 × 103 m/h)(1 h/3600 s)2 + 2(1.60 × 105 kg)
(45.0 × 103 m/h)2(1 h/3600 s)2
KEi = 1.58 × 107 J + 1.25 × 107 J = 2.83 × 107 J
1
KEf = 2(m1 + m2)vf2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
KEf = 2(4.00 × 105 kg + 1.60 × 105 kg)(35.7 × 103 m/h)2(1 h/3600 s)2
1
= 2(5.60 × 105 kg)(35.7 × 103 m/h)2(1 h/3600 s)2
KEf = 2.75 × 107 J
∆KE = KEf − KEi = 2.75 × 107 J − 2.83 × 107 J = −8 × 105 J
Section Two—Problem Workbook Solutions
II Ch. 6–9
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Givens
6. m1 = 21.3 kg
v1,i = 0 m/s
m2 = 1.80 × 10−1 kg
vf = 6.00 × 10−2 m/s
Solutions
(m1 + m2 )vf –m1v1,i
v2,i = 
m2
(21.3 kg + 1.80 × 10−1 kg)(6.00 × 10−2 m/s) – (21.3 kg)(0 m/s)
v2,i = 
1.80 × 10−1 kg
(21.5 kg)(6.00 ⫻ 10−2 m/s)
v2,i = 
⫽ 7.17 m/s
1.80 ⫻ 10–1 kg
1
1
KEi = 2m1v1,i 2 + 2m2 v2,i 2
KEi = 2(21.3 kg)(0 m/s)2 + 2(1.80 × 10−1 kg)(7.17 m/s)2
1
1
KEi = 0 J⫹ 4.63 J = 4.63 J
1
KEf = 2(m1 + m2)vf 2
KEf = 2(21.3 kg ⫹ 0.180 kg)(6.00 × 10−2 m/s)2 = 2(21.5 kg)(6.00 × 10−2 m/s)2
1
1
KEf = 3.87 × 10−2 J
∆KE = KEf − KEi = 3.87 × 10−2 J − 4.63 J = −4.59 J
II
7. m1 = 122 g
m2 = 96.0 g
v2,i = 0 m/s
Because v2,i = 0 m/s, m1v1,i = (m1 + m2)vf
(m1 + m2)vf
v1,i = 
m1
2 
2

∆KE KEf − KEi 2m1 + m2 vf − 2m1v1,i
fraction of KE dissipated =  =  = 
1
KEi
KEi
 m v 2
1
1
2
1 1,i
(m1 + m2)vf 2
(m1 + m2 )vf 2 − m1 
m1
fraction of KE dissipated = 
(m1 + m2)vf 2
m1 
m1
(m1vf )2 + 2m1m2vf 2 + (m2vf )2
m1vf 2 + m2vf 2 − 
m1
fraction of KE dissipated = 
(m1vf )2 + 2m1m2vf 2 + (m2vf )2

m1
m22
vf 2 m1 + m2 − m1 − 2m2 − 
m1
fraction of KE dissipated = 
2
m2
vf 2 m1 + 2m2 + 
m1
m22
(96.0 g)2
−m2 − 
−96.0 g − 
m1
122 g
fraction of KE dissipated =  = 
2
m2
(96.0 g)2
m1 + 2m2 + 
122 g + (2)(96.0 g) + 
m1
122 g
−171.5 g
−96.0 g − 75.5 g
fraction of KE dissipated =  = 
= −0.440
122 g + 192 g + 75.5 g 3.90 × 102 g
The fraction of kinetic energy dissipated can be determined without the initial velocity because this value cancels, as shown above. The initial velocity is needed to find
the decrease in kinetic energy.
II Ch. 6–10 Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Additional Practice 6G
Givens
Solutions
1. m2 = 0.500 m1
Momentum conservation
3
v1, i = 3.680 × 10 km/h
2
v1, f = −4.40 × 10 km/h
v2, f = 5.740 × 103 km/h
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m1v1, f + m2v2, f − m1v1, i m1v1, f + (0.500)m1v2, f − m1v1, i
v2, i = 
= 
(0.500)m1
m2
v2, i = (2.00)v1, f + v2, f − (2.00)v1, i = (2.00)(−4.40 × 102 km/h) + 5.740 × 103 km/h
− (2.00)(3.680 × 103 km/h) = −8.80 × 102 km/h + 5.740 × 103 km/h
− 7.36 × 103 km/h
v2, i = −2.50 × 103 km/h
Conservation of kinetic energy (check)
1
m v 2
2 1 1, i
+ 2m2v2, i2 = 2m1v1, f 2 + 2m2v2, f 2
1
1
1
m v 2
2 1 1, i
+ 2(0.500)m1v2, i2 = 2m1v1, f 2 + 2(0.500)m1v2, f 2
1
1
1
1
v1, i2 + (0.500)v2, i2 = v1, f 2 + (0.500)v2, f 2
(3.680 × 103 km/h)2 + (0.500)(−2.50 × 103 km/h)2 = (−4.40 × 102 km/h)2 + (0.500)
(5.740 × 103 km/h)2
1.354 × 107 km2/h2 + 3.12 × 106 km2/h2 = 1.94 × 105 km2/h2 + 1.647 × 107 km2/h2
7
2
2
7
2
II
2
1.666 × 10 km /h = 1.666 × 10 km /h
2. m1 = 18.40 kg
m2 = 56.20 kg
v2, i = 5.000 m/s to the left
= −5.000 m/s
v2, f = 6.600 × 10−2 m/s to
the left
= − 6.600 × 10−2 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v1, f = 10.07 m/s to the left
= −10.07 m/s
Momentum conservation
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m1v1, f + m2v2, f − m2v2, i
v1, i = 
m1
(18.40 kg)(−10.07 m/s) + (56.20 kg)(−6.600 × 10−2 m/s) − (56.20 kg)(−5.000 m/s)
v1, i = 
18.40 kg
−185.3 kg•m/s − 3.709 kg•m/s + 281.0 kg•m/s 92.0 kg•m/s
v1, i =  =  = 5.00 m/s
18.40 kg
18.40 kg
v1, i = 5.00 m/s to the right
Conservation of kinetic energy (check)
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2m1v1, f 2 + 2m2v2, f 2
1
1
(18.40 kg)(5.00 m/s)2 + (56.20)(−5.000
2
2
1
+ 2(56.20 kg)(−6.600 × 10−2 m/s)2
1
m/s)2 = 2(18.40 kg)(−10.07 ms)2
2.30 × 102 J + 702.5 J = 932.9 J + 0.1224 J
932 J = 933 J
The slight difference arises from rounding.
Section Two—Problem Workbook Solutions
II Ch. 6–11
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Givens
Solutions
3. m1 = m2
Momentum conservation
v1, i = 5.0 m/s to the right
= +5.0 m/s
v1, f = 2.0 m/s to the left
= −2.0 m/s
v2, f = 5.0 m/s to the right
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m1v1, f + m2v2, f − m1v1, i
v2, i = 
= v1, f + v2, f − v1, i
m2
v2, i = −2.0 m/s + 5.0 m/s − 5.0 m/s = −2.0 m/s
v2, i = 2.0 m/s to the left
Conservation of kinetic energy (check)
= +5.0 m/s
1
m v 2
2 i 1, i
1
1
1
+ 2m2v2, i2 = 2m1v1, f 2 + 2m2v2, f 2
v1,i2 + v2,i2 = v1, f 2 + v2, f 2
(5.0 m/s)2 + (−2.0 m/s)2 = (−2.0 m/s)2 + (5.0 m/s)2
25 m2/s2 + 4.0 m2/s2 = 4.0 m2/s2 + 25 m2/s2
29 m2/s2 = 29 m2/s2
4. m1 = 45.0 g
II
Momentum conservation
v1, i = 273 km/h to the right
= +273 km/h
v2, i = 0 km/h
v1, f = 91 km/h to the left
= −91 km/h
v2, f = 182 km/h to the right
= +182 km/h
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m1v1, f − m1v1, i (45.0 g)(−91 km/h) − (45.0 g)(273 km/h)
 = 
m2 = 
v2, i − v2, f
0 km/h − 182 km/h
−4.1 ⫻ 103 g •km/h − 12.3 ⫻ 103 g •km/h −16.4 ⫻ 103 g •km/h
m2 =  = 
−182 km/h
−182 km/h
m2 = 90.1 g
Conservation of kinetic energy (check)
1
m v 2
2 1 1,i
1
1
1
+ 2m2 v2,i2 = 2m1v1, f 2 + 2m2 v2, f
1
1
(45.0 g)(273 ⫻ 103 m/h)2(1 h/3600 s)2 + (90.1 g)(0 m/s)2
2
2
1
1
= 2(45.0 g)(−91 × 103 m/h)2(1 h/3600 s)2 + 2(90.1 g)(182 × 103
m/h)2 (1 h/3600 s)2
129 J + 0 J = 14 J + 115 J
5. v1,i = 185 km/h to the right
⫽ ⫹185 km/h
v2,i = 0 km/h
Momentum conservation
m1v1, i + m2v2,i = m1v1,f + m2v2,f
m v − mv = v –v
m
m [185 km/h − (−80.0 km/h)] = v
m
v =  (265 km/h) to the right
m m1
vi,f = 80.0 km/h to the left
= −80.0 km/h
–2
m1 = 5.70 ⫻ 10 kg
m1
1,i
2
1,f
2,f
2,i
2
1
2,f
− 0 km/h
2
1
2,f
2
Conservation of kinetic energy
1
1
1
KEi = 2m1v1,i2 + 2m2 v2,i2 = 2m1v1,i2
1
1
KEf = 2m1v1, f 2 + 2m2v2,f 2
1
 m v 2
2 1 1,i
1
1
= 2m1v1,f 2 + 2m2 v2,f 2
m1
m
 (v1,i)2 = 1 (v1,f)2 + v2,f 2
m2
m2
m
m
m (185 km/h) = m (−80.0 km/h) + v
1
2
II Ch. 6–12
Holt Physics Solution Manual
2
1
2
2
2
2,f
Copyright © by Holt, Rinehart and Winston. All rights reserved.
129 J = 129 J
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Givens
Solutions
mm(3.42×10km/h)−mm(6.40×10km/h) = v
1
4
2
1
2
2
v2,f =
3
2
2
2,f
2
mm (2.78×10k m/h) = mm 167 km/h
1
4
2
2
2
1
2
Equating the two results for v2,f yields the ratio of m1 to m2.
m (265 km/h) = m (167 km/h)
m1
m1
2
2
265 km/h =
m (167 km/h)
m2
1
m2
265 km/h 2
 =  = 2.52
m1
167 km/h
m2 = (2.52) m1 ⫽ (2.52)(5.70 ⫻ 10–2 kg)
m2 = 0.144 kg
6. m1 = 4.00 × 105 kg
II
Momentum conservation
m2 = 1.60 × 105 kg
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, i = 32.0 km/h to the right
m1v1, i + m2v2, i − m1v1, f
v2, f = 
m2
v2, i = 36.0 km/h to the right
v1, f = 35.5 km/h to the right
(4.00 × 105 kg)(32.0 km/h) + (1.60 × 105 kg)(36.0 km/h) − (4.00 × 105 kg)(35.5 km/h)
v2, f = 
1.60 × 105 kg
1.28 × 107 kg•km/h + 5.76 × 106 kg•km/h − 1.42 × 107 kg•km/h
v2, f = 
1.60 × 105 kg
6
4.4 × 10 kg•km/h
v2, f = 
1.60 × 105 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v2, f = 28 km/h to the right
Conservation of kinetic energy (check)
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2m1v1, f 2 + 2m2 v2, f 2
1
(4.00
2
1
× 105 kg)(32.0 × 103 m/h)2(1 h/3600 s)2 + 2(1.60 × 105 kg)(36.0 × 103 m/h)2
1
1
(1 h/3600 s)2 = 2(4.00 × 105 kg)(35.5 × 103 m/h)2(1 h/3600 s)2 + 2(1.60 × 105 kg)
3
2
2
(28 ⫻ 10 m/h) (1 h/3600 s)
1.58 × 107 J + 8.00 × 106 J = 1.94 × 107 J + 4.8 × 106 J
2.38 × 107 J = 2.42 × 107 J
The slight difference arises from rounding.
Section Two—Problem Workbook Solutions
II Ch. 6–13
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Givens
Solutions
7. m1 = 5.50 × 105 kg
Momentum conservation
m2 = 2.30 × 105 kg
m1v1,i + m2v2,i = m1v1,f + m2v2,f
v1, i = 5.00 m/s to the right
= +5.00 m/s
m1v1,i + m2v2,i − m2v2,f
v1,f = 
m1
v2, i = 5.00 m/s to the left
= −5.00 m/s
(5.50 × 105 kg)(5.00 m/s) + (2.30 × 105 kg)(−5.00 m/s) − (9.10 m/s)
v1,f = 
5.50 × 105 kg
v2, f = 9.10 m/s to the right
= +9.10 m/s
2.75 ⫻ 106 kg•m/s – 1.15 ⫻106 kg•m/s –2.09 × 106 kg•m/s
v1,f = 
= −0.89 m/s right
5.50 × 105 kg
v1,f = 0.89 m/s left
Conservation of kinetic energy (check)
1
m v 2
2 1 1, i
1
1
1
+ 2m2 v2, i2 = 2 m1v1, f 2 + 2m2v2, f 2
1
 (5.50
2
1
1
× 105 kg)(5.00 m/s)2 + 2(2.30 × 105 kg)(−5.00 m/s)2 = 2(5.50 × 105 kg)
1
(−0.89 m/s)2 + 2(2.30 × 105 kg)(9.10 × m/s)2
II
6.88 × 106 J + 2.88 × 106 J = 2.2 × 105 J + 9.52 × 106 J
9.76 × 106 J = 9.74 × 106 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
The slight difference arises from rounding.
II Ch. 6–14
Holt Physics Solution Manual
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Rotational Motion and the Law of Gravity
Chapter
7
Additional Practice 7A
Givens
Solutions
1. r = 10.0 km
∆s = r∆q = (10.0 km)(15.0 rad) = 1.50 × 102 km
∆q = +15.0 rad
The particle moves in the positive, or counterclockwise , direction around the
neutron star’s “north” pole.
2. ∆q = 3(2p rad)
∆s = r∆q = (6560 km)[(3)(2p rad)] = 1.24 × 105 km
r = 6560 km
1.40 × 105 km
3. r = 
2
a. ∆s = r∆q = (7.00 × 104 km)(1.72 rad) = 1.20 × 105 km
5
∆s (1.20 × 10 km)(1 rev/2p rad)
b. ∆qE =  = 
= 3.00 rev, or 3.00 orbits
6.37 × 103 km
rE
4
= 7.00 × 10 km
∆q = 1.72 rad
II
rE = 6.37 × 103 km
4. ∆q = 225 rad
∆s = 1.50 × 106 km
5. r = 5.8 × 107 km
∆s 1.5 × 108 km
∆q =  = 
= 2.6 rad
r
5.8 × 107 km
∆s = 1.5 × 108 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆s 1.50 × 106 km
r =  =  = 6.67 × 103 km
∆q
225 rad
6. ∆s = −1.79 × 104 km
r = 6.37 × 103 km
∆s −1.79 × 104 km
= −2.81 rad
∆q =  = 
r
6.37 × 103 km
Additional Practice 7B
1. r = 1.82 m
−1
wavg = 1.00 × 10
∆t = 60.0 s
2. ∆t = 120 s
wavg = 0.40 rad/s
3. r = 30.0 m
∆s = 5.0 × 102 m
∆q = wavg∆t = (1.00 × 10−1 rad/s)(60.0 s) = 6.00 rad
rad/s
∆s = r∆q = (1.82 m)(6.00 rad) = 10.9 m
∆q = wavg∆t = (0.40 rad/s)(120 s) = 48 rad
∆q
∆s
5.0 × 102 m
wavg =  =  =  = 0.14 rad/s
∆t r∆t (30.0 m)(120 s)
∆t = 120 s
Section Two—Problem Workbook Solutions
II Ch. 7–1
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Givens
Solutions
4. ∆q = 16 rev
∆q (16 rev)(2p rad/rev)
wavg =  =  = 0.37 s
(4.5 min)(60 s/min)
∆t
∆t = 4.5 min
5. wavg = 2p rad/24 h
0.262 rad
∆q
∆t =  =  = 1.00 h
wavg
2p rad

24 h
∆q = 0.262 rad
6. r = 2.00 m
∆s = 1.70 × 102 km
1.70 × 105 m
∆q
∆s
∆t =  =  =  = 1.44 × 104 s = 4.00 h
wavg rwavg (2.00 m)(5.90 rad/s)
wavg = 5.90 rad/s
Additional Practice 7C
w2 − w1

∆t = 
aavg
1. aavg = 2.0 rad/s2
w1 = 0 rad/s
II
9.4 rad/s − 0.0 rad/s
∆t = 
2.0 rad/s2
w2 = 9.4 rad/s
∆t = 4.7 s
2. ∆tJ = 9.83 h
aavg = −3.0 × 10−8 rad/s2
w2 = 0 rad/s
∆q
2p rad
w1 =  =  = 1.78 × 10−4 rad/s
∆tJ (9.83 h)(3600 s/h)
−4
w2 − w1 0.00 rad/s − 1.78 × 10 rad/s
 = 
∆t = 
−8
−3.0 × 10 rad/s2
aavg
w2 = 3.15 rad/s
w2 − w1 3.15 rad/s − 2.00 rad/s 1.15 rad/s
 =  = 
aavg = 
3.6 s
3.6 s
∆t
∆t = 3.6 s
aavg = 0.32 rad/s2
3. w1 = 2.00 rad/s
4. w1 = 8.0 rad/s
w2 = 3w1 = 24 rad/s
w2 − w1 24 rad/s − 8.0 rad/s 16 rad/s
 =  = 
aavg = 
∆t
25 s
25 s
∆t = 25 s
aavg = 0.64 rad/s2
5. ∆t1 = 365 days
2p rad
∆q
w1 = 1 =  = 1.99 × 10−7 rad/s
∆t2 (365 days)(24 h/day)(3600 s/h)
∆q1 = 2p rad
aavg = 6.05 × 10−13 rad/s2
∆t2 = 12.0 days
w2 = 1.99 × 10−7 rad/s + 6.27 × 10−7 rad/s = 8.26 × 10−7 rad/s
6. w1 = 0 rad/s
2
aavg = 0.800 rad/s
∆t = 8.40 s
w2 = w1 + aavg∆t2 = 1.99 × 10−7 rad/s + (6.05 × 10−13 rad/s2)(12.0 days)
(24 h/day)(3600 s/h)
w2 − w1

a avg = 
∆t
w2 = w1 + a avg ∆t
w2 = 0 rad/s + (0.800 rad/s2)(8.40 s)
w2 = 6.72 rad/s
II Ch. 7–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 5.9 × 103 s
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Additional Practice 7D
Givens
Solutions
w f = w i + a∆t
1. w i = 5.0 rad/s
a = 0.60 rad/s2
w f = 5.0 rad/s + (0.60 rad/s2)(0.50 min)(60.0 s/min)
∆t = 0.50 min
w f = 5.0 rad/s + 18 rad/s
w f = 23 rad/s
2. a = 1.0 × 10−10 rad/s2
∆t = 12 h
w f = w i + a∆t = 2.66 × 10−6 rad/s + (1.0 × 10−10 rad/s2)(12 h)(3600 s/h)
2p rad
w i = 
27.3 days
w f = 2.66 × 10−6 rad/s + 4.3 × 10−6 rad/s = 7.0 × 10−6 rad/s
∆s
∆q = 
r
43 m
3. r = 
2p rad
2a∆s
wf2 = wi2 + 2a∆q = wi2 + 
r
wi = 0 rad/s
∆s = 160 m
−2
a = 5.00 × 10
2
rad/s
wf =
w+2ar∆s =
2
i
wf = 1.5 rad/s
4. ∆s = 52.5 m
−5
a = −3.2 × 10
2
rad/s
wf = 0.080 rad/s
r = 8.0 cm
(2)(5.00 × 10−2 rad/s2)(160 m)
(0 rad/s)2 + 
43 m

2p rad
II
∆s
∆q = 
r
2a∆s
wf2 = wi2 + 2a∆q = wi2 + 
r
wi =
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2p rad 1 day 1 h
w i =    = 2.66 × 10−6 rad/s
27.3 days 24 h 3600 s
(2)(−3.2 × 10−5 rad/s2)(52.5 m)
(0.080 rad/s)2 − 
8.0 × 10−2 m
2a∆s
wf 2 −  =
r
wi = 6.
s2
+4.2
s2 = 4.
s2
4×10−3ra
d2/
×10−2ra
d2/
8×10−2ra
d2/
wi = 0.22 rad/s
5. r = 3.0 m
w i = 0.820 rad/s
w f = 0.360 rad/s
∆s = 20.0 m
∆s
∆q = 
r
w f 2 − w i 2 wf2 − wi2 (0.360 rad/s)2 − (0.820 rad/s)2
a =  =  = 
2∆q
20.0 m
2 ∆s
(2) 

3.0 m
r
0.130 rad2/s2 − 0.672 rad2/s2
−0.542 rad2/s2
a =  = 
20.0 m
20.0 m
(2) 
(2) 
3.0 m
3.0 m
a = −4.1 × 10−2 rad/s2
Section Two—Problem Workbook Solutions
II Ch. 7–3
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Givens
Solutions
6. r = 1.0 km
w i = 5.0 × 10−3 rad/s
∆t = 14.0 min
∆q = 2p rad
7. w i = 7.20 × 10−2 rad/s
1
∆q = w i ∆t + 2a∆t 2
−3
2(∆q − wi∆t) (2)[2p rad − (5.0 × 10 rad/s)(14.0 min)(60 s/min)]

a = 
=
[(14.0 min)(60 s/min)]2
∆t2
(2)(2.1 rad)
(2)(6.3 rad − 4.2 rad)
a = 2 = 2 = 6.0 × 10−6 rad/s2
[(14.0
min)(60 s/min)]
[(14.0 min)(60 s/min)]
∆t = (4 min)(60 s/min) + 22 s = 262 s
∆q = 12.6 rad
∆q = w i ∆t + 2a∆t 2
∆t = 4 min, 22 s
2(∆q − wi∆t) (2)[12.6 rad − (7.20 × 10−2 rad/s)(262 s)]
a = 
= 
(262 s)2
∆t2
1
(2)(12.6 rad − 18.9 rad)
(2)(−6.3 rad/s)
a = 
= 
= −1.8 × 10−4 rad/s2
(262 s)2
(262 s)2
II
w f = 32.0 rad/s
∆t = 6.83 s
9. a = 2.68 × 10−5 rad/s2
∆t = 120.0 s
2p rad
wi = 
12
10. wi = 6.0 × 10−3 rad/s
wf = 3wi = 18 × 10−3 rad/s
a = 2.5 × 10−4 rad/s2
w f − w i 32.0 rad/s − 27.0 rad/s 5.0 rad/s
aavg =  =  = 
6.83 s
6.83 s
∆t
aavg = 0.73 rad/s2
1
∆q = wi∆t + 2a∆t2
2p rad
1
∆q =  (1 h/3600 s)(120.0 s) + 2(2.68 × 10−5 rad/s2)(120.0 s)2
12 h
∆q = 1.7 × 10−2 rad + 1.93 × 10−1 rad = 0.210 rad
wf2 − wi2
∆q =  
2a
3.2 × 10−4 rad2/s2 − 3.6 × 10−5 rad2/s2
(18 × 10−3 rad/s)2 − (6.0 × 10−3 rad/s)2
∆q = 
= 
−4
2
(2)(2.5 × 10 rad/s )
(2)(2.5 × 10−4 rad/s2)
2.8 × 10−4 rad2/s2
∆q = 
= 0.56 rad
5.0 × 10−4 rad/s2
11. w i = 9.0 × 10−7 rad/s
w f = 5.0 × 10−6 rad/s
a = 7.5 × 10−10 rad/s2
wf − wi
∆t = 
a
4.1 × 10−6 rad/s
5.0 × 10−6 rad/s − 9.0 × 10−7 rad/s

∆t = 
=
7.5 × 10−10 rad/s2
7.5 × 10−10 rad/s2
∆t = 5.5 × 103 s = 1.5 h
II Ch. 7–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. w i = 27.0 rad/s
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Givens
Solutions
12. r = 7.1 m
∆s
1
∆q =  = w i ∆t + 2a∆t 2
r
∆s
1
a∆t 2 + w ∆t −  = 0
i
2
r
∆s = 500.0 m
w i = 0.40 rad/s
a = 4.0 × 10−3 rad/s2
Using the quadratic equation:
−∆s
w
−4a
r
∆t = 
− wi ±
i
1

2
2
22a 1
500.0 m
1
−0.40 rad/s ± (0.40 rad/s)2 + (4)2(4.0 × 10−3 rad/s2) 
7.1 m
∆t = 
(2)2(4.0 × 10−3 rad/s2)
1
s2
72
rad
2/
−0.40 rad/s ± 0.
s2
+0.5
s2 – 0.40 rad/s ± 0.
16
rad
2/
6ra
d2/
∆t = 
= 
−3
2
−3
2
4.0 × 10 rad/s
4.0 × 10 rad/s
Choose the positive value:
−0.40 rad/s + 0.85 rad/s 0.45 rad/s
∆t = 
= 
= 1.1 × 102 s
4.0 × 10−3 rad/s2
rad/s2
II
Additional Practice 7E
1. w = 4.44 rad/s
v
4.44 m/s
r = t =  = 1.00 m
w 4.44 rad/s
vt = 4.44 m/s
2. vt = 16.0 m/s
−5
Copyright © by Holt, Rinehart and Winston. All rights reserved.
w = 1.82 × 10
rad/s
v
16.0 m/s
r = t = 
= 8.79 × 105 m = 879 km
w 1.82 × 10−5 rad/s
circumference = 2pr = (2p)(879 km) = 5.52 × 103 km
3. w = 5.24 × 103 rad/s
vt = 131 m/s
4. vt = 29.7 km/s
8
r = 1.50 × 10 km
19.0 mm
5. r =  = 9.50 mm
2
v
131 m/s
= 2.50 × 10−2 m = 2.50 cm
r = t = 
w 5.24 × 103 rad/s
v
29.7 km/s
w = t = 
= 1.98 × 10−7 rad/s
r 1.50 × 108 km
vt = rw = (9.50 × 10−3 m)(25.6 rad/s) = 0.243 m/s
w = 25.6 rad/s
Section Two—Problem Workbook Solutions
II Ch. 7–5
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Additional Practice 7F
Givens
Solutions
1. r = 32 m
at = 0.20 m/s2
a 0.20 m/s2
a = t = 
r
32 m
a = 6.2 × 10−3 rad/s2
2. r = 8.0 m
at = −1.44 m/s2
3. ∆w = −2.4 × 10−2 rad/s
∆t = 6.0 s
at = − 0.16 m/s2
4. ∆q ′ = 14 628 turns
∆t′ = 1.000 h
II
at = 33.0 m/s2
wi = 0 rad/s
∆q = 2p rad
a −1.44 m/s2
a = t =  = −0.18 rad/s2
r
8.0 m
∆w −2.4 × 10−2 rad/s
a =  =  = −4.0 × 10−3 rad/s2
∆t
6.0 s
−0.16 m/s2
a
r = t = 
= 4.0 × 101 m
a −4.0 × 10−3 rad/s2
a
r =  t
a
wf 2 − wi2
a = 
2∆q
∆q ′
wf = 
∆t′
at
2
∆q ′
  − wi2
∆t′
2at∆q
r =  = 
2
2∆q
∆q ′
  − wi 2
∆t′
2
(2)(33.0 m/s )(2p rad)
5. r = 56.24 m
wi = 6.00 rad/s
wf = 6.30 rad/s
∆t = 0.60 s
2
− (0 rad/s) = 0.636 m
2
at = ra
wf − wi
a = 
∆t
6.30 rad/s − 6.00 rad/s
(56.24 m)(0.30 rad/s)
wf − wi
at = r  = (56.24 m)  = 
0.60 s
0.60 s
∆t
at = 28 m/s2
6. r = 1.3 m
∆q = 2p rad
∆t = 1.8 s
wi = 0 rad/s
at = ra
2(∆q − wi ∆t)
α = 
∆t2
(2)[2p rad − (0 rad/s)(1.8 s)]
2(∆q − wi∆t)
at = r 
= (1.3 m) 
(1.8 s)2
∆t2
at = 5.0 m/s2
II Ch. 7–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(14 628 turns)(2p rad/turn)
r = 
(1.000 h)(3600 s/h)
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Additional Practice 7G
Givens
Solutions
v 2 (0.17 m/s)2
r = t = 
= 0.10 m
ac
0.29 m/s2
1. vt = 0.17 m/s
ac = 0.29 m/s2
2. w = 2p rad/day
−7
ac = 2.65 × 10
ac = rw2
2
m/s
2.65 × 10−7 m/s2
a
r = c2 = 2
[(2p rad/day)(1 day/24 h)(1 h/3600 s)]
w
r = 50.1 m
58.4 cm
3. r =  = 29.2 cm
2
ac = 8.50 × 10−2 m/s2
12 cm
4. r =  = 6.0 cm
2
vt = ra
c = (2
9.
2×10−2m
)(
8.
50
×10−2m
/s
2
vt = 0.158 m/s
vt = ra
0−2m
m/s
c = (6
.0
×1
)(
0.
28
2) = 0.13 m/s
II
ac = 0.28 m/s2
5. r = 20.0 m
∆q
ac = rw 2 = r 
∆t
∆t = 16.0 s
(20.0 m)(2p rad)2
ac = 
= 3.08 m/s2
(16.0 s)2
∆q = 2p rad
6. ∆t = 1.000 h
∆s = 47.112 km
3
r = 6.37 × 10 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2
∆s 2

∆t
v2
∆s2
ac = t =  = 2
r
r∆t
r
(47 112 m)2
ac = 
= 2.69 × 10−5 m/s2
6
(6.37 × 10 m)[(1.000 h)(3600 s/h)]2
Additional Practice 7H
mtot = m1 + m2 = 235 kg + 72 kg = 307 kg
1. m1 = 235 kg
m2 = 72 kg
v2
Fc = mtot ac = mtot t
r
rFc
(25.0 m)(1850 N)
vt =  =  = 12.3 m/s
mtot
307 kg
r = 25.0 m
Fc = 1850 N
2. m = 30.0 g
r = 2.4 m
FT = 0.393 N
2
g = 9.81 m/s
v2
FT = Fg + Fc = mg + mt
r
(2.4 m)[0.393 N − (30.0 × 10−3 kg)(9.81 m/s2)]
r(FT − mg)

 =
vt =
30.0 × 10−3 kg
m
(2.4 m)(0.393 N − 0.294 N)
(2.4 m)(0.099 N)
 = 
v = 30.0 × 10 kg
30.0 × 10 kg
t
−3
−3
vt = 2.8 m/s
Section Two—Problem Workbook Solutions
II Ch. 7–7
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Givens
Solutions
Fg = Fc
3. vt = 8.1 m/s
m vt2
m1g = 2
r
r = 4.23 m
m1 = 25 g
m gr
m2 = 1
vt2
g = 9.81 m/s2
(25 × 10−3 kg)(9.81 m/s2)(4.23 m)
m2 = 
= 1.6 × 10−2 kg
(8.1 m/s)2
v2
Fc = m t
r
4. vt = 75.57 km/h
m = 92.0 kg
(92.0 kg)[(75.57 km/h)(103 m/km)(1 h/3600 s)]2
mvt2
r = 
= 
12.8 N
Fc
Fc = 12.8 N
r = 3.17 × 103 m = 3.17 km
5. m = 75.0 kg
II
mvt 2 (75.0 kg)(12 m/s)2
Fc = 
=  = 24 N
446 m
r
r = 446 m
vt = 12 m/s
FT = Fc + mg = 24 N + (75.0 kg)(9.81 m/s2)
g = 9.81 m/s2
FT = 24 N + 736 N = 7.60 × 102 N
Additional Practice 7I
1. r = 6.3 km
Fgr
(2.5 × 10−2 N)(6.3 × 103 m)2
m2 =  = 
Gm1
N • m2
6.673 × 10−11 
(3.0 kg)
kg 2
2
Fg = 2.5 × 10
N
m1 = 3.0 kg
2
N•m
G = 6.673 × 10−11 
kg2
2. m1 = 3.08 × 104 kg
m2 = 5.0 × 1015 kg
Fgr2
(2.88 × 10−16 N)(1.27 × 107 m)2
m2 =  = 
Gm1
N • m2
6.673 × 10−11 
(3.08 × 104 kg)
kg 2
7
r = 1.27 × 10 m
Fg = 2.88 × 10−16 N
2
N•m
G = 6.673 × 10−11 
kg2
3. m1 = 5.81 × 104 kg
r = 25.0 m
Fg = 5.00 × 10−7 N
N•m2
G = 6.673 × 10−11 
kg2
II Ch. 7–8
m2 = 2.26 × 104 kg
Fgr2
(5.00 × 10−7 N)(25.0 m)2
m2 =  = 
Gm1
N • m2
6.673 ×10−11 
(5.81 × 104 kg)
kg 2
m2 = 80.6 kg
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
−2
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Givens
Solutions
4. m1 = 621 g
r=
m2 = 65.0 kg
N•m2
G = 6.673 × 10−11 
kg2
5. m1 = m2 = 1.0 × 108 kg
Fg = 1.0 × 10
N
−11
G = 6.673 × 10
Gm1m2
g
Fg = 1.0 × 10−12 N
−3


F
N•m2

kg2
(6.673 × 10−11 N • m2/kg2)(0.621 kg)(65.0 kg)

= 52 m
1.0 × 10−12 N
r=
r=

F
Gm1m2
g
r=
(6.673 × 10−11 N • m2/kg2)(1.0 × 108 kg)2

1.0 × 10−3 N
r = 2.6 × 104 m = 26 km
6. ms = 25 × 109 kg
m1 = m2 =
1
 m
2 s
r = 1.0 × 103 km
N•m2
G = 6.673 × 10−11 
kg2
7. m1 = 318mE
m2 = 50.0 kg
VJ = 1323VE
mE = 5.98 × 1024 kg
rE = 6.37 × 106 m
N • m2 1
6.673 × 10−11 
(25 × 109 kg)2
kg2 2
Gm1m2
Fg = 
= 
= 1.0 × 10−2 N
(1.0 × 106 m)2
r2
II
If VJ = 1323 VE , then rJ = 3 1323
rE .
Gm m2 (6.673 × 10−11 N • m2/kg2)(318)(5.98 × 1024 kg)(50.0 kg)
Fg = 1
= 
rJ 2
[( 3 13
23
)(6.37 × 106 m)]2
Fg = 1.30 × 103 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
N•m2
G = 6.673 × 10−11 
kg2
Section Two—Problem Workbook Solutions
II Ch. 7–9
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Rotational Equilibrium and Dynamics
Chapter
8
Additional Practice 8A
Givens
1. m = 3.00 × 105 kg
q = 90.0° − 45.0° = 45.0°
t = 3.20 × 107 N • m
g = 9.81 m/s2
2. t net = 9.4 kN • m
m1 = 80.0 kg
m2 = 120.0 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
3. t net = 56.0 N • m
Solutions
t = Fd(sin q) = mgl (sin q)
l
t
= 
mg(sin q)
l
3.20 × 107 N • m
= 
(3.00 × 105 kg)(9.81 m/s2)(sin 45.0°)
l
= 15.4 m
t net = t 1 + t 2 = F1d1(sin q 1) + F2d2(sin q 2)
q 1 = q 2 = 90°, so
t net = F1d1 + F2d2 = m1g
l
2 + m gl
2
l
t net

=m
1g
 + m2 g
2
l
9.4 × 103 N • m
9.4 × 103 N•m
=  = 
392 N + 1.18 × 103 N
(80.0 kg)(9.81 m/s2)
 + (120.0 kg)(9.81 m/s 2)
2
l
9.4 × 103 N • m
= 
= 6.0 m
1.57 × 103 N
t net = t 1 + t 2 = F1d1(sin q 1) + F2d2(sin q 2)
m1 = 3.9 kg
q 1 = q 2 = 90°, so
m2 = 9.1 kg
t net = F1d1 + F2d2 = m1gd1 + m2 g(1.000 m − x)
t net − m1gd1

x = 1.000 m − 
m2 g
d1 = 1.000 m − 0.700 m =
0.300 m
g = 9.81 m/s2
II
56.0 N • m − (3.9 kg)(9.81 m/s2)(0.300 m)
x = 1.000 m − 
(9.1 kg)(9.81 m/s2)
56.0 N• m − 11 N• m
45 N• m
x = 
= 
= 1.000 m − 0.50 m
(9.1 kg)(9.81 m/s2)
(9.1 kg)(9.81 m/s2)
x = 0.50 m = 5.0 × 101 cm
Section Two—Problem Workbook Solutions
II Ch. 8–1
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Givens
Solutions
4. t = −1.3 × 104 N • m
l
= 6.0 m
t = Fd(sin q) = −Fg(l − d)(sin q)
Fg =
d = 1.0 m
q = 90.0° − 30.0° = 60.0°
76 m
5. R =  = 38 m
2
q = 60.0°
t = −1.45 × 106 N • m
−t

(l − d)(sin q)
−(−1.3 × 104 N • m)
1.3 × 104 N • m
=  = 
(6.0 m − 1.0 m)(sin 60.0°)
(5.0 m)(sin 60.0°)
Fg = 3.0 × 103 N
t = Fd(sin q) = −FgR(sin q)
−(−1.45 × 106 N • m)
−t
Fg =  = 
(38 m)(sin 60.0°)
R(sin q)
Fg = 4.4 × 104 N
6. m1 = 102 kg
II
tnet = t1 + t2 = F1d1(sin q1) + F2d2(sin q2)
m2 = 109 kg
q1 = q2 = 90°, so
l = 3.00 m
l 1 = 0.80 m
l 2 = 1.80 m
tnet = F1d1 + F2d2 = m1g
tnet
2
g = 9.81 m/s
2 − l + m g2 − l 3.00 m
3.00 m
= (102 kg)(9.81 m/s )  − 0.80 m + (109 kg)(9.81 m/s )  − 1.80 m
2
2
l
l
1
2
2
2
2
tnet = (102 kg)(9.81 m/s2)(1.50 m − 0.80 m) + (109 kg)(9.81 m/s2)(1.50 m − 1.80 m)
tnet = (102 kg)(9.81 m/s2)(0.70 m) + (109 kg)(9.81 m/s2)(−0.30 m)
tnet = 7.0 × 102 N • m − 3.2 × 102 N • m
tnet = 3.8 × 102 N • m
7. m = 5.00 × 102 kg
a. t ⬘ = Fd(sin q) = mg d1(sin q1)
d1 = 5.00 m
t ⬘ = (5.00 × 102 kg)(9.81 m/s2)(5.00 m)(sin 80.0°)
t = 6.25 × 105 N • m
t ⬘ = 2.42 × 104 N • m
g = 9.81 m/s2
d2 = 4.00 m
q2 = 90°
b. t net = Fd2 (sin q2) − t ⬘ = Fd2 (sin q2) − mg d1(sin q1)
t net + mgd1(sin q1)
F = 
d2(sin q2)
6.49 × 105 N • m
6.25 × 105 N • m + 2.42 × 104 N • m
F =  = 
4.00 m
4.00 m (sin 90°)
F = 1.62 × 105 N
II Ch. 8–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q1 = 90.0° − 10.0° = 80.0°
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Additional Practice 8B
Givens
Solutions
1. t1 = 2.00 × 105 N • m
5
t2 = 1.20 × 10 N • m
h = 24 m
Apply the second condition of equilibrium, choosing the base of the cactus as the
pivot point.
tnet = t1 − t2 − Fd(sin q) = 0
Fd(sin q) = t1 − t2
For F to be minimum, d and sin q must be maximum. This occurs when the force is
perpendicular to the cactus (q = 90°) and is applied to the top of the cactus (d = h =
24 m).
2.00 × 105 N • m − 1.20 × 105 N • m
t1 − t2
 = 
Fmin = 
24 m
h
8.0 × 104 N•m
Fmin =  = 3.3 × 103 N applied to the top of the cactus
24 m
2. m1 = 40.0 kg
Apply the first condition of equilibrium.
m2 = 5.4 kg
Fn − m1g − m2g − Fapplied = 0
d1 = 70.0 cm
Fn = m1g + m2g + Fapplied = (40.0 kg)(9.81 m/s2) + (5.4 kg)(9.81 m/s2) + Fapplied
d2 = 100.0 cm − 70.0 cm
= 30.0 cm
Fn = 392 N + 53 N + Fapplied = 455 N + Fapplied
g = 9.81 m/s2
II
Apply the second condition of equilibrium, using the fulcrum as the location for the
axis of rotation.
Fappliedd2 + m2gd2 − m1gd1 = 0
2
2
m1gd1 − m2gd2 (40.0 kg)(9.81 m/s )(0.700 m) − (5.4 kg)(9.81 m/s )(0.300 m)
 = 
Fapplied = 
0.300 m
d2
275 N•m − 16 N•m
259 N•m
Fapplied =  = 
0.300 m
0.300 m
Fapplied = 863 N
Substitute the value for Fapplied into the first-condition equation to solve for Fn.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fn = 455 N + 863 N = 1318 N
3. m = 134 kg
Apply the first condition of equilibrium in the x and y directions.
d1 = 2.00 m
Fx = Fapplied (cos q) − Ff = 0
d2 = 7.00 m − 2.00 m = 5.00 m
Fy = Fn − Fapplied (sin q) − mg = 0
q = 60.0°
To solve for Fapplied, apply the second condition of equilibrium, using the fulcrum as
the pivot point.
2
g = 9.81 m/s
Fapplied (sin q)d2 − mgd1 = 0
(134 kg)(9.81 m/s2)(2.00 m)
mgd1
Fapplied = 
= 
(5.00 m)(sin 60.0°)
d2(sin q)
Fapplied = 607 N
Substitute the value for Fapplied into the first-condition equations to solve for Fn and
Ff .
Fn = Fapplied(sin q) + mg = (607 N)(sin 60.0°) + (134 kg)(9.81 m/s2)
Fn = 526 N + 1.31 × 103 N = 1.84 × 103 N
Ff = Fapplied(cos q) = (607 N)(cos 60.0°) = 304 N
Section Two—Problem Workbook Solutions
II Ch. 8–3
Givens
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Solutions
Print
4. m = 8.8 × 103 kg
Apply the first condition of equilibrium in the x and y directions.
d1 = 3.0 m
Fx = Ffulcrum,x − F (sin q) = 0
d2 = 15 m − 3.0 m = 12 m
Fy Ffulcrum,y − F (cos q) − mg = 0
q = 20.0°
To solve for F, apply the second condition of equilibrium •, using the fulcrum as the
pivot point.
g = 9.81 m/s2
Fd2 − mg d1 (cos q) = 0
(8.8 × 103 kg)(9.81 m/s2)(3.0 m)(cos 20.0°)
mg d1 (cos q) 
 =
F= 
12 m
d2
F = 2.0 × 104 N
Substitute the value for F into the first-condition equations to solve for the components of Ffulcrum.
Ffulcrum,x = F (sin q) = (2.0 × 104 N)(sin 20.0°)
Ffulcrum,x = 6.8 × 103 N
Ffulcrum,y = F (cos q) + mg = (2.0 × 104 N)(cos 20.0°) + (8.8 × 104 kg)(9.81 m/s2)
Ffulcrum,y = 1.9 × 104 N + 8.6 × 105 N = 8.8 × 105 N
II
5. m1 = 64 kg
Apply the first condition of equilibrium to solve for Fapplied.
m2 = 27 kg
Fn − m1 g − m2 g − Fapplied = 0
3.00 m
d1 = d2 =  = 1.50 m
2
Fapplied = Fn − m1 g − m2 g = 1.50 × 103 N − (64 kg)(9.81 m/s2) − (27 kg)(9.81 m/s2)
Fn = 1.50 × 103 N
g = 9.81 m/s2
Fapplied = 1.50 × 103 N − 6.3 × 102 N − 2.6 × 103 N = 6.1 × 102 N
To solve for the lever arm for Fapplied, apply the second condition of equilibrium,
using the fulcrum as the pivot point.
Fapplied d + m2 g d2 − m1 g d1 = 0
kg)(9.81 m/s2)(1.50 m) − (27 kg)(9.81 m/s2)(1.50 m)
m1 g d1 − m2 g d2 (64
 = 
d= 
6.1 × 102 N
Fapplied
d = 0.89 m from the fulcrum, on the same side as the less massive seal
6. m1 = 3.6 × 102 kg
Apply the second condition of equilibrium, using the pool’s edge as the pivot point.
m2 = 6.0 × 102 kg
Assume the total mass of the board is concentrated at its center.
l = 15 m
l 1 = 5.0 m
m1 g d − m2 g l − l 1 = 0
2
g = 9.81 m/s2
m2 l − l 1
m2 g l − l 1
2
2
d =  = 
m1 g
m1
15m
(6.0 × 102 kg)  − 5.0 m
(6.0 × 102 kg)(7.5 m − 5.0 m)
2
d = 
= 
2
3.6 × 102 kg
3.6 × 10 kg
(6.0 × 102 kg)(2.5 m)
d = 
= 4.2 m from the pool’s edge
3.6 × 102 kg
II Ch. 8–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9.4 × 102 N • m − 4.0 × 102 N • m
5.4 × 102 N • m
d = 
= 
2
6.1 × 10 N
6.1 × 102 N
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Givens
Solutions
7. m = 449 kg
l
Apply the first condition of equilibrium to solve for F2.
F1 + F2 − mg = 0
= 5.0 m
3
F1 = 2.70 × 10 N
2
g = 9.81 m/s
F2 = mg − F1
F2 = (449 kg)(9.81 m/s2) − 2.70 × 103 N = 4.40 × 103 N − 2.70 × 103 N = 1.70 × 103 N
Apply the second condition of equilibrium, using the left end of the platform as the
pivot point.
F2 l − m g d = 0
F2 l (1.70 × 103 N)(5.0 m)
d =  = 
m g (449 kg)(9.81 m/s2)
d = 1.9 m from the platform’s left end
8. m1 = 414 kg
l
= 5.00 m
Apply the first condition of equilibrium to solve for F2.
F1 + F2 − m1 g − m2 g = 0
m2 = 40.0 kg
F2 = m1 g + m2 g − F1 = (m1 + m2) g − F1
F1 = 50.0 N
F2 = (414 kg + 40.0 kg)(9.81 m/s2) − 50.0 N = (454 kg)(9.81 m/s2) − 50.0 N = 4.45 ×
103 N − 50.0 N
g = 9.81 m/s2
II
F2 = 4.40 × 103 N
Apply the second condition of equilibrium, using the supported end (F1) of the stick
as the rotation axis.
2 − m g l = 0
m
414 kg
 + m g l
 + 40.0 kg (9.81 m/s )(5.0 m)
2
2
d =  = 
F2 d − m1 g
l
1
2
2
2
4.40 × 103 N
F2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(207 kg + 40.0 kg)(9.81 m/s2)(5.00 m)
(247 kg)(9.81 m/s2)(5.00 m)
d = 
= 
3
4.40 × 10 N
4.40 × 103 N
d = 2.75 m from the supported end
Additional Practice 8C
M = 1.20 × 106 kg
1.0 × 109 N • m
t
t
a =  = 2 = 
(1.20 × 106 kg)(50.02)
I MR
t = 1.0 × 109 N • m
a = 0.33 rad/s2
1. R = 50.0 m
2. M = 22 kg
R = 0.36 m
5.7 N•m
t
t
a =  = 2 = 2
(22 kg)(0.36 m)
I MR
t = 5.7 N • m
a = 2.0 rad/s2
Section Two—Problem Workbook Solutions
II Ch. 8–5
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Givens
Solutions
3. M = 24 kg
l
The force is applied perpendicular to the lever arm, which is half the pencil’s length.
= 2.74 m
Therefore,
F = 1.8 N
t = F d (sin q) = F
F
t
a =  =
I
2
l

1
 M l 2
12
2
l
2.74 m
(1.8 N) 
2
= 
1
 (24 kg)(2.74 m)2
12
a = 0.16 rad/s2
4. M = 4.07 × 105 kg
R = 5.0 m
t = 5.0 × 104 N • m
t
t
a =  = 
1
I
MR2
2
(5.0 × 104 N • m)
a = 
1
(4.07 × 105 kg)(5.0 m)2
2
a = 9.8 × 10−3 rad/s2
II
5. R = 2.00 m
The force is applied perpendicular to the lever arm, which is the ball’s radius.
F = 208 N
Therefore,
a = 3.20 × 10−2 rad/s2
t = F d (sin q) = F R
(208 N)(2.00 m)
t FR
T =  =  = 
3.20
× 10−2 rad/s2
a a
6. r = 8.0 m
3
t = 7.3 × 10 N• m
a = 0.60 rad/s2
t
I =  = mr 2
a
7.3 × 103 N•m
I = 
= 1.2 × 104 kg • m2
0.60 rad/s2
1.2 × 104 kg•m2
I
m = 2 = 
= 1.9 × 102 kg
(8.0 m)2
r
7. vt,i = 2.0 km/s
l
= 15.0 cm
∆t = 80.0 s
t = −0.20 N•m
vt,f = 0 m/s
t
v
−
t
t,f vt,i
t
I =  =  = 
a
wf − wi

d2 ∆t
∆t
l
I =
−0.20 N•m
3
0 m/s − 2.0 × 10 m/s

0.150 m
 (80.0 s)
2
I = 6.0 × 10−4 kg • m2
II Ch. 8–6
Holt Physics Solution Manual
=
−0.20 N•m
−2.0 × 103 m/s

(0.075 m)(80.0 s)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
I = 1.30 × 104 kg • m2
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Givens
Solutions
1.70 m
8. R =  = 0.85 m
2
t = 125 N • m
t
I =  = MR2
a
wi = 0 rad/s
125 N•m
125 N • m
t
t
I =  =  =  = 2
6.0 rad/s
a
12 rad/s − 0 rad/s
wf − wi


2.0 s
∆t
wf = 12 rad/s
I = 21 kg • m2
∆t = 2.0 s
21 kg•m2
I



M= 2 =
= 29 kg
(0.85 m)2
R
9. R = 3.00 m
M = 17 × 103 kg
wi = 0 rad/s
wf − wi
1

t = I a = 2 MR2 
∆t
3
wf = 3.46 rad/s
(17 × 10 kg)(3.00 m)2(3.46 rad/s − 0 rad/s)
t =  = 2.2 × 104 N•m
(2)(12 s)
∆t = 12 s
10. R = 4.0 m
8
M = 1.0 × 10 kg
wi = 0 rad/s
wf = 0.080 rad/s
wf − wi
1

t = Ia = 2 MR2 
∆t
II
(1.0 × 108 kg)(4.0 m)2 (0.080 rad/s − 0 rad/s)
t =  = 1.1 × 106 N •m
(2)(60.0 s)
∆t = 60.0 s
11. I = 2.40 × 103 kg•m2
∆q = 2(2p rad) = 4p rad
∆t = 6.00 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
wi = 0 rad/s
1
∆q = wi ∆t + 2 a ∆t2
Because wi = 0,
1
∆q = 2 a∆t2
2∆q
a = 2
∆t
3
2
2 I ∆q (2)(2.40 × 10 kg•m )(4p rad)

t = Ia = 
2
2 =
(6.00 s)
∆t
t = 1.68 × 103 N • m
12. m = 7.0 × 103 kg
r = 18.3 m
at = 25 m/s2
a
t = Ia = (mr 2)t = mrat
r
t = (7.0 × 103 kg)(18.3 m)(25 m/s2)
t = 3.2 × 106 N • m
Section Two—Problem Workbook Solutions
II Ch. 8–7
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Additional Practice 8D
Givens
Solutions
1. ri = 4.95 × 107 km
vi = 2.54 × 105 Km/h
vf = 1.81 × 105 km/h
Li = Lf
Ii wi = If wf
v
v
mri2 i = mrf2 f
ri
rf
ri vi = rf vf
(4.95 × 107 km)(2.54 × 105 km/h)
r v
rf = i i = 
1.81 × 105 km/h
vf
rf = 6.95 × 107 km
2. vi = 399 km/h
Li = Lf
vf = 456 km/h
Ii wi = If wf
R = 0.20 m
v
v
m ri2 i = mrf2 f
ri
rf
∆q = 20 rev
ri vi = rf vf
II
rf = ri − ∆s = ri − R ∆q
ri vi = (ri − R ∆q) vf
ri (vf − vi) = (R ∆q) vf
vf R ∆q (456 km/h)(0.20 m)(20 rev)(2p rad/rev)
 = 
ri = 
456 km/h − 399 km/h
vf − vi
(456 km/h)(0.20 m)(20 rev)(2p rad/rev)
= 
57 km/h
ri = 2.0 × 102 m
Li = Lf
R = 15.0 cm
Ii wi = If wf
wi = 4.70 × 10−3 rad/s
2
Ii wi 5MR wi
If = 
=
wf
wf
wf = 4.74 × 10−3 rad/s
2
(2)(25.0 kg)(0.150 m)2(4.70 × 10−3 rad/s)
If = 
= 0.223 kg•m2
(5)(4.74 × 10−3 rad/s)
2
2
Ii = 5 MR2 = 5 (25.0 kg)(0.150 m)2 = 0.225 kg•m2
∆I = If − Ii = 0.223 kg•m2 − 0.225 kg•m2 = −0.002 kg•m2
The moment of inertia decreases by 0.002 kg•m2.
II Ch. 8–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. M = 25.0 kg
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Givens
Solutions
4. vi = 395 km/h
ri = 1.20 × 102 m
∆r
 = 0.79 m/s
∆t
∆t = 33 s
∆r
rf = ri −  ∆t
∆t
Li = Lf
Ii wi = Lf wf
∆r
r v = r v = r −  ∆t v
∆t v
v
mri2 i = mrf2 f
ri
rf
i i
f f
i
f
(1.20 × 102 m)(395 km/h)
ri vi
vf = 
= 
1.20 × 102 − (0.79 m/s)(33 s)
∆r
[ri −  ∆t]
∆t
(1.20 × 102 m)(395 km/h)
(1.20 × 102 m)(395 km/h)

=
vf = 
1.20 × 102 m − 26 m
94 m
vf = 5.0 × 102 km/h
10.0 m
5. ri =  = 5.00 m
2
4.00 m
rf =  = 2.00 m
2
wi = 1.26 rad/s
II
Li = Lf
Iiwi = If w f
mri2wi = mrf 2wf
2
ri2wi (5.00 m) (1.26 rad/s)


wf = 
=
(2.00 m)2
rf2
wf = 7.88 rad/s
Li = Lf
6. R = 3.00 m
M = 1.68 × 10 kg
Iiwi = If wf
ri = 2.50 m
2MR2 + mri2wi = 2MR2 + mrf2wf
1
2MR2 + mri2wi
wf = 
1
2MR2 + mrf2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4
rf = 3.00 m
m = 2.00 × 102 kg
wi = 3.46 rad/s
1
1
2(1.68 × 104 kg)(3.00 m)2 + (2.00 × 102 kg)(2.50 m)2(3.46 rad/s)
wf =
1
2(1.68 × 104 kg)(3.00 m)2 + (2.00 × 102 kg)(3.00 m)2
1
(7.56 × 104 kg • m2 + 1.25 × 103 kg • m2)(3.46 rad/s)
wf = 
7.56 × 104 kg • m2 + 1.80 × 103 kg • m2
(7.68 × 104 kg • m2)(3.46 rad/s)
wf = 
7.74 × 104 kg • m2
wf = 3.43 rad/s
∆w = wf − wi = 3.43 rad/s − 3.46 rad/s = −0.03 rad/s
The angular speed decreases by 0.03 rad/s.
Section Two—Problem Workbook Solutions
II Ch. 8–9
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Additonal Practice 8E
Givens
Solutions
1. m = 407 kg
MEi = MEf
1
1

2
h = 57.0 m
mgh = 2 mvf 2 +
vf = 12.4 m/s
1

2
wf = 28.0 rad/s
2mgh − mvf2 m(2gh − vf2)
I = 
= 
wf2
wf2
g = 9.81 m/s2
I wf 2
1
I wf2 = mgh − 2 mvf 2
(407 kg)[(2)(9.81 m/s2)(57.0 m) − (12.4 m/s)2]
I = 
(28.0 rad/s)2
(407 kg)(1.12 × 103 m2/s2 − 154 m2/s2)
(407 kg)(9.7 × 102 m2/s2)
I = 
= 
2
(28.0 rad/s)
(28.0 rad/s)2
I = 5.0 × 102 kg • m2
2. h = 5.0 m
MEi = MEf
2
g = 9.81 m/s
II
1
1
mgh = 2mvf 2 + 2Iw f 2
vf 2
1
1
mgh = 2mvf 2 + 2(mr 2) 
r2
mgh = mvf 22 + 2 = mvf2
1
1
vf = gh
m)
= (9
.8
1m
/s
2)(5
.0
vf = 7.0 m/s
the mass is not required
MEi = MEf
2
g = 9.81 m/s
1
 mv 2
i
2
+ 2Iwi2 = mgh
1
1
mv 2
i
2
+
1 1
  mr 2
2 2
vi2
2 = mgh
r
mvi2 2 + 4 = 4 mvi2 = mgh
1
vi =
4. vf = 12.0 m/s
I = 0.80mr
2
g = 9.81 m/s2
1
3
(4)(9.81 m/s2)(1.2 m)
 = 4.0 m/s
3
4gh
 =
3
MEi = MEf
0.90 vf2
(0.90)(12.0 m/s)2
h = 
= 
g
9.81 m/s2
h = 13 m
II Ch. 8–10 Holt Physics Solution Manual
2
v
1
1
1
1
mgh = 2mvf 2 + 2Iw f 2 = 2mvf 2 + 2(0.80 mr2) f
r
0.
80
1
mgh = mvf 2 2 +  = 0.90 mvf 2
2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. h = 1.2 m
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Givens
Solutions
MEi = MEf
5. vi = 5.4 m/s
2
g = 9.81 m/s
q = 30.0°
1
mv 2
i
2
1
+ 2Iw i2 = mgh = mgd(sin q)
1
mgd(sin q) = 2mvi2 +
1
mvi2 2
mgd(sin q) =
1 2
  
2 5
=
1
+ 5
v2
mr 2 i2
r
7

10
mvi2
(7)(5.4 m/s)2
7 vi 2
d = 
= 
= 4.2 m
10 g (sin q) (10)(9.81 m/s2)(sin 30.0°)
MEi = MEf
6. r = 2.0 m
wf = 5.0 rad/s
g = 9.81 m/s2
m = 4.8 × 103 kg
1
1
mgh = 2mvf2 + 2Iwf 2
mgh = 2mr 2w f 2 + 25 mr 2 w f 2
1
1 2
mgh = mr 2w f 2 2 + 5 = 10 mr 2 w f 2
1
7
1
II
7
 r 2w 2
f
10
(7)(2.0 m)2(5.0 rad/s)2
h =  = 
(10)(9.81 m/s2)
g
h = 7.1 m
1
1
KE trans = 2mvf 2 = 2mr 2wf 2
1
KE trans = 2(4.8 × 103 kg)(2.0 m)2(5.0 rad/s)2
KE trans = 2.4 × 105 J
MEi = MEf
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. m = 5.55 kg
h = 1.40 m
2
g = 9.81 m/s
1
1
1
1 2
  
2 5
mgh = 2mvf 2 + 2Iwf 2
mgh = 2mvf 2 +
vf 2
mr 2 
r2
mgh = mvf 2 2 + 5 = 10 mvf 2
1
vf =
7
1
 = 4.43 m/s
107gh = 7
(10)(9.81 m/s2)(1.40 m)
1
1
KErot = 2Iw f 2 = 5mvf 2
(5.55 kg)(4.43 m/s)2
KErot =  = 21.8 J
5
Section Two—Problem Workbook Solutions
II Ch. 8–11
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Fluid Mechanics
Chapter
9
Additional Practice 9A
Givens
Solutions
FB = Fg
1. mp = 1158 kg
ρVg = (mp + mr)g
3
V = 3.40 m
ρ = 1.00 × 10 kg/m
mr = ρV − mp = (1.00 × 103 kg/m3)(3.40 m3) − 1158 kg = 3.40 × 103 kg − 1158 kg
g = 9.81 m/s2
mr = 2.24 × 103 kg
3
3
2. V = 4.14 × 10−2 m3
FB = Fg − apparent weight
ρswVg = mg − apparent weight
apparent weight =
3.115 × 103 N
ρsw = 1.025 × 10 kg/m
apparent weight
3.115 × 103 N
m = ρswV +  = (1.025 × 103 kg/m3)(4.14 × 10−2 m3) + 
g
9.81 m/s 2
g = 9.81 m/s2
m = 42.4 kg + 318 kg = 3.60 × 102 kg
3
3.
l
3
II
Fnet,1 = Fnet,2 = 0
= 3.00 m
FB,1 − Fg,1 = FB,2 − Fg,2
2
A = 0.500 m
rfw = 1.000 × 10 kg/m
rfwVg − mg = rsw Vg − (m + mballast)g
rsw = 1.025 × 103 kg/m3
mballast g = (rsw − rfw)Vg
3
3
mballast = (rsw − rfw)A l
m0 = (1.025 × 103 kg/m3 − 1.000 × 103 kg/m3)(0.500 m2)(3.00 m)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m0 = (25 kg/m3)(0.500 m2)(3.00 m) = 38 kg
4. A = 3.10 × 104 km2
h = 0.84 km
FB = rVg = rAhg
FB = (1.025 × 103 kg/m3)(3.10 × 1010 m2)(840 m)(9.81 m/s2) = 2.6 × 1017 N
r = 1.025 × 103 kg/m3
g = 9.81 m/s2
5. m = 4.80 × 102 kg
2
g = 9.81 m/s
FB = Fg − apparent weight = mg − apparent weight
FB = (4.80 × 102 kg)(9.81 m/s2) − 4.07 × 103 N = 4.71 × 103 N − 4.07 × 103 N
apparent weight = 4.07 × 103 N FB = 640 N
Section Two—Problem Workbook Solutions
II Ch. 9–1
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Givens
Solutions
Fg,i = FB
6. h = 167 m
riVi g = rswVswg
H = 1.50 km
3
3
rsw = 1.025 × 10 kg/m
ri(h + H)Ag = rswHAg
rswH
ri = 
h+H
3
3
3
(1.025 × 103 kg/m3)(1.50 × 103 m) (1.025 × 10 kg/m )(1.50 × 10 m)
ri = 
=  = 921 kg/m3
3
1670 m
167 m + 1.50 × 10 m
7.
l
Fnet = Fg − FB
= 1.70 × 102 m
13.9 m
r =  = 6.95 m
2
msuba = msubg − mswg
msw = 2.65 × 107 kg
ρsub(g − a)V = mswg
m g
m g
 = sw
ρsub = sw
(g − a)V (g − a)(πr2 l )
(2.65 × 107 kg)(9.81 m/s2)
ρsub = 
(9.81 m/s2 − 2.00 m/s2)(π)(6.95 m)2(1.70 × 102 m)
ρsubVa = ρsubVg − mswg
a = 2.00 m/s2
g = 9.81 m/s2
II
(2.65 × 107 kg)(9.81 m/s2)
ρsub = 
(9.81 m/s2)(π)(6.95 m)2(1.70 × 102 m)
ρsub = 1.29 × 103 kg/m3
Fg,1 = Fg,2
∆ apparent weight = 800 N
FB,1 + apparent weight in water = FB,2 + apparent weight in PEG solution
ρwater = 1.00 × 10 kg/m
ρwaterVg + apparent weight in water − apparent weight in PEG solution = ρsolnVg
3
2
g = 9.81 m/s
3
ρwaterVg + ∆ apparent weight
ρsoln = 
Vg
(1.00 × 103 kg/m3)(6.00 m3)(9.81 m/s2) + 800 N
ρsoln = 
(6.00 m3)(9.81 m/s2)
5.89 × 104 N + 800 N
5.97 × 104 N
ρsoln = 
3
2 = 
(6.00 m )(9.81 m/s ) (6.00 m3)(9.81 m/s2)
ρsoln = 1.01 × 103 kg/m3
Additional Practice 9B
1. P = 1.01 × 105 Pa
F = PA (1.01 × 105 Pa)(3.3 m2) = 3.3 × 105 N
A = 3.3 m2
II Ch. 9–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. V = 6.00 m3
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Givens
Solutions
2. P = 4.0 × 1011 Pa
F = PA = P(πr2)
F = (4.0 × 1011 Pa)(π)(50.0 m)2
r = 50.0 m
F = 3.1 × 1015 N
P 1 = P2
3. m 1 = 181 kg
m2 = 16.0 kg
F
F
1 = 2
A1 A2
2
A1 = 1.8 m
g = 9.81 m/s2
3m2gA1 3m2A1
F A
A2 = 2 1 =  = 
m1g
m1
F1
(3)(16.0 kg)(1.8 m2)
A2 =  = 0.48 m2
181 kg
4. m = 4.0 × 107 kg
4
F2 = 1.2 × 10 N
A2 = 5.0 m2
g = 9.81 m/s2
P1 = P2
F
F
1 = 2
A1 A2
II
A F A mg
A1 = 21 = 2
F2
F2
(5.0 m2)(4.0 × 107 kg)(9.81 m/s2)
A1 = 
= 1.6 × 105 m2
1.2 × 104 N
5. P = 2.0 × 1016 Pa
F = 1.02 × 1031 N
F 1.02 × 1031 N
= 5.1 × 1014 m2
A =  = 
P 2.0 × 1016 Pa
A = 4πr2
r=
5.1 × 1014 m2

4π
A
=
4π
Copyright © by Holt, Rinehart and Winston. All rights reserved.
r = 6.4 × 106 m
6. F = 4.6 × 106 N
38 cm
r =  = 19 cm
2
F
P = 
A
Assuming the squid’s eye is a sphere, its total surface area is 4πr 2. The outer half of
the eye has an area of
A = 2πr 2
F
4.6 × 106 N
P = 2 = 2 = 2.0 × 107 Pa
2πr
(2π)(0.19 m)
7. A = 26.3 m2
F = 1.58 × 107 N
Po = 1.01 × 105 Pa
F 1.58 × 107 N
P =  = 
= 6.01 × 105 Pa
26.3 m2
A
Pgauge = P − Po = 6.01 × 105 Pa − 1.01 × 105 Pa = 5.00 × 105 Pa
Section Two—Problem Workbook Solutions
II Ch. 9–3
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Additional Practice 9C
Givens
Solutions
1. h = (0.800)(16.8 m)
P = 2.22 × 105 Pa
Po = 1.01 × 105 Pa
g = 9.81 m/s2
2.22 × 105 Pa − 1.01 × 105 Pa
1.21 × 105 Pa
P−P
ρ = o = 
= 
2
(9.81 m/s )(0.800)(16.8 m)
(9.81 m/s2)(0.800)(16.8 m)
gh
ρ = 918 kg/m3
P = Po + ρgh
2. h = −950 m
4
P = 8.88 × 10 Pa
Po = 1.01 × 105 Pa
g = 9.81 m/s2
−1.2 × 104 Pa
8.88 × 104 Pa − 1.01 × 105 Pa
P − Po


ρ= 
=
=
2
gh
(9.81 m/s2)(−950 m)
(9.81 m/s )(−950 m)

1
ρ = 1.3 kg/m3
3. P = 13.6P0
P = P0 + rgh
5
P0 = 1.01 × 10 Pa
r = 1.025 × 103 kg/m3
g = 9.81 m/s2
13.6P − P
12.6P
h = 00 = 0
rg
rg
(12.6)(1.01 × 105 Pa)
h = 
= 127 m
(1.025 × 103 kg/m3)(9.81 m/s2)
4. P = 4.90 × 106 Pa
P0 = 1.01 × 105 Pa
r = 1.025 × 103 kg/m3
P = P0 + rgh
4.90 × 106 Pa − 1.01 × 105 Pa
P −P
h = 0 = 
(1.025 × 103 kg/m3)(9.81 m/s2)
rg
4.80 × 106 Pa
h = 
= 477 m
(1.025 × 103 kg/m3)(9.81 m/s2)
g = 9.81 m/s2
P = Po + ρgh
5. h = 245 m
Po = 1.01 × 10 Pa
P = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(245 m)
ρ = 1.025 × 103 kg/m3
P = 1.01 × 105 Pa + 2.46 × 106 Pa
g = 9.81 m/s2
P = 2.56 × 106 Pa
5
6. h = 10 916 m
P = P0 + rgh = 1.01 × 105 Pa + (1.025 × 103 kg/m3)(9.81 m/s2)(10 916 m)
5
P0 = 1.01 × 10 Pa
r = 1.025 × 103 kg/m3
P = 1.01 × 105 Pa + 1.10 × 108 Pa = 1.10 × 108 Pa
g = 9.81 m/s2
II Ch. 9–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
II
P = Po + ρgh
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Additional Practice 9D
Givens
Solutions
1. P1 = (1 + 0.12)P2 = 1.12 P2
v1 = 0.60 m/s
v2 = 4.80 m/s
ρ = 1.00 × 103 kg/m3
h1 = h2
1
1
P1 + 2ρv12 + ρgh1 = P2 + 2ρv22 + ρgh2
P1
h1 = h2, and P2 = ,
so the equation simplifies to
1.12
P1 1 2
1
P1 + 2ρv12 = 
+  ρv2
1.12 2
1
1
P1 1 −  = 2ρ(v22 − v12)
1.12
ρ(v2 − v12)
2
P1 =
1
(2) 1 − 
1.12
=
(1.00 × 103 kg/m3)[(4.80 m/s)2 − (0.60 m/s)2]
1
(2) 1 − 
1.12
(1.00 × 103 kg/m3)(23.0 m2/s2 − 0.36 m2/s2)
P1 = 
(2)(1 − 0.893)
(1.00 × 103 kg/m3)(22.6 m2/s2)
P1 =  = 1.06 × 105 Pa
(2)(0.107)
II
4.10 m
2. r1 =  = 2.05 m
2
v1 = 3.0 m/s
2.70 m
r2 =  = 1.35 m
2
P2 = 82 kPa
ρ = 1.00 × 103 kg/m3
h1 = h2
A1v1 = A2v2
Av
πr 2v1 r12v1
v2 = 11 = 1
= 
A2
πr22
r22
(2.05 m)2(3.0 m/s)
v2 = 
= 6.9 m/s
(1.35 m)2
1
1
P1 + 2 rv12 + ρgh1 = P2 + 2 rv22 + ρgh2
h1 = h2, so the equation simplifies to
1
1
P1 + 2rv12 = P2 + 2rv22
1
P1 = P2 + 2r(v22 − v12)
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P1 = 82 × 103 Pa + 2 (1.00 × 103 kg/m3)[(6.9 m/s)2 − (3.0 m/s)2]
1
P1 = 82 × 103 Pa + 2 (1.00 × 103 kg/m3)(48 m2/s2 − 9.0 m2s2)
1
P1 = 82 × 103 Pa + 2 (1.00 × 103 kg/m3)(39 m2/s2)
P1 = 82 × 103 Pa + 2.0 × 104 Pa = 10.2 × 104 Pa = 102 kPa
1
3. h2 − h1 = 2 h
∆x = 19.7 m
To find the horizontal speed of the cider, recall that for a projectile with no initial
vertical speed,
∆x = v∆t
1
1
∆y = − 2g ∆t2 = − 2h
∆t =
hg
∆x ∆x
v =  =  =
∆t
h

g
1

2
g ∆x2

h
1
P1 + ρv12 + ρgh1 = P2 + 2 ρv22 + ρgh2
Section Two—Problem Workbook Solutions
II Ch. 9–5
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Givens
Solutions
Assuming the vat is open to the atmosphere, P1 = P2.
Also assume v2 ≈ 0. Therefore, the equation simplifies to
1

2
ρv12 + ρgh1 = ρgh2
1 2
 v
2 1
1
= 2
2 2
g ∆x
h = g(h − h ) = g h
2
1
1

2
g ∆x2
 = gh
h
h2 = ∆x2
h = ∆x = 19.7 m
4. v1 = 59 m/s
1
2
g = 9.81 m/s
1
P1 + 2rv12 + rgh1 = P2 + 2rv22 + rgh2
Assume v2 ≈ 0 and P1 = P2 = P0 .
1
rv 2
1
2
+ rgh1 = rgh2
(59 m/s)2
v2
h2 − h1 = 1 = 
= 1.8 × 102 m
2g (2)(9.81 m/s2)
II
5. h2 − h1 = 66.0 m
2
g = 9.81 m/s
1
1
P1 + 2rv12 + rgh1 = P2 + 2rv22 + rgh2
Assume v2 ≈ 0 and P1 = P2 = P0 .
1
rv 2
1
2
+ rgh = rgh
1
2
v1 = 2g
)(9.
m/s
(h
2 −h
81
2)(6
6.
0m
) = 36.0 m/s
1) = (2
2
g = 9.81 m/s
1
1
P1 + 2rv12 + rgh1 = P2 + 2rv22 + rgh2
Assume v2 ≈ 0 and P1 = P2 = P0 .
1
rv 2
1
2
+ rgh1 = rgh2
v1 = 2g
)(9.
m/s
02
m) = 76.7 m/s
(h
2−h
81
2)(3
.0
0×1
1) = (2
7. h2 − h1 = 6.0 m
2
g = 9.81 m/s
1
1
P1 + 2rv12 + rgh1 = P2 + 2rv22 + rgh2
Assume v2 ≈ 0 and P1 = P2 = P0.
1
rv 2
1
2
+ rgh1 = rgh2
v1 = 2g
)(9.
m/s
m) = 11 m/s
(h
h
81
2)(6
.0
2 −
1) = (2
II Ch. 9–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. h2 − h1 = 3.00 × 102 m
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Additional Practice 9E
Givens
Solutions
1. V = 3.4 × 105 m3
T = 280 K
N = 1.4 × 1030 atoms
kB = 1.38 × 10−23 J/K
2. V = 1.0 × 10−3 m3
N = 1.2 × 1013 molecules
T = 300.0 K
kB = 1.38 × 10−23 J/K
3. V = 3.3 × 106 m3
N = 1.5 × 1032 molecules
T = 360 K
kB = 1.38 × 10−23 J/K
4. N = 1.00 × 1027 molecules
T = 2.70 × 102 K
P = 36.2 Pa
Copyright © by Holt, Rinehart and Winston. All rights reserved.
kB = 1.38 × 10−23 J/K
5. V1 = 3.4 × 105 m2
T1 = 280 K
P1 = 1.6 × 104 Pa
T2 = 240 K
4
P2 = 1.7 × 10 Pa
6. A = 2.50 × 102 m2
T = 3.00 × 102 K
P = 101 kPa
N = 4.34 × 1031 molecules
kB = 1.38 × 10−23 J/K
PV = NkBT
30
−23
NkBT (1.4 × 10 atoms)(1.38 × 10 J/K)(280 K)
P = 
= 
5
3
3.4 × 10 m
V
P = 1.6 × 104 Pa
PV = NkBT
13
−23
NkBT (1.2 × 10 molecules)(1.38 × 10 J/K)(300.0 K)
P = 
= 
−3
3
1.0 × 10 m
V
P = 5.0 × 10−5 Pa
P V = NkB T
32
−23
Nk T (1.5 × 10 molecules)(1.38 × 10 J/K)(360 K)
P = B = 
6 3
3.3 × 10 m
V
II
P = 2.3 × 105 Pa
P V = NkB T
27
−23
2
Nk T (1.00 × 10 molecules)(1.38 × 10 J/K)(2.70 × 10 K)
V = B = 
36.2 Pa
P
V = 1.03 × 105 m3
P1 V1 P2 V2
 = 
T1
T2
4
5 3
P1 V1 T2 (1.6 × 10 Pa)(3.4 × 10 m )(240 K)
V2 =  = 
4
(1.7 × 10 Pa)(280 K)
P2 T1
V2 = 2.7 × 105 m3
P V = NkB T
31
−23
2
Nk T (4.34 × 10 molecules)(1.38 × 10 J/K)(3.00 × 10 K)
V = B = 
3
101 × 10 Pa
P
V = 1.78 × 106 m3
V = lA
V
A
l =  =
1.78 × 106 m3

= 7.12 × 103 m
2.50 × 102 m2
Section Two—Problem Workbook Solutions
II Ch. 9–7
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Givens
Solutions
7. V = 7.36 × 104 m3
P = 1.00 × 105 Pa
N = 1.63 × 1030 particles
kB = 1.38 × 10−23 J/K
(1.00 × 105 Pa)(7.36 × 104 m3)
PV
T =  = 
30
−23
NkB (1.63 × 10 particles)(1.38 × 10 J/K)
T = 327 K
P V = NkBT
8. l = 3053 m
A = 0.040 m2
27
N = 3.6 × 10 molecules
P = 105 kPa
kB = 1.38 × 10−23 J/K
P Al
V
T = P  = 
NkB
NkB
(105 × 103 Pa)(0.040 m2)(3053 m)
T = 
= 260 K
(3.6 × 1027 molecules)(1.38 × 10−23 J/K)
V1 = 3.00 m3
P1 V1 P2 V2
 = 
T1
T2
5
3
P2 V2 T1 (1.01 × 10 Pa)(57.0 m )(495 K)
T2 =  = 
6
(2.50 × 10 Pa)(3.00 m3)
P1 V1
V2 = 57.0 m3
T2 = 3.80 × 102 K
9. P1 = 2.50 × 106 Pa
T1 = 495 K
II
PV = NkBT
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P2 = 1.01 × 105 Pa
II Ch. 9–8
Holt Physics Solution Manual
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Heat
Chapter 10
Additional Practice 10A
Givens
1. TC = 14°C
Solutions
T = TC + 273.15
T = (14 + 273.15) K
T = 287 K
9
TF = 5TC + 32.0
TF = 5(14) + 32.0°F = (25 + 32.0)°F
9
TF = 57°F
2. TF = (4.00 × 102)°F
II
5
TC = 9(TF − 32.0)
5
5
TC = 9[(4.00 × 102) − 32.0]°C = 9(368)°C
TC = 204°C
T = TC + 273.15
T = (204 + 273.15) K
T = 477 K
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. TC,1 = 117°C
TC,2 = −163°C
∆TC = TC,1 − TC,2 = 117°C − (−163°C)
∆TC = (2.80 × 102)°C
9
∆TF = 5∆TC
9
∆TF = 5(2.80 × 102)°F
∆TF = 504°F
4. TF = 860.0°F
5
TC = 9(TF − 32.0)
5
5
TC = 9(860.0 − 32.0)°C = 9(828.0)°C
TC = 460.0°C
Section Two—Problem Workbook Solutions
II Ch. 10–1
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Givens
Solutions
5. ∆TF = 49.0°F
TC,2 = 7.00°C
9
TF = 5TC + 32.0
5(7.00) + 32.0°F = (12.6 + 32.0)°F
9
TF,2 =
TF,2 = 44.6°F
TF,1 = TF,2 − ∆TF = 44.6°F − 49.0°F
TF,1 = −4.4°F
5
TC,1 = 9(TF,1 − 32.0)
5
5
TC,1 = 9(−4.4 − 32.0)°C = 9(−36.4)°C
TC,1 = −20.2°C
6. ∆TC = 56°C
TC,2 = −49°C
TC,1 = TC,2 + ∆TC
TC,1 = −49°C + 56°C
TC,1 = 7°C
II
T1 = TC,1 + 273.15
T1 = (7 + 273.15) K
T1 = 2.80 × 102 K
T2 = TC,2 + 273.15
T2 = (−49 + 273.15) K
T2 = 2.24 × 102 K
5
T = TC + 273.15 = 9(TF − 32.0) + 273.15
T = 9(116 − 32.0) + 273.15 K =
5
9(84) + 273.15 K = (47 + 273.15) K
5
T = 3.20 × 102 K
Additional Practice 10B
1. mH = 3.05 × 105 kg
vi = 120.0 km/h
vf = 90.0 km/h
∆T = 10.0°C
∆U
k =  =
mw ∆T
4186 J

(1.00 kg)(1.00°C)
∆PE + ∆KE + ∆U = 0
∆PE = 0
mH 2
∆KE =
−
= (vf − vi 2)
2
mH 2
∆U = −∆KE = (vi − vf 2)
2
1
m v 2
2 H f
mw ∆T

mw = 
∆U
1
m v 2
2 H i
∆T = k ∆T = 2k∆T (v
∆U
∆U
1
3.05 × 105 kg
mw = 
(2)(4186 J)(10.0°C)

(1.00 kg)(1.00°C)
mH
i
2
− vf 2)
2
mw = (3.64 kg • s2/m2)(1110 m2/s2 − 625 m2/s2)
mw = (3.64 kg • s2/m2)(480 m2/s2)
mw = 1.7 × 103 kg
II Ch. 10–2
Holt Physics Solution Manual
2
120.0 km
90.0 km
 − 
h
h
1h

3600 s
2
103 m

1 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. TF = 116 °F
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Givens
Solutions
∆PE = −mgh
2. h = 228 m
Ti = 0.0°C
When the ice lands, its kinetic energy is tranferred to the internal energy of the
ground and the ice. Therefore, ∆KE = 0 J.
2
g = 9.81 m/s
fraction of MEf converted
to U = 0.500
k = (∆U/m) = energy
needed to melt ice =
3.33 × 105 J/1.00 kg
∆U = (0.500)(MEf ) = −(0.500)(∆PE) = (0.500)(mgh)
∆U
 = (0.500)(gh)
m
∆U (0.500)(gh)
f =  =  
km
k
(0.500)(9.81 m/s2)(228 m)
f = 
(3.33 × 105 J/1.00 kg)
f = 3.36 × 10−3
3. vi = 2.333 × 103 km/h
3
h = 4.000 × 10 m
2
(0.0100)(∆PE + ∆KE) + ∆U = 0
∆PE = PEf − PEi = 0 − mgh = −mgh
g = 9.81 m/s
∆KE = KEf − KEi = 0 − 2mvi 2 = − 2mvi 2
fraction of ME converted
to U = 0.0100
∆U = −(0.0100)(∆PE + ∆KE) = −(0.0100)(m)−gh − 2 v 2 = (0.0100)(m)gh + 2 v 2
∆U
k =  =
m ∆T
(355 J)

(1.00 kg)(1.00°C)
1
1
1
m ∆U
1
II
(0.0100)gh + 2v 2
∆T =  = 
355 J
k

(1.00 kg)(1.00°C)
1
6
2
1 2.333 × 10 m/h
(0.0100) (9.81 m/s2)(4.000 × 103 m) + 2 
3600 s/h
∆T = 
2
2
355 m /s • °C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(0.0100)[(3.92 × 104 m2/s2) + (2.100 × 105 m2/s2)]
∆T = 
355 m2/s2 • °C
∆T = 7.02°C
4. h = 8848 m
∆PE = −mgh
2
g = 9.81 m/s
ME f = −∆PE
fraction of MEf converted
to U = 0.200
When the hook lands, its kinetic energy is transferred to the internal energy of the
hook and the ground. Therefore, ∆KE = 0 J.
Ti = −18.0°C
∆U = (0.200)(ME f ) = (−0.200)(∆PE) = (0.200)(mgh)
∆T
1.00°C
 = 
∆U/m 448 J/kg
∆U/m = (0.200)(gh)
1.00°C
∆T =  (0.200)(9.81 m/s )(8848 m) = 38.7°C
448 J/kg
∆T ∆U
1.00°C
∆T =   =  [(0.200)(gh)]
∆U/m m
448 J/kg
2
Tf = Ti + ∆T = −18.0°C + 38.7°C
Tf = 20.7°C
Section Two—Problem Workbook Solutions
II Ch. 10–3
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Givens
Solutions
5. hi = 629 m
Because hf = 0 m, PEf = 0 J. Further, vi = 0 m/s, so KE i = 0 J.
2
g = 9.81 m/s
∆U = −(0.050)(∆ME) = −(0.050)(∆PE + ∆KE)
vf = 42 m/s
∆PE = PEf − PEi = 0 J − PEi = −mgh
fraction of ME converted
to U = 0.050
∆KE = KE f − KE i = KE f − 0 J = 2 mvf 2
m = 3.00 g
∆U = (0.050)(3.00 × 10−3 kg)[(9.81 m/s2)(629 m) − (0.5)(42 m/s)2]
1
1
1
∆U = (0.050)(−∆PE − ∆KE) = (0.050)(mgh − 2mvf 2) = (0.050)(m)(gh − 2vf 2)
∆U = (0.050)(3.00 × 10−3 kg)(6170 m2/s2 − 880 m2/s2)
∆U = (0.050)(3.00 × 10−3 kg)(5290 m2/s2)
∆U = 0.79 J
6. h = 2.49 m
(0.500)(∆PE + ∆KE) + ∆U = 0
2
g = 9.81 m/s
∆PE = PEf − PEi = 0 − mgh = −mgh
m = 312 kg
−mv 2
1
∆KE = KEf − KEi = 0 − 2mv 2 = 
2
v = 0.50 m/s
II
fraction of ME converted
to U = 0.500
mv 2
mv 2
∆U = −(0.500)(∆PE + ∆KE) = −(0.500) −mgh −  = (0.500) mgh + 
2
2
∆U = (0.500)[(312 kg)(9.81 m/s2)(2.49 m) + (0.5)(312 kg)(0.50 m/s)2]
∆U = (0.500)[(7.62 × 103 J) + 39 J]
∆U = 3.83 × 103 J
g = 9.81 m/s
(0.100)(MEf ) (0.100)(∆PE)
∆U
∆U/m
 = ∆T  =  =  = (0.100)(gh)
m
m
∆T
m
fraction of MEf converted
to U = 0.100
Because the kinetic energy at the bottom of the falls is converted to the internal energy of the water and the ground, ∆KE = 0 J.
∆U/m 4186 J/kg
 =  
∆T
1.00°C
∆U/m
∆T 
∆T
h = 
(0.100)(g)
2
0.230°C
4186 J/kg

h = 
(0.100)(9.81 m/s2) 1.00°C
h = 981 m
II Ch. 10–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. ∆T = 0.230°C
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Additional Practice 10C
Givens
Solutions
1. Q = (2.8 × 109 W)(1.2 s)
cp,c = 387 J/kg • °C
Tc = 26°C
Q = mc cp,c (Tc − Tf )
Q
mc = 
cp,c(Tc − Tf )
(2.8 × 109 W)(1.2 s)
mc = 
(387 J/kg• °C)(26°C − 21°C)
Tf = 21°C
(2.8 × 109 W)(1.2 s)
mc =  = 1.7 × 106 kg
(387 J/kg• °C)(5°C)
2. mw = 14.3 × 103 kg
Tw = 20.0°C
Tx = temperature of
burning wood = 280.0°C
Tf = 100.0°C
cp,x mx (Tx − Tf ) = cp,w mw (Tf − Tw)
cp,w mw (Tf − Tw)
mx = 
cp,x(Tm − Tf )
 (143 × 10 kg)(100.0 − 20.0) °C
kg °C
m = 
4186 J
cp,x = specific heat
capacity of wood =
1.700 × 103 J/kg • °C
cp,w = 4186 J/kg • °C
3. ∆U = Q = (0.0100)
(1.450 GW)(1.00 year)
cp,x = specific heat capacity
of iron = 448 J/kg • °C
mx = mass of steel =
25.1 × 109 kg
3
•
x
(1.700 × 103 J/kg • °C)(280.0 − 100.0) °C
II
mx = 1.56 × 105 kg
Q = cp,x mx ∆T
Q
∆T = 
cp,x mx
(0.0100)(1.450 × 109 W)(1.00 year) 365.25 days 24 h 3600 s
  
∆T = 
1 year
1 day 1 h
(448 J/kg • °C)(25.1 × 109 kg)
∆T = 40.7°C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. ml = 2.25 × 103 kg
Tl,i = 28.0°C
cp,l = cp,w = 4186 J/kg • °C
mlcp,l (Tl,i − Tl,f ) = micp,i (Ti,f − Ti,i)
2
mi = 9.00 × 10 kg
Ti,i = −18.0°C
cp,i = 2090 J/kg • °C
Ti,f = 0.0°C
5. mw = 1.33 × 1019 kg
Tw = 4.000°C
cp,w = 4186 J/kg • °C
P = 1.33 × 1010 W
3
∆t = 1.000 × 10 years
9.00 × 10 kg 2090 J/kg °C
= (28.0°C) −   [(0.0°C) − (−18.0°C)]
2.25 × 10 kg 4186 J/kg °C
m cp,i
Tl,f = Tl,i − i  (Ti,f − Ti,i)
ml cp,l
Tl,f
2
•
3
•
Tl,f = 28.0°C − 3.59°C = 24.4°C
P∆t = Q = mw cp,w (Tf − Tw)
P∆t
Tf =  + Tw
mw cp,w
365.25 day 24 h 3600 s
(1.33 × 1010 W)(1.000 × 103 years)   
1 day 1 h
1 year
Tf = 
+ 4.000°C
19
(1.33 × 10 kg)(4186 J/kg • °C)
Tf = (7.54 × 10−3)°C + 4.000°C
Tf = 4.008°C
Section Two—Problem Workbook Solutions
II Ch. 10–5
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Givens
Solutions
Q = mx cp,x ∆T
6. Ti = 18.0°C
Tf = 32.0°C
Q
Q
cp,x =  = 
mx ∆T mx (Tf − Ti )
Q = 20.8 kJ
20.8 × 103 J
cp,x = 
(0.355 kg)(32.0°C − 18.0°C)
mx = 0.355 kg
cp,x = 4190 J/kg • °C
mmcp,m(Tf − Tm) = mw cp,w (Tw − Tf )
7. Tm = −62.0°C
mm = 180 g
Tw − Tf
mw
cp,m = 
(cp,w) 
mm
Tf − Tm
mw = 0.500 kg
cp,m
Tw = 38.0°C
Tf = 36.9°C
cp,w = 4186 J/kg • °C
0.500 kg
(38.0°C) − (36.9°C)
=  (4186 J/kg °C) 
0.18 kg (36.9°C) − (−62.0°C)
•
cp,m = 1.3 × 102 J/kg • °C
The metal could be gold (cp = 129 J/kg • °C) or lead (cp = 128 J/kg • °C).
II
Additional Practice 10D
1. mw,S = 1.20 × 1016 kg
QS = energy transferred by heat from Lake Superior = cp,w mw,S (TS − Tf )
mw,E = 4.8 × 10 kg
QE = energy transferred by heat to Lake Erie = mw,E Lf + mw,E cp,w (Tf − TE)
TE = 0.0°C
From the conservation of energy,
TS = 100.0°C
QS = QE
14
cp,w mw,S (TS − Tf ) = mw,E Lf + mw,E cp,w (Tf − TE)
cp,w = 4186 J/kg • °C
5
Lf of ice = 3.33 × 10 J/kg
(mw,E cp,w + cp,w mw,S)Tf = cp,w mw,STS + mw,E cp,wTE − mw,E Lf
cp,w mw,STS − mw,ELf
Tf = 
cp,w (mw,E + mw,S)
(4186 J/kg • °C)(1.20 × 1016 kg)(100.0°C) − (4.8 × 1014 kg)(3.33 × 105 J/kg)
Tf = 
(4186 J/kg•°C)[(4.8 × 1014 kg) + (1.20 × 1016 kg)]
( 5.02 × 1021 J) − (1.6 × 1020 J )
Tf = 
(4186 J/kg • °C)(1.25 × 1016 kg)
Tf = 92.9°C
2. Tf = −235°C
Tfreezing = 0.0°
mw = 0.500 kg
cp,w = 4186 J/kg • °C
cp,ice = cp,i = 2090 J/kg • °C
Lf of ice = 3.33 × 105 J/kg
Qtot = 471 kJ
Qtot = cp,w mw (Ti − 0.0°) + Lf mw + mw cp,i (0.0°− Tf ) = cp,w mwTi + Lf mw − cp,i mw Tf
Qtot − Lf mw + cp,i mwTf
Ti = 
cp,wmw
(4.71 × 105 J) − (3.33 × 105 J/kg)(0.500 kg) + (−235°C)(2090 J/kg • °C)(0.500 kg)
Ti = 
(4186 J/kg • °C)(0.500 kg)
(4.71 × 105 J) − (1.66 × 105 J) − (2.46 × 105 J)
Ti = 
2093 J/°C
Ti = 28°C
II Ch. 10–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
cp,w (mw,STS + mw,E TE ) − mw,ELf
Tf =  , where TE = 0.0°C
cp,w (mw,E + mw,S)
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Givens
Solutions
3. Ti = 0.0°C
ms Lv = miLf
6
mi = 4.90 × 10 kg
5
Lf of ice = 3.33 × 10 J/kg
Ts = 100.0°C
Lv of steam = 2.26 × 106 J/kg
miLf
ms = 
Lv
(4.90 × 106 kg)(3.33 × 105 J/kg)
ms = 
2.26 × 106 J/kg
ms = 7.22 × 105 kg
ms Lf,s = mi Lf,i
4. ms = 1.804 × 106 kg
Lf of silver = Lf,s =
8.82 × 104 J/kg
ms L f,s
mi = 
L f,i
Lf of ice = Lf,i =
3.33 × 105 J/kg
(1.804 × 106 kg)(8.82 × 104 J/kg)
mi = 
3.33 × 105 J/kg
mi = 4.78 × 105 kg
5. mg = 12.4414 kg
Q = mg cp,g (Tg,f − Tg,i) + mg Lf
Tg,i = 5.0°C
II
Q − mg cp,g (Tg, f − Tg,i)
Lf = 
mg
Q = 2.50 MJ
cp,g = 129 J/kg • °C
(2.50 × 106 J) − (12.4414 kg)(129 J/kg • °C)(1063°C − 5.0°C)
Lf = 
12.4414 kg
Tg,f = 1063°C
(2.50 × 106 J) − (1.70 × 106 J)
Lf = 
12.4414 kg
Lf = 6.4 × 104 J/kg
a. mc = rcVc
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. Vp = 7.20 m3
Vc = (0.800)(Vp )
3
mc = (8.92 × 103 kg/m3)(0.800)(7.20 m3)
3
rc = 8.92 × 10 kg/m
Lf = 1.34 × 105 J/kg
mc = 5.14 × 104 kg
Q
15
b. fraction of melted coins =  = 
mcLf 100
15(5.14 × 104 kg)(1.34 × 105 J/kg)
15
Q =  mc Lf = 
100
100
Q = 1.03 × 109 J
Section Two—Problem Workbook Solutions
II Ch. 10–7
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Givens
Solutions
7. mw = 3.5 × 1019 kg
Ti = 10.0°C
Tf = 100.0°C
cp,w = 4186 J/kg•°C
Lv of water = 2.26 × 106 J/kg
P = 4.0 × 1026 J/s
a. Q = mw cp,w (Tf − Ti ) + mw Lv
Q = (3.5 × 1019 kg)(4186 J/kg • °C)(100.0°C − 10.0°C)
+ (3.5 × 1019 kg)(2.26 × 106 J/kg)
Q = 1.3 × 1025 + 7.9 × 1025 J = 9.2 × 1025 J
b. Q = P∆t
9.2 × 1025 J
Q
∆t =  = 
= 0.23 s
P 4.0 × 1026 J/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
II
II Ch. 10–8
Holt Physics Solution Manual
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Thermodynamics
Chapter 11
Additional Practice 11A
Givens
Solutions
1. P = 5.1 kPa
W = P∆V
W
3.6 × 103 J
= 7.1 × 10−1 m3
∆V =  = 
P
5.1 × 103 Pa
3
W = 3.6 × 10 J
Vi = 0.0 m3
V = Vi + ∆V = 0.0 m3 + 7.1 × 10−1 m3 = 7.1 × 10−1 m3
W = mgh = −P∆V
2. m = 207 kg
2
g = 9.81 m/s
mgh
∆V = 
−P
h = 3.65 m
(207 kg)(9.81 m/s2)(3.65 m)
∆V = 
= −4.1 × 10−3 m3
−(1.8 × 106 Pa)
P = 1.8 × 106 Pa
II
W = P∆V
3. rf = 1.22 m
ri = 0.0 m
W
P = 
∆V
W = 642 kJ
4
∆V = 3 p(rf 3 − ri3)
W
642 × 103 J
P =  =  = 8.44 × 104 Pa
4
4
 p (1.22 m)3 − (0.0 m)3
 p(r 3 − r 3)
f
i
3 3
W = P∆V
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. rf = 7.0 × 105 km
4
8
= 7.0 × 10 m
V = 3 pr 3
ri = 0.0 m
4
∆V = Vf − Vi = 3 p(rf 3 − ri3)
W = 3.6 × 1034 J
4
W = (P)( 3 p)(rf 3 − ri3)
W
P = 
4
(3 p)(rf 3 − ri3)
(3.6 × 1034 J)
P = 
= 2.5 × 107 Pa
4
(3 p )[(7.0 × 108 m)3 − (0.0 m)3]
W = P∆V
5. P = 87 kPa
−3
3
∆V = −25.0 × 10 m
W = (87 × 103 Pa)(−25.0 × 10−3 m3) = −2.2 × 103 J
Section Two—Problem Workbook Solutions
II Ch. 11–1
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Givens
Solutions
W = P∆V
6. rf = 29.2 cm
ri = 0.0 m
4
P = 25.0 kPa
m = 160.0 g
∆V = 3 p(rf 3 − ri3)
1
W = 2 mv 2
1
 mv 2
2
= (P) 3 p (rf 3 − ri3)
4
8p P
v=
3m (r
v=
f
3
− ri3)
(8p)(25.0 × 103 Pa)

(29.2 × 10−2 m)3 − (0.0 m)3
(3)(160.0 × 10−3 kg)
v = 181 m/s
Additional Practice 11B
∆U = Uf − Ui = Q − W
1. m = 227 kg
II
W = mgh
h = 8.45 m
g = 9.81 m/s
Uf = Ui + Q − W = Ui + Q − mgh
Ui = 42.0 kJ
Uf = (42.0 × 103 J) + (4.00 × 103 J) − (227 kg)(9.81 m/s2)(8.45 m)
Q = 4.00 kJ
Uf = (42.0 × 103 J) + (4.00 × 103 J) − (18.8 × 103 J)
2
Uf = 27.2 × 103 J = 27.2 kJ
2. m = 4.80 × 102 kg
Q =0J
a. Assume that all work is transformed to the kinetic energy of the cannonball.
1
v = 2.00 × 102 m/s
W = 2mv 2
1
W = 2 (4.80 × 102 kg)(2.00 × 102 m/s)2
Uf = 12.0 MJ
b. ∆U = Q − W = −W
Uf − Ui = −W
Ui = Uf + W
Ui = (12.0 × 106 J) + (9.60 × 106 J)
Ui = 21.6 × 106 J = 21.6 MJ
3. m = 4.00 × 104 kg
a. Q = mcp ∆T
cp = 4186 J/kg • °C
Q = (4.00 × 104 kg)(4186 J/kg • °C)(−20.0°C)
∆T = −20.0°C
Q = −3.35 × 109 J
W = 1.64 × 109 J
b. Q of gas = −Q of jelly = −(−3.35 × 109 J) = 3.35 × 109 J
∆U = Q − W
∆U = (3.35 × 109 J) − (1.64 × 109 J)
∆U = 1.71 × 109 J
II Ch. 11–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
W = 9.60 × 106 J = 9.60 MJ
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Givens
Solutions
4. m = 1.64 × 1015 kg
W = mgh = (1.64 × 1015 kg)(9.81 m/s2)(75.0 m)
h = 75.0 m
W = 1.21 × 1018 J
g = 9.81 m/s2
Q = −mcp,w (Tf − Ti ) − mLv = −m[cp,w (Tf − Ti ) + Lv]
Ti = 6.0°C
Tf − Ti = 100.0°C − 6.0°C = 94.0°C
Tf = 100.0°C
Q = −(1.64 × 1015 kg)[(4186 J/kg • °C)(94.0°C) + (2.26 × 106 J/kg)]
cp,w = 4186 J/kg • °C
Q = −4.35 × 1021 J
Lv of water = 2.26 × 106 J/kg
∆U = Uf − Ui = (−0.900)(Ui ) = Q − W
∆U = (−0.900)(Ui )
Uf = (1 − 0.900)Ui = (0.100)(Ui )
Uf
Uf
−∆U = Ui − Uf =  − Uf = (0.900)  = W − Q
0.100
0.100
0.100
U =  [(1.21 × 10
0.900
0.100
Uf =  (W − Q)
0.900
18
f
J) − (−4.35 × 1021 J)]
Uf = 4.83 × 1020 J
II
(Note: Nearly all of the energy is used to increase the temperature of the water
and to vaporize the water.)
5. m = 5.00 × 103 kg
1
W = 2 mv 2
∆U = Uf − Ui = Q − W
v = 40.0 km/h
Ui = 2.50 × 10 J
Uf = 2Ui
Uf = 2Ui
∆U = 2Ui − Ui = Ui
5
1
Q = ∆U + W = Ui + 2 mv 2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2
40.0 km 1 h
103m
1
Q = (2.50 × 105 J) + 2(5.00 × 103 kg)   
3600 s 1 km
h
Q = (2.50 × 105 J) + (3.09 × 105 J)
Q = 5.59 × 105 J
Q
6. P =  = 5.9 × 109 J/s
∆t
∆t = 1.0 s
9
∆U = 2.6 × 10 J
∆U = Q − W = P∆t − W
W = P∆t − ∆U
W = (5.9 × 109 J/s)(1.0 s) − (2.6 × 109 J)
W = 3.3 × 109 J
Section Two—Problem Workbook Solutions
II Ch. 11–3
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Givens
Solutions
7. h = 1.00 × 102 m
v = 141 km/h
Ui = 40.0 ΜJ
m = 76.0 kg
a. ∆U = Q − W
All energy is transferred by heat, so W = 0 J
1
b. ∆U = Q = 2 mv 2
∆U
mv 2
 × 100 =  (100)
Ui
2Ui
2
2
2
1h
141 km
103 m
(76.0 kg)    (100)
3600 s 1 km
h
∆U
 × 100 = 
Ui
(2)(40.0 × 106 J)
U × 100 = 0.146 percent
∆U
i
Additional Practice 11C
1. eff = 8 percent = 0.080
II
Qh = 2.50 kJ
Qh − Qc
Q
 = 1 − c
eff = 
Qh
Qh
−Qc = Qh (eff − 1)
Qc = Qh (1 − eff )
Qc = (2.50 kJ)(1 − 0.08) = (2.50 kJ)(0.92)
Qc = 2.3 kJ
W = Qh − Qc
W = 2.50 kJ − 2.3 kJ
W = 0.2 kJ
eff = 16 percent = 0.16
Qh = 2.0 × 109 J
W et Pnet∆t
= 
eff = n
Qh
Qh
(eff )(Q h )
∆t = 
Pnet
(0.16)(2.0 × 109 J)
∆t = 
1.5 × 106 W
∆t = 2.1 × 102 s
3. Pnet = 19 kW
eff = 6.0 percent = 0.060
∆t = 1.00 h
Wnet = Pnet ∆t
W et
eff = n
Qh
W et Pnet ∆t
Qh = n
= 
eff
eff
3.6 × 103 s
(19 kW)(1.00 h) 
1h
Qh = 
(0.060)
Qh = 1.1 × 106 kJ = 1.1 × 109 J
II Ch. 11–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. Pnet = 1.5 MW
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Givens
Solutions
4. Pnet = 370 W
∆t = 1.00 min = 60.0 s
eff = 0.19
W et Pnet ∆t
eff = n
= 
Qh
Qh
Pnet ∆t
Qh = 
eff
(370 W)(60.0 s)
Qh = 
0.19
Qh = 1.2 × 105 J = 120 kJ
5. Wnet = 2.6 MJ
MJ
Qh/m = 32.6 
kg
m = 0.80 kg
Wnet
Wnet
= 
eff = 
Qh
Qh
 (m)
m
2.6 MJ
eff = 
MJ
32.6  (0.80 kg)
kg
eff = 0.10 = 10 percent
4
6. m = 3.00 × 10 kg
2
g = 9.81 m/s
II
Wnet = mgh = Qh − Qc
Qh = Wnet + Qc
2
h = 1.60 × 10 m
8
Qc = 3.60 × 10 J
mgh
W et
Wnet
eff = n
= 
= 
Wnet + Qc mgh + Qc
Qh
(3.00 ⫻ 104 kg)(9.81 m/s2)(1.60 ⫻ 102 m)
eff = 
(3.00 ⫻ 104 kg)(9.81 m/s2)(1.60 ⫻ 102 m) + 3.60 ⫻ 108 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
eff = 0.12
Section Two—Problem Workbook Solutions
II Ch. 11–5
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Vibrations and Waves
Chapter 12
Additional Practice 12A
Givens
Solutions
1. m = 0.019 kg
F mg
k =  = 
x
x
2
g = 9.81 m/s
(0.019 kg)(9.81 m/s2)
x = 
83 N/m
k = 83 N/m
x = 2.25 × 10−3 m
2. m = 187 kg
4
k = 1.53 × 10 N/m
(187 kg)(9.81 m/s2)
x = 
= 0.120 m
1.53 × 104 N/m
g = 9.81 m/s2
3. m1 = 389 kg
−3
x 2 = 1.2 × 10
F mg
k =  = 
x
x
m
m2 = 1.5 kg
II
F
F
1 = 2
x1
x2
Fx
m gx
x1 = 12 = 1 2
F2
m2 g
(389 kg)(1.2 × 10−3 m)
x1 =  = 0.31 m
(1.5 kg)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. m = 18.6 kg
x = 3.7 m
2
F mg (18.6 kg)(9.81 m/s )
k =  =  = 
(3.7 m)
x
x
g = 9.81 m/s2
k = 49 N/m
5. h = 533 m
x1 =
1
 h
3
m = 70.0 kg
2
x2 = 3 h
F
mg
3mg
k =  =  = 
x (x2 − x1)
h
3(70.0 kg)(9.81 m/s2)
k =  = 3.87 N/m
(533 m)
g = 9.81 m/s2
6. k = 2.00 × 102 N/m
x = 0.158 m
g = 9.81 m/s2
a. F = kx = (2.00 × 102 N/m)(0.158 m) = 31.6 N
F kx
b. m =  = 
g
g
(2.00 × 102 N/m)(0.158 m)
m = 
(9.81 m/s2)
m = 3.22 kg
Section Two—Problem Workbook Solutions
II Ch. 12–1
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Givens
Solutions
7. h = 1.02 × 104 m
L = 4.20 × 103 m
F = kx = k(h − L)
F = (3.20 × 10−2 N/m)(6.0 × 103 m) = 190 N
k = 3.20 × 10−2 N/m
F = kx = k(h − L) = mg
8. h = 348 m
2
L = 2.00 × 10 m
k = 25.0 N/m
(25.0 N/m)(148 m)
m = 
= 377 kg
(9.81 m/s2)
g = 9.81 m/s2
Additional Practice 12B
1. L = 6.7 m
g = 9.81 m/s2
2. L = 0.150 m
II
g = 9.81 m/s2
3. x = 0.88 m
2
g = 9.81 m/s
T = 2π
Lg = 2π 
(9(.861.7
m/)s) =
m
T = 2p
 = 0.777 s
Lg = 2p (9.81 m/s )
2
T = 2p
5.2 s
(0.150 m)
2

4gx = 2p (9.81 m/s )
4(0.88 m)
2
T = 3.8 s
4. f = 6.4 × 10−2 Hz
2
g = 9.81 m/s
1
T =  = 2p
f
Lg
L = 61 m
5. t = 3.6 × 103 s
N = 48 oscillations
g = 9.81 m/s2
T = 2p
Lg = Nt
2
t
 g
N
(3.6 × 103 s)2(9.81 m/s2)

L=
=
4p 2 (48)2
4p 2
L = 1.4 × 103 m
6. L = 1.00 m
T = 10.5 s
II Ch. 12–2
2
4p 2L 4p (1.00 m)
g = 
= 
= 0.358 m/s2
2
(10.5 s)2
T
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(9.81 m/s2)
g
L = 22 = 
2
4π (6.4 × 10−2 Hz)2
4p f
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Additional Practice 12C
Givens
Solutions
1. f1 = 90.0 Hz
2
k = 2.50 × 10 N/m
T = 2p

k = 
3.00 × 10 f
1
m
−2
1
(2.50 × 102 N/m)
k

m = 
=
2
4p 2(3.00 × 10−2)2f12 4p (3.00 × 10−2)2(90.0 Hz)2
m = 0.869 kg
2. m1 = 3.5 × 106 kg
f = 0.71 Hz
k = 1.0 × 106 N/m
1
T =  = 2p
f
m1 + m2

k
k
m2 = 22 − m1
4p f
(1.0 × 106 N/m)
m2 = 
− 3.5 × 104 kg = 1.5 × 104 kg
4p 2(0.71 Hz)2
3. m = 20.0 kg
42.7
f =  = 0.712 Hz
60 s
4p 2 m
k = 
= 4p 2mf 2
T2
II
k = 4p 2 (20.0 kg)(0.712 Hz)2
k = 4.00 × 102 N/m
4. m = 2.00 × 105 kg
T = 1.6 s
T = 2p
mk
2
5
4p 2m 4p (2.00 × 10 kg)

k = 
=
= 3.1 × 106 N/m
(1.6 s)2
T2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. m = 2662 kg
x = 0.200 m
g = 9.81 m/s2
6. m = 10.2 kg
k = 2.60 × 102 N/m
T = 2p
k = 2p mg
T = 2p
0 m)
= 0.897 s

((90.8.210
m/
s )
T = 2p
 = 1.24 s
mk = 2p (2.60 × 10 N/m)
m
mx
2
(10.2 kg)
2
Additional Practice 12D
1. f = 2.50 × 102 Hz
v = 1530 m/s
v
1530 m/s
= 6.12 m
l =  = 
f
2.50 × 102 Hz
Section Two—Problem Workbook Solutions
II Ch. 12–3
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Givens
Solutions
2. f = 123 Hz
v = 334 m/s
3. l = 2.0 × 10−2 m
v = 334 m/s
4. l = 2.54 m
v = 334 m/s
5. f = 73.4 Hz
l = 4.50 m
6. f = 2.80 × 105 Hz
v
(334 m/s)
l =  =  = 2.72 m
f
(123 Hz)
v
(334 m/s)
f =  = 
= 1.7 × 104 Hz
l
(2.0 × 10−2 m)
v (334 m/s)
f =  =  = 131 Hz
l
(2.54 m)
v = fl = (73.4 Hz)(4.50 m) = 3.30 × 102 m/s
v = fl = (2.80 × 105 Hz)(5.10 × 10−3 m)
l = 5.10 × 10−3 m
v = 1.43 × 103 m/s
∆x = 3.00 × 103 m
∆x
(3.00 × 103 m)
∆t =  = 
v
(1.43 × 103 m/s)
II
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 2.10 s
II Ch. 12–4
Holt Physics Solution Manual
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Sound
Chapter 13
Additional Practice 13A
Givens
Solutions
1. Intensity = 3.0 × 10−3 W/m2
r = 4.0 m
P
Intensity = 2
4p r
P = 4pr 2(Intensity) = 4p (4.0 m)2(3.0 × 10−3 W/m2)
P = 0.60 W
2. r = 8.0 × 103 m
−12
Intensity = 1.0 × 10
2
W/m
P
Intensity = 2
4p r
P = 4pr 2(Intensity)
P = 4p(8.0 × 103 m)2(1.0 × 10−12 W/m2) = 8.0 × 10−4 W
3. Intensity = 1.0 × 10−12 W/m2
−6
P = 2.0 × 10
W
P
Intensity = 2
4p r
r=
r=
4. Intensity = 1.1 × 10−13 W/m2
−4
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P = 3.0 × 10
W
r = 2.5 m
6. Intensity = 2.5 × 10−6 W/m2
r = 2.5 m


4p (I
n
te
n
si
ty)
P
2.0 × 10−6 W

= 4.0 × 102 m
4p (1.0 × 10−12 W/m2)
P
r 2 = 
4p Intensity
r =
5. P = 1.0 × 10−4 W
II
(3.0 × 10−4 W)

= 1.5 × 104 m
4p(1.1 × 10−13 W/m2)
P
 =
4p Intensity
P
Intensity = 2
4p r
(1.0 × 10−4 W)
Intensity = 
= 1.3 × 10−6 W/m2
4p(2.5 m)2
P = 4pr 2(Intensity)
P = 4p(2.5 m)2(2.5 × 10−6 W/m2)
P = 2.0 × 10−4 W
Section Two—Problem Workbook Solutions
II Ch. 13–1
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Additional Practice 13B
Givens
Solutions
1. f15 = 26.7 Hz
15v 15(334 m/s)
L =  = 
4(26.7 Hz)
4 f15
v = 334 m/s
n = 15
L = 46.9 m
2. l = 3.47 m
2
vs = 5.00 × 10 m/s
n =3
va = 334 m/s
3. n = 19
II
l vsn (3.47 m)(5.00 × 102 m/s)(3)
L = 
= 
2(334 m/s)
2va
L = 7.79 m
L = 86 m
nv
19(334 m/s)
f19 =  = 
2L
2(86 m)
v = 334 m/s
f19 = 37 Hz
4. L = 3.50 × 102 m
f 75 = 35.5 Hz
75 v
f75 = 
2L
2(3.50 × 102 m)(35.5 Hz)
2Lf75
v = 
= 
75
75
n = 75
v = 331 m/s
5. L = 4.7 × 10−3 m
4(4.7 × 10−3 m)
4L
n =  = 
= 5
(3.76 × 10−3 m)
ln
Copyright © by Holt, Rinehart and Winston. All rights reserved.
l = 3.76 × 10−3 m
4L
l n = 
n
II Ch. 13–2
Holt Physics Solution Manual
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Light and Reflection
Chapter 14
Additional Practice 14A
Givens
Solutions
1. f = 9.00 × 108 Hz
d = 60.0 m
c = 3.00 × 108 m/s
c = fl
d
d
df
 =  = 
l
c
c

f
8
d (60.0 m)(9.00 × 10 Hz)
 = 
8
(3.00 × 10 m/s)
l
d
 = 1.80 × 102 wavelengths
l
2. f = 5.20 × 1014 Hz
−4
d = 2.00 × 10
m
8
c = 3.00 × 10 m/s
II
c = fl
d
d
df
 =  = 
l
c
c

f
−4
14
d (2.00 × 10 m)(5.20 × 10 Hz)
 = 
(3.00 × 108 m/s)
l
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d
 = 347 wavelengths
l
3. f = 2.40 × 1010 Hz
8
c = 3.00 × 10 m/s
c = fl
c
l = 
f
(3.00 × 108 m/s)
l = 
(2.40 × 1010 Hz)
l = 1.25 × 10−2 m = 1.25 cm
4. l = 1.2 × 10−6 m
8
c = 3.00 × 10 m/s
c = fl
c
f = 
l
(3.00 × 108 m/s)
f = 
(1.2 × 10−6 m)
f = 2.5 × 1014 Hz = 250 TH
Section Two—Problem Workbook Solutions
II Ch. 14–1
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Givens
Solutions
5. l1 = 2.0 × 10−3 m
−3
l 2 = 5.0 × 10
m
c = 3.00 × 108 m/s
c = fl
(3.00 × 108 m/s)
c
f1 =  = 
(2.0 × 10−3 m)
l1
f1 = 1.5 × 1011 Hz = 15 × 1010 Hz
(3.00 × 108 m/s)
c
f2 =  = 
(5.0 × 10−3 m)
l2
f2 = 6.0 × 1010 Hz
6.0 × 1010 Hz < f < 15 × 1010 Hz
6. f = 10.0 Hz
c = fl
8
c = 3.00 × 10 m/s
c
l = 
f
(3.00 × 108 m/s)
l = 
(10.0 Hz)
II
l = 3.00 × 107 m = 3.00 × 104 km
Additional Practice 14B
5
f = 2.50 × 10 m
1 1 1
 =  − 
q
p
f
1
1
1
(4.00 × 10−6) (2.70 × 10−6)
 =  
− 
=  − 
5
5
q (2.50 × 10 m) (3.70 × 10 m)
1m
1m
1.30 × 10−6
q = 
1m
−1
= 7.69 × 105 m = 769 km
q −7.69 × 105 m
M = −  = 
p
3.70 × 105 m
M = −2.08
2. h = 8.00 × 10−5 m
f = 2.50 × 10−2 m
q = −5.9 × 10−1 m
1 1 1
 +  = 
p q f
1 1 1
 =  − 
p f
q
1
1
1
40.0 1.69 41.7
 = 
−  =  −  = 
p (2.50 × 10−2 m) (−5.9 × 10−1 m)
1m 1m 1m
p = 2.40 × 10−2 m
h⬘
q
(−5.9 × 10−1 m)
M =  = −  = − 
= 24.6
h
p
(2.40 × 10−2 m)
II Ch. 14–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. p = 3.70 × 105 m
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Givens
3. h⬘ = −28.0 m
h = 7.00 m
f = 30.0 m
Image is real, so q > 0 and
h⬘ < 0.
Solutions
q h⬘
M = −  = 
p h
ph⬘
q = − 
h
1 1
1
 +  = 
p
q
f
1
1
1
 +  = 
p
f
−ph⬘

h
h
p = f 1 − 
h⬘
7.00 m
p = (30.0 m) 1 + 
28.0 m
1
h
1
 1 −  = 
p
h⬘
f
II
p = 37.5 m
4. h⬘ = 67.4 m
h = 1.69 m
R = 12.0 m
(h⬘ > 0, q < 0)
q h⬘
M = −  = 
p h
ph⬘
q = − 
h
1 1 2
 +  = 
p q R
1
1
2
 +  = 
p
R
−ph⬘

h
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
h
2
 1 −  = 
p
h⬘
R
R
h
p =  1 − 
2
h⬘
(12.0 m)
1.69 m
p =  1 −  = (6.00 m)(0.975)
2
67.4 m
p = 5.85 m
Image is virtual and therefore upright.
Section Two—Problem Workbook Solutions
II Ch. 14–3
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Givens
Solutions
5. h = 32 m
f = 120 m
p = 180 m
1 1 1
 +  = 
p q
f
1 1 1
 =  − 
q f
p
1
1
1
0.0083 0.0056 0.0027
 =  −  =  −  = 
q (120 m) (180 m)
1m
1m
1m
q = 3.7 × 102 m
h⬘
q
M =  = − 
h
p
qh
h⬘ = − 
p
−(370 m)(32 m)
h⬘ = 
(180 m)
h⬘ = −66 m
The image is inverted (h⬘ < 1)
and real (q > 0)
II
6. h = 0.500 m
R = 0.500 m
p = 1.000 m
1 2 1
 =  − 
q R p
1
2
1
4.00 1.000 3.00
 =  −  =  −  = 
q (0.500 m) (1.000 m) 1 m
1m
1m
q = 0.333 m = 333 mm
qh −(0.333 m)(0.500 m)
h⬘ = −  = 
(1.000 m)
p
h⬘ = −0.166 m = −166 mm
The image is real (q > 0).
II Ch. 14–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q h⬘
M = −  = 
p h
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Givens
7. p = 1.00 × 105 m
h = 1.00 m
h⬘ = −4.00 × 10−6 m
(h⬘ < 0 because image is
inverted)
Solutions
q h⬘
M = −  = 
p h
ph⬘
q = − 
h
1 1 2
 +  = 
p q R
2
2p
R =  = 
1
h
h
 − 
1 − 
p ph⬘
h⬘
(2.00 × 105 m)
2(1.00 × 105 m)
R =  = 
(1 + 2.50 × 105)
(1.00 m)
1 + 
(4.00 × 10−6 m)
R = 0.800 m = 80.0 cm
8. h = 10.0 m
p = 18.0 m
h⬘ = −24.0 m
Image is real, so q > 0, and
h⬘ must be negative.
q h⬘
M = −  = 
p h
II
h⬘p
q = − 
h
1 1
2
 +  = 
p q R
2
2p
R=  = 
1
h
h
 − 
1 − 
p ph⬘
h⬘
2(18.0 m)
36.0 m
(36.0 m)
R =  =  = 
(1
+
0
.4
17)
(1.417)
10.0 m
1 + 
24.0 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
R = 25.4 m
Section Two—Problem Workbook Solutions
II Ch. 14–5
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Additional Practice 14C
Givens
Solutions
1. R = −6.40 × 106 m
p = 3.84 × 108 m
h = 3.475 × 106 m
1 1 2
 +  = 
p q R
1 2 1
 =  − 
q R p
1
2
1
(3.13 × 10−7) (2.60 × 10−9)
 =  


 − 
−
=
−
q (−6.40 × 106 m) (3.84 × 108 m)
1m
1m
3.16 × 10−7
q = − 
1m
−1
= −3.16 × 106 m = −3.16 × 103 km
q h⬘
M = −  = 
p h
qh
h⬘ = − 
p
−(−3.16 × 106 m)(3.475 × 106 m)
h⬘ = 
(3.84 × 108 m)
II
h⬘ = 2.86 × 104 m = 28.6 km
2. p = 553 m
2
R = −1.20 × 10 m
1 1 2
 +  = 
p q R
1 2 1
 =  − 
q R p
0.0167 0.00181
0.0185
1
2
1
 =  
−  = −  −  = − 
1m
1m
1m
q (−1.20 × 102 m)
(553 m)
q = −54.1 m
M = 9.78 × 10−2
3. R = −35.0 × 103 m
p = 1.00 × 105 m
1 1 2
 +  = 
p q R
1 2 1
 =  − 
q R p
1
2
1
(5.71 × 10−5) (1.00 × 10−5)
 =  


 − 
−
=
−
q (−35.0 × 103 m) (1.00 × 105 m)
1m
1m
6.71 × 10−5
q = − 
1m
II Ch. 14–6
Holt Physics Solution Manual
−1
= −1.49 × 104 m = −14.9 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q −(−54.1 m)
M = −  = 
(553 m)
p
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Givens
Solutions
h⬘
M = 
h
4. h = 1.4 × 106 m
h⬘ = 11.0 m
11.0 m
M = 
(1.4 × 106 m)
R = −5.50 m
M = 7.9 × 10−6
Scale is 7.9 × 10−6:1
q = −pM
1 1 2
 +  = 
p q R
R
1
p =  1 − 
2
M
−5.50 m
1−1
p =  
2 7.9 × 10 1
1
2
 1 −  = 
p
M
R
II
−6
p = 3.5 × 105 m = 3.5 × 102 km
5. scale factor = 1:1400
−3
f = −20.0 × 10
1
M = 
1400
m
q
M = − 
p
q = −Mp
1 1 1
 +  = 
p q
f
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
1
1
 1 −  = 
p
M
f
1
p = f 1 − 
M
p = (−20.0 × 10−3 m)(1 − 1400)
p = 28 m
Section Two—Problem Workbook Solutions
II Ch. 14–7
Menu
Print
Givens
Solutions
6. h = 1.38 m
q h⬘
M = −  = 
p
h
p = 6.00 m
h⬘ = 9.00 × 10−3 m
ph⬘
q = − 
h
1
1
2
 +  = 
p
q R
1
h
2
 −  = 
p ph⬘ R
2p
R= 
h
1 − 
h⬘
2(6.00 m)
12.0 m
R =  = 
(1
− 153)
1.38 m
1 − 
−3
9.00 × 10 m
R = −7.89 × 10−2 m = −7.89 cm
II
7. h⬘ = 4.78 × 10−3 m
−2
h = 12.8 × 10
m
−2
f = −64.0 × 10
m
q h⬘
M = −  = 
p h
qh
p = − 
h⬘
1 1 1
 +  = 
p q
f
1 −h⬘
1
  + 1 = 
q h f
h⬘
q = f 1 − 
h
4.78 × 10−3 m
q = (−64.0 × 10−2 m) 1 − 
= (−64.0 × 10−2 m)(0.963)
12.8 × 10−2 m
q = −61.6 × 10−2 m = −61.6 cm
II Ch. 14–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
1 1
 +  = 
q
f
−qh

h⬘
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Givens
Solutions
8. h = 0.280 m
−3
h⬘ = 2.00 × 10
m
−2
q = −50.0 × 10
m
h⬘
q
M =  = − 
h
p
qh
p = − 
h⬘
1 1 1
 +  = 
p q
f
1
1 1
 +  = 
q
f
−qh

h⬘
1 −h⬘
1
  + 1 = 
q h f
q
f = 
h⬘
1 − 
h
(−50.0 × 10−2 m)
(−50.0 × 10−2 m)
f =  = 
−3
(0.993)
(2.00 × 10 m)
1 − 
(0.280 m)
II
Copyright © by Holt, Rinehart and Winston. All rights reserved.
f = −50.4 × 10−2 m = −50.4 cm
Section Two—Problem Workbook Solutions
II Ch. 14–9
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Refraction
Chapter 15
Additional Practice 15A
Givens
1. qi = 72°
θr = 34°
Solutions
(sin θi)
(sin 72°)
= (1.00)  = 1.7
nr = ni 
(sin θr)
(sin 34°)
ni = 1.00
2. θi = 47.9°
θr = 29.0°
(sin θi)
(sin 47.9°)
nr = ni 
= (1.00)  = 1.53
(sin θr)
(sin 29.0°)
ni = 1.00
3. θr = 17°
ni = 1.5
II
glass to water:
nr = 1.33
nr
1.33
θi = sin−1 (sin
θr) = sin−1 (sin 17°) = 15°
ni
1.5
θr = 15°
air to glass:
nr = 1.5
1.5
nr
θi = sin−1 (sin
θr) = sin−1 (sin 15°) = 23°
1.00
ni
ni = 1.00
4. θi = 55.0°
θr = 53.8°
(sin θr)
(sin 53.8°)
= 1.33  = 1.31
ni = nr 
(sin θi)
(sin 55.0°)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
nr = 1.33
5. θi = 48°
ni = 1.00
air to glass 1:
nr = 1.5
ni
1.00
θr = sin−1 (sin
θi) = sin−1 (sin 48°) = 3.0 × 101 °
nr
1.5
θi = 3.0 × 101 °
glass 1 to glass 2:
ni = 1.5
nr = 1.6
ni
1.5
θr = sin−1 (sin
θi) = sin−1 (sin 3.0°) = 28°
nr
1.6
θi = 28°
glass 2 to glass 3:
ni = 1.6
ni
1.6
θr = sin−1 (sin
θi) = sin−1 (sin 28°) = 26°
nr
1.7
nr = 1.7
Section Two—Problem Workbook Solutions
II Ch. 15–1
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Additional Practice 15B
Givens
Solutions
1. f = 8.45 m
q = −25 m
1
1
1
 +  = 
p
q
f
1
1 1
 =  − 
p
f
q
1
1
1
0.118 0.040 0.158
 =  −  =  +  = 
p (8.45 m) (− 25)
1m
1m
1m
p = 6.3 m
q −(−25 m)
M = −  = 
p
6.3 m
M = 4.0
2. h⬘ = 1.50 m
q = −6.00 m
II
f = −8.58 m
1
 =
p
1
 =
p
1 1
1
1
 −  =  − 
f q −8.58 m −6.00 m
−0.117 0.167 0.050
 +  = 
1m
1m
1m
p = 20.0 m
−h′p
(1.50 m)(20.0 m)
h =  = −  = 5.00 m
q
(−6.00 m)
3. h = 7.60 × 10−2 m
h⬘ = 4.00 × 10−2 m
f = −14.0 × 10−2 m
h⬘
q
M =  = − 
h
p
ph⬘
q = − 
h
1
1
−1
 +  = 
p
f
−ph′

h
1
h
1
 1 −  = 
p
h⬘ f
h
p = f 1 − 
h⬘
7.60 × 10−2 m
p = (−14.0 × 10−2 m) 1 − 
= (−14.0 × 10−2 m)(0.90)
4.00 × 10−2 m
p = 1.30 × 10−1 m = 13.0 cm
(1.3 × 10−1 m)(4.00 × 10−2 m)
ph⬘
q = −  = − 
(7.60 × 10−2 m)
h
q = −6.84 × 10−2 m = −6.84 cm
II Ch. 15–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1 1
1
 +  = 
p q
f
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Givens
4. h = 28.0 m
h⬘ = 3.50 m
f = −10.0 m
Solutions
h⬘
q
M =  = − 
h
p
h⬘p
q = − 
h
1 1 1
 +  = 
p q f
1
1
1
 +  = 
p
f
−h⬘p

h
1
h
1
 1 −  = 
p
h⬘ f
h
p = f 1 − 
h⬘
28.0 m
p = (−10.0 m) 1 − 
3.50 m
p = 70.0 m
5. h⬘ = 1.40 cm
q = −19.0 cm
f = 20.0 cm
II
1 1 1
1
1
 =  −  =  − 
p f
q
−20.0 cm −19.0 cm
1 −0.0500 0.0526 2.60 × 103
 =  +  = 
1 cm
1 cm
1 cm
p
p = 385 cm = 3.85 m
ph⬘
(385 cm)(1.40 cm)
h = −  = −  = 28.4 cm
q
(−19.0 cm)
6. h = 1.3 × 10−3 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
h⬘ = 5.2 × 10−3 m
f = 6.0 × 10−2 m
h⬘
q
M =  = − 
h
p
ph⬘
q = − 
h
1 1 1
 +  = 
p q
f
1
1
1
 +  = 
p
f
ph⬘
− 
h
1
h
1
 1 −  = 
p h⬘ f
h
p = f 1 − 
h⬘
(1.3 × 10−3 m)
p = (6.0 × 10−2 m) 1 − 
(5.2 × 10−3 m)
p = 4.5 × 10−2 m = 4.5 cm
−(4.5 × 10−2 m)(5.2 × 10−3 m)
ph⬘
q = −  = 
(1.3 × 10−3 m)
h
q = −0.18 m = −18 cm
Section Two—Problem Workbook Solutions
II Ch. 15–3
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Print
Givens
Solutions
7. f = 26.7 × 10−2 m
p = 3.00 m
Image is real, so h⬘ < 0.
1 1
1
 +  = 
p q
f
1
1 1
 =  − 
q
f
p
1
1
1
3.75 0333 3.42
 = 
−  =  −  = 
q (26.7 × 10−2 m) (3.00 m) 1 m
1m
1m
q = 0.292 m = 29.2 cm
q
M = − 
p
(0.292 m)
M = − 
(3.00 m)
M = −9.73 × 10−2
8. h⬘ = 2.25 m
p = 12.0 m
II
f = −5.68 m
1
1
1 1 1
 =  −  =  − 
p −5.68 m 12.0 m
q f
1 −0.176 0.083 −0.259
 =  −  = 
q
1m
1m
1m
q = −3.86 m
h′p
(2.25 m)(12.0 m)
h = −  = −  = 6.99 m
q
−3.86 m
p = 4h = 0.432 m
f = −0.216 m
1 1 1
 +  = 
p q
f
1 1 1
 =  − 
p
q f
1
1
1
4.63
2.31
6.94
 =  −  = −  −  = − 
q (−0.216 m)
(0.432 m)
1m
1m
1m
q = − 0.144 m = −144 mm
h⬘
q
 = − 
h
p
qh
h⬘ = − 
p
− (−0.144 m)(0.108 m)
h⬘ = 
(0.432 m)
h⬘ = 0.0360 m = 36.0 mm
II Ch. 15–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. h = 0.108 m
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Givens
Solutions
10. p = 117 × 10−3 m
q
M = − 
p
M = 2.4
q = −pM = −(117 × 10−3 m)(2.4)
q = − 0.28 m
1 1 1
 +  = 
p q
f
1
1
5.0
1
8.55 3.6
 = 
−  =  −  = 
−3
(117 × 10 m) (0.28 m) 1 m 1 m 1 m
f
f = 0.20 m = 2.0 × 102 mm
11. Image is real, and therefore
inverted.
h⬘
 = M = −64
h
q = 12 m
q
p = − 
M
1 1
1
 +  = 
p q
f
M 1
1
−  +  = 
q
q
f
II
q
f = 
(1 − M)
(12 m)
f = 
[1 − (− 64)]
f = 0.18 m = 18 cm
12. h⬘ = −0.55 m
h = 2.72 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
p = 5.0 m
Image is inverted
q h⬘
−  = 
p h
ph⬘
q = − 
h
1 1
1
 +  = 
p q
f
1 1
h
 =  − 
f
p ph⬘
1 1
h
 =  1 − 
p
h⬘
f
p
f = 
h
1 − 
h⬘
5.0 m
5.0 m
f =  = 
5.9
(2.72 m)
1 − 
(−0.55 m)
f = 0.85 m
Section Two—Problem Workbook Solutions
II Ch. 15–5
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Givens
Solutions
13. p = 12.0 × 10−2 m
q
M = − 
p
M = 3.0
q = − Mp
1 1 1
 +  = 
p q
f
1
1
1
 −  = 
p Mp f
1
1
1
 1 −  = 
p
M
f
p
f = 
1
1 − 
M
(12.0 × 10−2 m)
f = 
1
1 − 
3.0
f = 1.8 × 10−1 m = 18 cm
II
14. h = 7.60 × 10−2 m
p = 16.0 × 10−2 m
f = −12.0 × 10−2 m
1 1 1
 +  = 
p q f
q h⬘
M = −  = 
p h
ph⬘
q = − 
h
1
1
1
 +  = 
p
f
−ph⬘

h
h
p
1 −  = 
h⬘ f
h
p
 = 1 − 
h⬘
f
h
h⬘ = 
p
1 − 
f
(7.60 × 10−2 m)
(7.60 × 10−2 m)
h⬘ = 
= 
−2
(2.33)
(16.0 × 10 m)
1 − 
−
2
(−12.0 × 10 m)
h⬘ = 3.26 × 10−2 m = 3.26 cm
II Ch. 15–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
h
1
 1 −  = 
p
h⬘ f
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Givens
Solutions
15. h = 48 m
−1
f = 1.1 × 10
m
p = 120 m
1 1
1
 +  = 
p q
f
1 1 1
 =  − 
q
f
p
1
1
1
9.1
(8.3 × 10−3)
9.1
 = 




  = 
−
=
−
q
(1.1 × 10−1 m)
(120 m) 1 m
1m
1m
q = 1.1 × 10−1m
h⬘
q
M =  = − 
h
p
qh
h⬘ = − 
p
−(1.1 × 10−1m)(48 m)
h⬘ = 
(120 m)
h⬘ = 4.4 × 10−2 m = − 4.4 cm
16. f = −0.80 m
h⬘ = 0.50 × 10−3 m
h = 0.280 m
h⬘
q
M =  = − 
h
p
II
ph⬘
q = − 
h
1 1 1
 +  = 
p q f
1
1
1
 +  = 
p
f
−p h⬘

h
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
h
1
 1 −  = 
p
h⬘ f
h
p = f 1 − 
h⬘
(0.280 m)
p = (− 0.80 m) 1 − 
(0.50 × 10−3 m)
p = 4.5 × 102 m
−(4.5 × 102 m)(0.50 × 10−3 m)
ph⬘
q = −  = 
(0.280 m)
h
q = −0.80 m
Additional Practice 15C
1. θc = 46°
ni = 1.5
nr = nisinθc = (1.5)(sin 46°) = 1.1
Section Two—Problem Workbook Solutions
II Ch. 15–7
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Givens
Solutions
2. ni = 1.00
qi = 75.0°
(sin θi)
(sin 75.0°)
nr = ni 
= (1.00)  = 2.44
(sin θr)
(sin 23.3°)
qr = 23.3°
ni = 2.44
nr = 1.00
3. θc = 42.1°
nr = 1.00
4. ni = 1.56
nr = 1.333
1.00
2.44 = 24.2°
n
θc = sin = −1 r
ni
θc = sin−1
nr
1.00
ni = 
=  = 1.49
sin θc sin 42.1°
n
sin qc = r
ni
n
1.333
θc = sin−1 r = sin−1  = 58.7°
ni
1.56
II
5. ni = 1.52
h = 0.025 mm
nr = 1.00
n
1.00
θc = sin−1 r = sin−1  = 41.1°
ni
1.52
∆x = h(tan qc)
n
where tan qc = r
ni
n
1.00
∆x = h r = (0.025 mm)  = 0.0160 mm
ni
1.52
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 2∆x = 2(0.0160 mm) = 0.0320 mm
II Ch. 15–8
Holt Physics Solution Manual
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Interference and Diffraction
Chapter 16
Additional Practice 16A
Givens
Solutions
1. d = 1.20 × 10−6 m
−9
l = 156.1 × 10
m
m = 5; constructive
interference
For constructive interference,
d(sin q ) = ml
ml
sin q = 
d
(5)(156.1 × 10 m)


(1.20 × 10 m) ml
q = sin−1 
d
−9
q = sin−1
−6
II
q = 40.6°
2. d = 6.00 × 10−6 m
−7
l = 6.33 × 10
m
m = 0; destructive
interference
For destructive interference,
d(sin q ) = m + 2l
1
m + 2l
1
sin q = 
d
HRW material copyrighted under notice appearing earlier in this book.
m + 2l
q = sin−1 
d
−1
q = sin
1
0 + 2(6.33 × 10−7 m)

(6.00 × 10−6 m
1
q = 3.02°
3. d = 0.80 × 10−3 m
m = 3; destructive
interference
q = 1.6°
For destructive interference,
d(sin q ) = m + 2l
1
d(sin q)
l= 
1
m + 2
(0.80 × 10−3 m)[sin (1.6°)]
l = 
1
3 + 2
l = 6.4 × 10−6 m = 6.4 mm
Section Two — Problem Workbook Solutions
II Ch. 16–1
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Givens
Solutions
4. d = 15.0 × 10−6 m
m = 2; constructive
interference
q = 19.5°
For constructive interference,
d(sin q ) = ml
d(sin q )
l = 
m
(15.0 × 10−6 m)[sin(19.5°)]
l = 
2
l = 2.50 × 10−6 m = 2.50 mm
5. l = 443 × 10−9 m
For destructive interference,
m = 4; destructive
interference
d(sin q ) = m + 2l
q = 2.27°
d= 
(sin q )
1
m + 2l
1
4 + 2(443 × 10−9 m)
1
d = 
[sin (2.27°)]
II
d = 5.03 × 10−5 m
6. f = 60.0 × 103 Hz
8
c = 3.00 × 10 m/s
m = 4; constructive
interference
q = 52.0°
For constructive interference,
mc
d(sin q ) = ml = 
f
mc
d = 
f (sin q )
(4)(3.00 × 108 m/s)
d = 
(60.0 × 103 Hz)[sin (52.0°)]
7. f = 137 × 106 Hz
8
c = 3.00 × 10 m/s
m = 2; constructive
interference
q = 60.0°
For constructive interference,
mc
d(sin q ) = ml = 
f
mc
d = 
f (sin q )
(2)(3.00 × 108 m/s)
d = 
(137 × 106 Hz)[sin(60.0°)]
d = 5.06 m
d[sin(90.0°)] d df
mmax =  =  = 
l
l
c
(5.06 m)(137 × 106 Hz)
mmax = 
= 2.31
(3.00 × 108 m/s)
The second-order maximum (m = 2) is the highest observable with this apparatus.
II Ch. 16–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 2.54 × 104 m = 25.4 km
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Additional Practice 16B
Givens
Solutions
1
1. d = 
1.00 × 102 lines/m
m =1
q = 30.0°
8
c = 3.00 × 10 m/s
d(sin q ) = ml
d(sin q)
l = 
m
[sin(30.0°)]
l = 
(1.00 × 102 lines/m)(1)
l = 5.00 × 10−3 m = 5.00 mm
c (3.00 × 108 m/s)
f =  = 
l
(5.00 × 10−3 m)
f = 6.00 × 1010 Hz = 60.0 Ghz
2. d = 2.0 × 10−8 m
d(sin q ) = ml
m =3
d(sin q)
l = 
m
q = 12°
II
−8
(2.0 × 10 m)[sin(12°)]
l = 
3
l = 1.4 × 10−9 m = 1.4 nm
3. l = 714 × 10−9 m
m =3
d(sin q ) = ml
ml
d = 
(sin q)
q = 12.0°
(3)(714 × 10−9 m)
d = 
[sin (12.0°)]
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 1.03 × 10−5 m between lines
or
9.71 × 104 lines/m
4. l = 40.0 × 10−9 m
−9
d = 150.0 × 10
m =2
m
d(sin q ) = ml
2(40.0 × 10 m)

(150.0
× 10 m)
ml
q = sin−1 
d
q = sin−1
−9
−9
q = 32.2°
Section Two — Problem Workbook Solutions
II Ch. 16–3
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Givens
Solutions
5. f = 1.612 × 109 Hz
8
c = 3.00 × 10 m/s
d = 45.0 × 10−2 m
m =1
mc
d(sin q ) = ml = 
f
(1)(3.00 × 10 m/s)

(45.0
× 10 m)(1.612 × 10 Hz)
mc
q = sin−1 
df
q = sin−1
8
−2
9
q = 24.4°
6. l = 2.2 × 10−6 m
1
d = 4
6.4 × 10 lines/m
q = 34.0°
d(sin q ) = ml
d(sin q )
m = 
l
[sin (34.0°)]
m = 
= 4.0
(6.4 × 104 lines/m)(2.2 × 10−6 m)
m = 4.0
II
1
7. d = 4
25 × 10 lines/m
−7
l = 7.5 × 10
q = 48.6°
m
d(sin q ) = ml
d(sin q )
m = 
l
[sin(48.6°)]
m = 
= 4.0
4
(25 × 10 lines/m)(7.5 × 10−7 m)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m = 4.0
II Ch. 16–4
Holt Physics Solution Manual
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Electric Forces and Fields
Chapter 17
Additional Practice 17A
Givens
Solutions
1. q1 = 0.085 C
q q2
Felectric = kC 1
r2
3
r = 2.00 × 10 m
Felectric = 8.64 × 10−8 N
kC = 8.99 × 109 N • m2/C2
Felectric r 2

q2 = 
kC q1
(8.64 × 10−8 N)(2.00 × 103 m)2
q2 = 
= 4.5 × 10−10 C
(8.99 × 109 N • m2/C2)(0.085 C)
2. q1 = q
q q2
3q 2


F = kC 1
=
k
C
r2
r2
q2 = 3q
Felectric = 2.4 × 10−6 N
q=
r = 3.39 m
II
(2.4 × 10−6 N)(3.39 m)2

3k = (3)(8.99 × 10 N m /C )
Fr 2
9
C
kC = 8.99 × 109 N • m2/C2
3. Felectric = 1.0 N
•
2
2
q = 3.2 × 10−8 C
N 2(qe)2
F = kC 
r2
r = 2.4 × 1022 m
kC = 8.99 × 109 N • m2/C2
qe =
Fkr = r kF
2
C
C

8.99 × 10 N m /C 1.0 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
qe = (2.4 × 1022 m)
9
•
2
2
qe = 2.5 × 1017 C
9
2 2
−9
−9
q q2 (8.99 × 10 N • m /C )(2.0 × 10 C)(2.8 × 10 C)
= 
Felectric = kC 1
2
2
(1034 m)
r
4. r = 1034 m
q1 = 2.0 × 10−9 C
q2 = −2.8 × 10−9 C
9
2
2
kC = 8.99 × 10 N • m /C
r2 = 2r
Felectric = 4.7 × 10−14 N
r2 = 2r = (2)(1034 m) = 2068 m
q=
(4.7 × 10−14 N)(2068 m)2

8.99 × 109 N • m2/C2
Felectricr22
 =
kC
q = 4.7 × 10−9 C
5. q1 = 1.0 × 105 C
q2 = −1.0 × 105 C
r = 7.0 × 1011 m
kC = 8.99 × 109 N • m2/C2
q q2
F = kC 1
r2
(1.0 × 105 C)2
F = (8.99 × 109 N • m2/C2) 
(7.0 × 1011 m)2
F = 1.8 × 10−4 N
Section Two—Problem Workbook Solutions
II Ch. 17–1
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Givens
Solutions
6. N = 2 000 744
−19
qp = 1.60 × 10
C
3
r = 1.00 × 10 m
kC = 8.99 × 109 N • m2/C2
Nqp (2 000 744)(1.60 × 10−19 C)
q =  =  = 1.60 × 10−13 C
2
2
q2
Felectric = kC 2
r
(1.60 × 10−13 C)2
Felectric = (8.99 × 109 N • m2/C2) 
(1.00 × 103 m)2
Felectric = 2.30 × 10−22 N
7. N1 = 4.00 × 103
N2 = 3.20 × 105
q = 1.60 × 10−19 C
r = 1.00 × 103 m
kC = 8.99 × 109 N • m2/C2
kC q1q2 kC N1N2 q 2
 = 
Felectric = 
r2
r2
(8.99 × 109 N • m2/C2)(4.00 × 103)(3.20 × 105)(1.60 × 10−19 C)2
Felectric = 
(1.00 × 103 m)2
Felectric = 2.95 × 10−25 N
9
2 2
5 2
−19
2
N 2q 2 (8.99 × 10 N • m /C )(3.20 × 10 ) (1.60 × 10 C)

Felectric = kC 2 
=
(1.00 × 103 m)2
r2
II
Felectric = 2.36 × 10−23 N
8. Felectric = 2.0 × 10−28 N
N = 111
qp = 1.60 × 10−19 C
N 2qp2
q2
Felectric = kC 2 = kC 
r2
r
r=
kC = 8.99 × 109 N • m2/C2
(8.99 × 109 N • m2/C2)(111)2(1.60 × 10−19 C)2

2.0 × 10−28 N
kCN 2qp2
 =
Felectric
r = 1.2 × 102 m
Felectric = 4.48 m × 104 N
r=
(8.99 × 109 N • m2/C2)(1.0 C)2

= 448 m
4.48 × 104 N
q2
kC  =
Felectric
kC = 8.99 × 109 N • m2/C2
10. Felectric = 1.18 × 10−11 N
q1 = 5.00 × 10−9 C
q2 = −2.50 × 10−9 C
kC = 8.99 × 109 N • m2/C2
q q2
Felectric = kC 1
r2
r=
(8.99 × 109 N • m2/C2)(5.00 × 10−9 C)(2.50 × 10−9 C)

1.18 × 10−11 N
k q2
C =
Felectric
r = 97.6 m
L = r cos q = (97.6 m)cos 45° = 69.0 m
II Ch. 17–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. q = 1.00 C
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Additional Practice 17B
Givens
Solutions
1. q1 = 2.80 × 10−3 C
q2 = −6.40 × 10−3 C
q3 = 4.80 × 10−2 C
r1,3 = 9740 m
r1,2 = 892 m
kC = 8.99 × 109 N • m2/C2
q q2
F = kC 1
r2
(8.99 × 109 N • m2/C2)(2.80 × 10−3 C)(6.40 × 10−3 C)
F1,2 = 
= 2.02 × 10−1 N
(892 m)2
(8.99 × 109 N • m2/C2)(2.80 × 10−3 C)(4.80 × 10−2 C)
F1,3 = 
= 1.27 × 10−2 N
(9740 m)2
F1,tot = F1,2 + F1,3 = −(2.02 × 10−1 N) + (1.27 × 10−2 N) = −0.189 N
F1,tot = 0.189 N downward
2. q1 = 2.0 × 10−9 C
−9
q2 = 3.0 × 10
C
q3 = 4.0 × 10−9 C
q4 = 5.5 × 10−9 C
2
r1,2 = 5.00 × 10 m
r1,3 = 1.00 × 103 m
q q2
F = kC 1
r2
(8.99 × 109 N • m2/C2)(2.0 × 10−9 C)(3.0 × 10−9 C)
F1,2 = 
= 2.2 × 10−13 N
(5.00 × 102 m)2
(8.99 × 109 N • m2/C2)(2.0 × 10−9 C)(4.0 × 10−9 C)
= 7.2 × 10−14 N
F1,3 = 
(1.00 × 103 m)2
9
2
−9
2
II
−9
r1,4 = 1.747 × 103 m
(8.99 × 10 N • m /C )(2.0 × 10 C)(5.5 × 10 C)
F1,4 = 
= 3.2 × 10−14 N
(1.747 × 103 m)2
kC = 8.99 × 109 N • m2/C2
F1,tot = F1,2 + F1,3 + F1,4 = (2.2 × 10−13 N) + (7.2 × 10−14 N) + (3.2 × 10−14 N)
F1,tot = 3.2 × 10−13 N down the rope
3. w = 7.00 × 10−2 m
−1
L = 2.48 × 10
−9
q = 1.0 × 10
m
C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
kC = 8.99 × 109 N • m2/C2
q q2
F = kC 1
r2
L

Fx = F1 + F2 (cos q) = F1 + F2 2
w +
L2
2.48 × 10 m
1
F = k q  + 
(2.48 × 10 m) [(7.00 × 10 m) + (2.48 × 10 m) ] 1
L

Fx = kC q 2 2 + 
L
(w 2 + L2)3/2
x
C
−1
2
−1
−2
2
−1
2
2 3/2
Fx = (8.99 × 109 N • m2/C2)(1.0 × 10−9 C)2(30.8/m2) = 2.8 × 10−7 N
w

Fy = F3 + F2 (sin q) = F3 + F2 2
w +
L2
7.00 × 10 m
1
F = k q  + 
(7.00 × 10 m) [(7.00 × 10 m) + (2.48 × 10 m) ] 1
w

Fy = kC q 2 2 + 
w
(w 2 + L2)3/2
y
C
−2
2
−2
−2
2
2
−1
2 3/2
Fy = (8.99 × 109 N • m2/C2)(1.0 × 10−9 C)2(2.00 × 102/m2) = 1.8 × 10−6 N
2 + F 2 = (2.8 × 10−7 N)2 + (1.8 × 10−6 N)2
Fnet = F
x y
Fnet = 1.8 × 10−6 N
Fy
1.8 × 10−6 N
q⬘ = tan−1  = tan−1 
= 81°
2.8 × 10−7 N
Fx
Fnet = 1.8 × 10−6 N, 81° above the positive x-axis
Section Two—Problem Workbook Solutions
II Ch. 17–3
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Givens
Solutions
4. L = 10.7 m
q q2
F = kC 1
r2
w = 8.7 m
Fx = F4 + F3(cos q)
q1 = −1.2 × 10−8 C
q2 = 5.6 × 10−9 C
Fy = F2 + F3(sin q)
q3 = 2.8 × 10−9 C
q
q3 L

Fx = kC q1 42 + 
2
L
(L + w 2)3/2
q4 = 8.4 × 10−9 C
9
2
2
kC = 8.99 × 10 N • m /C
(2.8 × 10−9 C)(10.7 m)
(8.4 × 10−9 C)

Fx = (8.99 × 109 N • m2/C2)(1.2 × 10−8 C) 
+
[(10.7 m)2 + (8.7 m)2]3/2
(10.7 m)2
−9
Fx = 9.1 × 10 N
q2
qw
Fy = kC q1 
+ 3
w 2 (L2 + w 2)3/2
(2.8 × 10−9 C)(8.7 m)
(5.6 × 10−9 C)
Fy = (8.99 × 109 N • m2/C2)(1.2 × 10−8 C) 
+ 
2
[(10.7 m)2 + (8.7 m)2]3/2
(8.7 m)
−9
Fy = 9.0 × 10 N
2 + F 2 = (9.1 × 10−9 N)2 + (9.0 × 10−9 N)2 = 1.28 × 10−8 N
Fnet = F
x y
Fy
9.0 × 10−9 N
q ⬘ = tan−1  = tan−1 
= 45°
Fx
9.1 × 10−9 N)
II
Fnet = 1.28 × 10−8 N, 45° above the positive x-axis
q1 = 1.6 × 10−2 C
q2 = 2.4 × 10−3 C
−3
q3 = −3.2 × 10
−3
q4 = −4.0 × 10
C
C
kC = 8.99 × 109 N • m2/C2
1.2 × 103 m
d
 = 
∆x = ∆y = = 8.5 × 102 m
2
2
kC q1q2

F=
r2
q2
q3(cos 45°)

Fx = −F2 + F3 (cos 45°) = kC q1 − 
+ 
2
∆x
d2
(3.2 × 10−3 C)(cos 45°)
2.4 × 10−3 C

Fx = (8.99 × 109 N • m2/C2)(1.6 × 10−2 C) − 
+
(1.2 × 103 m)2
(8.5 × 102 m)2
Fx = −0.24 N
q4
q3 (sin 45°)

Fy = −F4 − F3(sin 45°) = kC q1 
+
2
∆y
d2
(3.2 × 10−3 C)(sin 45°)
4.0 × 10−3 C
Fy = −(8.99 × 109 N • m2/C2)(1.60 × 10−2 C) 
+ 
2
2
(1.2 × 103 m)2
(8.5 × 10 m)
Fy = −1.0 N
2 + F 2 = (0.24 N)2 + (1.0 N)2 = 1.0 N
Fnet = F
x y
Fy
(1.0 N)
q⬘ = tan−1  = tan−1  = 77°
Fx
(0.24 N)
Fnet = 1.0 N, 77° below the negative x-axis
II Ch. 17–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. d = 1.2 × 103 m
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Givens
Solutions
228.930 × 103 m
6. d =  =
3
7.631 × 104 m
kC q1q2

F=
r2
Fx = F2 − F3(cos 60.0°)
q1 = 8.8 × 10−9 C
Fy = F3(sin 60.0°)
q2 = −2.4 × 10−9 C
9
2
2
kC = 8.99 × 10 N • m /C
q = 60.0°
q
q3(cos 60.0°)

Fx = kC q1 22 − 
d
d2
q3 = 4.0 × 10−9 C
(4.0 × 10−9 C)(cos 60.0°)
2.4 × 10−9 C


Fx = (8.99 × 109 N • m2/C2)(8.8 × 10−9 C) 
−
4
(7.631 × 104 m)2
(7.631 × 10 m)2
−18
Fx = 5.5 × 10
N
kC q1q3(sin 60.0°)

Fy = − 
r2
(8.99 × 109 N • m2/C2)(8.8 × 10−9 C)(4.0 × 10−9 C)(sin 60.0°)
Fy = − 
(7.631 × 104 m)2
Fy = −4.7 × 10−17 N
2 + F 2 = (5.5 × 10−18 N)2 + (4.7 × 10−17 N)2 = 4.7 × 10−17 N
Fnet = F
x y
Fy
4.7 × 10−17 N
q ⬘ = tan−1  = tan−1 
= 83°
Fx
5.5 × 10−18 N
II
Fnet = 4.7 × 10−18 N, 83° below the positive x-axis
Additional Practice 17C
1. q1 = 2.5 × 10−9 C
q3 = 1.0 × 10−9 C
r2,1 = 5.33 m
q q1
q q2
= kC 3
F3,1 = F3,2 = kC 3
(r3,1)2
(r3,2)2
r ,2
q2 = q1 3
r3,1
r3,1 = 1.90 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
r3,2 = r2,1 − r3,1 = 5.33 m − 1.90 m = 3.43 m
2
3.43 m 2
q2 = (2.50 × 10−9 C)  = 8.15 × 10−9 C
1.90 m
2. q1 = 7.5 × 10−2 C
q3 = 1.0 × 10−4 C
2
r2,1 = 6.00 × 10 km
r3,1 = 24 km
r3,2 = r2,1 − r3,1 = 6.00 × 102 km − 24 km = 576 km
q q1
q3q2
F3,1 = F3,2 = kC 3
2 = kC 2
(r3,1)
(r3,2)
r ,2
q2 = q1 = 3
r3,1
2
576 km 2
q2 = (7.5 × 10−2 C)  = 43 C
24 km
Section Two—Problem Workbook Solutions
II Ch. 17–5
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Givens
Solutions
3. mE = 6.0 × 1024 kg
mm = 7.3 × 1022 kg
G = 6.673 × 10−11 N • m2/kg2
kC = 8.99 × 109 N • m2/C2
Fg = Felectric
GmEmm kC q 2

= 
r2
r2
q=
(6.673 × 10−11 N • m2/kg2)(6.0 × 1024 kg)(7.3 × 1022 kg)

8.99 × 109 N • m2/C2
GmE mm
 =
kC
q = 5.7 × 1013 C
4. m = 17.23 kg
r = 0.800 m
Fnet = 167.6 N
g = 9.81 m/s2
kC = 8.99 × 109 N • m2/C2
Fnet = Fg − Felectric
kC q 2

Fnet = mg − 
r2
q=
q=
II
r 2(mg − Fnet )

kC
(0.800 m)2[(17.23 kg)(9.81 m/s2) − (167.6 N)]

8.99 × 109 N • m2/C2
q = 1.0 × 10−5 C
5. m1 = 9.00 kg
Fg,1 = Fg,2 + Felectric
m2 = 8.00 kg
r = 1.00 m
kC q 2

g(m1 − m2 ) = 
r2
kC = 8.99 × 109 N • m2/C2
q=
6. m = 9.2 × 104 kg
l1 = 1.00 m
g = 9.81 m/s2
l2 = 8.00 m
r = 2.5 m
kC = 8.99 × 109 N • m2/C2
7. q1 = 2.0 C
q2 = 6.0 C
q3 = 4.0 C
L = 2.5 × 109 m
q = 3.30 × 10−5 C
t1 = t2
kC q 2l2

mgl1 = 
r2

r kmlgl
(2.5 m) (9.2 × 10 kg)(9.81 m/s )(1.00 m)
 = 8.9 × 10
q = (8.99 × 10 N m /C )(8.00 m)
2
q=
1
C2
2
4
9
2
•
2
−3
2
Fnet = 0 = F1 + F2
kC q2 q3
q q3

kC 1
= 
(L − x)2
x2
q
q2
21 = 
x
(L − x)2
(L − x) q1 = x q2
L x
 −  =
x x
q
q2
L
 =
x
1
9
L
2.5 × 10 m
x =  =  = 9.3 × 108 m
q2
6.0 C
 + 1
 + 1
q1
2.0 C
II Ch. 17–6
Holt Physics Solution Manual
q + 1
q2
1
C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
(9.81 m/s2)(1.00 m)2(9.00 kg − 8.00 kg)

8.99 × 109 N • m2/C2
gr 2(m1 − m2 )
 =
kC
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Givens
Solutions
Fnet = 0 = F1 + F2
8. q1 = 55 × 10−6 C
q2 = 137 × 10−6 C
q q3
kC q2 q3

kC 1
= 
x2
(L − x)2
q
q2
21 = 
x
(L − x)2
q3 = 14 × 10−6 C
L = 87 m
(L − x) q1 = x q2
L x
 −  =
x x
L
 =
x
q
q2
1
q + 1
q2
1
87 m

L
x =  = 137 × 10−6 C
= 34 m

+1
q2
−6
55 × 10 C
 + 1
q1
kC q1q2

F=
r2
9. F = 1.00 × 108 N
4
q1 = 1.80 × 10 C
r=
4
q2 = 6.25 × 10 C
9
2
2
kC = 8.99 × 10 N • m /C
r=
II
kC q1q2

F
(8.99 × 109 N • m2/C2)(1.80 × 104 C)(6.25 × 104 C)

1.00 × 108 N
r = 3.18 × 105 m
10. m = 5.00 kg
Fg = Felectric
q = 4.00 × 10−2 C
9
2
2
kC = 8.99 × 10 N • m /C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
kC q 2

mg = 
h2
kC q 2

h=
mg
h=
11. m = 1.0 × 10−19 kg
r = 1.0 m
q = 1.60 × 10−19 C
kC = 8.99 × 109 N • m2/C2
(8.99 × 109 N • m2/C2)(4.00 × 10−2 C)2

= 542 m
(5.00 kg)(9.81 m/s2)
kC q 2

Fres = Felectric = 
r2
(8.99 × 109 N • m2/C2)(1.60 × 10−19 C)2
Fres = 
(1.0 m)2
Fres = 2.3 × 10−28 N
12. m = 5.0 × 10−6 kg
−15
q = 2.0 × 10
C
r = 1.00 m
kC = 8.99 × 109 N • m2/C2
G = 6.673 × 10−11 N • m2/kg2
Fnet = Felectric + Fg
9
2 2
−15
2
kC q 2 (8.99 × 10 N • m /C )(2.0 × 10 C)

Felectric = 
= 
2
2
(1.00 m)
r
−11
2
2
−6
2
Gm 2 (6.673 × 10 N • m /kg )(5.0 × 10 kg)
Fg = 
= 
2
2
(1.00 m)
r
Fnet = 3.6 × 10−20 N + 1.7 × 10−21 N = 3.8 × 10−20 N
Section Two—Problem Workbook Solutions
II Ch. 17–7
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Givens
Solutions
13. m = 2.00 × 10−2 kg
Felectric = Ffriction
q1 = 2.0 × 10−6 C
q2 = −8.0 × 10−6 C
r = 1.7 m
kC = 8.99 × 109 N • m2/C2
g = 9.81 m/s2
kC q1q2

= mk mg
r2
kC q1q2

mk = 
mgr 2
(8.99 × 109 N • m2/C2)(2.0 × 10−6 C)(8.0 × 10−6 C)
mk = 
= 0.25
(2.00 × 10−2 kg)(9.81 m/s2)(1.7 m)2
Additional Practice 17D
1. r = 3.72 m
k q
E = C
r2
E = 0.145 N/C
kC = 8.99 × 109 N • m2/C2
q = 60.0°
k q
k q
Ex = C
(cos 60.0°) − C
(cos 60.0°) = 0 N/C
2
r
r2
Because Ex = 0 N/C, the electric field points directly upward.
2kC q (sin 60.0°)

Ey = 
r2
II
Ey r 2
(0.145 N/C)(3.72 m)2
q =  = 
= 1.29 × 10−10 C
2kC (sin 60.0°) (2)(8.99 × 109 N • m2/C2)(sin 60.0°)
−8
q1 = 1.2 × 10
C
∆x = 120 m
Ex = 1.60 × 10−2 N/C
kC = 8.99 × 109 N • m2/C2
k q
E = C
r2
kC q 2 (∆x)
k q1
Ex = E1 + E2 (cos q) = C
+ 
2
2
∆x
x2+
(∆x + ∆y 2) ∆
∆y2
k q1 (∆x 2 + ∆y 2)3/2

q2 = Ex − C
kC ∆x
∆x 2
(8.99 × 109 N • m2/C2)(1.2 × 10−8 C)
k q1
−2

Ex − C
=
1.60
×
10
N/C
−
(120 m)2
∆x 2
= 8.5 × 10−3 N/C
[(120 m)2 + (190 m)2]3/2
(∆x 2 + ∆y 2)3/2
 = 
(8.99 × 109 N • m2/C2)(120 m)
kC ∆x
= 1.0 × 10−5 C2/N
q2 = (8.5 × 10−3 N/C)(1.0 × 10−5 C2/N) = 8.5 × 10−8 C
3. q1 = 1.80 × 10−5 C
q2 = −1.20 × 10−5 C
Enet = 22.3 N/C toward q2
k
Enet = C(q
1 + q2)
r2
r=
k
(qE
+q)
C
1
2
net
kC = 8.99 × 109 N • m2/C2
r=
(8.99 × 109 N • m2/C2)[(1.80 × 10−5 C) + (1.20 × 10−5 C)]

22.3 N/C toward q2
r = 1.10 × 102 m
II Ch. 17–8
k
r 2 = C(q1 + q2)
Enet
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. ∆y = 190 m
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Givens
Solutions
4. d = 86.5 m
Enet = E1 + E2 = 0
−9
q1 = 4.8 × 10
−8
q2 = 1.6 × 10
C
E1 = E2
C
q
q2
12 = 
x
(d − x)2
(d − x) q1 = x q2
x q1 + q2 = d q1
8×10−9C
d q1
(86.5 m) 4.

=
x = 
−
9
) + (1
.6
×10−8C
)
(4
.8
×10C
q2 q1 + x = 3.0 × 101 m
5. q = 3.6 × 10−6 C
L = 960 m
w = 750 m
kC = 8.99 × 109 N • m2/C2
k q
E = C
r2
kC qw
k q

Ey = E1 + E2 (sin q) = C
+
2
w2
+
w
L 2 (w 2 + L 2)
II
1
w

Ey = kC q 2 + 
2
w
(w + L2)3/2
750 m
1
Ey = (8.99 × 109 N • m2/C2)(3.6 × 10−6 C) 2 + 
[(750 m)2 + (960 m)2]3/2
(750 m)
Ey = 7.1 × 10−2 N/C
kC qL
k q

Ex = E3 + E2 (cos q) = C
+
2
L2
w +
L 2 (w 2 + L 2)
1
L

Ex = kC q 2 + 
2
L
(w + L2)3/2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
960 m
1
Ex = (8.99 × 109 N • m2/C2)(3.6 × 10−6 C) 2 + 
[(750 m)2 + (960 m)2]3/2
(960 m)
Ex = 5.2 × 10−2 N/C
2 + E 2 = (7.1 × 10−2 N/C)2 + (5.2 × 10−2 N/C)2 = 8.8 × 10−2 N/C
Enet = E
y x
Ey
7.1 × 10−2 N/C
= 54°
q ⬘ = tan−1  = tan−1 
Ex
5.2 × 10−2 N/C
Enet = 8.8 × 10−2 N/C, 54° above the horizontal
Section Two—Problem Workbook Solutions
II Ch. 17–9
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Givens
Solutions
h 2
+
w 2
6. w = 218 m
r =  =  = 112 m
2
2
h = 50.0 m
q = 6.4 × 10−9 C
9
2
2
kC = 8.99 × 10 N • m /C
q1 = q2 = q
q3 = 3q
q4 = 2q
0.
0m
)2+(21
8m
(5
)2
h
50.0 m
q = tan−1  = tan−1  = 12.9°
w
218 m
The electric fields of charges on opposite corners of the rectangle cancel to give 2q
on the lower left corner and q on the lower right corner.
k q
E = C
r2
kC 2q kC q
kC q(cos q)
 − 

Ex = 
(cos q) = 
r2
r2
r2
(8.99 × 10 N • m2/C2)(6.4 × 10−9 C)(cos 12.9°)
Ex = 
= 4.5 × 10−3 N/C
(112 m)2
9
kC 2q kC q
3kC q (sin q)


Ey = 
+ 
(sin q) = 
2
2
r
r
r2
II
(3)(8.99 × 109 N • m2/C2)(6.4 × 10−9 C)(sin 12.9°)
Ey = 
= 3.1 × 10−3 N/C
(112 m)2
2 + E 2 = (4.5 × 10−3 N/C)2 + (3.1 × 10−3 N/C)2
Enet = E
x y
Enet = 5.5 × 10−3 N/C
Ey
3.1 × 10−3 N/C
q ⬘ = tan−1  = tan−1 
= 35°
4.5 × 10−3 N/C
Ex
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Enet = 5.5 × 10−3 N/C, 35° above the positive x-axis
II Ch. 17–10 Holt Physics Solution Manual
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Electrical Energy and Capacitance
Chapter 18
Additional Practice 18A
Givens
Solutions
1. PEelectric = 0.868 J
r = 15.4 × 10 m
qq
PEelectric = kC 12
r
N • m2
kC = 8.99 × 109 
C2
(PEelectric )(r)

q1q2 = q 2 = 
kC
3
q=

k
(PEelectric )(r)
C
q=
(0.868 J)(15.4 × 103 m)

(8.99 × 109 N • m2/C2)
q = q1 = q2 = 1.22 × 10−3 C
2. r = 281 m
q1 = 2.40 × 10−7 C
PEelectric = −2.0 × 10−5 J
HRW material copyrighted under notice appearing earlier in this book.
N • m2
kC = 8.99 × 109 
C2
II
kC q1q2

PEelectric = 
r
(PEelectric)(r)

q2 = 
(kC)(q1)
(−2.0 × 10−5 J)(281 m)
q2 = 
N • m2
(8.99 × 109 )(2.40
× 10−7 C)
C2
q2 = −2.6 × 10−6 C
3. PEelectric = 4.80 × 10−4 J
d = 2365 m
E = −1.50 × 102 N/C
PEelectric = −qEd
−PE lectric
q = e
Ed
−(4.80 × 10−4 J)
q = 
(−1.50 × 102 N/C)(2365 m)
q = 1.35 × 10−9 C
4. q1 = 44 × 10−6 C
q2 = 44 × 10−6 C
PEelectric = 1.083 × 10−2 J
N • m2
kC = 8.99 × 109 
C2
qq
PEelectric = kC 12
r
kC q1q2
r = 
PEelectric
N • m2 (44 × 10−6 C)2

r = 8.99 × 109 
(1.083 × 10−2 J)
C2
r = 1.6 × 103 m = 1.6 km
Section Two—Problem Workbook Solutions
II Ch. 18–1
Print
Givens
Solutions
5. q1 = −16.0 × 10−3 C
−3
q2 = 24.0 × 10
−4
W = 2.8 × 10
C
J
rf = ∞
N • m2
kC = 8.99 × 109 
C2
6. d = 1410 m
W = ∆PEelectric = PEelectric,f − PEelectric,i
1 1
−kC q1q2

W = kC q1q2  −  = 
rf ri
ri
−kC q1q2
−(8.99 × 109 N • m2/C2)(−16.0 × 10−3 C)(24.0 × 10−3 C)
ri = 
= 
W
2.8 × 10−4 J
ri = 122 m
∆PEelectric = −qEd
E = 380 N/C
∆PEelectric = −(−1.60 × 10−19 C)(380 N/C)(1410 m)
q = −1.60 × 10−19 C
∆PEelectric = 8.6 × 10−14 J
7. d = 275 m
q = 12.5 × 10−9 C
E = 1.50 × 102 N/C
II
8. P = 1.5 × 105 W
vf = 2.50 × 102 km/h
vi = 0 km/h
PEelectric = −qEd
PEelectric = −(12.5 × 10−9 C)(1.50 × 102 N/C)(275 m)
PEelectric = −5.16 × 10−4 J
1
1
1
a. W = ∆KE = KEf − KEi = 2 mvf 2 − 2 mvi2 = 2 mvf 2
1
W = 2 (mc + mp)vf 2
mp = 2.000 × 103 kg
W = 6.03 × 106 J
N • m2
kC = 8.99 × 109 
C2
b. W = P∆t
(vf + vi )
∆x =  ∆t
2
(vf + vi ) W
vf W
∆x =   = 
2P
2
P
1 h 103 m
(2.50 × 102 km/h)   (6.03 × 106 J)
3600 s 1 km
∆x = 
2(1.5 × 105 W)
∆x = 1.4 × 103 m = 1.4 km
qq
c. PEelectric = kC 12
r
PEelectric = W
r = ∆x
q1 = q2 = q =

(P
Ek
)(r) = (W
k)(∆x)
electric
C
q=
(6.03 × 106 J)(1.4 × 103 m)

N • m2
8.99 × 109 
C2
q = 0.97 C
II Ch. 18–2
2
mc = 5.00 × 102 kg
km 1 h 103 m
1
W = 2[(5.00 × 102 kg) + (2.000 × 103 kg)] 2.50 × 102   
h 3600 s 1 km
Holt Physics Solution Manual
C
HRW material copyrighted under notice appearing earlier in this book.
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Additional Practice 18B
Givens
Solutions
1. q1 = −12.0 × 10−9 C
q2
q
q
q
V = kC 1 + 2 = kC 1 + 
r1 r2
r1 (d − r1)
q2 = −68.0 × 10−9 C
V q
q2
 − 1 + 
kC r1 (d − r1)
V = −25.3 V
r1 = 16.0 m
r2 = d − r1
2
N •m
kC = 8.99 × 109 
C2
q2
d=
+ r1
V q1
 − 
kC r1
−68.0 × 10−9 C
d =  +16.0 m
−25.3 V
(−12.0 × 10−9 C)
9
− 
2
2
8.99 × 10 N•m /C
16.0 m
d = 33.0 m + 16.0 m = 49.0 m
2. q1 = 18.0 × 10−9 C
q
q q
 = kC 1 + 2
r
r1 r2
V = kC
q2 = 92.0 × 10−9 C
V = 53.3 V
q1
q
V = kC 
+ 2
d − r2 r2
r1 = d − r2
d = 97.5 m
kC = 8.99 × 10 
C2
II

k = 
(d − r )(r )
V
V
−  r +  dr = (q − q )r + q d
k k V
Vd
kr + q − q − k r + q d = 0
(q1r2 + q2d − q2r2 )
V
2
9 N •m
2
C
2
2
2
2
C
1
2 2
2
C
2
2
1
2
C
2
2
C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Solve using the quadratic formula:
Vd
Vd 2 4Vq2 d
− q1 − q2 −  ±
q1 − q2 −  − 
kC
kC
kC
r2 = 
2V

kC
q − q − k = 18.0 × 10
Vd
1
−9
2
C
q − q − k = −652 × 10
Vd
1
−9
2
(53.3 V)(97.5 m)
C − 92.0 × 10−9 C − 
N • m2
8.99 × 109 
C2
C
C
4Vq 2 d 4(53.3 V)(92.0 × 10−9 C)(97.5 m)
 = 
= 2.13 × 10−13 C2
kC
(8.99 × 109 N • m2/C2)
2V
C
2(53.3 V )
 = 
= 11.9 × 10−9 
9
2
2
kC (8.99 × 10 N • m /C )
m
65
2×10−9C
13
×1
)2
−(2.
0−13
C2)
−(−652 × 10−9 C) ± (−
r2 = 
(11.9 × 10−9 C/m)
652 ± 460
r2 =  m
11.9
Section Two—Problem Workbook Solutions
II Ch. 18–3
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Givens
Solutions
Of the two roots, the one that yields the correct answer is
(652 − 460)
r2 =  m
11.9
r2 = 16.1 m
3. V = 1.0 × 106 V
−2
r = 12 × 10
m
N • m2
kC = 8.99 × 109 
C2
q
V = kC 
r
Vr
q = 
kC
(1.0 × 106 V)(12 × 10−2 m)
q = 
N • m2
8.99 × 109 
C2
q = 1.3 × 10−5 C
II
4. ME = 5.98 × 1024 kg
mVgravity = qVelectric
N • m2
G = 6.673 × 10−11 
kg2
mMEG qQEkC
 =  
r
r
N • m2
kC = 8.99 × 109 
C2
mMEG
QE = 
qkC
m = 1.0 kg
N • m2
(1.0 kg)(5.98 × 1024 kg) 6.673 × 10−11 
kg2
QE = 
N • m2
(1.0 C) 8.99 × 109 
C2
q = 1.0 C
5. msun = 1.97 × 1030 kg
mH = mass of hydrogen
atom = 1.67 × 10−27 kg
q1 = charge of proton
= +1.60 × 10−19 C
q2 = charge of electron
= −1.60 × 10−19 C
11
r1 = 1.1 × 10 m
r2 = 1.5 × 1011 m − 1.1 ×
1011 m = 4.0 × 1010 m
N•m2
kC = 8.99 × 109 
C2
msunq1

a. Q + = charge of proton cloud = (number of protons)q1 = 
mH
(1.97 × 1030 kg)(1.60 × 10−19 C)
Q + = 
(1.67 × 10−27 kg)
Q + = 1.89 × 1038 C
msun q2

Q − = charge of electron cloud = 
mH
Q − = −1.89 × 1038 C
b. V = kC
q
Q
Q
 = kC + + −
r
r1
r2
N • m2 1.89 × 1038 C 1.89 × 1038 C

V = 8.99 × 109 
− 
C2
1.1 × 1011 m
4.0 × 1010 m
V = −2.7 × 1037 V
II Ch. 18–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
QE = 4.44 × 104 C
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Givens
Solutions
6. r = r1 = r2 = r3 = r4 =
V = kC
x2 y
 + 
2
2
2
y = 276 m
q = 64 × 10−9 C
N • m2
8.99 × 109 
(64 × 10−9 C)(4.5)
C2
V = 
292 m 2 276 m 2
 + 
2
2
q1 = 1.0q
q2 = −3.0q
q3 = 2.5q
q4 = 4.0q
N • m2
kC = 8.99 × 109 
C2
7. q1 = q2 = q3 = q
= 7.2 × 10−2 C
V = 13 V
V = kC
7
= 1.6 × 10 m
l
r1 = r2 = 
2
r3 =
kC q(1.0 − 3.0 + 2.5 + 4.0)
V = 
x2 y 2
 + 
2
2
x = 292 m
l
q
q
q
q
q
 = kC 1 + 2 + 3 + 4
r
r1 r2 r3 r4
l − 2l 2
2
N • m2
kC = 8.99 × 109 
C2
q
q
q
q
 = kC 1 + 2 + 3
r
r1 r2 r3
q
q
1
q
kC q
V = kC  +  +  = 
2+2+ 
2
12
l
l
l
l
2
1 − 2


− 
l
2
2
2
N • m2
8.99 × 109 
(0.072 C)
1
C2
4+ 
3
V = 
4
(1.6 × 107 m)
II
V = 2.1 × 102 V
8. q1 = q2 = q3 = q
= 25.0 × 10−9 C
V = kC
Copyright © by Holt, Rinehart and Winston. All rights reserved.
r1 = r2 = l
r3 =
l
2+ 2
l
l
= 184 m
N • m2
kC = 8.99 × 109 
C2
q
q q q
 = kC 1 + 2 + 3
r
r1 r2 r3
1
1
1
1
kC q

 =
V = kC q  +  + 1+1+ l l
l
2
l 2+l 2
N • m2
8.99 × 109 
(25.0 × 10−9 C)
C2
V =  (2.707)
(184 m)
V = 3.31 V
Section Two—Problem Workbook Solutions
II Ch. 18–5
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Additional Practice 18C
Givens
Solutions
1. ∆V = 3.00 × 102 V
PEelectric = 17.1 kJ
1
PEelectric = 2 C(∆V )2
2PE ctric

C = ele
(∆V )2
2(17.1 × 103 J)
C = 
(3.00 × 102 V)2
C = 3.80 × 10−1 F
2. PEelectric = 1450 J
∆V = 1.0 × 104 V
1
PEelectric = 2 C(∆V )2
2PE ctric

C = ele
(∆V )2
2(1450 J)
C = 
(1.0 × 104 V)2
II
C = 2.9 × 10−5 F
3. Emax = 3.0 × 106 V/m
d = 0.2 × 10−3 m
A = 6.7 × 103 m2
e0 = 8.85 × 10−12 C2/N•m2
∆Vmax = Emaxd
Q
Qmax
=
∆Vmax = max
C
e0 A

d
Qmax
Emax d = 
e0 A

d
Qmax = Emax e0 A
Qmax = 0.18 C
Qmax = C∆Vmax = CEmax d
4. r = 3.1 m
−3
d = 1.0 × 10
m
6
Emax = 3.0 × 10 V/m
e0 = 8.85 × 10−12 C2/N • m2
e A
C = 0
d
Qmax = e0 AE max = e0 pr 2E max
Qmax = (8.85 × 10−12 C2/N • m2)(p)(3.1 m)2(3.0 × 106 V/m)
Qmax = 8.0 × 10−4 C = 0.80 mC
5. P = 5.0 × 1015 W
−12
∆t = 1.0 × 10
C = 0.22 F
s
1
PEelectric = 2C(∆V)2
PEelectric = P∆t
1
P∆t = 2C(∆V )2
2P∆t
∆V = 
C
∆V =
2(5.0 × 1015 W)(1.0 × 10−12 s)

(0.22 F)
∆V = 210 V
II Ch. 18–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Qmax = (3.0 × 106 V/m)(8.85 × 10−12 C2/N • m2)(6.7 × 103 m2)
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Givens
6. A = 2.32 × 105 m2
d = 1.5 × 10−2 m
Q = 0.64 × 10−3 C
e0 = 8.85 × 10−12 C2/N • m2
Solutions
2
1 Q
PEelectric = 2 
C
e0 A
C = 
d
2
1 Q d
PEelectric = 2 
e0 A
PEelectric =
1

2
(0.64 × 10−3 C)2(1.5 × 10−2 m)

(8.85 × 10−12 C2/N • m2)(2.32 × 105 m2)
PEelectric = 1.5 × 10−3 J
7. r = 18.0 m
∆V = 575 V
PEelectric = 3.31 J
1
PEelectric = 2C(∆V)2
2 PE ctric 2(3.31 J)
 = 2
C = ele
(575 V)
(∆V)2
C = 2.00 × 10−5 F
e A e0 πr2

d = 0 = 
C
C
II
(8.85 × 10−12 C2/N• m2)(π)(18.0 m)2
d = 
(2.00 × 10−5 F)
d = 4.5 × 10−4 m = 0.45 mm
8. di = 5.00 × 10−3 m
df = 0.30 × 10−3 m
e0 = 8.85 × 10−12 C2/N • m2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
A = 1.20 × 10−4 m2
e A e A
∆C = Cf − Ci = 0 − 0
df
di
1
1
∆C = e0 A  − 
df di
C2
1
1
∆C = 8.85 × 10−12 2 (1.20 × 10−4 m2) 
− 
N •m
0.30 × 10−3 m 5.00 × 10−3 m
∆C = 3.3 × 10−12 F = −3.3 pF
9. A = 98 × 106 m2
C = 0.20 F
e0 = 8.85 × 10−12 C2/N • m2
e A
C = 0
d
e A
d = 0
C
(8.85 × 10−12C2/N • m2)(98 × 106 m2)
d = 
(0.20 F)
d = 4.3 × 10−3 m = 4.3 mm
Section Two—Problem Workbook Solutions
II Ch. 18–7
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Givens
Solutions
10. A = 7.0 m × 12.0 m
e A
a. C = 0
d
d = 1.0 × 10−3 m
e0 = 8.85 × 10−12 C2/N • m2
PEelectric = 1.0 J
(8.85 × 10−12 C2/N • m2)(7.0 m)(12.0 m)
C = 
(1.0 × 10−3 m)
C = 7.4 × 10−7 F = 0.74 mF
1
b. PEelectric = 2 C(∆V )2
∆V =

C
∆V =
0 J)

(7.42(×1.1
0F)
2PEelectric
−7
∆V = 1.6 × 103 V = 1.6 kV
11. A = 44 m2
II
−12
e0 = 8.85 × 10
−6
Q = 2.5 × 10
∆V = 30.0 V
C
2
2
C /N • m
Q
a. C = 
∆V
(2.5 × 10−6 C)
C = 
(30.0 V)
C = 8.3 × 10−8 F = 83 nF
e A
b. C = 0
d
e A
d = 0
C
(8.85 × 10−12 C2/N • m2)(44 m2)
d = 
(8.3 × 10−8 F)
1
c. PEelectric = 2Q∆V
PEelectric = 2(2.5 × 10−6 C)(30.0 V)
1
PEelectric = 3.8 × 10−5 J
II Ch. 18–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 4.7 × 10−3 m
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Current and Resistance
Chapter 19
Additional Practice 19A
Givens
Solutions
1. I = 3.00 × 102 A
∆Q = I∆t
∆t = 2.4 min
60 s
∆Q = (3.00 × 102 A)(2.4 min) 
1 min
∆Q = 4.3 × 104 C
2. ∆t = 7 min, 29 s
∆Q = I∆t
I = 0.22 A
60 s
∆Q = (0.22 A) (7 min)  + 29 s = (0.22 A)(449 s)
1 min
∆Q = 99 C
3. ∆t = 3.3 × 10−6 s
I = 0.88 A
C
q = e = 1.60 × 10−19 
electron
II
∆Q = I∆t = nq
I∆t
n = 
q
(0.88 A)(3.3 × 10−6 s)
n = 
(1.60 × 10−19 C/electron)
n = 1.8 × 1013 electrons
4. ∆t = 3.00 h
4
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆Q = 1.51 × 10 C
∆Q
I = 
∆t
(1.51 × 104 C)
I = 
3.60 × 103 s
(3.00 h) 
1h
I = 1.40 A
5. ∆Q = 1.8 × 105 C
∆t = 6.0 min
∆Q
I = 
∆t
(1.8 × 105 C)
I = 
60 s
(6.0 min) 
1 min
I = 5.0 × 102 A
Section Two—Problem Workbook Solutions
II Ch. 19–1
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Givens
Solutions
6. I = 13.6 A
5
Q = 4.40 × 10 C
∆Q
∆t = 
I
(4.40 × 105 C)
∆t = 
(13.6 A)
∆t = 3.24 × 104 s = 9.00 h
Additional Practice 19B
1. ∆V = 440 V
∆V
R = 
I
I = 0.80 A
(440 V)
R = 
(0.80 A)
R = 5.5 × 102 Ω
II
2. ∆V = 9.60 V
∆V
R = 
I
I = 1.50 A
(9.60 V)
R = 
(1.50 A)
R = 6.40 Ω
3. ∆V = 312 V
5
∆Q = 2.8 × 10 C
∆Q
I = 
∆t
∆V
∆V
∆V∆t
R =  =  = 
I
∆Q
∆Q

∆t
∆t = 1.00 h
3.60 × 103 s
(312 V)(1.00 h) 
1h
R = 
5
(2.8 × 10 C)
R = 4.0 Ω
4. I = 3.8 A
∆V = IR
R = 0.64 Ω
∆V = (3.8 A)(0.64 Ω)
∆V = 2.4 V
5. R = 0.30 Ω
3
I = 2.4 × 10 A
II Ch. 19–2
∆V = IR = (2.4 × 103 A)(0.30 Ω) = 7.2 × 102 V
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Givens
Solutions
6. ∆V = 3.0 V
∆V
I = 
R
R = 16 Ω
(3.0 V)
I = 
(16 Ω)
I = 0.19 A
7. ∆V = 6.00 × 102 V
R = 4.4 Ω
∆V (6.00 × 102 V)
I =  =  = 1.4 × 102 A
R
(4.4 Ω)
Additional Practice 19C
1. P = 12 × 103 W
P = I 2R
R = 2.5 × 102 Ω
I=
R
I=
P
(12 × 103 W)

(2.5 × 102 Ω)
II
I = 6.9 A
2. P = 33.6 × 103 W
∆V = 4.40 × 102 V
P = I∆V
P
I = 
∆V
(33.6 × 103 W)
I = 
(4.40 × 102 V)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
I = 76.4 A
P = I∆V
3. P = 850 W
V = 12.0 V
P
I = 
∆V
850 W
I = 
12.0 V
I = 70.8 A
4.2 × 1010 J
4. P = 
1.1 × 103 h
R = 40.0 Ω
(∆V )2
P = 
R
∆V = P
R
∆V =
.2 × 10 J 1 h
  (40.0 Ω)
4
3600s
1
.1×10
h 10
3
∆V = 6.5 × 102 V
Section Two—Problem Workbook Solutions
II Ch. 19–3
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Givens
Solutions
5. P = 6.0 × 1013 W
6
∆V = 8.0 × 10 V
(∆V )2
P = 
R
(∆V )2
R = 
P
(8.0 × 106 V)2
R = 
(6.0 × 1013 W)
R = 1.1 Ω
6. I = 6.40 × 103 A
∆V = 4.70 × 103 V
P = I∆V
P = (6.40 × 103 A)(4.70 × 103 V)
P = 3.01 × 107 W = 30.1 MW
Additional Practice 19D
1. P = 8.8 × 106 kW
II
total cost = $1.0 × 106
cost of energy =
$0.081/kW• h
total cost of electricity = P∆t (cost of energy)
total cost of electricity
∆t = 
P(cost of energy)
$1.0 × 106
∆t = 
6
(8.8 × 10 kW)($0.081/kW• h)
∆t = 1.4 h
cost of energy =
$0.120/kW • h
purchase power = $18 000
purchase power
energy that can be purchased =  = P∆t
cost of energy
purchase power
∆t = 
(cost of energy)(P)
$18 000
∆t = 
($0.120/kW • h)(104 kW)
∆t = 1.4 × 103 h = 6.0 × 101 days
3. ∆t = 1.0 × 104 h
cost of energy =
$0.086/kW • h
total cost = $23
total cost of electricity = P∆t (cost of energy)
total cost of electricity
P = 
∆t(cost of energy)
$23
P = 
(1.0 × 104 h)($0.086 kW• h)
P = 2.7 × 10−2 kW
II Ch. 19–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. P = 104 kW
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Givens
4. ∆V = 110 V
R = 80.0 Ω (for maximum
power)
∆t = 24 h
cost of energy =
$0.086/kW • h
Solutions
(∆V)2
P = 
R
total cost of electricity = P∆t(cost of energy)
(∆V)2(∆t)
total cost =  (cost of energy)
R
(110 V)2(24 h) $0.086 1 kW
total cost =   
1 kW•h 1000 W
(80.0 Ω)
total cost = $0.31
5. 15.5 percent of solar energy
converted to electricity
cost of energy =
$0.080/kW • h
purchase power = $1000.00
purchase power
(0.155)Esolar = 
cost of energy
($1000.00)
Esolar = 
(0.155)($0.080/kW • h)
Esolar = 8.1 × 104 kW • h = 2.9 × 1011 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
II
Section Two—Problem Workbook Solutions
II Ch. 19–5
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Circuits and Circuit Elements
Chapter 20
Additional Practice 20A
Givens
Solutions
1. R = 160 kΩ
R1 = 2.0R
Req = R1 + R2 + R3 = 2.0R + 3.0R + 7.5R = 12.5R
Req = (12.5)(160 kΩ) = 2.0 × 103 kΩ
R2 = 3.0R
R3 = 7.5R
2. R = 5.0 × 108 Ω
1
R1 = 3 R
R2 =
2
 R
7
1
2
1
Req = R1 + R2 + R3 = 3 R + 7 R + 5 R
35 + 30 + 21
86
86
Req =  R = R =  (5.0 × 108 Ω) = 4.1 × 108 Ω
105
105
105
II
1
R3 = 5 R
3. R1 = 16 kΩ
R2 = 22 kΩ
R4 = Req − R1 − R2 − R3 = 82 kΩ − 16 kΩ − 22 kΩ − 32 kΩ = 12 kΩ
R3 = 32 kΩ
Req = 82 kΩ
4. R1 = 3.0 kΩ
R2 = 4.0 kΩ
Copyright © by Holt, Rinehart and Winston. All rights reserved.
R3 = 5.0 kΩ
P = (0.0100)(3.2 MW) =
0.032 MW
5. R1 = 4.5 Ω
R2 = 4.0 Ω
R3 = 16.0 Ω
6. R1 = 2.20 × 102 Ω
∆Vi = 1.20 × 102 V
∆Vf = 138 V
Req = R1 + R2 + R3 = 3.0 kΩ + 4.0 kΩ + 5.0 kΩ = 12.0 kΩ
(∆V )2
P = 
R
∆V = P
04
W)(
04
Ω) = 2.0 × 104 V
R
eq
= (3
.2
×1
1.
20
×1
R12 = R1 + R2 = 4.5 Ω + 4.0 Ω = 8.5 Ω
R13 = R1 + R3 = 4.5 Ω + 16.0 Ω = 20.5 Ω
R23 = R2 + R3 = 4.0 Ω + 16.0 Ω = 20.0 Ω
Because the current is unchanged, the following relationship can be written.
Vf
Vi
 = 
R1 R1 + R2
Vf R1 − Vi R1 (138 V)(220 Ω) − (120 V)(220 Ω)
R2 =  = 
Vi
120 V
30 400 V • Ω − 26 400 V • Ω 4000 V• Ω
R2 =  =  = 33 Ω
120 V
120 V
Section Two—Problem Workbook Solutions
II Ch. 20–1
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Givens
Solutions
7. R1 = 3.6 × 10−5 Ω
R2 = 8.4 × 10− 6 Ω
Req = R1 + R2 = 3.6 × 10−5 Ω + 8.4 × 10−6 Ω = 4.4 × 10−5 Ω
P = I 2Req = (280 A)2(4.4 × 10−5 Ω) = 3.4 W
I = 280 A
Additional Practice 20B
R2 = 5.0 Ω
R3 = 32 Ω
2. R = 450 Ω
R1 = R
R2 = 2.0R
R3 = 0.50R
II
3. R1 = 2.48 × 10−2 Ω
Req = 6.00 × 10−3 Ω
−1
−1
R4 = 11R
Req = 6.38 × 10−2 Ω
5. ratio = 1.22 × 10−2 Ω/m
l = 1813 km
1
 R
4
−1
1
−1
−1
−1
−1
−2
−1
= R + 3R + 7R + 11R
362
231 + 77 + 33 + 21
1.57
=  =  = 
231R 231R R 1
1
1
−1
1
−1
R = 1.57Req = 1.57(6.38 × 10−2 Ω) = 0.100 Ω
a. R = (ratio)(l ) = (1.22 × 10−2 Ω/m)(1.813 × 106 m) = 2.21 × 104 Ω
−1
−1
−1
31
= 
1.00 × 1010 Ω
= 3.23 × 108 Ω
(∆V)2
P = 
R
(∆V)2 (14.4 V)2
R =  =  = 0.922 Ω
P
225 W
−1
4
Req = 
R
R 0.922 Ω
=  =  = 0.230 Ω
4
4
∆V
14.4 V
I =  =  = 62.6 A
Req 0.230 Ω
Holt Physics Solution Manual
−1
2 4 5 20
=  +  +  + 
R R R R
1
II Ch. 20–2
= 1.3 × 102 Ω
= 0.012 Ω
R4 = 20 R
P = 225 W
−1
1
= 0.0078 
Ω
2
1
= 86 
Ω
6. ∆V = 14.4 V
−1
−3
31
Req = 
R
1
R3 = 5 R
−1
 − 
= 
6.00 × 10 Ω 2.48 × 10 Ω
1
2
R2 =  − 
Req R1
Req
= 1.3 Ω
1
1
1
Req = 0.0022  + 0.0011  + 0.0045 
Ω
Ω
Ω
1
1
1
1
b. Req =  +  +  + 
R1 R2 R3 R4
1
R1 = 2 R
−1
−1
R3 = 7R
1
1
1
1
=  +  + 
450 Ω 900 Ω 220 Ω
1
1
1
1
Req =  +  +  + 
R1 R2 R3 R4
R2 = 3R
1
1
1
1
Req =  +  + 
R1 R2 R3
R2 =
1
−1
Req
1
1
R2 = 167  − 80.6 
Ω
Ω
4. R1 = R
−1
= 1.8Ω + 5.0Ω + 32Ω
1
1
1
1
= 0.55  + 0.20  + 0.031  = 0.78 
Ω Ω Ω Ω
1
1
1
Req =  +  + 
R1 R2 R3
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. R1 = 1.8 Ω
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Givens
Solutions
7. L = 3.22 × 105 km
l
3
= 1.00 × 10 km
−2
ratio = 1.0 × 10
∆V = 1.50 V
Ω/m
1
Req = N 
R
L
where N =  and R = (ratio)l
l
L
Req = 2
(ratio)l
−1
3.22 × 108 m
= 
(1.0 × 10−2 Ω/m)(1.00 × 106 m)2
−1
= 31 Ω
∆V 1.50 V
I =  =  = 0.048 A
Req
31 Ω
Additional Practice 20C
1. R1 = 6.60 × 102 Ω
R2 = 2.40 × 102 Ω
R3 = 2.00 × 102 Ω
R4 = 2.00 × 102 Ω
R12 = R1 + R2 = 660 Ω + 240 Ω = 900 Ω
−1
−1
= 900Ω + 200Ω
1
1
1
= 0.00111  + 0.00500  = 0.00611 
Ω
Ω Ω
1
1
R123 =  + 
R12 R3
1
1
−1
R123
−1
= 164 Ω
Req = R123 + R4 = 164 Ω + 200 Ω = 364 Ω
2. ∆V = 24 V
R1 = 2.0 Ω
R2 = 4.0 Ω
R3 = 6.0 Ω
R4 = 3.0 Ω
R12 = R1 + R2 = 2.0 Ω + 4.0 Ω = 6.0 Ω
−1
−1
1
1
1
= 0.17  + 0.33  = 0.50 
Ω Ω Ω
1
1
1
1
=  +  =  + 
R R 6.0 Ω 2.0 Ω
1
1
1
= 0.17  + 0.50  = 0.67 
Ω Ω Ω
1
1
R34 =  + 
R3 R4
1
1
=  + 
6.0 Ω 3.0 Ω
−1
R34
−1
−1
Req
12
= 2.0 Ω
−1
34
−1
Req
Copyright © by Holt, Rinehart and Winston. All rights reserved.
II
−1
= 1.5 Ω
∆V 24 V
I =  =  = 16 A
Req 1.5 V
3. R1 = 2.5 Ω
R2 = 3.5 Ω
R3 = 3.0 Ω
R4 = 4.0 Ω
R5 = 1.0 Ω
∆V = 12 V
R12 = R1 + R2 = 2.5 Ω + 3.5 Ω = 6.0 Ω
−1
−1
1
1
1
= 0.17  + 0.33  = 0.50  = 2.0 Ω
Ω Ω Ω
1
1
1
1
=  +  =  + 
R R 4.0 Ω 1.0 Ω
1
1
1
= 0.25  + 1.0  = 1.2  = 0.83 Ω
Ω Ω Ω
1
1
R123 =  + 
R12 R3
1
1
=  + 
6.0 Ω 3.0 Ω
−1
−1
R123
−1
R45
4
−1
5
−1
R45
−1
Req = R123 + R45 = 2.0 Ω + 0.83 Ω = 2.8 Ω
∆V 12 V
I =  =  = 4.3 A
2.8 Ω
R
Section Two—Problem Workbook Solutions
II Ch. 20–3
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Givens
Solutions
4. ∆V = 1.00 × 103 V
R1 = 1.5 Ω
R2 = 3.0 Ω
R3 = 1.0 Ω
−1
1
1
−1
R12
−1
= 1.5Ω + 3.0Ω
1
1
1
= 0.67  + 0.33  = 1.00 
Ω Ω Ω
1
1
R12 =  + 
R1 R2
−1
= 1.00 Ω
Req = R12 + R3 = 1.00 Ω + 1.0 Ω = 2.0 Ω
(∆V )2 (1.00 × 103 V)2
P =  =  = 5.0 × 105 W
Req
2.0 Ω
5. ∆V = 2.00 × 103 V
I = 1.0 × 10−8 A
R1 = r
R12 = R1 + R2 = r + 3r = 4r
R34 = R3 + R4 = 2r + 4r = 6r
R2 = 3r
R4 = 4r
−1
5
3+2
=  = 
12r 12r
1
1
Req =  + 
R12 R34
R3 = 2r
II
∆V 2.00 × 103 V
Req =  = 
= 2.0 × 1011 Ω
I
1.0 × 10−8 A
−1
Req
−1
−1
1
1
=  + 
4r 6r
12
=  r
5
5
5
r = Req = (2.0 × 1011 Ω) = 8.3 × 1010 Ω
12
12
6. P = 6.0 × 105 W
∆V = 220 V
(∆V)2
(220 V)2
R =  = 
= 8.1 × 10−2 Ω
P
6.0 × 105 W
R12 = R45 = 2R = 2(0.081 Ω) = 0.16 Ω
−1
R12345
−1
= 0.042 Ω
Req = R12345 + R6 = 0.042 Ω + 0.081 Ω = 0.123 Ω
(∆V)2 (220 V)2
P =  =  = 3.9 × 105 W
Req
0.123 Ω
Additional Practice 20D
1. R = 8.1 × 10−2 Ω
Req = 0.123 Ω
∆V
220 V
a. I =  =  = 1800 A
Req 0.123 Ω
∆V = 220 V
∆V12345 = IR12345 = (1800 A)(0.042 Ω) = 76 V
R12 = R45 = 0.16 Ω
∆V3 = ∆V12345 = 76 V
R12345 = 0.042 Ω
∆V
76 V
I3 = 3 = 
= 9.4 × 102 A
8.1 × 10−2 Ω
R3
II Ch. 20–4
Holt Physics Solutions Manual
−1
1
1
1
=  +  + 
0.16 Ω 0.081 Ω 0.16 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
−1
1
1
1
1
= 6.2  + 12  + 6.2  = 24 
Ω Ω Ω Ω
1
1
1
R12345 =  +  + 
R12 R3 R45
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Givens
Solutions
b. ∆V12 = ∆V12345 = 76 V
∆V12
76 V
I12 = 
=  = 4.8 × 102 A
R12
0.16 Ω
I2 = I12 = 4.8 × 102 A
∆V2 = I2R2 = (4.8 × 102 A)(8.1 × 10−2 Ω) = 39 V
c. Same as part b:
I4 = 4.8 × 102 A
∆V4 = 39 V
2. ∆V = 12 V
R1 = 2.5 Ω
R3 = 3.0 Ω
R4 = 4.0 Ω
R5 = 1.0 Ω
R12 = 6.0 Ω
a. ∆V45 = IR45 = (4.3 A)(0.83 Ω) = 3.6 V
∆V5 = ∆V45 = 3.6 V
∆V
3.6 V
I5 = 5 =  = 3.1 A
R5
1.0 Ω
b. ∆V123 = IR123 = (4.3 A)(2.0 Ω) = 8.6 V
R123 = 2.0 Ω
∆V12 = ∆V123 = 8.6 V
R45 = 0.83 Ω
∆V12 8.6 V
I1 = I12 = 
=  = 1.4 A
R12
6.0 Ω
Req = 2.8 Ω
I = 4.3 A
II
∆V1 = I1R1 = (1.4 A)(2.5 Ω) = 3.5 V
c. I45 = I = 4.3 A
∆V45 = I45 R45 = (4.3 A)(0.83 Ω) = 3.6 V
Copyright © by Holt, Rinehart and Winston. All rights reserved.
V4 = ∆V45 = 3.6 V
∆V
3.6 V
I4 = 4 =  = 0.90 V
R4
4.0 Ω
d. ∆V3 = ∆V123 = 8.6 V
∆V
8.6 V
I3 = 3 =  = 2.9 A
R3
3.0 Ω
Section Two—Problem Workbook Solutions
II Ch. 20–5
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Givens
Solutions
3. R1 = 15 Ω
R2 = 3.0 Ω
R23 = R2 + R3 = 3.0 Ω + 2.0 Ω = 5.0 Ω
R4 = 5.0 Ω
R234
R5 = 7.0 Ω
R6 = 3.0 Ω
R7 = 3.0 × 101 Ω
∆V = 2.00 × 103 V
−1
1
= 0.40  = 2.5 Ω
Ω
R3 = 2.0 Ω
1
1
R234 =  + 
R23 R4
−1
1
1
=  + 
5.0 Ω 5.0 Ω
−1
R56 = R5 + R6 = 7.0 Ω + 3.0 Ω = 10.0 Ω
−1
1
1
=  + 
10.0 Ω 30 Ω
−1
R567
−1
1
1
1
= 0.100  + 0.033  = 0.133 
Ω Ω Ω
1
1
R567 =  + 
R56 R7
−1
= 7.52 Ω
Req = R1 + R234 + R567 = 15 Ω + 2.5 Ω + 7.52 Ω = 25 Ω
∆V 2.00 × 103 V
a. I =  =  = 80 A
Req
25 Ω
∆V234 = IR234 = (80 A)(2.5 Ω) = 2.0 × 102 V
II
∆V4 = ∆V234 = 2.0 × 102 V
∆V
200 V
I4 = 4 =  = 4.0 × 101 A
R4
5.0 Ω
b. ∆V23 = ∆V234 = 200 V
∆V23 200 V
I23 = 
=  = 40 A
R23
5.0 Ω
I3 = I23 = 4.0 × 101 A
∆V3 = I3R3 = (40 A)(2.0 Ω) = 8.0 × 101 V
V567 = I567R567 = (80 A)(7.52 Ω) = 600 V
∆V56 = ∆V567 = 600 V
∆V56 600 V
I56 = 
=  = 60 A
10.0 Ω
R56
I5 = I56 = 6.0 × 101 A
∆V5 = I5R5 = (60 A)(7.0 Ω) = 4.2 × 102 V
d. ∆V7 = ∆V567 = 6.0 × 102 V
∆V
600 V
I7 = 7 =  = 2.0 × 101 A
R7
30 Ω
II Ch. 20–6
Holt Physics Solutions Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
c. I567 = I = 80 A
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Magnetism
Chapter 21
Additional Practice 21A
Givens
Solutions
1. B = 45 T
Fmagnetic = qvB
6
v = 7.5 × 10 m/s
Fmagnetic = (1.60 × 10−19 C)(7.5 × 106 m/s)(45 T)
q = e = 1.60 × 10−19 C
Fmagnetic = 5.4 × 10−11 N
me = 9.109 × 10−31 kg
2. q = 12 × 10−9 C
Fmagnetic = qvB
B = 2.4 T
−6
Fmagnetic = 3.6 × 10 N
3. v = 350 km/h
−8
II
Fmagnetic = qvB = q[v (sin q)]B
B = 7.0 × 10−5 T
1h
103 m
Fmagnetic = (3.6 × 10−8 C)(350 km/h)   (sin 30.0°)(7.0 × 10−5 T)
3600 s 1 km
q = 30.0°
Fmagnetic = 1.2 × 10−10 N
q = 3.6 × 10
C
Fmagnetic = qvB
4. v = 2.60 × 102 km/h
−17
Fmagnetic = 3.0 × 10
−19
q = 1.60 × 10
Copyright © Holt, Rinehart and Winston. All rights reserved.
1h
103 m
Fmagnetic = (12 × 10−9 C)(450 km/h)   (2.4 T)
3600 s 1 km
v = 450 km/h
C
N
Fmagnetic
B = 
qv
(3.0 × 10−17 N)
B = 
1h
103 m
(1.60 × 10−19 C)(2.60 × 102 km/h)  
3600 s 1 km
B = 2.6 T
Section Two—Problem Workbook Solutions
II Ch. 21–1
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Givens
Solutions
5. q = 1.60 × 10−19 C
Fmagnetic = qvB
v = 60.0 km/h
−22
Fmagnetic = 2.0 × 10
N
Fmagnetic
B = 
qv
(2.0 × 10−22 N)
B = 
1h
103 m
(1.60 × 10−19 C)(60.0 km/h)  
3600 s 1 km
−5
B = 7.5 × 10
6. q = 88 × 10−9 C
T
Fmagnetic = qvB
B = 0.32 T
Fmagnetic = 1.25 × 10−6 N
Fmagnetic
v = 
qB
(1.25 × 10−6 N)
v = 
(88 × 10−9 C)(0.32 T)
v = 44 m/s = 160 km/h
II
7. q = 1.60 × 10−19 C
B = 6.4 T
Fmagnetic = 2.76 × 10−16 N
a. Fmagnetic = qvB
Fmagnetic
v = 
qB
(2.76 × 10−16 N)
v = 
(1.60 × 10−19 C)(6.4 T)
v = 2.7 × 102 m/s = 9.7 × 102 km/h
(vf + vi )
b. ∆x =  ∆t
2
∆x = 4.0 × 103 m
vi = 0 m/s
2(4.0 × 103 m)
∆t = 
(270 m/s + 0 m/s)
∆t = 3.0 × 101 s
8. B = 0.600 T
a. Fmagnetic = qvB
−19
q = 1.60 × 10
C
v = 2.00 × 105 m/s
m1 = 9.98 × 10−27 kg
m2 = 11.6 × 10−27 kg
II Ch. 21–2
Fmagnetic = (1.60 × 10−19 C)(2.00 × 105 m/s)(0.600 T)
Fmagnetic = 1.92 × 10−14 N
m v2
b. Fc,1 = 1 = Fmagnetic
r1
m2 v 2
Fc,2 =  = Fmagnetic
r2
m1v 2
r1 = 
Fmagnetic
m v2
r2 = 2
Fmagnetic
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
2∆x
∆t = 
(vf + vi )
vf = 270 m/s
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Givens
Solutions
(9.98 × 10−27 kg)(2.00 × 105 m/s)2
r1 = 
(1.92 × 10−14 N)
r1 = 2.08 × 10−2 m
(11.6 × 10−27 kg)(2.00 × 105 m/s)2
r2 = 
(1.92 × 10−14 N)
r2 = 2.42 × 10−2 m
r2 − r1 = 3.40 × 10−3 m = 3.4 mm
Additional Practice 21B
1. B = 22.5 T
l
Fmagnetic = BI l
−2
= 12 × 10
Fmagnetic = (22.5 T)(8.4 × 10−2 A)(12 × 10−2 m)
m
I = 8.4 × 10−2 A
2.
Fmagnetic = 0.23 N
l = 1066 m
Fmagnetic = BI l
−2
Fmagnetic = 6.3 × 10
N
I = 0.80 A
II
Fmagnetic
B= 
Il
(6.3 × 10−2 N)
B = 
(0.80 A)(1066 m)
B = 7.4 × 10−5 T
3.
l
= 5376 m
Fmagnetic = BI l = [B(sin q )]I l
Fmagnetic = 3.1 N
Fmagnetic
B= 
I l (sin q)
I = 12 A
Copyright © Holt, Rinehart and Winston. All rights reserved.
q = 38°
(3.1 N)
B = 
(12 A)(5376 m)(sin 38.0°)
B = 7.8 × 10−5 T
4.
l
Fmagnetic = BI l
= 21.0 × 103 m
B = 6.40 × 10−7 T
−2
Fmagnetic = 1.80 × 10
N
Fmagnetic
I= 
Bl
(1.80 × 10−2 N)
I = 
(6.40 × 10−7 T)(21.0 × 103 m)
I = 1.34 A
Section Two—Problem Workbook Solutions
II Ch. 21–3
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Givens
Solutions
5. B = 2.5 × 10−4 T
l
−2
= 4.5 × 10
Fmagnetic = BI l
m
−7
Fmagnetic = 3.6 × 10
N
Fmagnetic
I= 
Bl
(3.6 × 10−7 N)
I = 
(2.5 × 10−4 T)(4.5 × 10−2 m)
I = 3.2 ⫻ 10–2 A
6. Fmagnetic = 5.0 × 105 N
B = 3.8 T
I = 2.00 × 102 A
7. Fmagnetic = 16.1 N
−5
B = 6.4 × 10
T
I = 2.8 A
8. B = 0.040 T
Fmagnetic
= 
BI
l
(5.0 × 105 N)
= 
(3.8 T)(2.00 × 102 A)
l
= 6.6 × 102 m
Fmagnetic = BIl
l
Fmagnetic
= 
BI
l
(16.1 N)
= 
(6.4 × 10−5 T)(2.8 A)
l
= 9.0 × 104 m
Fmagnetic = BI l = [B(sin q)]Il
I = 0.10 A
Fmagnetic = (0.040 T)(sin 45°)(0.10 A)(0.55 m)
q = 45°
l = 55 cm ⫽ 0.55 m
9. B = 38 T
l
l
Fmagnetic = 1.6 × 10−3 Ν
Fmagnetic = BIl
= 2.0 m
Fg = mg
m = 75 kg
Fmagnetic = Fg
2
g = 9.81 m/s
BIl = mg
mg
I= 
Bl
(75 kg)(9.81 m/s2)
I = 
(38 T)(2.0 m)
I = 9.7 A
II Ch. 21–4
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
II
Fmagnetic = BI l
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Givens
10.
Solutions
l = 478 × 103 m
Fmagnetic = BIl
Fmagnetic = 0.40 N
Fmagnetic
I= 
Bl
−5
B = 7.50 × 10
T
(0.40 N)
I = 
(7.50 × 10−5 T)(478 × 103 m)
I = 1.1 × 10−2 A
Copyright © Holt, Rinehart and Winston. All rights reserved.
II
Section Two—Problem Workbook Solutions
II Ch. 21–5
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Induction and Alternating Current
Chapter 22
Additional Practice 22A
Givens
Solutions
1. Ai = 6.04 × 105 m2
Af =
1
 (6.04
2
5
2
× 10 m )
B = 6.0 × 10−5 T
emf = 0.80 V
N = 1 turn
q = 0.0°
−N∆[AB(cos q)]
emf = 
∆t
−NB(cos q)
∆t =  ∆A
emf
−NB(cos q)
∆t =  (Af − Ai )
emf
−(1)(6.0 × 10−5 T)(cos 0.0°)
1
∆t =  (6.04 × 105 m2)2 − 1
(0.80 V)
II
∆t = 23 s
100.0 m
2. r =  = 50.0 m
2
Bi = 0.800 T
Bf = 0.000 T
q = 0.00°
emf = 46.7 V
N = 1 turn
Copyright © Holt, Rinehart and Winston. All rights reserved.
3. emf = 32.0 × 106 V
Bi = 1.00 × 103 T
Bf = 0.00 T
A = 4.00 × 10−2 m2
N = 50 turns
q = 0.00°
−N∆[AB (cos q)]
emf = 
∆t
−N(pr 2)(cos q)(Bf − Bi )
∆t = 
emf
−(1)(p)(50.0 m)2(cos 0.0°)(0.000 T − 0.800 T)
∆t = 
(46.7 V)
∆t = 135 s
−N∆[AB(cos q)]
emf = 
∆t
−NA(cos q)(Bf − Bi )
∆t = 
emf
−(50)(4.00 × 10−2 m2)(cos 0.00°)[(0.00 T) − (1.00 × 103 T)]
∆t = 
(32.0 × 106 V)
∆t = 6.3 × 10−5 s
4. Af = 3.2 × 104 m2
2
Ai = 0.0 m
∆t = 20.0 min
B = 4.0 × 10−2 T
N = 300 turns
q = 0.0°
−N∆[AB (cos q)] −NB(cos q)
emf =  =  (Af − Ai )
∆t
∆t
−(300)(4.0 × 10−2 T)(cos 0.0°)
emf =  [(3.2 × 104 m2) − (0.0 m2)]
60 s
(20.0 min) 
1 min
emf = −3.2 × 102 V
Section Two—Problem Workbook Solutions
II Ch. 22–1
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Givens
Solutions
5. Bi = 8.0 × 10−15 T
−14
Bf = 10 Bi = 8.0 × 10
−2
∆t = 3.0 × 10
s
2
A = 1.00 m
emf = −1.92 × 10−11 V
q = 0.0°
T
−N∆[AB (cos q)]
emf = 
∆t
−(emf )(∆t)
N = 
A(Bf − Bi )(cos q)
−(−1.92 × 10−11 V)(3.0 × 10−2 s)
N = 
(1.00 m2)[(8.0 × 10−14 T) − (8.0 × 10−15 T)](cos 0.0°)
N = 8 turns
−N∆[AB(cos q )]
∆B
emf =  = −NA(cos q) 
∆t
∆t
6. Bi = 0.50 T
Bf = 0.00 T
−(emf )(∆t )
A = 
N(cos q )(Bf − Bi )
N = 880 turns
∆t = 12 s
−(147 V)(12 s)
A = 
(880)(cos 0.0°)(0.00 T − 0.50 T)
emf = 147 V
q = 0.0°
A = 4.0 m2
II
Additional Practice 22B
1. f = 833 Hz
maximum emf = NABw = NAB(2pf )
D = 5.0 cm ⫽ 0.050 m
B = 8.0 × 10−2 T
maximum emf ⫽ 330 V
2
2
D
0.05 m
A ⫽ pr 2 ⫽ p  ⫽ p  ⫽ 2.0 ⫻ 10⫺3 m2
2
2
330 V
maximum emf
N ⫽  ⫽ 
–3 2
(2.0 ⫻ 10 m )(2p)(833 Hz)(8.0 ⫻ 10–2 T)
AB(2p f)
N = 4.0 × 102 turns
maximum emf = 214 V
−2
B = 8.00 × 10
T
2
A = 0.400 m
maximum emf
N = 
ABw
214 V
N = 
(0.400 m2)(0.0800 T)(335 rad/s)
N = 20.0 turns
19.3 m
3. r =  = 9.65 m
2
w = 0.52 rad /s
maximum emf = 2.5 V
N = 40 turns
maximum emf = NABw
maximum emf
B = 
N(p r 2)w
2.5 V
B = 
(40)(p)(9.65 m)2(0.52 rad/s)
B = 4.1 × 10−4 T
II Ch. 22–2
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
maximum emf = NABw
2. w = 335 rad/s
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Givens
Solutions
4. maximum emf =
8.00 × 103 V
maximum emf = NABw
maximum emf
B = 
NAw
N = 236
8.00 × 103 V
B = 
(236)(6.90 m)2(57.1 rad/s)
A = (6.90 m)2
w = 57.1 rad/s
B = 1.25 × 10−2 T
5. N = 1000 turns
−4
A = 8.0 × 10
−3
B = 2.4 × 10
maximum emf = NABw
2
m
T
maximum emf = 3.0 V
maximum emf
w = 
NAB
3.0 V
w = 
(1000)(8.0 × 10−4 m2)(2.4 × 10−3 T)
w = 1.6 × 103 rad/s
6. N = 640 turns
maximum emf = NABw
2
A = 0.127 m
maximum emf =
24.6 × 103 V
−2
B = 8.00 × 10
T
II
maximum emf
w = 
NAB
24.6 × 103 V
w = 
(640)(0.127 m2)(8.00 × 10−2 T)
w = 3.78 × 103 rad/s
7. f = 1.0 × 103 Hz
B = 0.22 T
maximum emf = (250)(p)(12 × 10−2 m)2(0.22 T)(2p)(1.0 × 103 Hz)
N = 250 turns
maximum emf = 1.6 × 104 V = 16 kV
r = 12 × 10−2 m
Copyright © Holt, Rinehart and Winston. All rights reserved.
maximum emf = NABw = NAB(2pf ) = N(pr 3)Bw = N(pr 2)B(2pf )
Additional Practice 22C
1. ∆Vrms = 120 V
R = 6.0 × 10−2 Ω
1
 = 0.707
2
∆Vrms
a. Irms = 
R
(120 V)
Irms = 
(6.0 × 10−2 Ω)
Irms = 2.0 × 103 A
b. Imax = (Irms) 2
(2.0 × 103 A)
Imax = 
(0.707)
Imax = 2.8 × 103 A
c. P = (Irms)(∆Vrms)
P = (2.0 × 103 A)(120 V)
P = 2.4 × 105 W
Section Two—Problem Workbook Solutions
II Ch. 22–3
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Givens
Solutions
2. P = 10.0(Acoustic power)
Acoustic power =
30.8 × 103 W
∆Vrms = 120.0 V
1
 = 0.707
2
P = ∆Vrms Irms
P
Irms = 
∆Vrms
Imax

Irms = 
2
Imax
P
 = 

2 ∆Vrms
P 2
Imax = 
Imax
∆Vrms
(10.0)(30.8 × 103 W)
= 
(120.0 V)(0.707)
Imax = 3.63 × 103 A
3. P = 1.325 × 108 W
∆Vrms = 5.4 × 104 V
II
1
 = 0.707
2
(∆V )2

P = ∆Vrms I rms = (Irms )2R = rms
R
Imax

Irms = 
2
2 P
Imax = 2 Irms = 
∆Vrms
1.325 × 108 W
Imax = 
(5.4 × 104 V)(0.707)
Imax = 3.5 × 103 A
(∆V )2

R = rms
P
(5.4 × 104 V)2
R = 
(1.325 × 108 W)
4. ∆Vrms = 1.024 × 106 V
Irms = 2.9 × 10−2 A
1
 = 0.707
2
∆Vmax = ∆Vrms
2
(1.024 × 106 V)
∆Vmax = 
(0.707)
∆Vmax = 1.45 × 106 V = 1.45 MV
Imax = Irms 2
(2.9 × 10−2 A)
Imax = 
(0.707)
Imax = 4.1 × 10−2 A
II Ch. 22–4
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
R = 22 Ω
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Givens
5. ∆Vmax = 320 V
Imax = 0.80 A
1
 = 0.707
2
Solutions
∆Vmax

∆Vrms = 
2
∆Vrms = (320 V)(0.707)
∆Vrms = 2.3 × 102 V
Imax

Irms = 
2
Irms = (0.80 A)(0.707)
Irms = 0.57 A
∆V ax ∆Vrms
 = 
R = m
Imax
Irms
(320 V)
(230 V)
R =  = 
(0.80 A) (0.57 A)
R = 4.0 × 102 Ω
6. Imax = 75 A
R = 480 Ω
1
 = 0.707
2
II
∆Vmax

∆Vrms = 
2
∆Vmax = (Imax )(R)
ImaxR

∆Vrms = 
2
∆Vrms = (75 A)(480 Ω)(0.707)
∆Vrms = 2.5 × 104 V = 25 kV
7. Ptot = 6.2 × 107 W
Ptot = 24 P
Copyright © Holt, Rinehart and Winston. All rights reserved.
R = 1.2 × 105 Ω
1
 = 0.707
2
P ot
P = (Irms )2R = t
24
6.2 × 107 W
P = 
24
P = 2.6 × 106 W = 2.6 MW
RP
(2.6 × 10 W)
= 
(1.2×1
0Ω)
Irms =
6
Irms
5
Irms = 4.7 A
Imax = 2 Irms
4.7 A
Imax = 
0.707
Imax = 6.6 A
Section Two—Problem Workbook Solutions
II Ch. 22–5
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Additional Practice 22D
Givens
Solutions
1. N1 = 5600 turns
N2 = 240 turns
∆V2 = 4.1 × 102 V
N
∆V1 = ∆V2 1
N2
5600
∆V1 = (4.1 × 103 V) ⫽ 
240
∆V1 = 9.6 × 104 V = 96 kV
2. N1 = 74 turns
74
∆V = (650 V) 
403
N
∆V1 = ∆V2 1
N2
N2 = 403 turns
∆V2 = 650 V
1
∆V1 = 120 V
II
3. ∆V1 = 2.0 × 10−2 V
N1 = 400 turns
N2 = 3600 turns
N
∆V2 = ∆V1 2
N1
3600
∆V2 = (2.0 × 10−2 V) 
400
∆V2 = 0.18 V
∆V2 = 2.0 × 10−2 V
N
∆V1 = ∆V2 1
N2
400
∆V1 = (2.0 × 10−2 V) 
3600
4. ∆V1 = 765 × 103 V
3
∆V2 = 540 × 10 V
N1 = 2.8 × 103 turns
∆V
N
2 = 2
∆V1 N1
540 × 10 V
N =  (2.8 × 10 )
765 × 10 V
∆V
N2 = 2 N1
∆V1
3
2
3
N2 = 2.0 × 103 turns
II Ch. 22–6
Holt Physics Solution Manual
3
Copyright © Holt, Rinehart and Winston. All rights reserved.
∆V1 = 2.2 × 10−3 V
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Givens
Solutions
5. ∆V1 = 230 × 103 V
3
∆V2 = 345 × 10 V
N1 = 1.2 × 104 turns
N
∆V
2 = 2
N1
∆V1
∆V
N2 = N1 2
∆V1
345 × 103 V
N2 = (1.2 × 104) 
230 × 103 V
N2 = 1.8 × 104 turns
6. P = 20.0 W
a. P = (∆V1)(I1)
∆V1 = 120 V
P
(20.0 W)
I1 =  = 
∆V1
(120 V)
I1 = 0.17 A
N1
 = 0.36
N2
∆V
N
b. 2 = 2
∆V1 N1
1
∆V =  (120 V)
0.36
II
N
∆V2 = 2 ∆V1
N1
2
∆V2 = 3.3 × 102 V
7. ∆V1 = 120 V
∆V2 = 220 V
∆V1I1 = ∆V2I2
I2 = 30.0 A
∆V
I1 = 2 I2
∆V1
N2 = 660 turns
Copyright © Holt, Rinehart and Winston. All rights reserved.
P1 = P2
220 V
I =  (30.0 A)
120 V
1
I1 = 55 A
N
∆V
2 = 2
N1 ∆V1
120 V
N = (660) 
220 V
∆V
N1 = N2 1
∆V2
1
N1 = 360 turns
Section Two—Problem Workbook Solutions
II Ch. 22–7
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Atomic Physics
Chapter 23
Additional Practice 23A
Givens
Solutions
C = 3.00 × 108 m/s
(6.63 × 10−34 J• s)(3.00 × 108 m/s)
hc
l =  = 
1.29 × 10−15 J
E
h = 6.63 × 10−34 J• s
l = 1.54 × 10−10 m = 0.154 nm
1. E = 1.29 × 10−15 J
C = 3.00 × 108 m/s
(6.63 × 10−34 J• s)(3.00 × 108 m/s)
hc
l =  = 
6.6 × 10−19 J
E
h = 6.63 × 10−34 J• s
l = 3.0 × 10−7 m
2. E = 6.6 × 10−19 J
II
3. E = 5.92 × 10−6 eV
C = 3.00 × 108 m/s
(6.63 × 10−34 J• s)(3.00 × 108 m/s)
hc
l =  = 
(5.92 × 10−6 eV)(1.60 × 10−19 J/eV)
E
h = 6.63 × 10−34 J• s
l = 0.210 m
4. E = 2.18 × 10−23 J
h = 6.63 × 10−34 J • s
E = hf
E
f = 
h
2.18 × 10−23 J
f = 
6.63 × 10−34 J • s
Copyright © Holt, Rinehart and Winston. All rights reserved.
f = 3.29 × 1010 Hz
5. E = 1.85 × 10−23 J
h = 6.63 × 10−34 J • s
6. f = 9 192 631 770 s−1
h = 6.626 0755 × 10−34 J • s
1 eV = 1.602 117 33 × 10−19 J
1.85 × 10−23 J
E
f =  = 
= 2.79 × 1010 Hz
h 6.63 × 10−34 J/s
E = hf
(6.626 0755 × 10−34 J • s)(9 192 631 770 s−1)
E = 
1.602 117 33 × 10−19 J/eV
E = 3.801 9108 × 10−5 eV
7. l = 92 cm = 92 × 10−2 m
c = 3.00 × 108 m/s
h = 6.63 × 10−34 J • s
h = 4.14 × 10−15 eV • s
c
f = 
l
3.00 × 108 m/s
f = 
92 × 10−2 m
f = 3.3 × 108 Hz = 330 MHz
E = hf
Section Two—Problem Workbook Solutions
II Ch. 23–1
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Givens
Solutions
E = (6.63 × 10−34 J • s)(3.3 × 108 Hz)
E = 2.2 × 10−25 J
E = (4.14 × 10−15 eV • s)(3.3 × 108 Hz)
E = 1.4 × 10−6 eV
8. v = 1.80 × 10−17 m/s
∆t = 1.00 year
365.25 days 24 h 3600 s
∆x = (1.80 × 10−17 m/s)(1.00 year)   
1 day 1 h
1 year
l = ∆x
8
c = 3.00 × 10 m/s
h
∆x = v∆t
= 6.63 × 10−34 J • s
∆x = 5.68 × 10−10 m
hc hc
E = hf =  = 
l ∆x
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
E = 
5.68 × 10−10 m
II
E = 3.50 × 10−16 J
Additional Practice 23B
[KEmax + hft] [3.8 eV + 4.5 eV]
ᎏ = 
= 2.0 × 1015 Hz
f=ᎏ
h
4.14 × 10−15 eV• s
1. hft = 4.5 eV
KEmax = 3.8 eV
h = 4.14 × 10−15 eV• s
KEmax = hf − hft
KEmax = 3.2 eV
−15
h = 4.14 × 10
eV • s
KEmax + hft

f = 
h
3.2 eV + 4.3 eV
f = 
4.14 × 10−15 eV • s
f = 1.8 × 1015 Hz
3. hft ,Cs = 2.14 eV
hft,Se = 5.9 eV
h = 4.14 × 10−15 eV • s
c = 3.00 × 108 m/s
KEmax = 0.0 eV for both
cases
hc
a. KEmax = hf − hft = 0.0 eV =  − hft
l
hc
l = 
hft
lCs
(4.14 × 10−15 eV • s)(3.00 × 108 m/s)
hc


=
= 
2.14 eV
hft,Cs
lCs = 5.80 × 10−7 m = 5.80 × 102 nm
(4.14 × 10−15 eV • s)(3.00 × 108 m/s)
hc
b. lSe =  = 
5.9 eV
hft,Se
lSe = 2.1 × 10−7 m = 2.1 × 102 nm
II Ch. 23–2
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
2. hft = 4.3 eV
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Givens
Solutions
4. l = 2.00 × 102 nm =
2.00 × 10−7 m
v = 6.50 × 105 m/s
me = 9.109 × 10−31 kg
c = 3.00 × 108 m/s
h = 4.14 × 10−15 eV • s
1
KEmax = 2 me v 2 = hf − hft
hc
=  − hft
l
hc 1
hft =  − 2 me v 2
l
1
 m v 2
2 e
(4.14 × 10−15 eV • s)(3.00 × 108 m/s) (0.5)(9.109 × 10−31 kg)(6.50 × 105 m/s)2
hft = 
− 
2.00 × 10−7 m
1.60 × 10−19 J/eV
hft = 6.21 eV − 1.20 eV
hft = 5.01 eV
5.01 eV
ft = 
= 1.21 × 1015 Hz
4.14 × 10−15 eV• s
5. f = 2.2 × 1015 Hz
hft = hf − KEmax
KEmax = 4.4 eV
−15
h = 4.14 × 10
KEmax = hf − hft
eV • s
hft = (4.14 × 10−15 eV • s)(2.2 × 1015 Hz) − 4.4 eV
II
hft = 9.1 eV − 4.4 eV = 4.7 eV
4.7 eV
ft = 
= 1.1 × 1015 Hz
4.14 × 10−15 eV • s
6. l = 2.00 × 102 nm =
2.00 × 10−7 m
KEmax = 0.46 eV
h = 4.14 × 10−15 eV • s
8
c = 3.00 × 10 m/s
KEmax = hf − hft
hc
hft = hf − KEmax =  − KEmax
l
(4.14 × 10−15 eV • s)(3.00 × 108 m/s)
hft = 
− 0.46 eV
2.00 × 10−7 m
hft = 6.21 eV − 0.46 eV
Copyright © Holt, Rinehart and Winston. All rights reserved.
hft = 5.8 eV
5.8 eV
ft = 
= 1.4 × 1015 Hz
4.14 × 10−15 eV•s
7. l = 589 nm = 589 × 10−9 m
hft = 2.3 eV
c = 3.00 × 108 m/s
h = 4.14 × 10−15 eV • s
hc
KEmax = hf − hft =  − hft
l
(4.14 × 10−15 eV • s)(3.00 × 108 m/s)
KEmax = 
− 2.3 eV
589 × 10−9 m
KEmax = 2.11 eV − 2.3 eV
KEmax = −0.2 eV
No. The photons in the light produced by sodium vapor need 0.2 eV more energy
to liberate photoelectrons from the solid sodium.
Section Two—Problem Workbook Solutions
II Ch. 23–3
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Givens
Solutions
8. hft = 2.3 eV
l = 410 nm = 4.1 × 10−7 m
h = 4.14 × 10−15 eV • s
c = 3.00 × 108 m/s
9. hft ,Zn = 4.3 eV
hc
KEmax =  − hft
l
(4.14 × 10−15 eV • s)(3.00 × 108 m/s)
KE = 
− 2.3 eV
4.1 × 10−7 m
KE = 3.03 eV − 2.3 eV = 0.7 eV
KEmax = hf − hft
hft ,Pb = 4.1 eV
KEmax ,Pb = hf − hft,Pb = (KEmax,Zn + hft,Zn) − hft,Pb
KEmax ,Zn = 0.0 eV
KEmax ,Pb = 2me v 2
me = 9.109 × 10−31 kg
1
1
m v 2
2 e
v=
= (KEmax,Zn + hft,Zn) − hft,Pb
2(KEmax ,Zn + hft,Zn − hft,Pb )

me
(2)(0.0 eV + 4.3 eV − 4.1 eV) 1.60 × 10−19 J


9.109 × 10−31 kg
1 eV
v=
v=
(2)(0.2 eV)
1.60 × 10 J
  = 3 × 10 m/s

1 eV
9
.109
×1
0 k
g
II
−19
−31
5
Additional Practice 23C
l = 3.0 × 10−32 m
h = 6.63 × 10−34 J• s
v = 64 m/s
h
= 6.63 × 10−34 J • s
6.63 × 10−34 J• s
h
= 6.9 × 10−3 kg
m =  = 
−32
l v (3.0 × 10 m)(3.2 m/s)
2. l = 6.4 × 10−11 m
h
mv = 
l
h
m = 
lv
6.63 × 10−34 J • s

m=
(6.4 × 10−11 m)(64 m/s)
m = 1.6 × 10−25 kg
3. q = (2)(1.60 × 10−19 C) =
3.20 × 10−19 C
∆V = 240 V
h = 6.63 × 10−34 J • s
l = 4.4 × 10−13 m
1
KE = q∆V = 2mv 2
2q∆V
m = 
V2
6.63 × 10−34 J • s
h
v =  = 
= 1.0 × 105 m/s
−13
−26
l m (4.4 × 10 m)(1.5 × 10 kg)
2(3.20 × 10−19 C)(240 V)
m = 
(1.0 × 105 m/s)2
m = 1.5 × 10−26 kg
II Ch. 23–4
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
1. v = 3.2 m/s
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Givens
4. l = 2.5 nm = 2.5 × 10−9 m
mn = 1.675 × 10−27 kg
h = 6.63 × 10−34 J • s
Solutions
h
mv = 
l
h
v = 
lmn
6.63 × 10−34 J • s
v = 
(2.5 × 10−9 m)(1.675 × 10−27 kg)
v = 1.6 × 102 m/s
5. m = 7.65 × 10−70 kg
l = 5.0 × 1032 m
h = 6.63 × 10−34 J • s
h
mv = 
l
h
v = 
lm
6.63 × 10−34 J • s
v = 
(5.0 × 1032 m)(7.65 × 10−70 kg)
v = 1.7 × 103 m/s
6. m = 1.6 g = 1.6 × 10−3 kg
v = 3.8 m/s
h = 6.63 × 10−34 J • s
II
h
mv = 
l
h
l = 
mv
6.63 × 10−34 J • s
l = 
(1.6 × 10−3 kg)(3.8 m/s)
l = 1.1 × 10−31 m
7. ∆x = 42 195 m
Copyright © Holt, Rinehart and Winston. All rights reserved.
∆t = 3 h 47 min = 227 min
m = 0.080 kg
h = 6.63 × 10−34 J • s
42 195 m
∆x
v =  =  = 3.10 m/s
∆t
60 s
(227 min) 
1 min
h
 = mv
l
h
l = 
mv
6.63 × 10−34 J • s
l = 
(0.080 kg)(3.10 m/s)
l = 2.7 × 10−33 m
Section Two—Problem Workbook Solutions
II Ch. 23–5
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Subatomic Physics
Chapter 25
Additional Practice 25A, p. 173
Givens
1. E = 610 TW • h
Solutions
9
6
E (610 × 10 kW • h)(3.6 × 10 J/kW • h)
a. ∆m = 2 = 
(3.00 × 108 m/s)2
c
∆m = 24 kg
atomic mass of 12 H =
2.014 102 u
24 kg
∆m
= 
b. N = 
−27
atomic mass of 12 H (1.66 × 10 kg/u)(2.014 102 u)
N = 7.2 × 1027 12 H nuclei
II
56
Fe =
atomic mass of 26
55.934 940 u
24 kg
∆m
c. N = 
= 
−27 kg/u)(55.934 940 u)
56
(1.66
×
10
atomic mass of 26 Fe
N = 2.6 × 1026
atomic mass of 226
88 Ra =
226.025 402 u
24 kg
∆m
d. N = 
= 
−27 kg/u)(226.025 402 u)
(1.66
×
10
Ra
atomic mass of 226
88
HRW material copyrighted under notice appearing earlier in this book.
N = 6.4 × 1025
2. m = 4.1 × 107 kg
h = 10.0 cm
Z = 26
N = 56 − 26 = 30
mH = 1.007 825 u
mn = 1.008 665 u
56
atomic mass of 26
Fe =
55.934 940 u
56
26 Fe nuclei
226
88 Ra nuclei
a. E = mgh = (4.1 × 107 kg)(9.81 m/s2)(0.100 m)
E = 4.0 × 107 J
(4.0 × 107 J)(1 × 10−6 MeV/eV)
= 2.5 × 1020 MeV
b. Etot = 
(1.60 × 10−19 J/eV)
2.5 × 1020 MeV
∆mtot =  = 2.7 × 1017 u
931.50 MeV/u
56
∆m = Z(atomic mass of H) + Nmn − atomic mass of 26
Fe
∆m = 26(1.007 825 u) + 30(1.008 665 u) − 55.934 940 u
∆m = 0.528 46 u
MeV
Ebind = (0.528 46 u) 931.50 
u
Ebind = 492.26 MeV
2.5 × 1020 MeV
E ot
N = t
=  = 5.1 × 1017 reactions
492.26 MeV
Ebind
kg
mtot = (5.1 × 1017)(55.934 940 u) 1.66 × 10−27  = 4.7 × 10−8 kg
u
Section Two—Problem Workbook Solutions
II Ch. 25–1
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Givens
Solutions
3. E = 2.0 × 103 TW • h =
2.0 × 1015 W • h
atomic mass of 235
92 U =
235.043 924 u
atomic mass of H =
1.007 825 u
mn = 1.008 665 u
Z = 92
N = 235 − 92 = 143
∆m = Z(atomic mass of H) + Nmn − atomic mass of 235
92U
∆m = 92(1.007 825 u) + 143(1.008 665 u) − 235.043 924 u = 1.915 071 u
MeV
Ebind = (1.915 071 u) 931.50  = 1.7839 × 103 MeV = 1.7839 × 109 eV
u
J
Ebind = (1.7839 × 109 eV) 1.60 × 10−19  = 2.85 × 10−10 J
eV
3.60 × 103 s
E = (2.0 × 1015 W • h)  = 7.2 × 1018 J
h
E
7.2 × 1018 J
N =  = 
= 2.5 × 1028 reactions
Ebind 2.85 × 10−10 J
kg
mtot = (2.5 × 1028)(235.043 924 u) 1.66 × 10−27  = 9.8 × 103 kg
u
4. E = 2.1 × 1019 J
∆m = Z(atomic mass of H) + Nmn − atomic mass of 126C
atomic mass of 126C =
12.000 000 u
II
atomic mass of H =
1.007 825 u
mn = 1.008 665 u
∆m = 6(1.007 825 u) + 6(1.008 665 u) − 12.000 000 u
∆m = 9.894 × 10−2 u
J
= (92.16 × 10 eV) 1.60 × 10  = 1.47 × 10
eV
MeV
Ebind = (9.894 × 10−2 u) 931.50  = 92.16 MeV
u
−19
6
−11
Z =6
Ebind
N = 12 − 6 = 6
E
2.1 × 1019 J
= 1.4 × 1030 reactions
N =  = 
Ebind 1.47 × 10−11 J
J
kg
mtot = (1.4 × 1030)(12.000 000 u) 1.66 × 10−27  = 2.8 × 104 kg
u
Z =2
∆m = Z(atomic mass of H) + Nmn − atomic mass of 42He
∆m = (2)(1.007 825 u) + (2)(1.008 665 u) − 4.002 602 u
N =4−2=2
atomic mass of
4.002 602 u
∆m = 0.030 378 u
4
2He =
atomic mass of H =
1.007 825 u
mn =1.008 665 u
6. P = 42 MW = 42 × 106 W
atomic mass of 147N =
14.003 074 u
atomic mass of H =
1.007 825 u
mn = 1.008 665 u
Z =7
N = 14 − 7 = 7
∆t = 24 h
E = (0.030 378 u)(931.50 MeV/u)
E = 28.297 MeV
(3.9 × 1026 J/s)(1 × 10−6 MeV/eV)
N P
 = tot
 = 
(1.60 × 10−19 J/eV)(28.297 MeV)
∆t
E
N
 = 8.6 × 1037 reactions/s
∆t
∆m = Z(atomic mass of H) + Nmn − atomic mass of 147N
∆m = 7(1.007 825 u) + 7(1.008 665 u) − 14.003 074 u
∆m = 0.112 356 u
MeV
Ebind = (0.112 356 u) 931.50  = 104.66 MeV
u
J
Ebind = (104.66 × 106 eV) 1.60 × 10−19  = 1.67 × 10−11 J
eV
(42 × 106 W)(24 h)(3600 s/h)
P∆t
N =  = 
= 2.2 × 1023 reactions
1.67 × 10−11 J
Ebind
kg
mtot = (2.2 × 1023)(14.003 074 u) 1.66 × 10−27  = 5.1 × 10−3 kg = 5.1 g
u
II Ch. 25–2
Holt Physics Solution Manual
HRW material copyrighted under notice appearing earlier in this book.
5. Ptot = 3.9 × 1026 J/s
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Givens
Solutions
7. P = 3.84 × 107 W
∆m = Z(atomic mass of H) + Nmn − atomic mass of 126 C
atomic mass of 126 C =
12.000 000 u
∆m = 6(1.007 825 u) + 6(1.008 665 u) − 12.000 000 u
∆m = 9.894 × 10−2 u
atomic mass of H =
1.007 825 u
J
eV
Ebind = (9.894 × 10−2 u) 931.50 × 106  1.60 × 10−19 
eV
u
mn = 1.008 665 u
−11
Ebind = 1.47 × 10
Z =6
J
N
3.84 × 107 W
P
 =  = 
= 2.61 × 1018 reactions/s
∆t Ebind 1.47 × 10−11 J
N = 12 − 6 = 6
mtot
kg
 = (2.61 × 1018 s−1)(12.000 000 u) 1.66 × 10−27  = 5.20 × 10−8 kg/s
∆t
u
Additional Practice 25B, pp. 174–175
1.
238
1
92 U + 0 n
mass number of X = 238 + 1 = 239
→X
X → 939
93Np +
0
−1e
atomic number of X = 92 + 0 = 92 (uranium)
+ v
239
239
0
93Np → 94 Pu + −1e
+ v
X = 239
92 U
II
The equations are as follows:
238
1
92 U + 0 n
→ 239
92 U
239
939
0
92 U → 93Np + −1e
+ v
239
239
0
93Np → 94 Pu + −1e
mass number of Z = 212 + 0 = 212
2. X → Y + 42 He
atomic number of Z = 83 − 1 = 82 (lead)
HRW material copyrighted under notice appearing earlier in this book.
Y → Z + 42 He
Z → 212
83 Bi +
+ v
Z = 212
82Pb
0
−1e
+ v
mass number of Y = 212 + 4 = 216
atomic number of Y = 82 + 2 = 84 (polonium)
Y = 216
84 Po
mass number of X = 216 + 4 = 220
atomic number of X = 84 + 2 = 86 (radon)
X = 220
86 Rn
The equations are as follows:
220
216
4
86 Rn → 84 Po + 2 He
216
212
4
84 Po → 82 Pb + 2 He
212
212
0
82 Pb → 83 Bi + −1e
3. X → 135
56Ba +
0
−1e
+ v
+ v
mass number of X = 135 + 0 = 135
atomic number of X = 56 + (−1) = 55 (cesium)
X = 135
55 Cs
135
135
0
55 Cs → 56Ba + −1e
+ v
Section Two—Problem Workbook Solutions
II Ch. 25–3
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Givens
4.
Solutions
235
1
144
92 U + 0 n → 56 Ba +
89
36 Kr + X
mass number of X = 235 + 1 − 144 − 89 = 3
atomic number of X = 92 + 0 − 56 − 36 = 0 (neutron)
X = 3 01n
235
1
144
89
1
92 U + 0 n → 56 Ba + 36 Kr + 30 n
235
1
140
92 U + 0 n → 54 Xe + Y +
1
2 0n
mass number of Y = 235 + 1 − 140 − 2 = 94
atomic number of Y = 92 + 0 − 54 − 0 = 38 (strontium)
Y = 94
38 Sr
235
1
140
94
1
92 U + 0 n → 54 Xe + 38Sr + 2 0 n
5.
228
4
90 Th → X + 2 He + g
mass number of X = 228 − 4 = 224
atomic number X = 90 − 2 = 88 (radium)
X = 224
88 Ra
228
224
4
90 Th → 88 Ra + 2 He + g
II
6. 11p + 73 Li → X + 42 He
mass number of X = 1 + 7 − 4 = 4
atomic number of X = 1 + 3 − 2 = 2 (helium)
X = 42 He
1
1p
7.
217
4
85At → X + 2 He
+ 73 Li → 42 He + 42 He
mass number of X = 217 − 4 = 213
atomic number of X = 85 − 2 = 83 (bismuth)
217
213
4
85 At → 83 Bi + 2 He
Additional Practice 25C, pp. 176–177
1. T1/2 = 26 min, 43.53 s
0.693
0.693
l =  = 
(26 min)(60 s/min) + 43.53 s
T1/2
l = 4.32 × 10−4 s−1
5 times the run time = 5 half-lives
percent of sample remaining = (0.5)5(100)
percent decayed = 100 − percent remaining = 100 − (0.5)5(100) = 96.875 percent
II Ch. 25–4
Holt Physics Solution Manual
HRW material copyrighted under notice appearing earlier in this book.
X = 213
83Bi
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Givens
Solutions
2. T1/2 = 1.91 years
decrease = 93.75 percent =
0.9375
3. T1/2 = 11.9 s
Ni = 1.00 × 1013 atoms
Nf = 1.25 × 1012 atoms
If 0.9375 of the sample has decayed, 1.0000 − 0.9375 = 0.0625 of the sample remains.
0.0625 = (0.5)4, so 4 half-lives have passed.
∆t = 4T1/2 = (4)(1.91 years) = 7.64 years
13
12
∆N 1.00 × 10 − 1.25 × 10
 = 
= 0.875
13
1.00 × 10
Ni
If 0.875 of the sample has decayed, 1.000 − 0.875 = 0.125 of the sample remains.
0.125 = (0.5)3, so 3 half-lives have passed.
∆t = 3T1/2 = (3)(11.9 s) = 35.7 s
4. ∆t = 4800 years
T1/2 = 1600 years
5. ∆t = 88 years
∆t
4800 years
 =  = 3 half-lives
T1/2 1600 years
amount remaining after 3T1/2 = (0.5)3 = 0.125 = 12.5 percent
0.0625 = (0.5)4, so 4 half-lives have passed.
amount of sample
1
remaining = 16 = 0.0625
II
88 years
1
T1/2 = 4∆t =  = 22 years
4
0.693
0.693
l =  = 
T1/2
22 years
l = 3.2 × 10−2 years−1
6. ∆t = 34 days, 6 h, 26 min
amount of sample
 = 1.95 × 10−3
remaining = 512
HRW material copyrighted under notice appearing earlier in this book.
1
24 h 60 min
60 min
∆t = (34 days)   + (6 h)  + 26 min
day
h
h
∆t = 4.9346 × 104 min
1.95 × 10−3 = (0.5)9, so 9 half-lives have passed.
4.9346 × 104 min
1
T1/2 = 9 ∆t =  = 5482.9 min
9
0.693
0.693
l =  = 
5482.9 min
T1/2
l = 1.26 × 10−4 min−1 = 0.182 days−1
7. T = 4.4 × 10−22 s
1
0.693
 = l = 
T
T1/2
T /2
T = 1
0.693
T1/2 = (0.693)(T ) = (0.693)(4.4 × 10−22 s) = 3.0 × 10−22 s
Section Two—Problem Workbook Solutions
II Ch. 25–5
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Section
Section Review
Worksheet Answers
III
Holt Physics
III
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The Science of Physics
Chapter
1
Section 1-1, p. 1
2. a. No. Scientist do not vote about their knowledge.
They use evidence to support or disprove scientific
arguments
1. a. mechanics (laws of motion)
b. vibrations and waves (sound or acoustics)
c. optics
b. No. Speed of light is determined in nature. We can
only measure it.
d. thermodynamics
c. Yes, by sharing their scientific arguments. Science is
a body of knowledge about the universe. Scientists
around the world work together to make it grow.
e. electricity
f. nuclear physics
Section 1-2, p. 2
c. 5.3657 × 10−5 s
1. 1018
−3
9
d. 5.32 × 10
2. 10
7
g
c. 53.236 kV
10
d. 4.62 ms
e. 8.8900 × 10 Hz
3. 10
12
4. a. 3.582 × 10 bytes
−7
b. 9.2331 × 10
W
−9
f. 8.3 × 10
b. 452 nm
m
6. 4.2947842; 4.29478; 4.295; 4.3
5. a. 36.582472 Mgrams
Section 1-3, p. 3
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. a. 6.0 × 108
b. 6 × 105
b. 1.5 × 102
c. 8 × 10−9
c. 1.5 × 10−3
d. 7 × 10−5
d. 6.0 × 103
e. 7 × 106
e. 1.5 × 103
f. 7 × 10−4
f. 6.0 × 10−7
4. a. about 10 cm by 25 cm
b. Check student responses,
which should indicate that
volume = (width)2 × (height).
c. Check student responses for
consistency with a and b.
III
3. a. 104
b. 10−1
2. a. 4 × 105
Chapter 1 Mixed Review
1. a. 2.2 × 105 s
b. 4
b. 3.5 × 107 mm
−4
c. 4.3 × 10
km
d. 2.2 × 10−5 kg
11
e. 6.71 × 10 mg
−5
f. 8.76 × 10
c. 10
d. 3
b. 4.597 × 107; 3.866 × 107;
1.546 × 1014; 11.58
e. 2
5. 15.9 m2
f. 4
6. The graph should be a straight line.
GW
3. a. 4
g. 1.753 × 10−1 ps
b. 5
2. a. 3
4. a. 1.0054; −0.9952; 5.080 × 10−3;
5.076 × 10−3
c. 3
Section Three—Section Review Worksheet Answers
III–1
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Motion In One Dimension
Chapter
2
Section 2-1, p. 6
1. Yes, from t1 to t4 and from t6 to t7.
5. Yes, from 0 to t1 and from t5 to t6.
2. Yes, from t4 to t5.
6. Yes, from t1 to t2 , from t2 to t4 , from t4 to t5, and from
t6 to t7.
3. greater than
7. −5.0 m (or 5.0 m to the west of where it started)
4. greater than
Section 2-2, p. 7
−v
3. a = i
∆t
1. vf = 0. The car is stopped.
2∆x
2. vi = 
∆t
1
5. vi = −a∆t ∆x = 2 vi ∆t
−vi2
4. a = 
2∆x
Section 2-3, p. 8
1. a. −g
d. height = g∆t 2/8
b. initial speed = g(∆t/2)
c. 1.2 s
2
2. a. −9.81 m/s
c. elapsed time = ∆t/2
b. 12 m/s
Chapter 2 Mixed Review
2. a. vf = a(∆t)
b. total distance = d1 + d2 + d3
1
b. vf = vi + a(∆t); ∆x = 2(vi + vf )∆t or ∆x = vi (∆t) +
1
a(∆t)2
2
c. total time = t1 + t2 + t3
3.
Time interval
Type of motion
v(m/s)
a(m/s2)
A
B
C
D
E
speeding up
speeding up
constant velocity
slowing down
slowing down
+
+
+
+
+
+
+
0
−
−
4. a.
III–2
b. 1 s
2
Time (s)
Position (m)
v(m/s)
a(m/s )
1
2
3
4
4.9
0
−14.7
−39.2
0
−9.8
−19.6
−29.4
−9.81
−9.81
−9.81
−9.81
Holt Physics Solution Manual
c. 2 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
III
1. a. t1 = d1/v1; t2 = d2 /v2 ; t3 = d3 /v3
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Two-Dimensional Motion and Vectors
Chapter
3
Section 3-1, p. 11
1. {A, C, E, H, I}; {D, G}, {B, F, J}
2. {A, D, H}, {B, C, G}, {I, J}
3. {A, H}
4. Both diagrams should show a vector A that is twice as
long as the original vector A, but still pointing up. The
first diagram should have the tip of 2A next to the tail of
B. The second diagram should have the tip of B next to
the tail of 2A. The resultant vectors should have the
same magnitude and direction, slanting towards the
upper right.
5. Both diagrams should show a vector B that is half as
long as the original vector B. The first diagram should
have the tip of A next to the tail of −B/2, and −B/2
should be pointing to the left. The second diagram
should have the tip of B/2 next to the tail of −A, and
−A should be pointing down. The resultant vectors
should have the same magnitude but opposite directions. The first will slant towards the upper left. The
second will slant towards the lower right.
Section 3-2, p. 12
1. Check students’ graph for accuracy.
Shot 2: 110 m; 64 m
2. Shot 1: 45 m; 45 m
Shot 3: 65 m; 33 m
Shot 4: 0 m; 14.89 m
3. 220 m
Section 3-3, p. 13
1. ∆t = vi sin q/g
2.
h = vi2(sin q )2/g
3. x = vi (cos q )(∆t)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2vi2 sin q
cos q
4. R = 
g
5.
Launch angle
Maximum height (m)
Range (m)
15°
30°
45°
60°
75°
17
64
130
190
240
130
220
250
220
130
III
Section 3-4, p. 14
1. vBL = vBW + vWL
2. Student diagrams should show vBW twice as long as
vWL but both are in the same direction as vBL, which is
long as both together.
3. Student diagrams should show vWL and vBW, longer
and opposite in direction. The vector vBL should be as
long as the difference between the two, and in the same
direction and in the same direction as vBW.
4. Student diagrams should show vWL and vBW at a right
angle with vBL forming the hypotenuse of a right triangle.
5. a. 6.0 km/h, due east
b. 2.0 km/h, due west
c. 4.5 km/h, q = 26.6°
Section Three—Section Review Worksheet Answers
III–3
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Chapter 3 Mixed Review
1. a. The diagram should indicate the relative distances
and directions for each segment of the path.
b. 5.0 km, slightly north of northwest
c. 11.0 km
2. a. The same
b. Twice as large
c. 1.58
b. 1.0 m/s, in the direction of the sidewalk’s motion
c. 4.5 m/s, in the direction of the sidewalk’s motion
d. 2.5 m/s, in the direction opposite to the sidewalk’s
motion
e. 4.7 m/s, q = 32°
4. a. 4.0 × 101 seconds
b. 6.0 × 101 seconds
3. a. 2.5 m/s, in the direction of the sidewalk’s motion
III–4
Copyright © by Holt, Rinehart and Winston. All rights reserved.
III
Holt Physics Solution Manual
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Forces and the Laws of Motion
Chapter
4
Section 4-1, p. 17
1. The diagram should show two forces: 1) Fg (or mg)
pointing down; 2) an equal and opposite force of the
floor on the box pointing up.
2. The diagram should show four forces: 1) Fg (or mg)
pointing down; 2) an equal and opposite force of the
floor on the box pointing up; 3) F pointing to the right,
parallel to the ground; 4) Fresistance pointing to the left,
parallel to the ground.
3. The diagram should show four forces: 1) Fg (or mg)
pointing down; 2) F pointing to the right at a 50° angle
to the horizontal; 3) a force equal to Fg minus the vertical
component of the force F being applied at a 50° angle;
and 4) Fresistance to the left, parallel to the ground.
Section 4-2, p. 18
1. Fnet = F1 + F2 + F3 = 0
2. String 1: 0, −mg
4. F1 = 20.6 N
String 3: F3 cos q2 , F3 sin q2
3. Fx net = −F2 cos q1 + F3 cos q2 = 0
F2 = 10.3 N
Fy net =
−F2 sin q1 + F3 sin q2 + F1 = 0
F3 = 17.8 N
String 2: −F2 cos q1, F2 sin q1
Section 4-3, p. 19
1. Fs on b and Fb on s ; Fg on s and Fs on g; Ffr,1 and −Ffr,1;
Ffr,2 and −Ffr,2.
2. Fs on b, Fb on s , −Ffr,1
5. Fy,box = Fs on b − mg = 0
6. Fx,sled = Ma = F cos q − Ffr,1 − Ffr,2
3. Fg on s , Fs on g ; Fb on s , Ffr,1, F, Ffr,2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. Fx,box = ma = −Ffr,1
7. Fy,sled = Fg on s + F sin q − Fb on s − Mg = 0
III
Section 4-4, p. 20
1. 44 N
3. a. 21 N, up the ramp
2. 31 N
4. a. 18 N, down the ramp
b. yes
b. yes
Chapter 4 Mixed Review
1. a. at rest, moves to the left, hits back wall
b. m2 a
b. moves to the right (with velocity v), at rest, neither
c. F − m2 a = m1a
c. moves to the right, moves to the right, hits front wall
m1
d. 
F
m1 + m2
2. a. mg, down
b. mg, up
c. no
d. yes
F
3. a. a = 
m1 + m2
F − Fk
4. a. a = 
m1 + m2
b. m2 a − Fk
c. F − m2 a − Fk = m1a − Fk
m1
d. 
(F − Fk)
m1 + m2
Section Three—Section Review Worksheet Answers
III–5
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Work and Energy
Chapter
5
Section 5-1, p. 23
1. Fd
3. 0 J
5. 0 N
−mgd
2. 
2
4. Fk d
6. 0 J
Section 5-2, p. 24
1
1
1. a. 2mvi2
1
c. 2kx12
1
1
c. 2mv 2 + 2kx12
b. 0
1
c. 2mvi2
2. a.
1
b. 2kx12
4. a. 2mvi2
3. a. 0
1
mv 2
2
b.
b. 0
1
kx 2
2 2
1
c. 2mvi2
Section 5-3, p. 25
1. a. 0
d. mghB
2. a. vA = 0
b. mghA
c.
1
mv 2
B
2
b. vB = 2g
(h
A−hB)
3.
C
D
E
F
G
III
KEA
0
0
0
0
0
PEA
KElocation
4
1.9 × 10 J
1.9 × 104 J
1.9 × 104 J
1.9 × 104 J
1.9 × 104 J
3
9 × 10 J
1.3 × 104 J
1.6 × 104 J
3 × 103 J
6 × 103 J
PElocation
3
9.6 × 10 J
6.4 × 103 J
3.2 × 103 J
1.6 × 104 J
1.3 × 104 J
vlocation
17 m/s
2.0 × 101 m/s
22 m/s
10 m/s
14 m/s
4. The sums are the same.
Section 5-4, p. 26
1. v = −gt
2. d =
1
− 2gt 2
3. F = mg
5. The graph should be a curved line.
4. W = Fd
6. 4.20 × 102 W
Chapter 5 Mixed Review
1. a. 60 J
b. −60 J
3. a. 2.9 J
1
1
4. a. 2mvi2 + mghi = 2mvf 2 + mghf +
Fkd
2. a. mgh
b. 1.8 J
b. Fk = mmg(cos 23°)
b. mgh
c. 1.2 J
c. vf =
c. vB = vA2+
2gh
d. no
III–6
e. no
Holt Physics Solution Manual
d. a, b: different; c: same
m
vi2+2g(d
sin
23°
−mcos23
°)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Location
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Momentum and Collisions
Chapter
6
Section 6-1, p. 29
1. Student drawings should show a vector with a length of
9.5 squares to the right.
2. Student drawings should show a vector with a length of
5.0 squares pointing down.
3. 10.7 squares, angle −28°
4. 11 kg • m/s
5. 12 m/s
6. use a protractor, or use tan−1(5.0/9.5)
7. Student drawings should show one vector with a length
of 6.0 squares to the right and another with a length of
12.5 squares to the right. Final momentum is about
6.5 kg • m/s with a final speed of about 43 m/s.
Section 6-2, p. 30
vsmall
 = 50
4. 
vbig
1. 0 kg • m/s
2. 0 kg • m/s
5. The ratio of velocities is the inverse ratio of the masses.
3. The vectors have equal length and opposite direction.
Section 6-3, p. 31
1. vector A added head-to-tail with
vector K
2. F
3. F
5. J
4. vector F subtracted (tail-to-tail)
with vector H
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 6 Mixed Review
1. a. The change due to the bat is greater than the change
due to the mitt.
b. The impulse due to the bat is greater than the impulse due to the mitt.
c. Check student diagrams. Bat: vector showing initial
momentum and a larger vector in the opposite direction showing impulse of bat, result is the sum of
the vectors. Mitt: vector showing initial momentum
and an equal length vector showing impulse of mitt,
result is the sum, which is equal to zero.
b. The total force on the bowling ball is the sum of
forces on pins. The force on the pins is equal but opposite of total force on ball.
III
3. m1v1i + m2v2i = (m1 + m2)vf ;
m1v1i /(m1 + m2) + m2v2i /(m1 + m2) = vf
4. a. M(6 m/s)
b. 2 m/s
c. objects trade momentum; if masses are equal, objects trade velocities
2. a. The impulses are equal, but opposite forces, occurring during the same time interval.
Section Three—Section Review Worksheet Answers
III–7
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Rotational Motion and the Law of Gravity
Chapter
7
Section 7-1, p. 34
1. a. 0.297 rad
2. a. 57.3°
3. a. 29 rad
b. 2.967 rad
b. 237°
b. 19 rad/s
c. 0.873 rad
c. −143°
c. 25 rad/s2
d. 4.014 rad
d. 217°
d. 38 rad/s
2
e. −0.349 rad
e. (1.8 × 10 )°
f. 5.934 rad
f. 90.0°
4. w = v/r; ∆q = v∆t/r; ∆t = T if
∆q = 2p; 2p = vT/r; 2pr/v = T
Section 7-2, p. 35
1. a. 0.10 rad/s
3. 0.35 m/s2
2. a. 0.035 m/s
b. 0.50 rad/s
b. 0.18 m/s
c. 1.0 rad/s
c. 0.35 m/s
b. 0.5
d. 2.0 rad/s
d. 0.70 m/s
c. 2
e. 5.0 rad/s
e. 1.8 m/s
f. 1.0 × 101 rad/s
f. 3.5 m/s
4. a. 4
5. a. 18.8 m/s2
b. friction between tires and road
Section 7-3, p. 36
b. 4
III
c.
1

4
d. 1
1
c. double the radius, decrease the force to 4
d. If measured in the opposite direction, the force will
be in the opposite direction.
3. Because of inertia, objects tend to go in a straight line.
A force is needed to change the direction of travel.
2. a. double one mass, double the force
b. double both masses, quadruple the force
Chapter 7 Mixed Review
1. a. 3.0, 3.0, 9.0, 27
b. quadrupled
b. 4.3, 1.0, 4.3, 37
c. reduced to 4
c. 16, 0.28, 11, 6.0 × 102
d. quadrupled
d. 630, 0.11,74, 8.7
e. reduced to 9
e. 5.0, 44, 0.11, 9.9
2. a. friction
b. gravitational force
c. tension in string
3. a. doubled
1
1
4. 190 m
5. Student diagrams should show vectors for weight and
normal force from elevator; descent should show normal
force less than weight; stopping should show normal
force greater than weight; “weightlessness” feeling is due
to acceleration.
6. 1050 s (17.5 min)
III–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. a. 2
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Rotational Equilibrium and Dynamics
Chapter
8
Section 8-1, p. 39
2. a. 1.20 × 102 N • m
1. a. Fd , Fe , Ff , Fg
b. 96.8 N • m
b. Student diagrams should show only forces Fd , Fe,
Ff , Fg .
c. The door rotates toward Sherry because she exerts
the larger torque.
c. Fe exerts the largest torque because it has the largest
lever arm.
Section 8-2, p. 40
1. point 5
d. no change
2. a. point 9
3. a. point 3
b. point 6
b. point 2
c. point 2
c. point 6
d. no change
4. a. point 5, point 4
b. at point 7, to the left
Section 8-3, p. 41
1. a. 79 rad/s
2. a. 47 J
2
2
b. 22 kg • m , 14 kg • m
b. 0.042 kg • m2
c. 1700 kg • m2/s, 1100 kg • m2/s
c. 3.0 m/s
d. −4.5 × 10−3 rad/s2, −7.1 × 10−3 rad/s2
d. The ball loses energy to external force, the loss of
energy reduces the speed of the ball.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
e. hollow
III
Section 8-4, p. 42
1. Simple machines reduce the force
required for task at the expense of
distance.
2. a. 1.2 × 104 J
b. 120 N
c. 110 m
c. 0.94
d. greater
4. Friction is always present.
5. lubrication and careful
manufacturing
3. a. 0.92
b. 0.90
Chapter 8 Mixed Review
1. a. If the knob is farther from the
hinge, torque is increased
torque for a given force.
b. twice as much
2. a. Rotational inertia is reduced.
b. Angular momentum remains
the same.
c. Angular speed increases.
c. 2.1 × 108 J
3. a. 2.0 kg
d. 3.1 × 103 m/s
b. 0.67 kg
4. a. 6.2 N • m, 0.016 kg • m2,
390 rad/s2
2
e. 2.2 × 108 J
2
b. 12 N • m, 0.062 kg • m , 190 rad/s
16
2
5. a. 8.1 × 10 kg • m
b. 5.9 × 1012 kg • m2/s
6. a. 4.0 × 104 J
b. 4.4 × 104 J
c. 4.9 × 104 J
d. 0.81
Section Three—Section Review Worksheet Answers
III–9
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Fluid Mechanics
Chapter
9
Section 9-1, p. 45
1. V = 30.0 m3
5. Fb = 1.91 × 105 N
2. 1.95 × 104 kg
6. 1.95 × 104 kg
3. Fg = 1.91 × 105 N
7. 19.5 m3
4. 0
8. 19.5 m3; 10.5 m3
9. Ethanol: Fb = 1.91 × 105 N; 1.95 ×
104 kg; 24.2 m3; 24.2 m3; 5.8 m3
Section 9-2, p. 46
1. P = 6.94 × 103 Pa
3. P = 6.94 × 103 Pa
2. P = 6.94 × 103 Pa
4. 12.5 N
5. a. V = 1.44 × 10−5 m3(14.4 cm3)
b. 0.02 m
Section 9-3, p. 47
1. 1.20 m3/s; 1.20 m3/s; 1.20 m3/s
3. 1 s, 1 s, 1 s
2. 6.00 m; 2.00 m; 12.0 m
4. 6.00 m/s; 2.00 m/s; 12.0 m/s
5. Speed increases in order to keep
the flow rate constant.
Section 9-4, p. 48
−4
2. V = 4.00 × 10
3
m
3. T2/T1 = 1/2; P2/P1 = 3/1; V2/V1 = 1/6
III
4. Increasing the pressure reduced the volume. The decrease in temperature reduced the volume.
5. There was no change in mass since the container was
sealed.
6. d = 1.08 kg/m3; The density increased 6 times when
volume of the mass was reduced to 1/6 of the original
volume.
Chapter 9 Mixed Review
1. a. 2.01 × 105 N/m2 (top); 2.51 × 105 N/m2 (bottom)
b. 3.02 × 105 N/m2; 3.52 × 105 N/m2
c. Ftop = 1.81 × 106 N; Fbottom = 2.11 × 106 N
d. Ftop is downward; Fbottom is upward and greater
h. Fb = 3.00 × 105 N. The buoyant force is equal to the
weight of water displace by the crate.
1
1
1
b. Both have the same depth. P1 + 2 rv12 = P2 + 2 rv22
e. net force = 3.0 × 105 N; Fbottom
c. based on the continuity equation: if A1 >> A2 , then
v1 << v2
f. The crate will sink because the buoyant force is less
than the weight of the crate.
d. P1 = P2 + 2 rv22
g. V = 30.0 m3
1
e. P2 = P0 ; P2 < P1 (by 1.00 × 106 N/m2)
f. 44.7 m/s
III–10
1
2. a. P1 + rgh1 + 2 rv12 = P2 + rgh2 + 2 rv22
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. m = 4.32 × 10−4 kg
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Heat
Chapter 10
Section 10-1, p. 51
1. 183 K to 268 K
2
3. a. no—tub is 36°C
2
2. a. 6.30 × 10 K; 2.34 × 10 K
b. no; yes
b. cold
4. a. 77.4 K; 90.2 K
b. The nitrogen is a gas because
the temperature is above its
boiling point. The oxygen is a
liquid because the temperature
is below its boiling point.
Section 10-2, p. 52
1. a. 3.12 × 105 J
b. 5.00 × 104 J
c. increase, 2.62 × 105 J
d. yes; 2.62 × 105 J
2. a. 3.92 × 104 J; 2.50 × 103 J;
4.17 × 104 J
c. decreased by 3.92 × 104 J
d. increase by 3.92 × 104 J;
melting the ice
b. 0 J; 2.50 × 103 J; 2.50 × 103 J
Section 10-3, p. 53
1. 1.04 × 106 J
4. 3-part graph with energy in joules on horizontal axis
and temperature in degrees celsius on the vertical axis:
graph goes up from {0 J, −25°C to 1.04 × 106 J, 0°C}, is
horizontal until {7.70 × 106 J, 0°C}, then goes up to
8.12 × 106 J, 0°C}
6
2. 6.66 × 10 J
3. 4.19 × 105 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section 10-4, p. 54
1. reflect radiation inside the cavity
3. radiation
2. conduction through pan, convection inside water, conduction by contact water to spaghetti
4. convection
III
5. evaporation extracts energy from the body
Chapter 10 Mixed Review
1. a. 78.5 J
c. 3.62°C
b. 78.5 J
d. 19.4°C
c. 51.2 J; less than loss in PE
d. 27.3 J
2. a. 2.26 × 109 J
b. 1.49 × 105 kg
3. a. They are at thermal equilibrium.
b. (100.0 − x)°C; (y − 20.0)°C
c. (2.000 kg)(4.19 × 103 J/kg • °C)
(100.0 − x)°C
d. (5.000 kg) (8.99 × 102 J/kg • °C)
(y − 20.0)°C
e. all of the energy was transferred from the water to the
pipe, no loss and no other
source of energy
f. 72°C
Section Three—Section Review Worksheet Answers
III–11
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Thermodynamics
Chapter 11
Section 11-1, p. 57
1. a. 0.020 m3
2. a. yes, marble to water
3
b. no, ∆U by heat only
3
c. decrease; temperature dropped
b. 7.0 × 10 J
c. 2.0 × 10 J increase
d. increase; more water, less ice
e. no change, the cup is insulated
Section 11-2, p. 58
1. a. −320 J
2. a. 0
b. The gas lost energy because ∆U was less than 0.
b. 540 J out
c. Student diagrams should show the W arrow and the
Q arrow pointing OUT of the container.
c. Student diagrams should show the W arrow
pointing IN and the Q arrow pointing OUT.
Section 11-3, p. 59
1. a. 8.0 × 103 J
2. a. 7.00 × 103 J
3. a. 5.0 × 102 J
b. 20%
b. 1.30 × 104 J
b. 3.4 × 102 J
c. 3.2 × 102 N
c. 4.0 × 101 m
c. 1.9 × 102 J
Section 11-4, p. 60
III
2. a. 1, 4, 6, 4, 1
d. [2-2]
b. 4
b. 16
c. [1-1] has probability 2/4
c. [2-2] has probability 6/16
3. Equal distribution states are more
likely than any other arrangement.
Chapter 11 Mixed Review
1. ∆U = 700 J increase
3
2. a. 0.005 m
b. 1.5 × 103 J
c. 1.5 × 103 J
3. a. 5.00 × 104 J
b. 1.40 × 104 J
III–12
Holt Physics Solution Manual
4. a. ∆U (compressed air) = W (added by person) −
Q (things warm up)
b. Disorder is increased by increasing internal energy
through heat.
5. Graph bars should convey that: PE1 = max, KE1 = 0,
1
U1 = 0 or U1 is any amount. Then, PE2 = 0, KE2 ≤ 2 PE1,
1
1
U2 ≥ U1 + 2 PE1. Then, PE3 ≤ 2 PE1, KE3 = 0, U3 ≈ U2.
3
1
Last: PE4 = 0, KE4 ≤ 4 PE1, and U4 ≥ 4PE1.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. a. 1; 2; 1
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Vibrations and Waves
Chapter 12
Section 12-1, p. 63
1. a. 0.21 m
b. 2.0 s
d. 0.50 m, 2.0 s, 0.5 Hz
c. 41.6 N
2. a. 49.0 N
d. 15.9 cm
2
b. 4.90 × 10 N
c. 0.5 Hz
Section 12-2, p. 64
1. 0.1 s, 10 Hz
3. a. 4.0 Hz, 0.25 s
2. a. 5.0 Hz
b. 10, 70
5. a. 1267 kg, 5066 kg
b. 4.0 Hz, 0.25 s, 5.0 cm
b. increase
4. 0.500 Hz, 2.00 s, 0.0621 m
Section 12-3, p. 65
2. a. 0.02 s, 5 × 101 Hz
1. 37.5 m, 250 m
b. 40.00 m, 2.000 × 103 m/s
Section 12-4, p. 66
1. a. Students’ drawings of amplitudes should have
magnitudes corresponding to 0.25 and 0.35.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
b. Students’ drawings should indicate constructive
interference, with a net amplitude of 0.60.
2. a. 1.5 s
b. 10.0 m
c. yes
III
Chapter 12 Mixed Review
b. PE: 0 s at A, 1 s at C, 2 s at A, 3 s at C, 4 s at A; KE:
0.5 s, 1.5 s, 2.5 s, 3.5 s at B
1. a. 0.20 s; 5.0 Hz
b. same, same, increase, increase
c. 0.5 s, 2.5 s at B to the right 1.5 s, 3.5 s at B to the left;
0 s, 2 s, 4 s at A to the right, 1 s, 3 s at C to the left
2. a. 60.0 N/m
b. 0.574 seconds; 1.74 Hz
5. 3.00 × 102 m/s
2
3. 6.58 m/s ; no
6. 3.0 s; 6.0
4. a. A: 0 s, 2 s, 4 s; B: 0.5 s, 1.5 s, 2.5 s, 3.5 s; C: 1, 3 s
Section Three—Section Review Worksheet Answers
III–13
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Sound
Chapter 13
Section 13-1, p. 69
1. 336 m/s
c. 3.51 s; 0.234 s
2. 1030 m
d. 1.14 × 104 Hz (no Doppler effect because the train
was stationary)
3. a. 3.00 cm
e. pitch decrease; same; increase
b. 1.50 cm
Section 13-2, p. 70
1. a. 9.95 × 10−3 to 2.49 × 10−3 W/m2
b. 6.22 × 10−4 to 2.76 × 10−4 W/m2
c. 1.59 × 10−5 W/m2, about 70
2. a. 1.00 × 10−2 W/m2
b. 3.14 W
c. 5000 m
Section 13-3, p. 71
1. a. 462 m/s
b. Student diagrams should show antinodes, nodes at
both ends; first has one antinode, second has two,
third has three.
c. 69.0 cm
2. a. 880 Hz, 1320 Hz, 1760 Hz
b. Check student graphs for accuracy. Wavelength of
first harmonic should be two wavelengths of second
harmonic, three wavelengths of third harmonic. The
second and third harmonics should have half the
amplitude. The resultant will be a wave with a large
maximum, a smaller peak, a small minimum, and a
large minimum.
III
1. a. 2.19 m; 2.27 m
b. wavelength increases when temperature increases
2. a. arrows pointing East on ambulance, police, and
truck, West on van.
b. police and ambulance (equal), truck, small car, van
3. These objects had the same natural frequency of
330 Hz, so resonance occurred.
III–14
Holt Physics Solution Manual
4. a. 1460 Hz, 2440 Hz
b. 70.8 cm, 23.6 cm, 14.1 cm
c. 0.177 m
d. 974 Hz, 1460 Hz; 70.8 cm, 35.4 cm, 23.6 cm; 0.354 m
5. a. 5
b. 435 Hz, because it will also provide a difference
of 5 Hz.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 13 Mixed Review
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Light and Reflection
Chapter 14
Section 14–1, p. 74
2. a. 7.1 × 1014 Hz; 6.7 × 1014 Hz;
5.5 × 1014 Hz; 5.0 × 1014 Hz;
4.3 × 1014 Hz
1. a. 499 s
b. 193 s
c. 1.97 × 104 s
b. Frequency decreases when
wavelength increases.
c. No, no
Section 14-2, p. 75
1. a. Check student drawings for accuracy. Angles of
reflection should be equal.
b. Extensions intersect on the normal through A, 25 cm
inside the mirror.
e. The person will see the image by receiving reflected
Ray from C.
f. angle at A close to 50°, angle at B close to 60°
g. The eraser’s image is 15 cm inside.
c. 50 cm
d. No, but the person will see image by receiving the
reflection of some other ray.
Section 14-3, p. 76
d. Image locations: B at 3.33 m inside the mirror; C at
2.00 m outside the mirror
1. a. midpoint between mirror and O
b. markings should be at scale: 1 cm for 1 m
2. 2.60 m; 3.33 m; −2.00 m
c. A’s image is 2.6 m inside.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section 14-4, p. 77
1. a. all but green because green is
reflected
III
2. a. white
b. red, because it lets the type of
light best absorbed by plants to
be transmitted
3. black
b. blue
c. black
d. black
Chapter 14 Mixed Review
1. 4.07 × 1016 m
−5
2. a. 3.33 × 10
−4
b. 1.00 × 10
6. a. Check student drawings for accuracy.
b. B is 4 m from A horizontally, C is 2 m below B
vertically
s
m
3. 3.84 × 10 m
c. D is 2 m below A vertically, E coincides with C
4. 3.00 × 1011 Hz
d. they will overlap the existing images or objects
8
5. Diffuse reflection: (nonshiny surfaces) table top, floor,
walls, car paint, posters (answers will vary)
Specular reflection: metallic surfaces, water, mirrors
(answers will vary)
Section Three—Section Review Worksheet Answers
III–15
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7. a. 9.00 cm
8. p = 30.0 cm; q = −6.92 cm; virtual; upright; 1.38 cm tall
b. p = 30.0 cm; q = 12.9 cm; real; inverted; 2.58 cm tall
p = 24.0 cm; q = −6.55 cm; virtual, upright; 1.64 cm tall
p = 24.0 cm; q = 14.4 cm; real, inverted; 3.60 cm tall
p = 18.0 cm; q = −6.00 cm; virtual; upright; 2.00 cm tall
p = 18.0 cm; q = 18.0 cm; real; inverted; 6.00 cm tall
p = 12.0 cm; q = −5.14 cm; virtual; upright; 2.57 cm tall
p = 12.0 cm; q = 36.0 cm; real; inverted; 2.00 cm tall
p = 6.0 cm; q = −3.6 cm; virtual; upright; 3.6 cm tall
p = 6.0 cm; q = −18 cm; virtual; upright; 18 cm tall
III–16
Copyright © by Holt, Rinehart and Winston. All rights reserved.
III
Holt Physics Solution Manual
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Refraction
Chapter 15
Section 15-1, p. 80
1. a. n = c/v
c. Angles inside glass: 25°, 35°, 40°; Angles coming out
of glass: 40°, 60°, 80°
8
b. 2.25 × 10 m/s
d. Student sketches should indicate that the rays exiting the glass are parallel to the rays entering it.
2. a. 13.0°
b. 13.0°, 20.0°
Section 15-2, p. 81
1. a. Check student diagrams. Rays should be drawn
straight, according to rules for ray tracing.
c. B is real, inverted, and smaller; C is virtual, upright,
and larger
2. A: 4.80 cm; B: 7.5 cm; C: −6.00 cm
b. A is real, inverted, and smaller.
Section 15-3, p. 82
1. a. qr = 55.8°
d. qr = 38.5°; qr = 74.5°; qr = 33.4°
b. sin qr = 1.28 > 1: internal reflection
2. qr = 48.8°, the angle is too large, light with 45° incident
angle will be refracted and exit
c. qr = 24.4°
Chapter 15 Mixed Review
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. a. Ray 1 at 45°; Ray 2 at 14.9°
3. a. 9.00 cm
b. Rays should intersect inside the aquarium.
b. 12.9 cm, 14.4 cm, 18.0 cm, 36.0 cm, −18.0 cm
c. Because the rays are no longer parallel, they will intersect in the water.
2.58 cm, 3.6 cm, 6.00 cm, 18.0 cm, −18.0 cm
2. a. First boundary: 70.0°, 45.0°
III
real, real, real, real, virtual
4. 18.0 cm, with all images virtual and on the left of the lens
Second boundary: 45.0°, 40.4°
−11.2, −10.3, −9.00, −7.20, −4.50
Third boundary: 40.3°, 36.8°
5. a. 6.00 cm in front of the lens
b. Incoming rays get closer and closer to the normal.
Reflected rays get farther away from the normal with
the same angles.
b. 0.857 cm
Section Three—Section Review Worksheet Answers
III–17
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Interference and Diffraction
Chapter 16
Section 16-1, p. 85
1. a. First: 1.6°, Second: 3.2°, Third: 4.8°
b. Bright: 16.2°, 34.0°, 4.01°
2. a. 475 nm
b. 7.80°, 11.7°, 15.7°
c. A smaller slit results in more separation between
fringes. With 2 cm, fringes would be so close they
would not be distinguishable.
Section 16-2, p. 86
1. a. 1.25 × 10−6 m
b. 18° spacing for 400 nm light and 34° for 700 nm
light. More lines per centimeter will give better
resolution
2. a. 1250 lines/cm
b. 4.4°, 8.9°, 13°
3. 565 nm
4. 4.38 × 10−6 m
III
1. Coherent light is individual light waves of the same wavelength that have the properties of a single light wave.
3. Lasers convert light, electrical energy, or chemical
energy into coherent light.
2. Student diagrams should show a coherent light source
with light waves moving in the same direction. The
incoherent light should have a light source with waves
radiating out in different directions.
4. Answers will vary. Examples are CD players, laser
scalpels, laser range finders.
Chapter 16 Mixed Review
1. a. 6.74 × 10−6 m
b. 47.9°
c. The maximum angle for light to reach the screen in
this arrangement is 45°.
2. a. Longer wavelengths are diffracted with a greater angle.
b. First order group of lines: blue, green, red; second
order: the same
III–18
Holt Physics Solution Manual
c. White
3. a. A = 5.0 × 10−6 m, B = 1.1 × 10−7 m, C = 3.3 × 10−8 m
b. visible: A; x-ray: A, B, or C; IR: none
4. a. Neither would work because they would act as different sources, so even with the same frequency, they
should not be in phase.
b. Interference is occurring.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section 16-3, p. 87
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Electric Forces and Fields
Chapter 17
Section 17-1, p. 90
1. a. Experiment A, no charges were transferred. Experiment B, charges were transferred between the sphere
and the ground. Experiment C, charges were transferred between the sphere and the rod
b. Student diagrams should show: Sphere A, negative
charges (−) on the left, positive (+) on the right; Sphere
B, excess (−) all over; Sphere C, excess (+) all over.
c. Sphere B has excess (−); Sphere C has excess (+)
d. Experiment A
e. no change in Experiment A or Experiment B; reduced charge in Experiment C
Section 17-2, p. 91
1. a. 20.0 cm
f. 36.9°
b. 0.899 N (attraction along the line q1 − q3)
g. F1x = −0.719 N; F2x = 0.719 N; F1y = −0.540 N;
F2y = −0.540 N
c. 0.899 N (attraction along the line q1 − q2)
h. −1.08 N pointing down
d. 1.40 N repulsion pulling to the right
e. Student diagrams should show F1 pointing from q3
toward q1 and F2 pointing from q3 toward q2 .
i. downward along the y-axis
Section 17-3, p. 92
1. a. 21.2 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
b. all same strength of 1.60 × 10−6 N/C along the diagonal lines, with E1 pointing away from q1, E2 from
q2 , E3 from q3, and E4 from q4
c. Resultant electric field E = 0
2. a. 4.61 × 10−14 N down
b. 4.61 × 10−14 N up
c. 1.44 × 10−18 C
III
d. 9 electrons
Chapter 17 Mixed Review
1. a. A; 1.87 × 1013 electrons; B: 3.12 × 1013 electrons
3. a. 1.92 × 1016 N
b. 2.87 × 1010 m/s2
b. the forces are equal and opposite, no
2. a. Resultant = 1.49 N, left; F(A-C) = 1.35 N, left;
F(B-C) = 0.140 N, left
b. Resultant = 0.788 N, right; F(A-C) = 1.35 N, right;
F(B-C) = 0.562 N, left
c. Resultant = 0.400 N, left; F(A-C) = 0.599 N, right;
F(B-C) = 0.999 N, left
c. 9.81 m/s2; this is negligible in comparison with the
acceleration a; alpha particles will move horizontally
4. a. Check students diagrams for accuracy.
b. 1.53 × 10−2 N
c. 7.65 × 103 N/C
5. 1 C = 6.25 × 1018; 1 mC = 6.25 × 1012
Section Three—Section Review Worksheet Answers
III–19
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Electrical Energy and Capacitance
Chapter 18
Section 18-1, p. 95
1. a. −5.62 × 10−7 J, yes, 5.62 × 10−7 J of work was done
b. 1.40 × 10−6 J; 8.43 × 10−7 J of work was done on the
charges
2. a. 9.60 × 10−18 J; Potential energy decreases
b. 9.60 × 10−18 J
c. 5.36 × 104 m/s
Section 18-2, p. 96
1. a. 8.99 × 105 V
y = 10.0 cm; V = 3.32 × 105 V
b. y = −10.0 cm; V = 3.33 × 105 V
c. x = −10.0 cm; V = 4.28 × 105 V
y = −2.00 cm; V = 8.08 × 105 V
x = −2.00 cm; V = 1.20 × 106 V
5
y = 2.00 cm; V = 8.08 × 10 V
6
x = 2.00 cm; V = 1.20 × 10 V
x = 10.0 cm; V = 4.28 × 105 V
2. a. 2.16 × 106 V
b. 0
c. 0
Section 18-3, p. 97
1. pF = 10−12 F; nF = 10−9 F; mC = 10−6 C; Farads measure
the ratio of charge to potential difference. Coulombs
measure the amount of charge.
2. 1 pF < 1 nF. The 1 pF capacitor has a higher potential
difference (1000 times) because ∆V = Q/C
3. a. 4.00 × 10−7 F = 4.00 × 102 nF
b. Capacitance does not change. Charge doubles (Q is
proportional to ∆V, ∆V doubled and C was the same)
c. 5.00 × 10−2 J; 2.00 × 101 J
III
1. a. 4.50 × 10−7 J for all cases
b. PE does not change
c. All force vectors should have same magnitude and
point toward the center
2. a. −1.28 × 10−15 J; decreases
b. 1.28 × 10−15 J; increases
−7
c. 5.3 × 10 m/s
III–20
Holt Physics Solution Manual
3. a. 5.000 × 103 V/m; yes, the field is constant
b. ∆V(+plate, A) = 50.0 V; ∆V(+plate, B) = 1.50 ×
102 V; ∆V(+plate, C) = 2.50 × 102 V
c. PE at positive plate = 4.80 × 10−17 J; PEA = 4.00 ×
10−17 J; PEB = 2.40 × 10−17 J; PEC = 8.00 × 10−18 J;
PE at negative plate = 0 J
4. a. 2.00 × 102 V
b. 4.00 × 10−3 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 18 Mixed Review
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Current and Resistance
Chapter 19
Section 19-1, p. 100
1. 2.50 × 102 A
c. 4.69 × 1018 electrons
3
2. a. 15.0 C; 225 C; 5.40 × 10 C
3. a. 320 s
b. 320 s
d. Electrons are in the wires and
the filament.
18
b. 4.69 × 10 electrons
c. 320 s
Section 19-2, p. 101
1. a. 25. 6 Ω
3. 134.7 V
4. a. 343 Ω to 286 Ω
b. 4.70 A; 8.61 A; 2.34 A; 0.391 A
2. a. 1.80 × 10−3 A; 1.80 A; 1.80 × 102 A
b. R > 255 Ω
c. R < 387 Ω
b. C (smaller resistor)
Section 19-3, p. 102
1. a. 932 W
3. a. 5.1 Ω
2. a. 417 W
7
b. 1.68 × 10 J = 4.66 kWh
b. 3.5 A
b. 24 A
c. 32.6 ¢
Chapter 19 Mixed Review
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. a. I increases because R decreases
(shorter)
b. no change
c. I decreases because R increases
with temperature
d. I decreases
b. 144 Ω; 96.0 Ω; 57.6 Ω
2. a. 4.8 A
4
b. 8.64 × 10 J
c. 580 W
c. 70.0 ¢; $1.05; $1.75
III
4. a. 144 V
7
d. 1.0 × 10 J
3. a. 0.833 A; 1.25 A; 2.08 A
b. 864 W
c. 104 seconds
Section Three—Section Review Worksheet Answers
III–21
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Circuits and Circuit Elements
Chapter 20
Section 20-1, p. 105
1. a. Check student diagrams, which should contain 2
bulbs, 2 resistors, 3 switches, and 1 battery, in a
closed cirucit.
b. Check student diagrams to be certain that the switches
labeled S1 and S2 cause short circuits when closed.
c. Check student diagrams to be certain that switch S3
causes a short circuit when closed.
b. Students should connect one end of B to the battery,
the other to the switch, then the other end of the
switch to the battery. Bulb A should simply be left
out with no connections.
c. Students should connect each end of B to one end of
the battery, the other to the switch, then the other
end of the switch to the battery. Also each end of A
should be connected to an end of the battery.
2. a. Students should connect one end of bulb A to the
battery, the other to the switch, then the other end of
the switch to the battery. Also connect one end of B
to the battery, and the other end of B to the switch.
Section 20-2, p. 106
1. a. 16.0 Ω
2. a. 3.00 Ω
b. 0.750 A for both
c. 4 A, I1 = 1 A; I2 = 3 A
b. 12 V
d. 12.0 V
c. 12.0 V; 9.0 V; 3.0 V
Section 20-3, p. 107
b. Ia = Ib = Ic = 0.600 A; Id = Ie = If = 0.200 A;
∆Va = ∆Vb = ∆Vc = 7.20 V; ∆Vd = ∆Ve = ∆Vf = 2.40 V
III
2. a. Check diagram
b. 54 Ω; Ia = Ib = Ic = If = 0.444 A; Id = Ie = 0.222 A;
∆Va = ∆Vb = ∆Vc = ∆Vf = 5.33 V; ∆Vd = ∆Ve = 2.67 V
Chapter 20 Mixed Review
1. a. D
b. switch 5
c. • switches 1 and 3 open, switches 2, 4, and 5 closed
c. 6 Ω
3. a. Check students diagrams.
• switches 1 and 4 open, switches 2, 3, and 5 closed
b. 12.0 V, 12.0 V
• switch 2 open, switches 1, 3, 4, and 5 closed; or
switches 3 and 4 open, switches 1, 2, and 5 closed;
or switches 2, 3, and 4 open, switches 1 and 5 closed
c. 0.25 A, 2.25 A
2. a. Check students’ diagrams, which should show a
bulb and a resistor in series with a battery.
III–22
b. 15 Ω
Holt Physics Solution Manual
d. 5.33 Ω
4. a. R = 6.15 Ω
b. R = 30.4 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. a. 40 Ω
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Magnetism
Chapter 21
Section 21-1, p. 110
2. Arrows should point away from S, toward N, building a
composite picture of the magnetic field.
1. a. No
b. No
3. Arrows should point away from S, toward N, mostly in
the area between the ends of the magnet and around it.
c. Magnet: A; Iron: B and C.
Section 21-2, p. 111
1. a. the field at A, B, C is pointing out (dot symbol); the
field at D, E, F is pointing in (× symbol).
b. all reversed: the field at A, B, C is pointing in (× symbol); the field at D, E, F is pointing out (dot symbol)
2. the strength at point A is weaker than B, C, D or E, and
about equal to that at F.
3. All directions of field are opposite to the answers in
questions 1. The relative strengths remain the same.
Section 21-3, p. 112
1. a. v-arrow to the right, B-arrow upward
−14
b. •; F = 4.8 × 10
3. a. v-arrow to the right, B-arrow upward
N, upward, out of the page
c. 0
b. •; F = 9.6 × 10−14 N, upward, out of the page
c. 0
2. a. v-arrow to the left, B-arrow upward
b. ×; F = 4.8 × 10−14 N, downward, into the page
4. No. When the force is not zero, it acts perpendicular to
velocity. They move in a circle perpendicular to the
magnetic field.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
c. 0
III
Chapter 21 Mixed Review
1. a. The magnetic field from the leftmost segment is •
and stronger. The magnetic field from the rightmost
segment is × and weaker.
b. At A, both horizontal segments contribute a ×
magnetic field of equal strength
c. B; ×; × weaker; ×; × same
e. inside
2. a. F = 4.3 N into the page
b. F = 0
3. a. Diagrams should show clockwise current.
b. Starting from the left side: F = 1.1 N into the page;
F = 0; F = 1.1 N out of the page; F = 0
C; ×; × same; ×; × same
c. Forces are equal and opposite, so no translational
motion will occur, but it could rotate around a vertical axis.
D; ×; × stronger; ×; × same
E; ×; • stronger; ×; × same
d. No. They reinforce each other in the same direction.
Section Three—Section Review Worksheet Answers
III–23
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Induction and Alternating Current
Chapter 22
Section 22-1, p. 115
3. a. 2.56 × 10−2 m2
1. side a: none, down, down, none, none
side b: none, none, none, none, none
b. 2.0 s
side c: none, none, down, down, none
c. 2.0 × 10−2 V
side d: none, none, none, none, none
d. 5.7 × 10−2 A
2. none, clockwise, none, counterclockwise, none
Section 22-2, p. 116
1. A to B
3. a. horizontal
2. increase, increase, increase
c. 0.25 s
d. 1.9 × 10−3 V
b. vertical
Section 22-3, p. 117
1. down through primary coil, and up elsewhere, including through the secondary coil
2. a −, b +
4. no change in field
5. disappearing field is a change which secondary coil
opposes
3. 24 V
Chapter 22 Mixed Review
b. 7.1 × 10−2 m2
1. e
2. a. 0.50 s
c. 110 V
2
b. 0.26 m
c. 2.6 V
3. a. magnetic field, conductor, relative motion
b. answers may vary, but could include the following:
water wheel, windmill, electric motor, combustion
engine
4. a. 6.28 rad/s
III–24
Holt Physics Solution Manual
d. 78 V
5. A motor converts electric energy to rotational energy;
generators converts rotational energy to electric energy.
6. a. increases
b. induces current while change occurs
c. It decreases magnetic field which will induce a current while the change occurs.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
III
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Atomic Physics
Chapter 23
Section 23-1, p. 120
1. a. This implies that there is an infinite energy output.
b. quantization of energy
c. As wavelength gets shorter, energy in photon gets smaller.
2. a. 2.9 × 10−31 J
b. 1.8 × 10−12 eV
3. a. hft = hf − KEmax
b. 2.30 eV
Section 23-2, p. 121
1. small positively charged nucleus and electrons in
planetary orbits
2. He expected diffuse positive charge with no scattering.
3. Most atoms went through.
4. As electrons radiated energy, they would spiral in toward nucleus.
Section 23-3, p. 122
b. 5.41 × 10−40 m
1. a. light radiating from the sun to Earth
c. 8.4 × 10−37 m
b. light scattering off electrons
2. The precision of measurements for very small objects is
relatively less than the precision of measurements of
very large objects.
3. a. 1.47 × 10−38 m
d. 3.7 × 10−35 m
4. It allowed for electron uncertainty and gave electrons
probable but not definite orbits.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 23 Mixed Review
1. There is not enough energy in any individual photon to
liberate the electron.
2. Some energy is used in liberating the electron.
3. a. Atoms contained areas of dense positive charge.
b. The foil is mostly empty space.
4. a. 1.5 × 10−8 m
b. 5.3 × 10−34 m
c. The wavelength is too small to detect.
5. a. Simultaneous measurements of position and momentum cannot be completely certain.
III
b. A theory of distinct orbits would require precise
knowledge of their location at any given time.
6. A photon does not measurably deflect a planet.
7. 1.16 × 1015 m
8. No electrons were ejected.
9. It is absorbed by atoms into vibrational motion, etc.
10. Energy is observed in increased temperature.
Section Three—Section Review Worksheet Answers
III–25
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Modern Electronics
Chapter 24
Section 24-1, p. 125
1. a. conductor
b. insulator
c. conductor
d. insulator
e. semiconductor
2. Semiconductors have a small energy gap in which
electrons can pass.
3. Thermal excitation and electromagnetic fields can provide
the energy to excite electrons into the conduction band.
4. Properties of materials are based on many atoms
together.
Section 24-2, p. 126
1. a. It is easier for the neighboring electron to move into
the hole in valence band.
b. The hole increases conductivity.
b. No, neutral atoms are added.
c. 5
d. No, neutral atoms are added.
2. a. 3
Section 24-3, p. 127
1. lattice imperfections
c. It is transferred via lattice to the second electron.
2. a. They distort toward the electron.
d. No, pairs are constantly formed, broken, and
reformed.
b. It increases the force on the electron.
III
1. Check student diagrams; conductor should have overlap, semiconductor have a small gap, insulator have a
large gap.
2. They have small or no gap to conduction band.
3. Many atoms are located near each other.
4. They are thermal excitation and application of an
electromagnetic field.
5. They are weakly bound through lattice interaction.
6. Transistors have two p-n junctions instead of one,
which makes three leads instead of two.
III–26
Holt Physics Solution Manual
7. a. valence electrons
b. It increases the number of charge carriers available.
8. The n-type are doped with extra valence electron (majority carrier); p-type are doped with one less valence
electron (holes are majority carrier).
9. The p-n junction creates an electric potential barrier,
which allows current to pass one way but resists flow in
other direction.
10. Superconductors have zero resistance.
11. The conducting ring dissipates energy as heat.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 24 Mixed Review
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Subatomic Physics
Chapter 25
Section 25-1, p. 130
1. a. 16
e. Energy is required to separate the nucleus.
b. 8
f. No, it is the same element but a different isotope.
c. 16
2. a. strong interaction
d. 2.81 MeV
b. decreases
Section 25-2, p. 131
1. alpha—helium nucleus; beta—
electron or positron; gamma—
photons
2. a. O-17
b. Th-231
c. Np-238
5. half-life = 0.693/decay constant
d. U-235
6. 0.050 s−1
3. It is the time required for half of
the sample to decay.
7. 3.15 × 107 s
8. 25.0% or 1/4
4. It gives decay rate for sample.
Section 25-3, p. 132
1. a. fission
b. neutron and uranium nucleus
c. barium, krypton, and 3 neutrons
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d. yes
e. more fission
b. proton and helium-3 nucleus
f. It has high energy output. In a
nuclear reactor, the high heat
leads to a meltdown.
c. alpha (He-4), positron and
neutrino
d. yes
2. a. fusion
Section 25-4, p. 133
III
1. Strong: 1, hold nucleons, 10−15 m; electromagnetic:
10−2, charged particles, 1/r2; weak: 10−13, fission,
10−18 m; gravitational: 10−38, all mass, 1/r2
3. a. It can unify weak and electromagnetic interactions
at high energy.
b. It requires very high energy interaction (1 TeV).
2. a. graviton; W and Z bosons; photons; gluons
b. graviton
Chapter 25 Mixed Review
1. a. 143
2. a. atomic number
b. 146
b. number of neutrons
c. 146
c. same number of neutrons
d. 1
d. different atomic numbers
e. 2
e. Both pairs increase mass by
one amu.
f. 8
g. 10
h. 22
f. First pair are isotopes; second
pair are different elements.
b. one higher
c. New one is higher; otherwise, it
wouldn’t decay.
d. new one
4. gravitational interaction
5. No, there are not enough nucleons
to form an alpha particle.
6. mass and charge
3. a. almost the same
Section Three—Section Review Worksheet Answers
III–27
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Section
Interactive Tutor
Worksheets Answers
IV
Holt Physics
IV
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Interactive Tutor Worksheets Answer Key
Modules 1-20
Module 1 One-Dimensional Motion
1. air resistance
12. negative
2. displacement or ∆x, initial velocity or vi , final velocity
or vf , acceleration or i, time interval or t
13. positive x, or +x, direction
3. displacement or net change in position
14. The apple falls towards the negative y-axis.
15. Zero is neither positive nor negative.
4. The net change in position is in a positive direction.
5. direction
16. vf 2 = vi 2 + 2a∆y
17. 0.508 s
6. slowing down
18. the landing airplane
7. positive
19. any three
8. one
20. +6.13 m/s; 0.508 s
9. –1.85 m
21. 117.0 m/s
10. this makes the calculation easier
11. slow down
Module 2 Vectors
1. magnitude (or amount); direction
9. origin; original
2. a small arrow over the variable
10. axis system (or coordinate system)
3. resultant; sum
11. vectors
4. vector magnitude
12. In the Detroit/Indianapolis flight, the x component is
the side adjacent to the known angle, so the cosine
function is used. For the Indianapolis/Chicago flight,
the x component is the side opposite the known angle,
so the sine function must be used.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. sum
6. coordinate
7. The sine and cosine functions both have the length of
the hypotenuse in their denominator. The sine has a
numerator equal to the side opposite the known angle,
and the numerator for the cosine function is the side
adjacent to the known angle.
13. +x; −y
14. 263 m, 61.9° to the x axis
15. 28.3 m, 45° to the positive x-axis
8. right triangle
Module 3 Two-Dimensional Motion
1. two
7. 9.81 m/s2 (or free fall acceleration, or −g)
2. Student answers will vary. Be certain that examples are
projectiles, such as a launched arrow or a thrown ball.
8. tangent
3. horizontal and vertical (or x and y)
9. initial; so that horizontal and vertical motion can be
treated separately
4. constant velocity
10. The arrow is moving in both directions simultaneously.
5. No. Velocity is constant, so acceleration, the change in
velocity, is zero.
11. The arrow begins by moving upward, but ends by
falling downward. Labeling on the coordinate axes
indicates that downward is negative.
6. constant; changing
Section Four—Interactive Tutor Worksheets Answers
IV
IV–1
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12. This equation is only associated with constant velocity.
For projectile motion, the velocity in the y-direction is
not constant.
13. 20.2 m
14. 25.9 m
Module 4 Net Force
1. motion
2. friction, Ffriction ; gravity, Fg ; normal, Fn ; friction slows
the skateboard
3. smaller; the mass has decreased, but the acceleration is
the same
4. It is not aligned with either axis.
5. changing
6. slowing down
7. zero
8. positive
9. The answers should be the same as in the program, but
the axis system has been rotated to describe a displacement along the y-axis and no movement along the x-axis.
10. The equal but opposite forces in the y direction indicate
that there is no net force in that direction. As a result,
there is no motion in the y direction.
11. Because there is no motion in the direction of the normal force, it must be equal and opposite to the other
forces in the y direction.
12. 330 N
13. 1.2 × 103 N
Module 5 Work
1. sum
9. opposite
2. the force is in the same direction as the displacement
10. normal force
3. magnitude; displacement
11. it is applied perpendicular to the displacement
4. the relative direction of the force and displacement
12. 3.6 × 103 J
5. negative; displacement
13. 2.2 × 104 J
6. force; displacement
14. 1.35 × 103 J
7. W = Fd cos q
15. 98.6 N; 1.73 × 103 J
Module 6 Work–Kinetic Energy Theorem
1. gravitational potential energy
7. Fg ; Ffriction ; Fn
2. In the pulley system, the block and brick move separately, but in the skateboard example, the board and
person move together.
8. Work done by gravity is accounted for in the mechanical energy term.
3. mechanical energy
9. The platform with the brick moves downward (−y),
and the block moves upward (+y).
4. kinetic energy; gravitational potential energy (students
may also include elastic potential energy)
10. The final velocity would increase.
5. Is anything moving?
12. −3.7 × 10−12 N
11. 18.2 J
6. zero level
IV
Module 7 Conservation of Momentum
1. mass; speed
4. constant; momentum
2. velocity
5. inelastic
3. vector
IV–2
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8. frictional force, force of gravity, normal force, force applied by the truck; the force applied by the truck
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6. The problem would be the same, but the coordinate
system would be rotated and the y axis would be
parallel to the direction of motion.
7. It is sitting still.
10. conservation
11. The more-massive object has a smaller velocity, and the
less-massive object has a greater velocity.
12. 0.19 m/s to the west
8. They are stuck together.
13. 6.16 m/s to the north
9. vector
14. 0.72 m/s
Module 8 Angular Kinematics
1. linear
9. counterclockwise
2. radian
10. 3.14 rad/s
3. 2p; circumference
11. angular acceleration
4. 2 times p (or 2 times 3.14)
12. spinning clockwise and slowing down
5. angular displacement; radians; rotation
13. 5.0 s; 25 rad; counterclockwise
6. positive
14. 4.5 s.
7. same
15. 201 rad
8. angular speed; rad/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Module 9 Torque
1. force; rotate
7. clockwise
2. magnitude; angle
8. negative, because the rotation now would be clockwise
3. perpendicular; rotation; force
9. the 25 cm wrench, because the lever arm would be
largest
4. As the lever arm increases, so does the distance between
the axis of rotation and the applied force.
10. 0.075 N • m
5. no effect
11. 0.17 N • m
6. magnitude; length
12. 460 N; 360 N
Module 10 Rotational Inertia
1. torque; rotational inertia (or moment of inertia)
8. more
2. rotational
9. To speed up, move toward the center so that I decreases.
3. torque, t
10. I = mr 2; I = 0.5 mr 2
4. rotational inertia (or moment of inertia or I )
11. 0.84 rad/s2
5. axis of rotation
12. 4.05 × 10−2 kg • m2
6. increases
13. 6.4 rad/s2; 3.1 s
7. increases
Module 11 Hooke’s Law
IV
1. restoring force
5. Hooke’s law; equilibrium position
2. equilibrium position
6. opposing (or opposite)
3. 0 N
7. spring constant; N/m
4. magnitude; displacement
8. increases; compress
Section Four—Interactive Tutor Worksheets Answers
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9. The first spring needs half as much force; the second
spring is stiffer because it has the larger force constant
10. lesser spring constants for softer beds
11. 15.0 cm
12. 6.8 × 10−4 m
13. 102 g; 102 g
Module 12 Frequency and Wavelength
1. transverse
9. compression
2. perpendicular; motion (or direction)
10. rarefactions; longitudinal
3. simple harmonic motion
11. compressions; rarefactions
4. oscillating mass in the stapler
12. wavelength; transverse
5. longitudinal
13. meters; period
6. sound waves
14. The frequency is the inverse of the period, and vice versa.
7. Each molecule of air moves back and forth in the direction of wave motion as the wave travels.
15. v = l/T
8. A crest is the point farthest from the equilibrium point
of the wave in the positive (or up) direction.
A trough is the point farthest from the equilibrium
point of the wave in the negative (or down) direction.
16. pitch
17. transverse; longitudinal
18. 3.91 × 10−3 s
19. 5.5 m/s
Module 13 Doppler Effect
2. period
8. The distance the source moves is added to the normal
spacing between wavefronts, or the speed of the source
is added to the speed of sound.
3. less than
9. receding; frequency
4. speed (or velocity); period (or frequency)
5. the second fly; the distance between wavefronts for the
second fly’s buzz is smaller than for the first fly’s buzz;
hence, the value of d is greater, producing a shorter
wavelength and high frequency.
10. The speed of the source was added to the speed of
sound instead of subtracted.
11. longer
12. 31.4 m/s
6. higher (or greater than)
7. The speed of the source is subtracted from the speed of
sound. This difference appears in the denominator of
the fapp equation, so the ratio of speeds is always
greater than 1 and the perceived frequency is always
greater than the source frequency.
Module 14 Reflection
1. ray; perpendicular
2. intersect
3. image
IV
4. incidence
5. law of reflection
6. ray tracing
7. A real image has a positive value of q and is located in
front of the mirror.
IV–4
Holt Physics Solution Manual
A virtual image has a negative value of q and is located
behind the mirror.
8. bisects
9. principal axis
10. radius of curvature
11. mirror
12. focal point
13. The image is real and located 0.5 m in front of the mirror.
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1. perceived
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14. three (or two, with one to check)
19. behind; mirror
15. parallel
20. positive
16. center; curvature
21. the sign of q is negative (or the image is upright.)
17. principal axis
22. any value greater than 0.563 m
18. base; principal
Module 15 Refraction
1. converging lens; refraction
9. image; visible
2. speed
10. diverge; same
3. perpendicular; interface (or boundary)
11. positive; whether the image is real or virtual.
4. increases; decreases; toward; away from
5. center; right
12. The ray is refracted the same amount upon entering
and leaving the lens but in opposite directions: first
toward the normal, then away from it.
6. focal point
13. 0.262 m in front of the lens
7. center; focal point
14. 20.0 cm
8. focal point
15. virtual, 15.0 cm
Module 16 Force Between Charges
1. positive; electrons
10. signs
2. protons; electrons
11. Answers may vary but should indicate a natural source
of charge. Lightning is one possible answer.
3. net charge
4. force
12. The spheres have net charges of opposite sign and are
drawn together by the force of attraction.
5. repulsion; unlike (or opposite)
13. 2.7 N; attractive
6. magnitudes; distance
14. no effect, because charges are not included in calculations using Coulomb’s law
7. N • m2/C2
15. 2.0 N; repulsive
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. the centers of the charges
2
9. F = (kC q )/r
2
Module 17 Electrical Circuits
1. separate; potential difference
2. negative charges moving one way; positive charges
moving the other way; simultaneous movement of
both positive and negative charges
3. positive
11. Each side of each resistor is attached to the same point
and thus is at the same electrical potential energy.
12. reciprocal
13. resistors
14. Any device that draws current is a resistor.
4. resistance; volts; current
15. fuse (or circuit breaker); current
5. greater (or more)
16. No; the current branches prior to passing through Rb.
6. potential energy
17. series
7. equivalent resistance
8. series (or parallel); parallel (or series)
9. There are no branches (or divisions) in the circuit.
IV
18. When the reciprocals of individual resistors connected
in parallel are summed, the answer is in Ω−1. Taking the
reciprocal gives Ω, the units of resistance. Looking at
the units keeps the student from using the wrong value.
10. added together (or summed)
Section Four—Interactive Tutor Worksheets Answers
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Module 18 Magnetic Field of a Wire
1. motion
2. Student answers will vary but should include examples
of moving charges. One possible answer is current in a
conducting wire.
3. close to the wire
4. tangent; concentric circles
5. It is an electric shock hazard.
6. moving charges (or electric current); magnetic field
7. it has direction and magnitude
8. tesla-meters per ampere, or (T • m)/A
9. The magnitude of the wire’s magnetic field at the compass’ location is small compared with the magnitude of
Earth’s magnetic field at the same point.
10. distance between wire and compass
11. less than; distance is in the denominator of the equation
used to calculate the magnitude of a magnetic field.
12. The value of B would decrease because current, I, is in
the numerator of the equation for calculating magnetic
field strength.
13. The magnetic field measured is at a point 5.0 × 10−3 m
from the conducting wire.
Module 19 Magnetic Force on a Wire
1. no
7. 0°, 180°, and their multiples
2. current is the movement of charges
8. equals
3. 0 N
9. fingers, current, palm
4. the 100 mm wire
10. magnitude, direction
5. movement of positive charges that would be equivalent
to the current
11. The current equals 0.011 A.
6. The arrow representing current to the left is oriented
directly opposite the arrow representing the current to
the right. The angle between these two rays is 180°. The
line representing the magnetic field cuts across the current line so the angle between the left current and the
field plus the angle between the field and the current to
the right must equal 180°.
12. The force is 1.4 × 10−3 N.
13. 71°
1. electromagnetic induction or induced current
8. decreases or steps down
2. primary; secondary
9. The value of the current is halved.
3. no, the primary coil is the one that induces a current in
the secondary coil
4. root mean square
5. number of turns in the primary and secondary coils
11. 4 × 105 V or 400 kV
6. The currents are also equal.
12. 1:2
7. steps up or increases
IV
IV–6
10. The potential difference is 0 V. A transformer does not
respond to direct current produced by a battery because there are no changes in the magnitude or direction of the magnetic field.
Holt Physics Solution Manual
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Module 20 Induction and Transformers
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Problem Answers
Modules 1-20
Module 1 One-Dimensional Motion
1. 82.0 m/s
5. 66.1 m/s
2. 3.0 s
6. 36 m/s
3. 9.98 m/s
7. −11 m/s
4. 6.82 s; The sandwich would require slightly more time
to fall because the downward acceleration would be reduced by the force of air resistance.
8. 8.38 m
9. 0.04 s
Module 2 Vectors
1. 216.5 m, 30.01° north of east
3. 77.6 s
3
2. total gain in altitude = ∆ytot = 13.58 × 10 m = 13.58 km,
up total horizontal displacement = ∆xtot = 23.90 × 103 m
= 23.90 km, horizontally
4. 1.36 × 106 m, 4° east of north
5. 2.5 × 105 m, east
Module 3 Two-Dimensional Motion
1. 60°
4. 16.6 m/s
6. 55.2°
2. 45.8 m/s
5. a. 21 m/s
7. 21.0°
3. 68.2 m/s
b. 22 m
8. 4.1 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Module 4 Net Force
1. 1.89 × 103 N, up
4. 1.72 × 103 N, up
2. 1.36 × 103 N
5. 4.59 N, forward
3
2.72 × 10 N
3. 1.75 × 103 N, up; An increase in the separation of the
cables would cause cos q to decrease and thus the
tension in the cables to increase.
Module 5 Work
1. 1.15 × 105 J = 115 kJ done against gravity
3. 1.5 J
2. 352 m
IV
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Module 6 Work–Kinetic Energy Theorem
1. 58.9 m/s
4. 71 m/s = 2.5 × 103 km/h
6. 7.0 × 101 kg
2. 2.83 × 103 J
5. 6.60 × 102 J
7. 2.70 × 102 m
3. 1.89 × 10−2 N, up
Module 7 Conservation of Momentum
1. 1.7 m/s
4. 5.3 km/h, away from the plane
2. 1.6 m
5. 66 kg
6. 5.3 km/h, away from the
skateboarder
7. 54 kg
3. 14 km/h, southeast
Module 8 Angular Kinematics
1. 23 rad/s
2. ∆q = 6.7 rad; a = −4.0 × 10−2 rad/s2
3. 6.0 × 10−6 rad/s2
2. 3.8 × 102 N • m, counterclockwise
3. 15.4 m
Module 9 Torque
1. 1.3 × 104 N • m, clockwise
Module 10 Rotational Inertia
1. 1.0 rad/s
2. 2.0 rad/s2
3. 210 N
Module 11 Hooke’s Law
2. 192 N
4. 3.22 kg
6. 49 N/m
2
5. 0.31 m = 3.1 × 10 mm
7. 3.73 N/m
3. 15.3 kN/m
Module 12 Frequency and Wavelength
1. 1780 wavelengths
2. 2.72 m
3. 16.7 kHz; This frequency is
audible to the human ear.
4. v = 1430 m/s; ∆t = 2.1 s
5. 131 Hz
Module 13 Doppler Effect
1. 1.21 × 102 Hz, 1.36 × 102 Hz
3. 5.06 × 102 Hz, 4.83 × 102 Hz
5. 1.19 × 107 m/s
2. 1.26 × 102 Hz, 1.30 × 102 Hz
4. 2180 Hz
6. 685 nm, 685 nm (no change)
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1. 377 kg
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Module 14 Reflection
1. 0.800 m
4. M = 2.08; The image is inverted.
7. p = 5.85 m; The image is upright.
2. q = 360 m; h⬘ = −64 m; The image
is inverted and real.
5. q = 333 mm; h⬘ = 167 mm; The
image is real.
8. h = 2.00 mm; The image is virtual.
3. 20.6 m
6. 37.5 m
9. 6.00 cm
Module 15 Refraction
1. 6.3 m
4. q = 0.293 m = 29.3 cm; M = −0.098
7. f = 2.0 × 102 mm
2. 0.044 m = 44 mm
5. q = −20.4 m; M = 3.40
8. f = 9.0 cm
3. f = 0.185 m = 18.5 cm; The image
is inverted.
6. p = 4.5 cm; q = −18 cm
Module 16 Force Between Charges
1. 3.2 × 10−8 C = 32 nC
6. 2.95 × 10−25 N; 2.36 × 10−23 N
2. 1.6 × 1032 electrons in each cloud
7. 4.7 × 10−14 N; 4.7 × 10−9 C
3. 1.60 × 10−13 C; 2.30 × 10−22 N
8. 4.6 × 104 N
4. 120 m
9. 1.2 × 10−11 N
5. 5 × 10−10 C
Module 17 Electrical Circuits
1. a. ∆V3 = 73 V, I3 = 91 A
2. a. ∆V = 3.4 V, I = 3.4 A
b. ∆V2 = 37 V, I2 = 46 A
b. ∆V = 8.6 V, I = 2.9 A
c. ∆V = 4.9 V, I = 1.4 A
d. ∆V = 3.4 V, I = 0.85 A
Copyright © by Holt, Rinehart and Winston. All rights reserved.
c. ∆V4 = 37 V, I4 = 46 A
Module 18 Magnetic Field of a Wire
1. 2.35 × 10−11 T clockwise
3. 1.67 × 10−4 T, downward
5. 1.82 × 10−6 T
2. 9.00 A
4. south, 2.23 × 10−5 m
6. 0.512 A
Module 19 Magnetic Force on a Wire
3. 1.3 × 10−4 T
1. 0.20 N
−5
2. 7.4 × 10
4. 1.34 A
T
Module 20 Induction and Transformers
1. 4.1 × 103 V
2. 6.5 × 102 V
Section Four—Interactive Tutor Worksheets Answers
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Section
Problem Bank
Solutions
V
Holt Physics
V
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The Science of Physics
Chapter
1
Additional Practice 1A
Givens
1. mass = 6.0 × 103 kg
Solutions
103 g 1 mg
a. mass = 6.0 × 103 kg ×  × −3
= 6.0 × 109 mg
1 kg 10 g
103 g 1 Mg
b. mass = 6.0 × 103 kg ×  × 6 = 6.0 Mg
1 kg 10 g
2. Volume = 6.4 × 104 cm3
1m
a. volume = 6.4 × 104 cm3 × 2 
10 cm
3
1 m3
 = 6.4 × 10−2 m3
= 6.4 × 104 cm3 × 
106 cm3
10 mm
b. volume = 6.4 × 104 cm3 × 
1 cm
3
103 mm3
= 6.4 × 104 cm3 × 
= 6.4 = 107 mm3
1 cm3
3. energy = 4.2 × 109 J
1 MJ
a. energy = 4.2 × 109 J × 
= 4.2 × 103 MJ
106 J
1 GJ
b. energy = 4.2 × 109 J × 
= 4.2 GJ
109 J
4. distance = 1 parsec
= 3.086 × 1016 m
1 km
a. distance = 1 parsec = 3.086 × 1016 m × 3 = 3.086 × 1013 km
10 m
1 Em
b. distance = 1 parsec = 3.086 × 1016 m × 18 = 3.086 × 10–2 Em
10 m
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5. area = 1 acre
= 4.0469 × 103 m2
1 km
a. area = 1 acre = 4.0469 × 103 m2 × 3
10 m
2
1 km2
area = 4.0469 × 103 m2 × 6 2 = 4.0469 × 10−3 km2
10 m
102 cm
b. area = 1 acre = 4.0469 × 103 m2 × 
1m
2
104 cm2
area = 4.0469 × 103 m2 × 
= 4.0469 × 107 cm2
1 m2
6. electric charge = 15 C
103 mC
a. electric charge = 15 C ×  = 1.5 × 104 mC
1C
1 kC
b. electric charge = 15 C × 3 = 1.5 × 10–2 kC
10 C
Section Five—Problem Workbook Solutions
V
V Ch. 1–1
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Givens
Solutions
7. depth = 1.168 × 103 cm
1m
a. depth = 1.168 × 103 cm × 2  = 1.168 × 101 m = 11.68 m
10 cm
1 mm
1m
b. depth = 1.168 × 103 cm × 2  × −6 = 1.168 × 107 mm
10 cm 10 m
8. area = 0.344 279 km2
103 m
a. area = 0.344 279 km2 × 
1 km
2
103 m
b. area = 0.344 279 km2 × 
1 km
2
= 0.344 279 × 106 m2 = 3.442 79 × 105 m3
103 mm
× 
1m
2
= 0.344 279 × 1012 mm2
area = 3.442 79 × 1011 mm2
9. time = 4.50 × 109 years ×
365.25 days 24 h 3600 s
 ×  × 
1 year
1 day
1h
= 1.42 × 1017 s
10. time = 6.7 × 10−17 s
1 Gs
= 1.42 × 108 Gs
a. time = 1.42 × 1017 s × 
109 s
1 Ps
b. time = 1.42 × 1017 s × 1
= 1.42 × 102 Ps = 142 Ps
10 5 s
106 ms
a. time = 6.7 × 10−17 s ×  = 6.7 × 10−11 ms
1s
Copyright © Holt, Rinehart and Winston. All rights reserved.
1018 as
b. time = 6.7 × 10−17 s ×  = 6.7 = 101 as = 67 as
1s
V
V Ch. 1–2
Holt Physics Solution Manual
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Motion in One Dimension
Chapter
2
Additional Practice 2A
Givens
Solutions
1. ∆x = 3.33 km forward
∆t = 30.0 s
∆x 3.33 × 103 m
vavg =  =  = 111 m/s forward
∆t
30.0 s
vavg = (111 m/s)(3600 s/h)(10−3 km/m) = 4.00 × 102 km/h forward
2. ∆x = 15.0 km west
∆t = 15.3 s
15.0 km
15.0 km
∆x
= 3.53 × 103 km/h west
vavg =  =  = 
−3
1
h
4.2
5
×
1
0
h
∆t
(15.3 s) 
3600 s
3. ∆x = 4.0 m
∆t = 5.0 min
4.0 m
∆x
vavg =  =  = 48 m/h
1h
∆t
(5.0 min) 
60 min
4. ∆x = 3.20 × 104 km south
∆t = 122 days
∆x 3.20 × 104 km
vavg =  =  = 262 km/day south
∆t
122 days
5. ∆x1 = 1.70 × 104 km south
= +1.70 × 104 km
d = total distance traveled = magnitude ∆x1 + magnitude ∆x2 + magnitude ∆x3
2
∆x2 = 6.0 × 10 km north
= −6.0 × 102 km
∆x3 = 1.44 × 104 km south
= +1.44 × 104 km
∆t = 122 days
Copyright © Holt, Rinehart and Winston. All rights reserved.
d = 1.70 × 104 km + 6.0 × 102 km + 1.44 × 104 km
d = (1.70 + 0.060 + 1.44) × 104 km
d = 3.20 × 104 km
3.20 × 104 km
d
average speed =  =  = 262 km/day
122 days
∆t
∆xtot ∆x1 + ∆x2 + ∆x3
vavg = 
= 
∆t
∆t
(1.70 × 104 km) + (−6.0 × 102 km) + (1.44 × 104 km)
vavg = 
122 days
(1.70 − 0.060 + 1.44) × 104 km
vavg = 
122 days
3.08 × 104 km
vavg =  = +252 km/day = 252 km/day south
122 days
6. ∆x1 = 20.0 km east
= + 20.0 km
∆x2 = 20.0 km west
= − 20.0 km
∆x3 = 0 km
∆x4 = 40.0 km east
= +40.0 km
∆t = 60.0 min
∆x1 + ∆x2 + ∆x3 + ∆x4
∆xtot
a. vavg =  = 
∆t
∆t
(20.0 km) + (−20.0 km) + (0 km) + (40.0 km)
vavg = 
60.0 min
40.0 km
vavg =  = + 40.0 km/h = 40.0 km/h east
1h
(60.0 min) 
60 min
Section Five—Problem Bank
V
V Ch. 2–1
Givens
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Solutions
Print
b. d = total distance traveled
d = magnitude ∆x1 + magnitude ∆x2 + magnitude ∆x3 + magnitude ∆x4
d = 20.0 km + 20.0 km + 0 km + 40.0 km = 80.0 km
d
80.0 km
average speed =  =  = 80.0 km/h
∆t
1h
(60.0 min) 
60 min
7. v = 89.5 km/h north
∆x = vavg ∆t = v(∆t − ∆trest)
vavg = 77.8 km/h north
∆trest = 22.0 min
∆t(vavg − v) = −v∆trest
1h
1h
(89.5 km/h)(22.0 min) 
(89.5 km/h)(22.0 min) 
60
min
60 min
∆t = v∆rest

=  = 
11.7 km/h
89.5 km/h − 77.8 km/h
v − vavg
∆t = 2.80 h = 2 h, 48 min
8. v = 6.50 m/s downward
= −6.50 m/s
∆x = v∆t = (−6.50 m/s)(34.0 s) = −221 m = 221 m downward
∆t = 34.0 s
9. vt = 10.0 cm/s
∆xt = vt∆tt
2
vh = 20 vt = 2.00 × 10 cm/s
∆xh = vh∆th = vh (∆tt − 2.00 min)
∆trace = ∆tt
∆xt = ∆xrace = ∆xh + 20.0 cm
∆th = ∆tt − 2.00 min
vt ∆tt = vh (∆tt − 2.00 min) + 20.0 cm
∆xt = ∆xh + 20.0 cm = ∆xrace
∆tt (vt − vh) = −vh (2.00 min) + 20.0 cm
20.0 cm − (2.00 × 102 cm/s)(2.00 min)(60 s/min)
∆trace = ∆tt = 
10.0 cm/s − 2.00 × 102 cm/s
−2.40 × 104 cm
20.0 cm − 2.40 × 104 cm
∆trace = 
= 
2
−1.90 × 10 cm/s
−1.90 × 102 cm/s
∆trace = 126 s
10. ∆xrace = ∆xt
vt = 10.0 cm/s
∆xrace = ∆xt = vt∆tt = (10.0 cm/s)(126 s) = 1.26 × 103 cm = 12.6 m
∆tt = 126 s
Additional Practice 2B
1. ∆t = 6.92 s
vf = 17.34 m/s
∆v vf − vi
17.34 m/s − 0 m/s
aavg =  =  =  = 2.51 m/s2
∆t
∆t
6.92 s
vi = 0 m/s
V
2. vi = 0 m/s
2
vf = 7.50 × 10 m/s
∆t = 2.00 min
V Ch. 2–2
vf − vi
∆v
7.50 × 102 m/s − 0 m/s
aavg =  =  = 
= 6.25 m/s2
60 s
∆t
∆t
(2.00 min) 
1 min
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
20.0 cm − vh (2.00 min)
∆tt = 
vt − vh
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Givens
Solutions
3. vi = 0 m/s
vf = 0.85 m/s forward
vf − vi
∆v
0.85 m/s − 0 m/s
aavg =  =  =  = 0.23 m/s2 forward
∆t
3.7 s
∆t
∆t = 3.7 s
4. vi = 13.7 m/s forward
= +13.7 m/s
vf = 11.5 m/s backward
= −11.5 m/s
vf − vi
∆v
(− 11.5 m/s) − (13.7 m/s) −25.2 m/s
aavg =  =  =  = 
∆t
0.021 s
0.021 s
∆t
aavg = −1200 m/s2, or 1200 m/s2 backward
∆t = 0.021 s
5. vi = +320 km/h
1h
103 m
(0 km/h − 320 km/h)  
3600 s 1 km
∆v vf − vi
aavg =  =  = 
0.18
s
∆t
∆t
vf = 0 km/h
∆t = 0.18 s
−89 m/s
aavg =  = −490 m/s2
0.18 s
1h
103 m
(386.0 km/h − 0 km/h)  
vf − vi
3600 s 1 km
∆v
∆t =  =  = 
2
16.5 m/s
aavg
aavg
6. aavg = 16.5 m/s2
vi = 0 km/h
vf = 386.0 km/h
107.2 m/s
∆t = 2 = 6.50 s
16.5 m/s
7. vi = −4.0 m/s
vf = aavg ∆t + vi
2
aavg = −0.27 m/s
vf = (−0.27 m/s2)(17 s) + (−4.0 m/s) = −4.6 m/s − 4.0 m/s = −8.6 m/s
∆t = 17 s
8. vi = 4.5 m/s
vf − vi
∆v
6.3 m/s
10.8 m/s − 4.5 m/s
∆t =  =  = 
= 2 = 7.4 s
2
aavg
aavg
0.85 m/s
0.85 m/s
Copyright © Holt, Rinehart and Winston. All rights reserved.
vf = 10.8 m/s
aavg = 0.85 m/s2
9. vf = 296 km/h
1h
103 m
(296 km/h − 0 km/h)  
vf − vi
3600 s 1 km
∆v
82.2 m/s
∆t =  =  = 
= 2 = 51.4 s
2
1.60 m/s
aavg
aavg
1.60 m/s
vi = 0 km/h
2
aavg = 1.60 m/s
10. aavg = − 0.87 m/s2
∆v = aavg ∆t = (−0.87 m/s2)(3.85 s) = –3.35 m/s
∆t = 3.85
Additional Practice 2C
1. ∆t = 0.910 s
∆x = 7.19 km
2∆x
(2)(7.19 km)
vf =  − vi =  − 0 km/s = 15.8 km/s
∆t
0.910 s
vi = 0 km/s
V
Section Five—Problem Bank
V Ch. 2–3
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Givens
Solutions
2. vi = 4.0 m/s
∆t = 18 s
∆x = 135 m
2∆x
(2)(135 m)
vf =  − vi =  − 4.0 m/s = 15 m/s − 4.0 m/s = 11 m/s
∆t
18 s
10 m
3600 s
vf = (11 m/s) 
1h
1 km
3
vf = 4.0 × 101 km/h
3. ∆x = 55.0 m
∆t = 1.25 s
2∆x
(2)(55.0 m)
vi =  − vf =  − 43.2 m/s = 88.0 m/s − 43.2 m/s = 44.8 m/s
∆t
1.25 s
vf = 43.2 m/s
4. ∆x = 38.5 m
∆t = 5.5 s
2∆x
(2)(38.5 m)
vf =  − vi =  − 0 m/s = 14 m/s
∆t
5.5 s
vi = 0 m/s
5. ∆x = 478 km
∆vi = 72 km/h
∆t = 5 h, 39 min
6. ∆x = 4.2 m
∆t = 3.0 s
(2)(478 km)
2∆x
(2)(478 km)
vf =  − vi =  − 72 km/h =  − 72 km/h
1h
∆t
5 h + 0.65 h
5 h + 39 min 
60 min
956 km
vf =  − 72 km/h = 169 km/h − 72 km/h = 97 km/h
5.65 h
2∆x
(2)(4.2 m)
vi =  − vf =  − 1.3 m/s = 2.8 m/s − 1.3 m/s = 1.5 m/s
∆t
3.0 s
vf = 1.3 m/s
7. vi = 25 m/s west
vf = 35 m/s west
2∆x
(2)(250 m)
5.0 × 102 m
= 8.3 s
∆t =  =  = 
vi + vf 25 m/s + 35 m/s 6.0 × 101 m/s
8. vi = 755.0 km/h
vf = 777.0 km/h
∆t = 63.21 s
9. vi = 0 m/s
vf = 30.8 m/s
1
∆x =  (1532.0 km/h)(1.756 × 10−2 h) = 13.45 km
2
2∆x
(2)(493 m)
986 m
∆t =  =  =  = 32.0 s
vi + vf 0 m/s + 30.8 m/s
30.8 m/s
∆x = 493 m
10. ∆x = 1220 km
vi = 11.1 km/s
2∆x
2440 km
(2)(1220 km)
∆t =  =  =  = 107 s
vi + vf 11.1 km/s + 11.7 km/s 22.8 km/s
vf = 11.7 km/s
V
V Ch. 2–4
1h
1
1
∆x =  (vi + vf)∆t =  (755.0 km/h + 777.0 km/h)(63.21 s) 
3600 s
2
2
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
∆x = 250 m west
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Additional Practice 2D
Givens
1. ∆x = 12.4 m upward
∆t = 2.0 s
Solutions
2∆x (2)(12.4 m)
Because vi = 0 m/s, a = 2 = 
= 6.2 m/s2 upward
∆t
(2.0 s)2
vi = 0 m/s
2. ∆t = 1.5 s
vi = 2.8 km/h
vf = 32.0 km/h
1h
103 m
(32.0 km/h − 2.8 km/h)  
vf − vi
3600 s km
a =  = 
∆t
1.5 s
1h
103 m
(29.2 km/h)  
3600 s 1 km
a =  = 5.4 m/s2
1.5 s
3. ∆x = 18.3 m
∆t = 2.74 s
2∆x (2)(18.3 m)
Because vi = 0 m/s, a = ᎏᎏ
= 
= 4.88 m/s2
∆t 2
(2.74 s)2
vi = 0 m/s
4. vi = 2.3 m/s
vf = 46.7 m/s
vf − vi
46.7 m/s − 2.3 m/s 44.4 m/s
a =  =  =  = 6.3 m/s2
∆t
7.0 s
7.0 s
∆t = 7.0 s
5. vi = 6.23 m/s
∆x = 255 m
∆t = 82 s
6. vi = 11 km/h
vf = 55 km/h
Copyright © Holt, Rinehart and Winston. All rights reserved.
∆ = 4.1 s
2(∆x − vi ∆t) (2)[255 m − (6.23 m/s)(82 s)]
= 
a = 
(82 s)2
∆t2
(2)(255 m − 510 m)
(2)(−255 m)
a = 
= ᎏᎏ
= −7.6 × 10−2 m/s2
6.7 × 103 s2
6.7 × 103 s2
1h
103 m
(55 km/h − 11 km/h)  
vf − vi
3600 s 1 km
a =  = 
∆t
4.1 s
1h
103 m
(44 km/h)  
3600 s 1 km
a =  = 3.0 m/s2
4.1 s
7. vi = 42.0 m/s southeast
∆t = 0.0090 s
∆x = 0.020 m/s southeast
2(∆x − vi ∆t)
(2)[0.020 m − (42.0 m/s)(0.0090 s)]
= 
a = 
2
∆t
(0.0090 s)2
(2)(−0.36 m)
(2)(0.020 m/s − 0.38 m)
a = 
= 
8.1 × 10−5 s2
8.1 × 10−5 s2
a = −8.9 × 103 m/s2, or 8.9 × 103 m/s2 northwest
8. ∆t = 28 s
a = 0.035 m/s2
vf = a∆t + vi = (0.035 m/s2)(28.0 s) + 0.76 m/s = 0.98 m/s + 0.76 m/s = 1.74 m/s
vi = 0.76 m/s
V
Section Five—Problem Bank
V Ch. 2–5
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Givens
Solutions
9. vi = 0 m/s
vf = 72.0 m/s north
a = 1.60 m/s2 north
∆t = 45.0 s
vf − vi 72.0 m/s − 0 m/s
a. ∆t =  = 
= 45.0 s
a
1.60 m/s2
1
1
b. ∆x = vi∆t + a∆t2 = (0 m/s)(45.0 s) +  (1.60 m/s2)(45.0 s)2 = 0 m + 1620 m
2
2
∆x = 1.62 km
10. vi = +4.42 m/s
vf = 0 m/s
a = −0.75 m/s2
∆t = 5.9 s
vf − vi 0 m/s − 4.42 m/s
−4.42 m/s
a. ∆t =  = 
= 2 = 5.9 s
−0.75 m/s
a
−0.75 m/s2
1
1
b. ∆x = vi∆t + a∆t2 = (4.42 m/s)(5.9 s) + (−0.75 m/s2)(5.9 s)2
2
2
∆x = 26 m − 13 m = 13 m
Additional Practice 2E
1. vi = 1.8 km/h
vf = 24.0 km/h
∆x = 4.0 × 102 m
1 h 2 103 m 2
[(24.0 km/h)2 − (1.8 km/h)2]  
1 km
3600 s
a =  = 
(2)(4.0 × 102 m)
2∆x
vf2 − vi2
1 h 2 103 m 2
(576 km2/h2 − 3.2 km2/h2)  
3600 s
1 km
a = 
2
8.0 × 10 m
1 h 2 103 m 2
(573 km2/h2)  
3600 s
1 km
a = 
= 5.5 × 10−2 m/s2
8.0 × 102 m
2. vf = 0 m/s
vf = 8.57 m/s
vf2 − vi2
73.4 m2/s2
(8.57 m/s)2 − (0 m/s)2
a =  =  =  = 1.88 m/s2
2∆x
39.06 m
(2)(19.53 m)
3. vi = 7.0 km/h
vf = 34.5 km/h
∆x = 95 m
1 h 2 103 m 2
[(34.5 km/h)2 − (7.0 km/h)2]  
1 km
3600 s
a =  = 
(2)(95 m)
2∆x
vf2 − vi2
1 h 2 103 m 2
(1190 km2/h2 − 49 km2/h2)  
3600 s
1 km
a = 
190 m
1 h 2 103 m 2
(1140 km2/h2)  
3600 s
1 km
a =  = 0.46 m/s2
190 m
4. ∆x = 2.00 × 102 m
vi = 9.78 m/s
V
vf = 10.22 m/s
V Ch. 2–6
vf2 − vi2
(10.22 m/s)2 − (9.78 m/s)2
104.4 m2/s2 − 95.6 m2/s2
 = 
= 
a= 
2
2∆x
(2)(2.00 × 10 m)
4.00 × 102 m
8.8 m2/s2
a = 
= 2.2 × 10−2 m/s2
4.00 × 102 m
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
∆x = 19.53 m
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Givens
5. ∆x = +42.0 m
vi = +153.0 km/h
vf = 0 km/h
Solutions
1 h 2 103 m 2
2
2

 
[(0
km/h)
−
(153.0
km/h)
]
3600 s
1 km
vf2 − vi2

a =  =
(2) (42.0 m)
2∆x
1 h 2 103 m 2
−(2.34 × 104 km2/h2)  
3600 s
1 km
a =  = −21.5 m/s2
(84.0 m)
6. vi = 50.0 km/h forward
= +50.0 km/h
vf = 0 km/h
a = 9.20 m/s2 backward
= −9.20 m/s2
1 h 2 103 m 2
[(0 km/h)2 − (50.0 km/h)2]  
3600 s
1 km
∆x =  = 
2a
(2)(−9.20 m/s2)
vf2 − vi2
1 h 2 103 m 2
−(2.50 × 103 km2/h2)  
3600 s
1 km
∆x = 
2
−18.4 m/s
∆x = 10.5 m = 10.5 m forward
7. a = 7.56 m/s2
∆x = 19.0 m
vi = 0 m/s
8. vi = 1.8 m/s
vf = 9.4 m/s
a = 6.1 m/s2
9. vi = 1.50 m/s to the right
= +1.50 m/s
Copyright © Holt, Rinehart and Winston. All rights reserved.
vf = 0.30 m/s to the right
= +0.30 m/s
a = 0.35 m/s2 to the left
= −0.35 m/s2
10. a = 0.678 m/s2
vf = 8.33 m/s
∆x = 46.3 m
vi2+
m/s
)2
+(2)
vf = 2a∆
x = (0
(7
.5
6m
/s
2)(1
9.
0m
)
vf = 28
s2 = ±16.9 m/s = 16.9 m/s
7m
2/
vf2 − vi2
88 m2/s2 − 3.2 m2/s2
(9.4 m/s)2 − (1.8 m/s)2
∆x =  = 
= 
2
2a
(2)(6.1 m/s )
(2)(6.1 m/s2)
85 m2/s2
∆x = 
= 7.0 m
(2)(6.1 m/s2)
vf2 − vi2
(0.30 m/s)2 − (1.50 m/s)2
∆x =  = 
(2)(−0.35 m/s2)
2a
9.0 × 10−2 m2/s2 − 2.25 m2/s2
∆x = 
−0.70 m/s2
−2.16 m2/s2
= +3.1 m = 3.1 m to the right
∆x = 
−0.70 m/s2
vf2−
)2
−(2)
m/s
vi = 2a∆
x = (8
.3
3m
/s
(0
.6
78
2)(4
6.
3m
)
vi = 69
m2/
s2
−62.
s2 = 6.
s2 = ±2.6 m/s = 2.6 m/s
.4
8m
2/
6m
2/
Additional Practice 2F
1. vi = 0 m/s
vf = 49.5 m/s downward
= 49.5 m/s
a = −9.81 m/s2
∆tot = −448 m
(−49.5 m/s)2 − (0 m/s)2
vf2 − vi2
2450 m2/s2
∆xi =  = 
= 
= −125 m
2
(2)(−9.81 m/s )
2a
(2)(−9.81 m/s2)
∆x2 = ∆xtot − ∆x1 = (−448 m) − (−125 m) = −323 m
distance from net to ground = magnitude ∆x2 = 323 m
Section Five—Problem Bank
V
V Ch. 2–7
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Givens
Solutions
2. vi = 0 m/s
a = −9.81 m/s2
∆t1 = 1.00 s
1
1
Because vi = 0 m/s, ∆x1 = a∆t12 = (−9.81 m/s2)(1.00 s)2 = −4.90 m
2
2
1
1
∆x2 = a∆t22 = (−9.81 m/s2)(2.00 s)2 = −19.6 m
2
2
∆t2 = 2.00 s
∆t3 = 3.00 s
1
1
∆x3 = a∆t32 = (−9.81 m/s2)(3.00 s)2 = −44.1 m
2
2
3. vi = 0 m/s
∆t = 2.0 s
1
1
Because vi = 0 m/s, ∆x = a∆t2 = (−9.81 m/s2)(2.0 s)2 = −2.0 × 101 m
2
2
a = −9.81 m/s2
distance of bag below balloon = 2.0 × 101 m
4. vi = +17.5 m/s
2
a = −9.81 m/s
∆ttot = 3.60 s
To find the ball’s maximum height, calculate ∆xup. The time for this distance to be
1
traveled is ∆tup = 2 ∆xtot
2
1
1
1 1
∆xup = vi ∆tup + a∆tup2 = vi ∆ttot + a ∆ttot
2
2
2 2
3.60 s 1
3.60 s 2
∆xup = (17.5 m/s)  + (−9.81 m/s2)  = (31.5 m) + (−15.9 m) = +15.6 m
2
2
2
∆xtot = 0 m
softball’s maximum height = 15.6 m
vf = 11.4 m/s downward
= −11.4 m/s
a = 3.70 m/s2 downward
= −3.70 m/s2
6. ∆ttot = 5.10 s
1
∆tdown =  ∆ttop
2
1.30 × 102 m2/s2
vf2 − vi2
(−11.4 m/s)2 − (0 m/s)2

 = 
∆x = 
=
= −17.6 m
−7.40 m/s2
2a
(2)(−3.70 m/s2)
∆x = 17.6 m downward
To find the ball’s maximum height, calculate the displacement from that height to its
1
original position. The time interval for this free-fall is 2 ∆ttot, and vi = 0 m/s.
2
vi = 0 m/s
1
1 1
∆xdown =  a∆tdown 2 = a ∆ttot
2
2 2
a = −9.81 m/s2
ball’s maximum height = 31.9 m
7. ∆ttot = 5.10 s
2
a = −9.81 m/s
∆xtot = 0 m
8. vi = 85.1 m/s upward
= + 85.1 m/s
a = −9.81 m/s2
∆x = 0 m
= −31.9 m
2
1
∆ttot = vi ∆ ttot + a∆ttot2
2
Because ∆xtot = 0,
1
1
vi = − a∆ttot = − (−9.81 m/s2)(5.10 s) = +25.0 m/s = 25.0 m/s upward
2
2
1
Because ∆x = 0 m, vi ∆t + a∆t 2 = 0
2
2v
(2)(85.1 m/s)
∆t = − −i = − −
= 17.3 s
a
(−9.81 m/s2)
V
V Ch. 2–8
1
5.10 s
= (−9.81 m/s2) 
2
2
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
5. vi = 0 m/s
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Givens
Solutions
9. ∆td = 4.00 s
∆tp = ∆td + 3.00 s
= 7.00 s
a = −9.81 m/s2
∆xd = ∆xp
1
∆xd = vi, d ∆td + a∆td2
2
1
∆xp = vi, p ∆tp + a∆tp2
2
1
1
vi, d ∆td + a∆td2 = vi, p∆tp + a∆tp2
2
2
vi, p = 0 m/s
vi, d
1
a(∆tp2 − ∆td2)
vi, p ∆tp
2

=  + 
∆td
∆td
Because vi, p = 0 m/s,
1
a(∆tp2 − ∆td2)
2
vi, d =  =
∆td
1
(−9.81 m/s2)[(7.00 s)2 − (4.00 s)2]
2

4.00 s
(9.81 m/s2)(49.0 s2 − 16.0 s2)
(9.81 m/s2)(33.0 s2)
vi, d = −  = −  = −40.5 m/s
8.00 s
8.00 s
vf, d = a ∆td + vi, d = (−9.81 m/s2)(4.00 s) + (−40.5 m/s) = (−39.2 m/s) + (−40.5 m/s)
vf, d = −79.7 m/s
10. aup = 3.125 m/s2, upward
= +3.125 m/s2
∆tup = 4.00 s
vf = aup ∆tup + vi = (3.125 m/s2) (4.00 s) + 0 m/s = 2.5 m/s
When the cable brakes, the upward-moving elevator undergoes free-fall acceleration.
vi = 0 m/s
vi = +12.5 m/s
2
a = −9.81 m/s
∆t1 = 0.00 s
Copyright © Holt, Rinehart and Winston. All rights reserved.
∆t2 = 1.00 s
∆t3 = 2.00 s
∆t4 = 3.00 s
vf, 1 = a∆t1 + vi = (−9.81 m/s2)(0.00 s) + 12.5 m/s = +12.5 m/s
vf, 2 = a∆t2 + vi = (−9.81 m/s2)(1.00 s) + 12.5 m/s = −9.81 m/s + 12.5 m/s = +2.7 m/s
vf, 3 = a∆t3 + vi = (−9.81 m/s2)(2.00 s) + 12.5 m/s = −19.6 m/s + 12.5 m/s = −7.1 m/s
vf, 4 = a∆t4 + vi = (−9.81 m/s2)(3.00 s) + 12.5 m/s = −29.4 m/s + 12.5 m/s = − 16.9 m/s
V
Section Five—Problem Bank
V Ch. 2–9
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Two -Dimensional Motion and Vectors
Chapter
3
Additional Practice 3A
Givens
1. ∆x1 = 8 m to the left = +8 m
∆x2 = 8 m to the right
= −8 m
Solutions
a. distance traveled = 8 m + 8 m + 8 m = 24 m
b. d = ∆x1 + ∆x2 + ∆x3 = 8 m + (−8 m) + 8 m = 8 m
∆x3 = 8 m to the left = +8 m
2. hi = 0.91 m
∆y = hf − hi = (0.90 − 1.00)hi
hf = (0.90)hi
∆y = (−0.10)(0.91 m) = −9.1 × 10−2 m
∆x = 0.11 m
d= ∆
∆y2 = (0
)2
+(−
)2 = 1.
0−2m
0−3m
x2+
.1
1m
9.
1×10−2m
2×1
2+8.3
×1
2
d = 2.0 × 10−2 m2 = 0.14 m
∆y
−9.1 × 10−2 m
q = tan−1  = tan−1 
∆x
0.11 m
q = −4.0 × 101° = 4.0 × 101° below the horizontal
3. ∆x = 165 m
∆y = −45 m
d= ∆
∆y2 = (1
m
)2
+(−
m
)2 2.
04
m2+
03
m2 = 2.
04
m2
x2+
65
45
72
×1
2.0
×1
92
×1
d = 171 m
∆y
−45 m
q tan−1  = tan−1 
∆x
165 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = −15° = 15° below the horizontal
4. ∆y = −13.0 m
∆x = 9.0 m
d= ∆
∆y2 = (9
m
)2
+(−
m
)2 = 81
m2+
m2 = 2.
02
m2
x2+
.0
13
.0
169
50
×1
d = 15.8 m
∆y
−13.0 m
q = tan−1  tan−1 
∆x
9.0 m
q = −55° = 55° below the horizontal
5. ∆x = 36.0 m, east
∆y = 42.0 m, north
∆z = 17.0 m, up
d= ∆
∆y2+
∆
z 2 = (3
)2
+(42
m
)2
+(17
m
)2
x2+
6.
0m
.0
.0
d = 1.
03
m2+
03
m2+
m2 = 3.
03
m2
30
×1
1.7
6×1
289
35
×1
d = 57.9 m
∆y
42.0 m
horizontal direction = qh = tan−1  = tan−1 
∆x
36.0 m
qh = 49.4° north of east
∆z

vertical direction = qv = tan−1 
∆x 2 + ∆y 2
Section Five—Problem Bank
V
V Ch. 3–1
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Givens
Solutions
17.0 m


3.
0
m
06
×1
17.0 m
17.0 m
 = tan−1 
qv = tan−1 3 2
2
2
1.
30
×
1
0
m+
m2
(3
6.
0
m
)
+
(
42
.0
m
)
1.7
6×103
= tan−1
3
2
qv = 17.1° above the horizontal
6. d = 599 m
d 2 = ∆x 2 + ∆y 2
∆y = 89 m north
∆x = d 2−
∆y2 = (5
m
)2
−(89
m
)2 = 3.
05
m2−
03
m2
99
59
×1
7.9
×1
∆x = 3.
05
m2
51
×1
∆x = 592 m, east
∆y
89 m
q = sin−1  = sin−1 
d
599 m
q = 8.5 ° north of east
7. d = 478 km
∆y = 42 km, south = −42 km
d 2 = ∆x 2 + ∆y 2
∆x = d 2−
∆
y 2 = (4
)2
−(−
)2 2.
05km
03km
78
km
42
km
28
×1
2−1.8
×1
2
∆x = 2.
05km
26
×1
2 = −475 km
∆x = 475 km, west
∆y
−42 km
q = sin−1  = sin−1 
d
478 km
q = 5.0° south of west
q = 26° south of west
∆y = 3200 km, south =
−3200 km
d2 = ∆x 2 = ∆y 2
∆x = d2−
∆
y 2 = (7
40
0km
)2−(−32
00
km
)2 = 5.
5×107km
2−1.0
×107km
2
∆x = 4.
07km
5×1
2 = −6700 km
∆x = 6700 km, west
9. d = 3.88 km
∆x = 3.45 km
h1 = 0.8 km
d 2 = ∆x 2 + ∆y 2
∆y = 1.8 km
height of mountain = h = ∆y + h1 = 1.8 km + 0.8 km
h = 2.6 km
V
V Ch. 3–2
∆y = d 2−
∆x2 = (3
)2
−(3.
)2 = 15
.8
8km
45
km
.1
km
2−11.
9km
2 = 3.
2km
2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. d = 7400 km
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Givens
10. d = 2.9 × 103 km
∆x = 2.8 × 103 km, west
= −2.8 × 103 km
Solutions
d 2 = ∆x 2 + ∆y 2
∆y = d 2−
∆x2 = (2
03km
)2
−(−
03
)2
.9
×1
2.
8×1
∆y = 8.
4×106km
2−7.8
×106km
2 = 0.
6×106km
2 = −800 km
∆y = 800 km, south
∆x
−2.8 × 103 km
q = cos−1  = cos−1 
d
2.9 × 103 km
q = 15° south of west
Additional Practice 3B
1. d = 5.3 km
q = 8.4° above horizontal
∆y = d(sin q) = (5.3 km)(sin 8.4°)
∆y = 0.77 km = 770 m
the mountain’s height = 770 m
2. d = 19.1 m
q = 3.0° to the left
∆y = d(sin q) = (19.1 m)(sin 3.0°)
∆y = 1.0 m to the left
the lane’s width = 1.0 m
3. d = 113 m
q = 82.4° above the
horizontal south
4. v = 55 km/h
q = 37° below the horizontal
= −37°
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. d = 2.7 m
q = 13° from the table’s
length
∆x = d(cos q) = (113 m)(cos 82.4°)
∆x = 14.9 m, south
vy = v(sin q) = (55 km/h)[sin(−37°)]
vy = −33 km/h = 33 km/h, downward
∆x = d(cos q) = (2.7 m)(cos 13°)
∆x = 2.6 m along the table’s length
∆y = d(sin q) = (2.7 m)(sin 13°)
∆y = 0.61 m along the table’s width
6. v = 1.20 m/s
q = 14.0° east of north
vx = v(sin q) = (1.20 m/s)(sin 14.0°)
vx = 0.290 m/s, east
vy = v(cos q) = (1.20 m/s)(cos 14.0°)
vy = 1.16 m/s, north
7. d = 31.2 km
q = 30.0° west of south
∆x = d (sin q) = (31.2 km)(sin 30.0°)
∆x = 15.6 km, west
∆y = d (cos q) = (31.2 km)(cos 30.0°)
∆y = 27.0 km, south
V
Section Five—Problem Bank
V Ch. 3–3
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Givens
Solutions
8. v = 165.2 km/s
q = 32.7°
vforward = v(cos q) = (165.2 km/s)(cos 32.7°)
vforward = 139 km/s, forward
vside = v(sin q) = (165.2 km/s)(sin 32.7°)
vside = 89.2 km/s to the side
9. v = 55.0 km/h
q = 13.0° above horizontal
vy = v(sin q) = (55.0 km/h)(sin 13.0°)
vy = 12.4 km/h, upward
vx = v(cos q) = (55.0 km/h)(cos 13.0°)
vx = 53.6 km/h, forward
10. v = 13.9 m/s
vz = v(sin qv) = (13.9 m/s)(sin 26.0°)
qh = 24.0° east of north
vz = 6.09 m/s, upward
qv = 26.0° above the
horizontal
horizontal velocity = vh = v(cos qv)
vy = vh(cos qh) = v(cos qv)(cos qh) = (13.9 m/s)(cos 26.0°)(cos 24.0°)
vy = 11.4 m/s, north
vx = vh(sin qh) = v(cos qv)(sin qh) = (13.9 m/s)(cos 26.0°)(sin 24.0°)
vx = 5.08 m/s, east
Additional Practice 3C
1. d1 = 55 km
∆x1 = d1(cos q1) = (55 km)(cos 37°) = 44 km
q1 = 37 north of east
∆y1 = d1(sin q1) = (55 km)(sin 37°) = 33 km
d2 = 66 km
∆x2 = d2(cos q2) = (66 km)(cos 0.0°) = 66 km
q2 = 0.0° (due east)
∆y2 = d2(sin q2) = (66 km)(sin 0.0°) = 0 km
∆ytot = ∆y1 + ∆y2 = 33 km + 0 km = 33 km
d = (∆
xt
)2
+(∆
yt
)2 = (1
)2
+(33
)2
10
km
km
ot
ot
= 1.
04km
03km
04km
21
×1
2+1.1
×1
2= 1.
32
×1
2
d = 115 km
∆ytot
33 km
q = tan−1 
= tan−1 
∆xtot
110 km
q = 17° north of east
V
V Ch. 3–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆xtot = ∆x1 + ∆x2 = 44 km + 66 km = 110 km
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Givens
2. d1 = 4.1 km
Solutions
∆x1 = d1(cos ∆1) = (4.1 km)(cos 180°) = −4.1 km
q1 = 180° (due west)
∆y1 = d1(sin q1) = (4.1 km)(sin 180°) = 0 km
d2 = 17.3 km
∆x2 = d2(cos q2) = (17.3 km)(cos 90.0°) = 0 km
q2 = 90.0° (due north)
∆y2 = d2(sin q2) = (17.3 km)(sin 90.0°) = 17.3 km
d3 = 1.2 km
q3 = 24.6° west of north
= 90.0° + 24.6° = 114.6°
∆x3 = d3(cos q3) = (1.2 km)(cos 114.6°) = −0.42 km
Dy3 = d3(sin q3) = (1.2 km)(sin 114.6°) = 1.1 km
∆xtot = ∆x1 + ∆x2 + ∆x3 = −4.1 km + 0 km + (−0.42 km) = −4.5 km
∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 km + 17.3 km + 1.1 km = 18.4 km
d = (∆
xtot
)2
+(∆
ytot
)2 = (−
)2
+(18
)2
4.
5km
.4
km
= 2.
01km
0×1
2+339
km
2 = 35
9km
2
d = 18.9 km
∆ytot
18.4 km
= tan−1  = −76° = 76° north of west
q = tan−1 
∆xtot
−4.5 km
3. d1 = 850 m
∆x1 = d1(cos q1) = (850 m)(cos 0.0°) = 850 m
q1 = 0.0°
∆y1 = d1(sin q1) = (850 m)(sin 0.0°) = 0 m
d2 = 640 m
∆x2 = d2(cos q2) = (640 m)(cos 36°) = 520 m
q2 = 36°
∆y2 = d2(sin q2) = (640 m)(sin 36°) = 380 m
∆xtot = ∆x1 + ∆x2 = 850 m + 520 m = 1370 m
∆ytot = ∆y1 + ∆y2 = 0 m + 380 m = 380 m
d=
=
(∆
xt
)2
+(∆
yt
)2 =
ot
ot
37
(1
0m
)2
+(38
0m
)2 =
9×1
1.
06
m2+
1.4
×1
05
m2
0×1
2.
06
m2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 1400 m
∆ytot
380 m
q = tan−1 
= tan−1 
∆xtot
1370 m
q = 16° to the side of the initial displacement
V
Section Five—Problem Bank
V Ch. 3–5
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Givens
Solutions
4. d1 = 2.00 × 102 m
∆x1 = d1(cos q1) = (2.00 × 102 m)(cos 0.02) = 2.0 × 102 m
q1 = 0.0°
∆y1 = d1(sin q1) = (2.00 × 102 m)(sin 0.0°) = 0 m
d2 = 3.00 × 102 m
∆x2 = d2(cos q2) = (3.00 × 102 m)(cos 3.0°) = 3.0 × 102 m
q2 = 3.0°
∆y2 = d2(sin q2) = (3.00 × 102 m)(sin 3.0°) = 16 m
2
d3 = 2.00 × 10 m
q3 = 8.8°
∆x3 = d3(cos q3) = (2.00 × 102 m)(cos 8.8°) = 2.0 × 102 m
∆y3 = d3(sin q3) = (2.00 × 102 m)(sin 8.8°) = 31 m
∆xtot = ∆x1 + ∆x2 + ∆x3 = 2.0 × 102 m + 3.0 × 102 m + 2.0 × 102 m = 7.0 × 102 m
∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 m + 16 m + 31 m = 47 m
d = (∆
xtot
m)2
+(47
m)2 = 4.
m2+
)2+(∆
ytot)2 = (7
.0
×102
9×105
2.2
×10
3 m2
= 4.
05
m2
9×1
d = 7.0 × 102 m
∆ytot
47 m
q = tan−1 
= tan−1 
∆xtot
7.0 × 102 m
q = 3.8° above the horizontal
5. d1 = 46 km
q1 = 15° south of east
= −15°
d2 = 22 km
∆x1 = d1(cos q1) = (46 km)[cos(−15°)] = 44 km
∆y1 = d1(sin q1) = (46 km)[sin(−15°)] = −12 km
∆x2 = d2(cos q2) = (22 km)[cos(−77°)] = 4.9 km
∆y2 = d2(sin q2) = (22 km)[sin(−77°)] = −21 km
q2 = 13° east of south
= −77°
∆x3 = d3(cos q3) = (14 km)[cos(−104°)] = −3.4 km
d3 = 14 km
∆y3 = d3(sin q3) = (14 km)[sin(−104°)] = −14 km
q3 = 14° west of south
= −90.0° − 14° = −104°
∆xtot = ∆x1 + ∆x2 + ∆x3 = 44 km + 4.9 km + (−3.4 km) = 46 km
∆ytot = ∆y1 + ∆y2 + ∆y3 = −12 km + (−21 km) + (−14 km) = −47 km
= 4.
03km
3×1
2
d = 66 km
∆ytot
−47 km
= tan−1  = −46°
q = tan−1 
∆xtot
46 km
q = 46° south of east
V
V Ch. 3–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = (∆
xt
)2
+(∆
yt
)2 = (4
)2
+(−
)2 = 2.
03km
03km
6km
47
km
1×1
2+2.2
×1
2
ot
ot
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Givens
Solutions
6. d1 = 6.3 × 108 km
∆x1 = d1(cos q1) = (6.3 × 108 km)(cos 0.0°) = 6.3 × 108 km
q1 = 0.0°
∆y1 = d1 (sin q1) = (6.3 × 108 km)(sin 0.0°) = 0 km
d2 = 9.4 × 108 km
∆x2 = d2(cos q2) = (9.4 × 108 km)(cos 68°) = 3.5 × 108 km
q2 = 68°
∆y2 = d2 (sin q2) = (9.4 × 108 km)(sin 68°) = 8.7 × 108 km
9
d3 = 3.4 × 10 km
q3 = 94° + 68° = 162°
∆x3 = d3 (cos q3) = (3.4 × 109 km)(cos 162°) = −3.2 × 109 km
∆y3 = d3 (sin q3) = (3.4 × 109 km)(sin 162°) = 1.1 × 109 km
∆xtot = ∆x1 + ∆x2 + ∆x3 = 6.3 × 108 km + 3.5 × 108 km + (−3.2 × 109 km) = −2.2 × 109 km
∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 km + 8.7 × 103 km + 1.1 × 109 km = 2.0 × 109 km
2
d = (x
ytot
to
(
)2 = (−
2.
2×109km
)2+(2.
0×109km
)2
t)+
= 4.
018km
018km
018km
8×1
2+4.0
×1
2 = 8.
8×1
2
d = 3.0 × 109 km
∆ytot
2.0 × 109 km
= tan−1 
= − 42°
q = tan−1 
∆xtot
−2.2 × 109 km
180.0° − 42° = 138°
q = 138° from the probe’s initial direction
7. di = 2.50 × 103 m
∆x1 = d1(cos q1) = (2.50 × 103 m)(cos 58.5°) = 1310 m
q1 = 58.5° north of east
∆y1 = d1(sin q1) = (2.50 × 103 m)(sin 58.5°) = 2130 m
d2 = 375 m
∆x2 = d2(cos q2) = (375 m)(cos 21.8°) = 348 m
q2 = 21.8° north of east
∆y2 = d2(sin q2) = (375 m)(sin 21.8°) = 139 m
d3 = 875 m
q3 = 21.5° east of north
∆x3 = d3(sin q3) = (875 m)(sin 21.5°) = 321 m
∆y3 = d3(cos q3) = (875 m)(cos 21.5°) = 814 m
∆xtot = ∆x1 + ∆x2 + ∆x3 = 1310 m + 348 m + 321 m = 1.98 × 103 m
∆ytot = ∆y1 + ∆y2 + ∆y3 = 2130 m + 139 m + 814 m = 3.08 × 103 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = (∆
xtot
)2
+(∆
ytot
)2 = (1
03
m
)2
+(3.
03
m
)2
.9
8×1
08
×1
= 3.
06
m2+
06
m2 = 13
06
m2
92
×1
9.4
9×1
.4
1×1
d = 3.66 × 103 m
3.08 × 103 m
∆ytot
= tan−1 
q = tan−1 
∆xtot
1.98 × 103 m
q = 57.3° north of east
V
Section Five—Problem Bank
V Ch. 3–7
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Givens
Solutions
8. d1 = 5.0 km
q1 = 36.9° south of east
= −36.9°
d2 = 1.5 km
q2 = 90.0° due south
= −90.0°
d3 = 8.5 km
q3 = 42.2° south of east
= −42.2°
d4 = 0.8 km
q4 = 0° (due east)
∆x1 = d1(cos q1) = (5.0 km)[cos(−36.9°)] = 4.0 km
∆y1 = d1(sin q1) = (5.0 km)[sin(−36.9°)] = −3.0 km
∆x2 = d2(cos q2) = (1.5 km)[cos(−90.0°)] = 0 km
∆y2 = d2(sin q2) = (1.5 km)[sin(−90.0°)] = −1.5 km
∆x3 = d3(cos q3) = (8.5 km)[cos(−42.2°)] = 6.3 km
∆y3 = d3(sin q3) = (8.5 km)[sin(−42.2°)] = −5.7 km
∆x4 = d4(cos q4) = (0.8 km)(cos 0°) = 0.8 km
∆y4 = d4(sin q4) = (0.8 km)(sin 0°) = 0 km
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = 4.0 km + 0 km + 6.3 km + 0.8 km = 11.1 km
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = (− 3.0 km) + (−1.5 km) + (−5.7 km) + 0 km = −10.2 km
d = (∆
xtot
)2
+(∆
ytot
)2 = (1
)2
+(−
)2 = 12
1.
1km
10
.2
km
3km
2+104
km
2
= 22
7km
2
d = 15.1 km
∆ytot
−10.2 km
q = tan−1 
= tan−1  = −42.6°
∆xtot
11.1 km
q = 42.6° south of east
q1 = 45.0° west of north
= 90.0° + 45.0° = 135.0°
d2 = 1.98 m
q2 = 45.0° east of north
= 45.0°
∆x1 = d1(cos q1) = (1.41 m)(cos 135.0°) = −0.997 m
∆y1 = d1(sin q1) = (1.41 m)(sin 135.0°) = 0.997 m
∆x2 = d2(cos q 2) = (1.98 m)(cos 45.0°) = 1.40 m
∆y2 = d2(sin q 2) = (1.98 m)(sin 45.0°) = 1.40 m
∆x3 = d3(cos q3) = (0.42 m)(cos 135.0°) = −0.30 m
d3 = 0.42 m
∆y3 = d3(sin q3) = (0.42 m)(sin 135.0°) = 0.30 m
q3 = 45.0° west of north
= 135.0°
∆x4 = d4(cos q4) = (1.56 m)(cos 225.0°) = −1.10 m
d4 = 1.56 m
∆y4 = d4(sin q4) = (1.56 m)(sin 225.0°) = −1.10 m
q4 = 45.0° south of west
= 180.0° + 45.0° = 225.0°
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (−0.997 m) + 1.40 m + (−0.30 m) + (−1.10 m)
= −0.997 m
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = 0.997 m + 1.40 m + 0.30 m + (−1.10 m) = 1.60 m
2
d = (∆
xt
)2
+(∆
yt
)2
+(1.
m
)2 = 0.
0.
99
7m
60
99
4m
2+2.5
6m
2
ot
ot = (−
= 3.
m2
55
d = 1.88 m
∆ytot
1.60 m
= tan−1  = −58.1°
q = tan−1 
∆xtot
−0.997 m
q = 58.1° north of west
V
V Ch. 3–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. d1 = 1.41 m
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Givens
Solutions
10. d1 = 79 km
∆x1 = d1(cos q1) = (790 km)(cos 162°) = −750 km
q1 = 18° north of west
180.0° − 18° = 162°
d2 = 150 km
q2 = 180.0° due west
∆y1 = d1(sin q1) = (790 km)(sin 162°) = 24 km
∆x2 = d2(cos q2) = (150 km)(cos 180.0°) = −150 km
∆y2 = d2(sin q2) = (150 km)(sin 180.0°) = 0 km
d3 = 470 km
∆x3 = d3(cos q3) = (470 km)(cos 90.0°) = 0 km
q3 = 90.0° due north
∆y3 = d3(sin q3) = (470 km)(sin 90.0°) = 470 km
d4 = 240 km
∆x4 = d4(cos q4) = (240 km)(cos 75°) = 62 km
q4 = 15° east of north
90.0° − 15° = 75°
∆y4 = d4(sin q4) = (240 km)(sin 75°) = 230 km
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (−750 km) + (−150 km) + 0 km + 62 km = −840 km
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = 240 km + 0 km + 470 km + 230 km = 940 km
d = (∆
xtot
)2
+(∆
ytot
)2 = (−
)2
+(94
)2 = 7.
05km
05km
84
0km
0km
1×1
2+8.8
×1
2
= 15
.9
×105km
2
d = 1260 km
∆ytot
940 km
= tan−1  = −48°
q = tan−1 
∆xtot
−840 km
q = 48° north of west
Additional Practice 3D
1. vx = 430 m/s
∆x = 4020 m
g = 9.81 m/s2
∆x
∆t = 
vx
1
∆y = − 2g∆t 2
2
2
4020 m
1 ∆x
1
∆y = − 2g  = − 2(9.81 m/s2)  = −430 m
vx
430 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
height of ridge = 430 m
2. ∆x = 101 m
vx = 14.25 m/s
2
g = 9.81 m/s
∆x
∆t = 
vx
1
∆y = − 2g∆t 2
2
2
101 m
1 ∆x
1
y = − 2g  = − 2(9.81 m/s2)  = 246 m
14.25 m/s
vx
height of building = 246 m
3. vx = 1.30 × 102 km/h
∆x = 135 m
∆x
∆t = 
vx
1
2
g = 9.81 m/s
∆y = − 2g∆t 2
2
2
135 m
1 ∆x
1
∆y = − 2g  = − 2(9.81 m/s2) 
vx
1.30 × 102 km/h
2
3600 s/h

 = −68.6 m
103 m/km
airship’s altitude = 68.6 m
V
Section Five—Problem Bank
V Ch. 3–9
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Givens
Solutions
4. vx = 9.37 m/s
∆x = 85.0 m
g = 9.81 m/s2
∆x
∆t = 
vx
1
∆y = − 2g∆t 2
2
2
85.0 m
1 ∆x
1
∆y = − 2g  = − 2(9.81 m/s2)  = − 404 m
vx
9.37 m/s
mountain’s height = 404 m
5. vx = 6.32 cm/s
∆x = 1.00 m
g = 9.81 m/s2
∆x
∆t = 
vx
1
∆y = − 2g∆t 2
2
2
1.00 m
1 ∆x
1
∆y = − 2g  = − 2(9.81 m/s2) 
= 1230 m
vx
6.32 × 10−2 m/s
building’s height = 1230 m
6. vx = 10.0 cm/s
∆x = 18.6 cm
g = 9.81 m/s2
∆x
∆t = 
vx
1
∆y = − 2g∆t 2
2
2
2
18.6 cm
1 ∆x
1
∆y = − 2g  = − 2(9.81 m/s2)  = −17.0 m
vx
10.0 cm/s
squirrel’s height = 17.0 m
∆x = 3.50 m
g = 9.81 m/s2
∆x
∆t = 
vx
1
∆y = − 2g∆t 2
2
3.50 m
1 ∆x
1
∆y = − 2g  = − 2(9.81 m/s2)  = −26.7 m
vx
1.50 m/s
the lunch pail falls 26.7 m
8. ∆y = −2.50 × 102 m
vx = 1.50 m/s
g = 9.81 m/s2
∆x = vx∆t
1
∆y = − 2g∆t2
2∆y
−g
2∆y
(2)(−2.50 × 10 m)
∆x = v  = (1.50 m/s) 
−9.81
−g
m/
s ∆t =
2
x
∆x = 10.7 m
V
V Ch. 3–10
Holt Physics Solution Manual
2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. vx = 1.50 m/s
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Givens
Solutions
9. vx = 1.50 m/s
∆y = −2.50 × 102 m
2
g = 9.81 m/s
vy,f 2 = −2g∆y + vy,i2
vy,i = 0 m/s, so
vy,f = vy = −
m/s
02
m)
2g
∆
y = −(2
)(
9.
81
2)(−
2.
50
×1
vy = 70.0 m/s
2
)2
+(70
m/s
)2 = 2.
m2/
s2
+4.9
03
m2/
s2
v = vx2+
v
.5
0m
/s
.0
25
0×1
y = (1
= 4.
03
m2/
s2
90
×1
v = 70.0 m/s
v
1.50 m/s
q = tan−1 x = tan−1 
vy
70.0 m/s
q = 1.23° from the vertical
10. vx = 85.3 m/s
∆t =
x
∆y = −1.50 m
g = 9.81 m/s2
2∆y ∆x
−g = v
∆x = vx
2∆y
 = (85.3 m/s)
−g
(2)(−1.50 m)

−9.81
m/s = 47.2 m
2
range of arrow = 47.2 m
Additional Practice 3E
1. vi = 15.0 m/s
∆x = 17.6 m
∆x
∆t = 
vi(cos q)
1
∆y = vi(sin q)∆t − 2g∆t 2 = 0
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆x
1
1
vi(sin q) = 2g∆t = 2g 
vi(cos q)
g∆x
2(sin q)(cos q) = 
vi 2
Using the identity 2 (sin q)(cos q) = sin (2 q),
g∆x
sin (2 q) = 
vi 2
g∆x
(9.81 m/s2)(17.6 m)

sin−1 
sin
−
vi 2
(15.0 m/s)2
q = 
= 
2
2
q = 25.1°
2. vi = 23.1 m/s
∆ymax = 16.9 m
2
g = 9.81 m/s
vy,f 2 − vy,i2 = −2g∆y
At maximum height, vy,f
vy,i = vi (sin q) = 2g
ymax
∆
−1
q = sin
2g
∆
y
m
ax
(2
m/s
)(
9.
81
2)(1
6.
9m
)

= sin 
23.1 m/s
v
−1
i
q = 52.0°
V
Section Five—Problem Bank
V Ch. 3–11
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Givens
Solutions
3. ∆x = 7.49 m
vi = 9.50 m/s
g = 9.81 m/s2
Using the form of the equation derived in problem 1,
g∆x
2(sin q)(cos q) = sin(2 q) = 
vi2
(9.81 m/s2)(7.49 m)
g∆x


2
(9.50 m/s2
−1 vi
−1 
q = sin  = sin
2
2
q = 27.3°
4. vi = 141 cm/s
∆x = 18.5 cm
g = 9.81 m/s2
Using the form of the equation derived in problem 1,
g∆x
2(sin q)(cos q) = sin(2 q) = 
vi 2
g∆x

vi2
q = sin−1
2
(9.81 m/s2)(18.5 × 10−2 m)
sin−1 
(141 × 10−2 m/s)2
2
q = 33.0°
5. vi = 6.03 m/s
vy,f 2 − vy,i 2 = −2g∆y
hi = 10.0 m
At maximum height, vy,f = 0.
hmax = hf = 11.7 m
vy,i = vi(sin q) = 2g
ymax
∆
∆x = 3.62 m
For the diver, hf is the maximum height above the diving board.
2
g = 9.81 m/s
∆y = hf − hi
2g
(hf
−
hi)
(2
)(
9.
81
)m
/s
2)(1
1.
7m
−10.
0m
)
q = sin−1  = sin−1 
vi
6.03 m/s
6. vi = 10.0 m/s
∆x = vi(cos q)∆t = (10.0 m/s)(cos 37.0°)(2.5 s)
q = 37.0°
∆x = 2.0 ⫻ 101 m
∆t = 2.5 s
∆y = vi(sin q)∆t − 2g∆t 2 = (10.0 m/s)(sin 37.0°)(2.5 s) − 2(9.81 m/s2)(2.5 s)2
1
g = 9.81 m/s2
= 15 m − 31 m
∆y = −16 m
7. vi = 250 m/s
At the maximum height
q = 35°
vy,f = vy,i − g∆t = 0
2
g = 9.81 m/s
vy,i = vi(sin q) = g∆t
vi(sin q) (250 m/s)(sin 35°)
 = 
∆t = 
g
9.81 m/s2
∆t = 15 s
V
V Ch. 3–12
Holt Physics Solution Manual
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 73.3°
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Givens
Solutions
8. ∆x = 73.0 m
∆y = −52.8 m
q = −8.00°
2
g = 9.81 m/s
∆x
∆t = 
vi(cos q)
2
∆x
∆x
1
1
∆y = vi(sin q)∆t − 2g∆t 2 = vi (sin q)  − 2g 
vi (cos q)
vi(cos q)
2
g∆x

∆y = ∆x(tanq) − 
2 vi2(cos q)2
g∆x2

 = ∆x(tan q) − ∆y
2 vi2(cos q)2
g∆x2
2vi2(cos q)2 = 
[∆x(tan q) − ∆y]
g∆x2

2
2(cos q) [∆x(tan q) − ∆y]
vi =
vi =

(2)[cos(−8.00°)] [(73.0 m)(tan[−8.00°]) − (−52.8 m)]
vi =
vi =
(9.81 m/s2)(73.0 m)2
2
(9.81 m/s2)(73.0 m)2

(2)[cos(−8.00°)]2 (−10.3 m + 52.8 m)
(9.81 m/s2)(73.0 m)2

(2)[−cos(–8.00°)]2 (42.5 m)
vi = 25.0 m/s
9. q = −30.0°
vi = 2.0 m/s
∆y = −45 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
1
∆y = vi (sin q)∆t − 2g∆t 2
2 ∆t 2 − [vi(sin q)]∆t + ∆y = 0
g
Solving for ∆t using the quadratic equation,
vi(sin q) ± [−vi(sin q)]2 − 42(∆y)
∆t = 
g
22
g
(2.0 m/s)[sin(−30.0°)] ± [(
−
2.
0
m
/s
)[
si
n
(−
30
.0
°)
]2
−(
2)
(9
.8
1
m
/s2)
(−
45
m)
∆t = 
2
9.81 m/s
−1.0 m/s ± 1.
m2/s2 −1.0 m/s ± 8.
m2/s2
0m
2/s2+8.8
×102
8×102

∆t = 
=
9.81 m/s2
9.81 m/s2
−1.0 m/s ± 3.0 × 101 m/s
∆t = 
9.81 m/s2
∆t must be positive, so the positive root must be chosen.
29 m/s
∆t = 2 = 3.0 s
9.81 m/s
V
Section Five—Problem Bank
V Ch. 3–13
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Givens
Solutions
10. ∆x1 = 0.46 m
∆xtot = ∆x1 + ∆x2
∆x
∆t = 
vi(cos q)
∆x2 = 4.00 m
∆xi + ∆x2 1 ∆x1 + ∆x2
1
 − 2g 
∆y = vi(sin q)∆t − 2g∆t 2 = vi(sin q) 
vi(cos q)
vi(cos q)
∆y = − 0.35 m
g(∆x1 + ∆x2 )2

∆y = (∆x1 + ∆x2)(tan q) − 
2 vi2(cos q)2
g = 9.81 m/s2
vi =
vi =
vi =
vi =
2
q = 41.0°
g(∆x1 + ∆x2)2

2
2(cos q) [(∆x1 + ∆x2)(tan q) − ∆y]
(9.81 m/s2)(0.46 m + 4.00 m)2

2
(2)(cos 41.0°) [(0.46 m + 4.00 m)(tan 41.0°) − (−0.35 m)]
(9.81 m/s2)(4.46 m)2

(2)(cos 41.0°)2(3.88 m + 0.35 m)
(9.81 m/s2)(4.46 m)2

(2)(cos 41.0°)2 (4.23 m)
vi = 6.36 m/s
Additional Practice 3F
1. vbw = 58.0 km/h, forward
= +58.0 km/h
vwe = 55.0 km/h, backward
= −55.0 km/h
vbe = vbw + vwe = + 58.0 km/h + (−55.0 km/h) = +3.0 km/h
∆x 1.4 km
∆t =  = 
vbe 3.0 km/h
∆t = 0.47 h = 28 min
∆x = 1.4 km
vmw = 4.20 m/s, west
2
∆x = 8.50 × 10 m
vme = vmw + vwe = 4.20 m/s + 1.50 m/s = 5.70 m/s, west
time of travel with walkway:
∆x 8.50 × 102 m
∆t1 =  =  = 149 s
vme
5.70 m/s
time of travel without walkway:
∆x 8.50 × 102 m
∆t2 =  =  = 202 s
vmw
4.20 m/s
time saved = ∆t2 − ∆t1 = 202 s − 149 s = 53 s
3. v1e = 286 km/h, forward
v12 + v2e = v1e
v2e = 252 km/h, forward
v12 = v1e − v2e
∆x = 0.750 km
v12 = v1e − v2e = 286 km/h − 252 km/h = 34 km/h
∆x 0.750 km
∆t =  −  = 2.2 × 10−2 h
v12 34 km/h
3600 s
∆t = (2.2 × 10−2 h)  = 79 s
1h
V
V Ch. 3–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. vwe = 1.50 m/s, west
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Givens
4. vma = 1.10 m/s, east
vae = 5.0 km/h at 35°
west of south
∆x = 540 m
Solutions
vme = vma + vae
Find the mosquito’s speed with respect to Earth in the x direction.
vx, me = vx, ma + vx, ae = vma + vae(cos qae)
qae = –90.0° − 35° = −125°
1h
vx, me = 1.10 m/s + (5.0 km/h)(103 m/km)  [cos(−125°)] = 1.10 m/s
3600 s
+ (−0.80 m/s) = 0.30 m/s
∆x
540 m
∆t =  = 
vx, me 0.30 m/s
∆t = 1800 s = 3.0 × 101 min
5. vga = 150 km/h at 7.0°
below horizontal
vae = 15 km/h upward
∆y = −165 m
vge = vga + vae
Find the glider’s speed with respect to Earth in the y (vertical) direction.
vy, ge = vga + vy, ae = vga(sin qga) + vae
qga = −7.0°
vy, ge = (150 km/h)[sin(−7.0°)] + 15 km/h = −18 km/h + 15 km/h = −3 km/h
Time of descent with updraft:
−166 m
∆y
∆t =  = 
vy, ge (−3 km/h)(103 m/km)(1 h/3600 s)
∆t = 200 s
Time of descent without updraft:
−166 m
∆y
∆t′ =  = 
vy, ga (−18 km/h)(103 m/km)(1h/3600 s)
∆t′ = 33 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. vfc = 87 km/h, west
vfe = vfc + vce
vce = 145 km/h, north
vx, fe = vx, fc + vx, ce = vfc = −87 km/h
∆t = 0.45 s
vy, fe = vy, fc + vy, ce = vce = +145 km/h
∆x = vx, fe ∆t = (−87 km/h)(103 m/km)(1 h/3600 s)(0.45 s) = −11 m
∆x = 11m, west
∆y = vy, fe ∆t = (145 km/h)(103 m/km)(1 h/3600 s)(0.45 s)
∆y = 18 m, north
V
Section Five—Problem Bank
V Ch. 3–15
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Givens
Solutions
7. vaw = 55.0 km/h, north
vwe = 40.0 km/h at 17.0°
north of west
vae = vaw + vwe
vx, ae = vx, aw + vx, we = vwe (cos qwe)
vy, ae = vy, aw + vy, we = vaw + vwe(sin qwe)
qwe = 180.0° − 17.0° = 163.0°
vx, ae = (40.0 km/h)(cos 163.0°) = −38.3 km/h
vy, ae = 55.0 km/h + (40.0 km/h)(sin 163.0°) = 55.0 km/h + 11.7 km/h = 66.7 km/h
2
2
vy
)2
+(66
)2
vae = vx
(
38
.3
km
/h
.7
km
/h
, a
e)+
, a
e0 = (−
vae = 1.
03km
03km
03km
47
×1
2/h2+
4.4
5×1
2/h2 = 5.
92
×1
2/h2
vae = 76.9 km/h
66.7 km/h
vy,ae
q = tan−1  = tan−1  = −60.1°
−38.3
km/h
vx,ae
q = 60.1° west of north
8. vae = 76.9 km/h at 29.9°
west of north
∆t = 15.0 min
∆x = vae(cos qae)∆t
∆y = vae(sin qae)∆t
qae = 90.0° + 29.9° = 119.9°
∆x = (76.9 km/h)(cos 119.9°)(15.0 min)(1 h/60 min) = −9.58 km
∆y = (76.9 km/h)(sin 119.9°)(15.0 min)(1 h/60 min) = 16.7 km
∆x =
9.58 km, west
∆y = 16.7 km, north
9. vtc = 51 km/h, east
vte = vtc + vce
vce = 4.0 km/h, south
vx, te = vx, tc + vx, ce = vtc = 51 km/h
∆t = 14 s
vy, te = vy, tc + vy, ce = vce = −4.0 km/h
the target is 2.0 × 102 m away
∆y = vy, te ∆t = (4.0 km/h)(103 m/km)(1 h/3600 s)(14 s) = 16 m
the torpedo must be fired 16 m north of the target
V
V Ch. 3–16
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆x = vx, tc∆t = (51 km/h)(103 m/km)(1 h/3600 s)(14 s) = 2.0 × 102 m
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Givens
Solutions
10. vbw = 12.0 km/h, south
vbe = vbw + vwe
vwe = 4.0 km/h at 15.0°
south of east
vx, be = vx, bw + vx, we = vwe(cos qwe)
vy, be = vy, bw + vy, we = vbw + vwe(sin qwe)
qwe = −15.0°
vx, be = (4.0 km/h)[cos(−15.0°)] = 3.9 km/h
vy, be = (−12.0 km/h) + (4.0 km/h)[sin(−15.0°)] = (−12.0 km/h) + (−1.0 km/h)
= −13.0 km/h
2
2
vbe = (v
x,b
v
.9
km
/h
)2+(−13
.0
km
/h
)2
e)+
y, be) = (3
vbe = 15
km
2/h2+169
km
2/h2 = 18
4km
2/h2
vbe = 13.6 km/h
vy, be
−13.0 km/h
q = tan−1  = tan−1  = −73°
vx, be
3.9 km/h
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 73° south of east
V
Section Five—Problem Bank
V Ch. 3–17
Print Forces and the Laws of Motion
Chapter
4
Additional Practice 4A
Givens
1. F1 = 7.5 × 104 N north
Solutions
Fx,net = Fx = F1(cos q1) + F2(cos q2)
F2 = 9.5 × 104 N at 15.0°
north of west
Fx,net = (7.5 × 104 N)(cos 90.0°) + (9.5 × 104 N)(cos 165.0°)
q1 = 90.0°
Fy,net = Fy = F1(sin q1) + F2(sin q2)
q2 = 180.0° − 15.0° = 165.0°
Fy,net = (7.5 × 104 N)(sin 90.0°) + (9.5 × 104 N)(sin 165.0°)
Fx,net = −9.2 × 104 N
Fy,net = 7.5 × 104 N + 2.5 × 104 N = 10.0 × 104 N
Fy,net
q = tan−1 
Fx,net
q=
2. F1 = 6.00 × 102 N north
F3 = 6.75 × 102 N at 30.0°
south of east
q1 = 90.0°
10.0 × 104 N
= tan−1 
−9.2 × 104
= −47°
47° north of west
Fx,net = Fx = F1(cos q1) + F2(cos q2) + F3(cos q3) = (6.00 × 102 N)(cos 90.0°)
F2 = 7.50 × 102 N east
+ (7.50 × 102 N)(cos 0.00°) + (6.75 × 102 N)[cos(−30.0°)]
Fx,net = 7.50 × 102 N + 5.85 × 102 N = 13.35 × 102 N
Fy,net = Fy = F1(sin q1) + F2(sin q2) + F3(sin q3) = (6.00 × 102 N)(sin 90.0°)
+ (7.50 × 102 N)(sin 0.00°) + (6.75 × 102 N)[sin (−30.0°)]
q2 = 0.00°
Fy,net = 6.00 × 102 N + (−3.38 × 102 N) = 2.62 × 102 N
q3 = −30.0°
2.62 × 102 N
Fy,net
q = tan−1  = tan−1 
13.35 × 102 N
Fx,net
q=
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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3. F1 = 2280.0 N upward
F2 = 2250.0 N downward
F3 = 85.0 N west
F4 = 12.0 N east
F2 = 8.0 N
11.1° north of east
Fy,net = Fy = F1 + F2 = 2280.0 N + (−2250.0 N) = 30.0 N
Fx,net = Fx = F3 + F4 = −85.0 N + 12.0 N = −73.0 N
Fy,net
30.0 N
q = tan−1  = tan−1  = −22.3°
Fx,net
−73.0 N
q=
4. F1 = 6.0 N
22.3° up from west
Fmax = F1 + F2 = 6.0 N + 8.0 N
Fmax = 14.0 N
Fmin = F2 − F1 = 8.0 N − 6.0 N
Fmin =
2.0 N
V
Section Five—Problem Bank
V Ch. 4–1
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Print
Givens
Solutions
5. F1 = 3.0 N east
F2 = 4.0 N south
Fx,net = F1 + F3(cos q) = 0
F3(cos q) = −F1 = −3.0 N
Fy,net = F2 + F3(sin q) = 0
F3(sin q) = −F2 = −(−4.0 N) = 4.0 N
2
2
2
F3 = [F
co
s
q
)]
si
n
q
)]2 = (−3
.0
N
)2
+(4
.0
N
)2 = 9.
0
N
6
N
= 25
N2
+[F
+1
3(
3(
F3 =
5.0 N
F3(sin q)
4.0 N
q = tan−1  = tan−1  = −53°
−3.0 N
F3(cos q)
q=
6. F1 = 4.00 × 103 N east
F2 = 5.00 × 103 N north
3
F3 = 7.00 × 10 N west
F4 = 9.00 × 103 N south
53° north of west
Fx,net = F1 + F3 = 4.00 × 103 N + (−7.00 × 103 N) = −3.00 × 103 N
Fy,net = F2 + F4 = 5.00 × 103 N + (−9.00 × 103 N) = −4.00 × 103 N
2
)2 = (−
03
N
)2
+(−
03
N
)2
Fnet = (F
x,ne
(Fy,n
et
3.0
0×1
4.
00
×1
t )+
Fnet = 9
.00
×1
06
N2+
06
N2 = 25
06
N2
16.
0×1
.0
×1
Fnet =
5.00 × 103 N
Fy,net
−4.00 × 103 N
q = tan−1  = tan−1 
−3.00 × 103 N
Fx,net
q = 53.1° south of west
7. F1 = 15.0 N
Fy = F(sin q) = (15.0 N)(sin 55.0°)
q = 55.0°
Fy = 12.3 N
Fx = F(cos q) = (15.0 N)(cos 55.0°)
Fx = 8.60 N
8. F = 76 N
q = 40.0°
Fx = 58 N
Fy = F(sin q) = (76 N)(sin 40.0°)
Fy = 49 N
9. F1 = 350 N
Fy,net = F1(sin q1) + F2(sin q2) = (350 N)(sin 58°) + (410 N)(sin 43)
q1 = 58.0°
Fy,net = 3.0 × 102 N + 2.8 × 102 N
F2 = 410 N
Fy,net = 580 N
q2 = 43°
10. F1 = 7.50 × 102 N
q1 = 40.0°
Fy,net = Fg = F1(cos q1) + F2(cos q2)
Fy,net = (7.50 × 102 N)(cos 40.0°) + (7.50 × 102 N) [cos(−40.0°)]
2
F2 = 7.50 × 10 N
Fg = 575 N + 575 N = 1.150 × 103 N
q2 = −40.0°
V
V Ch. 4–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fx = F(cos q) = (76 N)(cos 40.0°)
Additional Practice 4B
PrintGivens
Solutions
15 cm/s − 0 cm/s
vf − v
anet = i =  = 3.0 cm/s2 = 3.0 × 10−2 m/s2
5.0 s
∆t
Fnet = m anet
1. Fnet = 2850 N
vf = 15 cm/s
vi = 0 cm/s
Fnet
2850 N
m =  = 
anet
3.0 × 10−2 m/s2
∆t = 5.0 s
m=
9.5 × 104 kg
1.0 m/s
∆v
anet =  =  = 0.20 m/s2
5.0 s
∆t
Fnet = m anet = Fdownhill − Fuphill = 18.0 N − 15.0 N = 3.0 N
Fnet
3.0 N
m =  = 2 = 15 kg
anet
0.20 m/s
2. ∆t = 1.0 m/s
∆t = 5.0 s
Fdownhill = 18.0 N
Fuphill = 15.0 N
Fnet = m anet = Fmax − mg
3. Fmax = 4.5 × 104 N
m(anet + g) = Fmax
anet = 3.5 m/s2
Fmax
4.5 × 104 N
4.5 × 104 N

m =  = 
=
= 3.4 × 103 kg
anet + g 3.5 m/s2 + 9.81 m/s2
13.3 m/s2
2
g = 9.81 m/s
1
∆y = vi ∆t + anet ∆t2
2
4. m = 2.0 kg
∆y = 1.9 m
2∆y
(2)(1.9 m)
Because vi = 0 m/s, anet = 
= 
= 0.66 m/s2
∆t2
(2.4 s)2
Fnet = m anet = (2.0 kg)(0.66 m/s) = 1.32 N
∆t = 2.4 s
vi = 0 m/s
Fnet = 1.32 N upward
1
∆y = vi ∆t + anet ∆t2
2
5. m = 8.0 kg
∆y = 20.0 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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2∆y
(2)(20.0 × 10−2 m)
Because vi = 0 m/s, anet = 
= 1.6 m/s2
2 = 
∆t
(0.50 s)2
Fnet = m anet = (8.0 kg)(1.6 m/s2) = 13 N
∆t = 0.50 s
vi = 0 m/s
g = 9.81 m/s2
Fnet = 13 N upward
Fnet = Fupward − mg
Fupward = Fnet + mg = 13 N + (8.0 kg)(9.8 m/s2) = 13 N + 78 N = 91 N
Fupward = 91 N upward
anet = aforward − abackward = 0.15 m/s2 − 2 × 10−2 m/s2
6. m = 75 kg
2
aforward = 0.15 m/s west
−2
abackward = 2 × 10
east
anet = 0.13 m/s2 west
Fnet = m anet = (75 kg)(0.13 m/s) = 9.8 N
Fnet = 9.8 N west
V
Section Five—Problem Bank
V Ch. 4–3
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Givens
Solutions
Fnet
−65.0 N
anet =  =  =
m
0.145 kg
7. Fnet = −65.0 N
m = 0.145 kg
8. m = 214 kg
−448 m/s2
Fnet = Fbuoyant − mg = 790 N − (214 kg)(9.81 m/s2)
Fbuoyant = 790 N
Fnet = 790 N − 2.10 × 103 N = −1310 N
Fnet
−1310 N
anet =  =  = −6.12 m/s2
m
214 kg
g = 9.81 m/s2
9. m = 0.080 kg
Fnet = m anet = m g(sin q)
q = 37.0°
anet = g(sin q) = (9.81 m/s2)(sin 37.0°) = 5.90 m/s2
g = 9.81 m/s2
anet = 5.90 m/s2 down the incline (37.0° below horizontal)
10. m = 0.080 kg
Fnet = Fupward − m adownward = 1.40 N − (0.080 kg)(5.90 m/s2) = 1.40 N − 0.47 N = 0.93 N
Fupward = 1.40 N
adownward = 5.90 m/s2
Fnet = 0.93 N up the incline (37.0° above the horizontal)
0 .93 N
F
 =   = 12 m/s2
anet = net
0.080 kg
m
Additional Practice 4C
1. Fdownward = 4.26 × 107 N
mk = 0.25
Fnet = Fdownward − Fk = 0
Fk = mk Fn = Fdownward
Fdownward
4.26 × 107 N
Fn =  =  = 1.7 × 108 N
0.25
mk
q = 10.0°
2
g = 9.81 m/s
3. Fs,max = 2400 N
ms = 0.20
Fn = mg (cos q)
1.7 × 108 N
Fn
m =  = 
= 1.8 × 107 kg
g (cos q)
(9.81 m/s2)(cos 10.0°)
Fs,max = ms Fn
2400 N
Fs,max
Fn =  =  = 1.2 × 104 N
0.20
ms
q = 30.0°
g = 9.81 m/s2
Fn = 1.2 × 104 N perpendicular to and away from the incline
Fn = mg(cos q )
Fn
1.2 × 104 N
m =  = 
= 1400 kg
g (cos q)
(9.81 m/s2)(cos 30.0°)
4. m = 60.0 kg
For the passenger to remain standing without sliding,
a = 3.70 m/s2
Fs,max ≥ F = ma
ms = 0.455
Fs,max = ms Fn = ms mg
2
g = 9.81 m/s
V
ms mg ≥ ma
ms g ≥ a
(0.455)(9.81 m/s2) = 4.46 m/s2 > 3.70 m/s2
The passenger will be able to stand without sliding.
V Ch. 4–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. Fn = 1.7 × 108 N
Givens
Print
Fnet = mg(sin q) − Fk = 0
5. m = 90.0 kg
q = 17.0°
Fk = mg(sin q) = (90.0 kg)(9.81 m/s2)(sin 17.0°) = 258 N
2
g = 9.81 m/s
6. msled = 47 kg
msupplies = 33 kg
mk = 0.075
q = 15°
Fk = 258 N up the slope
Fk = mk Fn = mk(msled + msupplies)g = (0.075)(47 kg + 33kg)(9.81 m/s2)
Fk = (0.075)(8.0 × 101 kg)(9.81 m/s2)
Fk = 59 N
Fk = mk Fn = mk(msled + msupplies)g(cos q) = (0.075)(47 kg + 33 kg)(9.81 m/s2)(cos 15°)
Fk = (0.075)(8.0 × 101 kg)(9.81 m/s2)(cos 15°) = 57 N
7. m = 1.8 × 103 kg
q = 15.0°
Fs,max = 1.25 × 104 N
g = 9.81 m/s2
Fs,max = ms Fn = ms mg(cos q)
Fs,max
1.25 × 104 N
 = 
ms = 
mg(cos q)
(1.8 × 103 kg)(9.81 m/s2)(cos 15.0°)
ms = 0.73
8. m = 15.0 g
Fnet = mg(sin q) − Fk = 0
q = 2.3°
Fk = mk Fn = mk mg(cos q)
g = 9.81 m/s2
m g(sin q)
mk =   = tan q = tan 2.3°
mg (c os q)
mk = 0.040
9. vf = 88.0 km/h
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Solutions
Fnet = Fapplied − Fs,max = 0
vi = 0 km/h
Fs,max = ms Fn = ms mg
∆t = 3.07 s
Fapplied = m a
g = 9.81 m/s2
m a = ms mg
(88.0 km/h − 0 km/h)(103 m/km)(1 h/3600 s)
a vf − vi
ms =  =  = 
(9.81 m/s2)(3.07 s)
g
g∆t
ms = 0.812
10. q = 5.0°
Fnet = mg(sin q) − Fk = 0
Fk = mk Fn = mk mg(cos q)
mg(sin q )
mk =  = tan q = tan 5.0°
mg(cos q )
mk = 0.087
V
Section Five—Problem Bank
V Ch. 4–5
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Additional Practice 4D
Givens
Solutions
1. anet = 1.22 m/s2
q = 12.0°
Fnet = m anet = mg(sin q) − Fk
Fk = mk Fn = mk mg(cos q)
g = 9.81 m/s2
m anet + mk mg(cos q) = mg(sin q)
2.04 m/s2 − 1.22 m/s2
(9.81 m/s2)(sin 12.0°) − 1.22 m/s2
g(sin q ) − anet
mk = 
= 
= 
2
(9.81 m/s )(cos 12.0°)
(9.81 m/s2)(cos 12.0°)
g (cos q )
0.82 m/s2
= 0.085
mk = 
(9.81 m/s2)(cos 12.0°)
2. Fapplied = 1760 N
Fnet = Fapplied − mg(sin q) − Fs,max = 0
q = 17.0°
Fs,max = ms Fn = ms mg(cos q)
m = 266 kg
ms mg(cos q ) = Fapplied − mg(sin q)
g = 9.81 m/s2
2
Fapplied − mg(sin q) 1760 − (266 kg)(9.81 m/s )(sin 17°)
ms =  = 
2
(266 kg)(9.81 m/s )(cos 17°)
mg(cos q )
1760 − 760 N
1.00 × 103 N

ms = 
=
(266 kg)(9.81 m/s2)(cos 17°)
(266 kg)(9.81 m/s2)(cos 17°)
ms = 0.40
3. m = 5.1 × 102 kg
Fnet = Fapplied − mg(sin q) − Fs,max = 0
Fs,max = ms Fn = ms mg(cos q)
q = 14°
Fapplied = 4.1 × 103 N
2
g = 9.81 m/s
ms mg(cos q ) = Fapplied − mg(sin q)
4.1 × 103 N − (5.1 × 102 kg)(9.81 m/s2)(sin 14°)
Fapplied − mg(sin q)
ms =  = 
(5.1 × 102 kg)(9.81 m/s2)(cos 14°)
mg(cos q)
4.1 × 103 N − 1.2 × 103 N
2.9 × 103 N
= 
ms = 
2
2
2
(5.1 × 10 kg)(9.81 m/s )(cos 14°)
(5.1 × 10 kg)(9.81 m/s2)(cos 14°)
4. Fapplied = 5.0 N to the left
Fnet = m anet = Fapplied − Fk
Fk = Fapplied − m anet
m = 1.35 kg
2
anet = 0.76 m/s to the left
Fk = 5.0 N − (1.35 kg)(0.76 m/s2) = 5.0 N − 1.0 N = 4.0 N
Fk = 4.0 N to the right
Fnet = m anet = Fk
5. mk = 0.20
2
g = 9.81 m/s
Fk = mkFh = mkmg
m mg
anet = k  = mk g = (0.20)(9.81 m/s2)
m
anet = 2.0 m/s2
V
V Ch. 4–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ms = 0.60
Givens
Print
6. Fapplied = 2.50 × 102 N
m = 65.0 kg
Fnet = m anet = Fapplied − mg(sin q) − Fk
Fk = Fapplied − mg(sin q) − manet
q = 18.0°
Fk = 2.50 × 102 N − (65.0 kg)(9.81 m/s2)(sin 18.0°) − (65.0 kg)(0.44 m/s2)
2
anet = 0.44 m/s
Fk = 2.50 × 102 N − 197 N − 29 N = 24 N = 24 N downhill
Fnet = m anet = mg(sin q) − Fk
7. m = 65.0 kg
2
24 N
F
anet = g(sin q) − k = (9.81 m/s2)(sin 18.0°) −   = 3.03 m/s2 − 0.37 m/s2 = 2.66 m/s2
65.0 kg
m
g = 9.81 m/s
Fk = 24 N
anet = 2.66 m/s2 downhill
q = 18.0°
8. Fapplied = 3.00 × 102 N
Fx,net = Fapplied(cos q) − Fk = 0
q = 20.0°
Fy,net = Fn − mg + Fapplied(sin q) = 0
mk = 0.250
Fk = mkFn
2
Fapplied(cos q) (3.00 × 10 N)[cos(−20.0°)]
Fn =  =  = 1130 N
0.25°
mk
1130 N + (3.00 × 102 N)[sin(−20.0°)]
Fn + Fapplied(sin q)
 = 
m= 
9.81 m/s2
g
103 0 N
1130 N − 103 N
=  2 = 105 kg
m = 
9.81 m /s
9.81 m/s2
Fnet = Fapplied + Fs,max − Fdownhill = 0
9. Fapplied = 590 N
Fdownhill = 950 N
Fs,max = ms Fn = ms mg(cos q)
ms = 0.095
ms Fn = Fdownhill − Fapplied
q = 14.0°
Fdownhill − Fapplied 950 N − 590 N 360 N
 =  =  = 3800 N
Fn = 
ms
0.095
0.095
Fn = 3800 N perpendicular to and up from the ground
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Solutions
3800 N
Fn
m = 
= 
= 4.0 × 102 kg
(9.81 m/s2)(cos 14.0°)
g(cos q)
10. anet = 1.20 m/s2
Fx,net = Fapplied(cos q) − Fs,max = 0
3
Fapplied = 1.50 × 10 N
Fs,max = msFn
q = −10.0°
(1.50 × 103 N)[cos(−10.0°)]
Fapplied(cos q)
Fn =  =  = 2.27 × 103 N
0.650
ms
ms = 0.650
g = 9.81 m/s2
Fn = 2.27 × 103 N, upward
Fy,net = m anet = Fn − mg + Fapplied(sin q)
m(anet + g) = Fn + Fapplied(sin q)
Fn + Fapplied(sin q) 2.27 × 103 N + (1.50 × 103 N)[sin(−10.0°)]
 = 
m= 
1.20 m/s2 + 9.81 m/s2
anet + g
2.27 × 103 N − 2.60 × 102 N 2.01 × 103 N
m = 
= 
= 183 kg
11.01 m/s2
11.01 m/s2
V
Section Five—Problem Bank
V Ch. 4–7
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Work and Energy
Chapter
5
Additional Practice 5A
Givens
1. d = 3.00 × 102 m
W = 2.13 × 106 J
Solutions
2.13 × 106 J
W
= 7.10 × 3 N
F =  = 
(3.00 × 102 m)(cos 0°)
d(cos q)
q = 0°
2. d = 76.2 m
Wnet = 1.31 × 103 J
W et
1.31 × 103 J
Fnet = n
=  = 17.2 N
d(cos q )
(76.2 m)(cos 0°)
q = 0°
3. W = 1800 J
d1 = 1.5 m
d2 = 5.0 m
q = 0°
4. Wnet = 4.27 × 103 J
d = 17 m
W = F1 d1 = (cos q) = F2 d2 (cos q)
W
1800 J
F1 =  =  = 1.2 × 103 N
d1 (cos q)
(1.5 m)(cos 0°)
W
1800 J
F2 =  =  = 3.6 × 102 N
d2 (cos q)
(5.0 m)(cos 0°)
W et
4.27 × 103 J
Fnet = n
=  = 2.5 × 102 N
d(cos q)
(17 m) (cos 0°)
q = 0°
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. F = 1.6 N
d = 1.2 m
W = Fd(cos q) = (1.6 N) (1.2 m) (cos 180°) = −1.9 J
q = 180°
6. d = 15.0 m
Wnet = Fapplied d (cos q1) + Fk d (cos q2)
Fapplied = 35.0 N
Wnet = (35.0 N) (15.0 m) (cos 20.0°) + (24.0 N) (15.0 m) (cos 180°)
q1 = 20.0°
Wnet = 493 J + (−3.60 × 102 J)
Fk = 24.0 N
Wnet = 133J
q2 = 180°
Alternatively,
Wnet = Fnet d (cos q ′)
q ′ = 0°
Fnet = Fapplied (cos q1) − Fk
Wnet = [Fapplied (cos q1) − Fk] d (cos q ′)
Wnet = [(35.0 N) (cos 20.0°) − 24.0 N] (15.0 m) (cos 0°)
Wnet = (32.9 N − 24.0 N) (15.0 m) = (8.9 N)(15.0 m) = 130 J
Section Five—Problem Bank
V
V Ch. 5–1
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Givens
Solutions
7. v1 = 88.9 m/s
W = Fd(cos q)
vf = 0 m/s
F = ma
∆v vf − v
a =  = i
∆t
∆t
∆t = 0.181 s
d = 8.05 m
m = 70.0 kg
q = 180°
(70.0 kg)(0 m/s − 88.9 m/s)
m (vf − vi)
W = 
d (cos q) =  (8.05 m)(cos 180°)
(0.181 s)
∆t
(70.0 kg)(88.9 m/s)(8.05 m)
W = 
(0.181 s)
W = 2.77 × 105 J
8. F = 715 N
W = 2.72 × 104 J
W
2.72 × 104 J
d =  =  = 38.0 m
F (cos q)
(715 N)(cos 0°)
q = 0°
9. Fnet = 7.25 × 10−2 N
Wnet = 4.35 × 10−2 J
W et
4.35 × 10−2 J
d = n
= 
= 0.600 m
Fnet (cos q)
(7.25 × 10−2 N)(cos 0°)
q = 0°
10. W = 6210 J
F = 2590 N
W
6210 J
d =  =  = 2.398 m
F (cos q)
(2590 N)(cos 0°)
q = 0°
Additional Practice 5B
v = 57 km/h
2. v = 15.8 km/s
m = 0.20 g
3. v = 35.0 km/h
m = 9.00 × 102 kg
4. v1 = 220.0 km/h
1
KE = 9.4 × 109 J
KE = 2 mv 2 = 2 (0.20 × 10−3 kg)(15.8 × 103 m/s)2
1
1
KE = 2.5 × 104 J
1
1
KE = 2 mv 2 = 2 (9.00 × 102 kg) [(35.0 km/h)(103 m/km)(1 h/3600 s)]2
KE = 4.25 × 104 J
1
1
KE1 = 2 m1 v12 = 2 (8.84 × 105 kg)[(220.0 km/h)(103 m/km)(1 h/3600 s)]2
m1 = 8.84 × 105 kg
KE1 = 1.65 × 109 J
v2 = 320.0 km/h
KE2 = 2 m2 v22 = 2 (4.80 × 105 kg)[(320.0 km/h)(103 m/km)(1h/3600 s)]2
m2 = 4.80 × 105 kg
KE2 = 1.90 × 109 J
5. KE = 2.78 × 109 J
V
1
KE = 2 mv 2 = 2 (7.5 × 107 kg) [(57 km/h)(103 m/km)(1h/3600 s)]2
v = 275 km/h
V Ch. 5–2
1
1
(2)(2.78 × 109 J)
2KE

m = 
=
[(275 km/h)(103 m/km)(1h/3600 s)]2
v2
m = 9.53 × 105 kg
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. m = 7.5 × 107 kg
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Givens
Solutions
6. v = 850 km/h
KE = 9.76 × 109 J
(2)(9.76 × 109 J)
2 KE
m = 
2 = 
[(850 km/h)(103 m/km)(1 h/3600 s)]2
v
m = 3.50 × 105 kg
7. v = 9.78 m/s
KE = 6.08 × 104 J
8. KE = 7.81 × 104 J
m = 55.0 kg
9. KE = 1433 J
m = 47.0 g
1
10. KEA,i = 2KEB
2 KE (2)(6.08 × 104 J)
m = 
= 
= 1.27 × 103 kg
v2
(9.78 m/s)2
v=
v=
(2)(1433 J)
KE
 = 
 =
2
47.0 × 10 kg
m
247 m/s
−3
1
KEA,i = 2KEB
= 22 mBvB2
vA,f = vA,i + 1.3 m/s
1
m v 2
2 A A,i
KEA,f = KEB
1
(2m )v 2
B A,i
2
mA = 2.0 mB
(2)(7.81 × 104 J)
 = 53.3 m/s
55.0 kg
2 KE
 =
m
11
1
= 4 mBvB2
1
vA,i2 = 4 vB2, or vB2 = 4vA,i2
1
vA,i = 2 vB
KEA,f = KEB
1

2
mAvA,f 2 = 2 mBvB2
1
1

2
(2mB)(vA,i + 1.3 m/s)2 = 2 mB vB2.
1
1
1
(vA,i + 1.3 m/s)2 = 2 vB2 = 2 (4vA,i2) = 2vA,i2
vA,i2 + (2.6 m/s) vA,i + 1.7 m2/s2 = 2vA,i2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vA,i2 − (2.6 m/s) vA,i − 1.7 m2/s2 = 0
Using the quadratic equation,
s2
+6.8
m2/
s2
)2
−4(−
s2) 2.6 m/s ± 6.
2.6 m/s ± (−
8m
2/
2.
6m
/s
1.
7m
2/
vA,i =  = 
2
2
vA,i =
vA,i =
13.6
2.6 m/s ± m2/
s2
2.6 m/s ± 3.69 m/s
2
6.3 m/s
2
 =  =  = 3.2 m/s
2
1
 v
2 B
vB = vA,i = (2)(3.2 m/s) = 6.4 m/s
Additional Practice 5C
1. vi = 8.0 m/s
1
vf = 0 m/s
Wnet = Fnetd(cos q) = Fkd(cos q)
d = 45 m
1

2
Fk = 0.12 N
q = 180°
1
Wnet = ∆KE = KEf − KEi = 2 mvf2 − 2 mvi2
m(vf 2 − vi2) = Fkd(cos q)
(2)(0.12 N)(45 m)(cos 180°)
2 Fkd(cos q)
−(2)(0.12 N)(45 m)
= 
= 
m= 
2
2
2 
2
(0
m/s)
−
(8.0
m/s)
vf − vi
−64 m2/s2
V
m = 0.17 kg
Section Five—Problem Bank
V Ch. 5–3
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Givens
Solutions
2. vi = 15.00 km/s
vf = 14.97 km/s
–2
Fr = 9.00 × 10 N
d = 500.0 km
q = 180°
1
1
Wnet = ∆KE =KEf − KEi = 2 mvf 2 − 2 mvi2
Wnet = Fnetd(cos q) = Fr d (cos q)
1

2
m(vf 2 − vi2) = Fr d (cos q)
(2)(9.00 × 10−2 N)(500.0 × 103 m)(cos 180°)
2 Fr d (cos q)
m= 
= 
2 
2
(14.97 × 103 m/s)2 − (15.00 × 103 m/s)2
vf − vi
−(2)(9.00 × 10−2 N)(500.0 × 103 m)
−9.00 × 104 J
m = 
8 2 2
8 2 2 = 
2.241 × 10 m /s − 2.250 × 10 m /s
−9 × 108 m2/s2
m = 1.00 × 10−4 kg
3. vi = 48.0 km/h
vf = 59.0 km/h
d = 100.0 m
m = 1100 kg
q = 0°
1
1
Wnet = ∆KE = KEf − KEi = 2 mvf 2 − 2 mvi2
Wnet = Fnet d (cos q)
1
Fnet d (cos q) = 2 m(vf 2 − vi2)
2
2
(1100 kg)[(59.0 km/h)2 − (48.0 km/h)2](103 m/km)2 (1h/3600 s)2
m(vf − vi )
 = 
Fnet = 
(2)(100.0 m)(cos 0°)
2 d (cos q)
(1100 kg)(3.48 × 109 m2/h2 − 2.30 × 109 m2/h2)(1h/3600 s)2
Fnet = 
(2)(100.0 m)
(1100 kg)(1.18 × 109 m2/h2)(1h/3600 s)2
Fnet = 
(2)(100.0 m)
Fnet = 5.01 × 102 N
d = 7.0 m
vf = 1.1 m/s
vi = 0 m/s
1
Wnet = ∆KE = KEf − KEi = 2 mvf 2 − vi2
Wnet = Fnetd(cos q)
1
Fnetd(cos q) = 2 m(vf2 − vi2)
2
2
(450 kg)[(1.1 m/s)2 − (0 m/s)2]
m(vf − vi )
(450 kg)(1.2 m2/s2)
 =  = 
Fnet = 
(2)(7.0 m)(cos 0°)
2 d (cos q)
(2)(7.0 m)
q = 0°
Fnet = 39 N
5. vi = 2.40 × 102 km/h
vf = 0 km/h
anet = 30.8 m/s2
m = 1.30 × 104 kg
q = 180°
1
Wnet = ∆KE = KEf − KEi = 2 mvf 2 − vi2
Wnet = Fnet d (cos q) = manet d(cos q)
1
manetd(cos q) = 2 m(vf2 − vi2)
[(0 km/h)2 − (2.40 × 102 km/h)2] (103 m/km)2(1h/3600 s)2
vf 2 − vi2
d = 
= 
(2)(30.8 m/s2)(cos 180°)
2 anet (cos q)
(−5.76 × 104 km2/h2)(103 m/km)2 (1h/3600 s)2
d = 
= 72.2 m
−(2)(30.8 m/s2)
V
V Ch. 5–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. m = 450 kg
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Givens
6. m = 50.0 kg
vi = 47.00 m/s
vf = 5.00 m/s
Solutions
1
1
Wnet = ∆KE = KEf − KEi = 2 mvf 2 − 2 mvi2
1
1
Wnet = 2 m(vf2 − vi2) = 2 (50.0 kg)[(5.00 m/s)2 − (47.00 m/s)2]
1
1
Wnet = 2 (50.0 kg)(25.0 m2/s2 − 2209 m2/s2) = 2 (50.0 kg)(−2184 m2/s2)
Wnet = −5.46 × 104 J
7. m = 2.0 × 106 kg
d = 7.5 m
Wnet = ∆KE = KEf − KEi = KEf
Wnet = Fnet d (cos q) = manetd(cos q)
anet = 7.5 × 10−2 m/s2
KEf = manet d (cos q) = (2.0 × 106 kg)(7.5 × 10−2 m/s2)(7.5 m)(cos 0°)
KEi = 0 J
KEf = 1.1 × 106 J
8. Fapplied = 92 N
Wnet = ∆KE = KEf − KEi = KEf
m = 18 kg
Wnet = Fnet d (cos q)
mk = 0.35
Fnet = Fapplied − Fk = Fapplied − mk mg
d = 7.6 m
KEf = (Fapplied − mk mg) d (cos q)
= [92 N − (0.35)(18 kg)(9.81 m/s2)](7.6 m)(cos 0°)
g = 9.81 m/s2
q = 0°
KEf = (92 N − 62 N)(7.6 m) = (3.0 × 101 N)(7.6 m)
KEi = 0 J
KEf = 228 J
9. m = 2.00 × 102 kg
1
1
Wnet = ∆KE = KEf − KEi = 2 mvf2 − 2 mvi2
Fwind = 4.00 × 102 N
Wnet = Fnet d (cos q) = Fwind d(cos q)
d = 0.90 km
1

2
vi = 0 m/s
q = 0°
1
mvf2 − 2 mvi2 = Fwind d (cos q)
(2)(4.00 × 102 N)(0.90 × 103 m)(cos 0°)
 + (0 m/s)
2F
md
(cosq)+v = 2.00 × 10 kg
(2)(4.00 × 10 N)(9.0 × 10 m)

v = 2.00 × 10 kg
wind
vf =
2
i
2
2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
f
2
2
2
vf = 6.0 × 101 m/s
10. m = 20.0 g
d = 2.5 m
Fforward = 7.3 × 10−2 N
mk = 0.20
vi = 0 m/s
g = 9.81 m/s2
q = 0°
1
1
Wnet = ∆KE = KEf − KEi = 2 mvf 2 − 2 mvi2
Wnet = Fnetd(cos q) = (Fforward − Fk) d(cos q) = (Fforward − mk mg)d(cos q)
1

2
1
mvf 2 − 2 mvi2 = (Fforward − mk mg)d(cos q)
2[(Fforward − mk mg) d (cos q) + vi2]

m
(2)[7.3 × 10 N − (0.20)(20.0 × 10 kg)(9.81 m/s )](2.5 m)(cos 0°) + (0 m/s)

v = 20.0 × 10 kg
(2)(7.3 × 10 N − 3.9 × 10 N)(2.5 m)
(2)(3.4 × 10 N)(2.5 m)
 = 
v = 20.0 × 10 kg
2.00 × 10 kg
vf =
−2
−3
−2
f
2
2
−3
f
−2
−3
−2
−2
vf = 2.9 m/s
V
Section Five—Problem Bank
V Ch. 5–5
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Additional Practice 5D
Givens
Solutions
1. x = 5.00 cm
Assuming all of the kinetic energy becomes stored elastic potential energy,
−4
KEcar = 1.09 × 10
J
1
KEcar = PEelastic = 2 kx2
2 PE lastic
(2)(1.09 × 104 J)
k = e
= 
2
x
(5.00 × 10−2 m)2
k = 8.72 × 106 N/m
2. PEelastic = 5.78 × 107 J
x = 102 m for each spring
PEelastic = 2 2 kx2
1
PEelastic = kx2
PE astic
5.78 × 107 J
k = el
= 
2
x
(102 m)2
k = 5.56 × 103 N/m
elastic potential energy stored = decrease in gravitational potential energy
3. m = 0.76 kg
PEelastic = mgx
x = 2.3 cm
2
g = 9.81 m/s
1
PEelastic = 2 kx2 = mgx
2 mgx 2 mg (2)(0.76 kg)(9.81 m/s2)
k = 
=  = 
2.3 × 10−2 m
x2
x
k = 6.5 × 102 N/m
4. m = 5.0 kg
PEg = mgh = mgd(sin q)
PEg = 2.4 × 102 J
2.4 × 102 J
PEg
d =  = 
(5.0 kg)(9.81 m/s2)(sin 25.0°)
mg (sin q)
d=
5. k = 1.5 × 104 N/m
PEelastic = 120 J
12 m
1
PEelastic = 2 kx2
x=±

k
= ± 
1.5 ×10
N/m
2 PEelastic
(2)(120 J)
4
Spring is compressed, so negative root is selected.
x = −0.13 m = −13 cm
PEg = mgh
6. m = 1750 kg
10
PEg = 1.69 × 10 J
2
g = 6.44 m/s
1.69 × 1010 J
PEg
h =  = 
(1750 kg)(6.44 m/s2)
mg
h = 1.50 × 106 m = 1.50 × 103 km
PEg = mgh
7. h = 7.0 m
4
PEg = 6.6 × 10 J
V
g = 9.81 m/s2
6.6 × 104 J
PE
m = g = 
gh (9.81 m/s2)(7.0 m)
m = 9.6 × 102 kg
V Ch. 5–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 25.0°
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Givens
Solutions
8. PEg = 3.36 × 109 J
PEg = mgh
3.36 × 109 J
PEg
m =  = 
(9.81 m/s2)(1.45 × 103 m)
gh
h = 1.45 km
m = 2.36 × 105 kg
9. k = 550 N/m
PEelastic = 2kx2 = 2(550 N/m)(–1.2 × 10−2 m)2 =
1
4.0 × 10−2 J
1
x = −1.2 cm
10. h = 5334 m
PEg = mgh = (64.0 kg)(9.81 m/s2)(5334 m) =
m = 64.0 kg
3.35 × 106 J
g = 9.81 m/s2
Additional Practice 5E
PEi = KEf
1. m = 0.500 g
h = 0.250 km
2
g = 9.81 m/s
2. d = 96.0 m
mgh = KEf
KEf = (0.500 × 10−3 kg)(9.81 m/s2)(0.250 × 103 m) =
1.23 J
a. PE1 = mgh = mgd(sin q)
q = 18.4°
PE2 = 0 J
m = 70.0 kg
MEi = PE1 + PE2 = mgd(sin q)
2
g = 9.81 m/s
MEi = (70.0 kg)(9.81 m/s2)(96.0 m)(sin 18.4°) = 2.08 × 104 J
b. PE1 = 0 J
PE2 = mgh = mgd(sin q)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
MEf = PE1 + PE2 = mgd(sin q) = MEi
MEf = 2.08 × 104 J
v1 = v2 = 1.0 m/s
h1,f = 20.0 m
c. MEi = MEf
PE1,i + PE2,i + KE1 + KE2 = PE1,f + PE2,f + KE1 + KE2
The kinetic energy of each passenger remains unchanged during the trip once the
cars are in motion, so
PE1,i + PE2,i = PE1,f + PE2,f
mgh1,i + 0 J = mgh1,f + PE2,f
PE2,f = mgh1,f − mgh1,f = mgd(sin q) − mgh1,f
PE2,f = (70.0 kg)(9.81 m/s2)(96 m)(sin 18.4°) − (70.0 kg)(9.81 m/s2)(20.0 m)
PE2,f = 2.1 × 104 J − 1.37 × 104 J
PE2,f = 7.0 × 103 J
V
Section Five—Problem Bank
V Ch. 5–7
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Givens
Solutions
PEi + KEi = PEf + KEf
3. hi = 75.0 m
vi = 1.2 m/s + 3.5 m/s
= 4.7 m/s
vf = 0 m/s
1
hf =
g = 9.81 m/s2
1
mghi + 2 mvi2 = mghf + 2mvf2
1
m(vi2
2
− vf2)
mg
vi2 − vf2
+ hi
+ hi = 
2g
(4.7 m/s)2 − (0 m/s)2
hf = 
+ 75.0 m = 1.1 m + 75.0 m = 76.1 m
(2)(9.81 m/s2)
m = 20.0 g
PEi = KEf
4. m = 25.0 kg
1
v = 12.5 m/s
mgh = 2 mv 2
g = 9.81 m/s2
v2
(12.5 m/s)2
h =  = 
= 7.96 m
2g
(2)(9.81 m/s2)
MEi + ∆ME = MEf
5. m = 50.0 g
2
vi = 3.00 × 10 m/s
MEi = KEi = 2mvi2
vf = 89.0 m/s
MEf = KEf = 2mvf2
1
1
1
∆ME = MEf − MEi = 2m(vf 2 − vi2)
∆ME = 2 (50.0 × 10−3 kg)[(89.0 m/s)2 − (3.00 × 102 m/s)2]
1
∆ME = 2 (5.00 × 10−2 kg)(7.92 × 103 m2/s2 − 9.00 × 104 m2/s2)
1
∆ME = 2 (5.00 × 10−2 kg) (−8.21 × 104 m2/s2)
1
∆ME = −2.05 × 103 J
6. m = 50.0 g
For upward flight,
vi = 3.00 × 102 m/s
PE1,f − KE1,i = ∆ME1
vf = 89.0 m/s
where
For downward flight,
KE2,f − PE2,i = ∆ME2
Where
∆ME2 = Wnet,2 = Fnet,2 h (cos 0°) = (mg − Fresist) h
Solving for h,
∆ME2 − ∆ME1 = (mg − Fresist) h − [−(mg + Fresist)h] = 2 mgh
∆ME2 − ∆ME1 (KE2, f − PE2,i) − (PE1, f − KE1,i)
 = 
h= 
2 mg
2 mg
1
KE2,f = 2 mvf 2
1
KE1,i = 2 mvi2
PE1,f = PE2,i = mgh
1

2
1
mvf2 − mgh − mgh + 2 mvi2
vf2 + vi2
h =  = 
−h
2 mg
4g
V
V Ch. 5–8
(89.0 m/s)2 + (3.00 × 102 m/s)2
vf2 + vi2
h = 
= 
(8)(9.81 m/s2)
8g
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆ME1 = Wnet,1 = Fnet,1 h (cos 180°) = (mg + Fresist) h
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Givens
Solutions
7.92 × 103 m2/s2 + 9.00 × 104 m2/s2
9.79 × 103 m2/s2
h = 
= 
2
(8)(9.81 m/s )
(8)(9.81 m/s2)
h = 1.25 × 103 m = 1.25 km
PEg,i = PEelastic,f + PEg,f
7. m = 50.0 kg
4
k = 3.4 × 10 N/m
mghi = 2 kx2 + mghf
x = 0.65 m
(3.4 × 104 N/m)(0.65 m)2
kx 2
hi = hf +  = 0.35 m + 
= 0.35 m + 15 m
(2)(50.0 kg)(9.81 m/s2)
2 mg
hf = 1.00 m − 0.65 m
= 0.35 m
1
hi = 15 m
PEi = KEf
8. h = 3.0 m
2
g = 9.81 m/s
1
mgh = 2 mvf 2
vf = 2g
m/s
m)
h = (2
)(
9.
81
2)(3
.0
vf =
9. m = 100.0 g
x = 30.0 cm
k = 1250 N/m
7.7 m/s
PEelastic = KE
1

2
1
kx 2 = 2 mv2
v=
−2
(1250 N/m)(30.0 × 10 m)

kmx = 100.0 × 10 kg
2
2
−3
v = 33.5 m/s
10. mw = 546 kg
h = 5.64 m
g = 9.81 m/s2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
mflyer = 273 kg
PEw = KEflyer
1
mw gh = 2 mflyer vflyer2
vflyer =
(2)(546 kg)(9.81 m/s2)(5.64 m)

273 kg
2mw gh
 =
mflyer
vflyer = 14.9 m/s
Additional Practice 5F
1. P = 380.3 kW
W = 4.5 × 106 J
2. P = 331 W
h = 442 m
m = 55 kg
4.5 × 106 J
W
= 12 s
∆t =  = 
380.3 × 103 W
P
W = mgh
W mgh
(55 kg)(9.8 1 m/s2)
∆t =  =  =  = 720 s = 12 min
P
P
331 W
g = 9.81 m/s2
3. F = 334 N
d = 50.0 m
q = 0°
W = Fd(cos q)
(334 N)(50.0 m)(cos 0°)
W Fd(cos q)
∆t =  =  =  = 7.95 s
2100 W
P
P
V
P = 2100 W
Section Five—Problem Bank
V Ch. 5–9
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Givens
Solutions
4. P = 13.0 MW
∆t = 15.0 min
5. P = 1 hp = 745.7 W
∆t = 0.55 s
6. P = (4)(300.0 kW)
∆t = 25 s
7. ∆t = 39 s
P = 158 kW
8. W = 1.4 × 1013 J
∆t = 8.5 min
9. W = 2.82 × 107 J
∆t = 30.0 min
W = P ∆t = (13.0 × 106 W)(15.0 min)(60 s/min) = 1.17 × 1010 J
W = P ∆t = (745.7 W)(0.55 s) = 4.1 × 102 J
W = P ∆t = (4)(300.0 × 103 W)(25 s) =
W = P ∆t = (158 × 103 W)(39 s) =
6.2 × 106 J
W
1.4 × 1013 J
P =  =  = 2.7 × 1010 W = 27 GW
∆t (8.5 min)(60 s/min)
2.82 × 107 J
W
P =  =  = 1.57 × 104 W = 15.7 kW
∆t (30.0 min)(60 s/min)
P = (1.57 × 104 W)(1 hp/745.7 W) =
10. W = 3.0 × 106 J
21.1 hp
3.0 × 106 J
W
P =  =  = 1.0 × 104 W
∆t (5.0 min)(60 s/min)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 5.0 min
3.0 × 107 J
V
V Ch. 5–10
Holt Physics Solution Manual
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Momentum and Collisions
Chapter
6
Additional Practice 6A
Givens
Solutions
1. m = 1.46 × 105 kg
p = 9.73 × 105 kg•m/s to the
south
2. m = 25 kg
4
p = 6.8 × 10 kg•m/s
p 9.73 × 105 kg•m/s
v =  = 
m
1.46 × 105 kg
v = 6.66 m/s to the south
p 6.8 × 104 kg•m/s
v =  =  = 2.7 × 103 m/s
m
25 kg
v = (2.7 km/s)(3600 s/h) = 9.7 × 103 km/h
v = 2.7 × 103 m/s = 9.7 × 103 km/h
3. m = 5.00 × 102 kg
p = 8.22 × 103 kg•m/s to the
west
4. mc = 177.4 kg
p 8.22 × 103 kg•m/s
v =  = 
m
5.00 × 102 kg
v = 16.4 m/s to the west
md = 61.5 kg
4.416 × 103 kg•m/s
4.416 × 103 kg•m/s
p
p
v =  =  =  = 
177.4 kg + 61.5 kg
238.9 kg
m mc + md
p = 4.416 × 103 kg•m/s
v = 18.48 m/s = (18.48 m/s)(3600 s/h)(1 km/103 m) = 66.5 km/h
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v = 18.48 m/s = 66.53 km/h)
5. ∆x = 200.0 m
∆t = 19.32 s
∆x 200.0 m
vavg =  =  = 10.35 m/s
∆t
19.32 s
m = 77 kg
p = mv
pavg = mvavg = (77 kg)(10.35 m/s) = 7.8 × 102 kg•m/s
6. ∆x = 274 m to the north
∆t = 8.65 s
∆x 274 m
vavg =  =  = 31.7 m/s to the north
∆t
8.65 s
m = 50.0 kg
p = mv
pavg = mvavg = (50.0 kg)(31.7 m/s) = 1.58 × 103 kg•m/s to the north
7. m = 7.10 × 105 kg
v = 270 km/h
p = mv = (7.10 × 105 kg)(270 km/h)(103 m/km)(1 h/3600 s)
p = 5.33 × 107 kg•m/s
V
Section Five—Problem Bank
V Ch. 6–1
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Givens
Solutions
8. v = 50.0 km/h
p = 0.278 kg•m/s
0.278 kg•m/s
p
m =  = 
v (50.0 km/h)(103 m/km)(1 h/3600 s)
m = 2.00 × 10−2 kg = 20.0 g
9. vavg = 96 km/h to the
southeast
pavg = 4.8 × 104 kg•m/s to
the southeast
10. v = 255 km/s
p = 8.62 × 1036 kg•m/s
4.8 × 104 kg•m/s
p p vg
= 
m =  = a
(96 km/h)(103 m/km)(1 h/3600 s)
v vavg
m = 1.8 × 103 kg
p 8.62 × 1036 kg•m/s
m =  = 
= 3.38 × 1031 kg
v
255 × 103 m/s
Additional Practice 6B
∆p = mvf − mvi = F∆t
∆t = 5.0 s
(10.0 N)(5.05) + (3.0 kg)(0 m/s)
F∆t + mv
vf =  i =  = 17 m/s
3.0 kg
m
vi = 0 m/s
vf = 17 m/s to the right
m = 3.0 kg
2. m = 60.0 g
F = −1.5 N
∆p = mvf − mvi = F∆t
∆t = 0.25 s
mvf − F∆t (60.0 × 10−3 kg)(0 m/s) − (−1.5 N)(0.25 s)
(1.5 N)(0.25 s)
vi =   = 
= 
−3
m
60.0 × 10−3 kg
60.0 × 10 kg
vf = 0 m/s
vi = 6.2 m/s
3. F = 75 N
m = 55 kg
∆p = mvf − mvi = F∆t
∆t = 7.5 s
(75 N)(7.5 s) + (55 kg)(0 m/s)
F∆t + mv
vf = i = 
55 kg
m
vi = 0 m/s
vf = 1.0 × 101 m/s
4. m = 0.195 kg
vi = 0.850 m/s to the right
= +0.850 m/s
F = 3.50 N to the left
= −3.50 N
∆t = 0.0750 s
5. m = 5.00 g
vi = 255 m/s to the right
∆p = mvf − mvi = F∆t
(−3.50 N)(0.0750 s) + (0.195 kg)(0.850 m/s)
F∆t + mv
vf =  i = 
0.195 kg
m
−0.262 kg•m/s + 0.166 kg•m/s
−0.096 kg•m/s
vf =  =  = −0.49 m/s
0.195 kg
0.195 kg
vf = 0.49 m/s to the left
∆p = mvf − mvi = F∆t
vf = 0 m/s
mvf − mvi (5.00 × 10−3 kg)(0 m/s) − (5.00 × 10−3 kg)(255 m/s)
 =  = −0.879 N
F=
1.45 s
∆t
∆t = 1.45 s
F = 0.879 N to the left
V
V Ch. 6–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. F = 10.0 N to the right
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Givens
6. m = 1.1 × 103 kg
vf = 9.7 m/s to the east
vi = 0 m/s
∆t = 19 s
Solutions
∆p mvf − mvi

F =  = 
∆t
∆t
(1.1 × 103 kg)(9.7 m/s) − (1.1 × 103 kg)(0 m/s)
F =  = 560 N
19 s
F = 560 N to the east
7. m = 3.00 × 103 kg
vi = 0 m/s
vf = 8.9 m/s to the right
∆t = 5.5 s
8. m = 0.17 kg
∆v = −9.0 m/s
g = 9.81 m/s2
mk = 0.050
∆p mvf − mvi

F =  = 
∆t
∆t
(3.00 × 103 kg)(8.9 m/s) − (3.00 × 103 kg)(0 m/s)
(3.00 × 103 kg)(8.9 m/s)
F =  = 
5.5 s
5.5 s
F = 4.9 × 103 N to the right
F∆t = ∆p = m∆v
F = Fk = −mgmk
m∆v
∆v
−9.0 m/s
∆t =  =  = 
−mgmk −gmk
−(9.81 m/s2)(0.050)
∆t = 18 s
9. m = 12.0 kg
Fapplied = 15.0 N
q = 20.0°
Ffriction = 11.0 N
mvf − mvi
(12.0 kg)(4.50 m/s) − (12.0 kg)(0 m/s)
∆p
∆t =  =  = 
F
(cos
q)
−
F
(15.0 N)(cos 20.0°) − 11.0 N
F
applied
friction
vi = 0 m/s
5.40 kg•m/s
5.40 kg•m/s − 0 kg•m/s
∆t =  = 
3.1 N
14.1 N − 11.0 N
vf = 4.50 m/s
∆t = 1.7 s
10. vf = 15.8 km/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
F = Fapplied(cos q) − Ffriction
vi = 0 km/s
F = 12.0 N
m = 0.20 g
∆p mvf − mvi

∆t =  = 
F
F
(0.20 × 10−3 kg)(15.8 × 103 m/s) − (0.20 × 10−3 kg)(0 m/s)
∆t = 
12.0 N
(0.20 × 10−3 kg)(15.8 × 103 m/s)
∆t = 
12.0 N
∆t = 0.26 s
Additional Practice 6C
1. vi = 382 km/h to the right
vf = 0 km/h
mc = 705 kg
md = 65 kg
∆t = 12.0 s
∆p (mc + md)vf − (mc + md)vi
F =  = 
∆t
∆t
[(705 kg + 65 kg)(0 km/h) − (705 kg + 65 kg)(382 km/h)](103 m//km)(1 h/3600 s)
F = 
12.0 s
−(7.70 × 102 kg)(382 km/h)(103 m/km)(1 h/3600 s)
F =  = − 6.81 × 103 N
12.0 s
F = 6.81 × 103 N to the left
1
V
1
∆x = 2(vi + vf)∆t = 2(382 km/h + 0 km/h)(103 m/km)(1 h/3600 s)(12.0 s)
∆x = 637 m to the right
Section Five—Problem Bank
V Ch. 6–3
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Givens
Solutions
2. vi = 7.82 × 103 m/s
vf = 0 m/s
m = 42 g
∆t = 1.0 × 10−6 s
∆p mvf − mv
F =  = i
∆t
∆t
(42 × 10−3 kg)(0 m/s) − (42 × 10−3 kg)(7.82 × 103 m/s)
F = 
1.0 × 10−6 s
−(42 × 10−3 kg)(7.82 × 103 m/s)
F = 
1.0 × 10−6 s
F = −3.3 × 108 N
∆x = 2(vi + vf)∆t = 2(7.82 × 103 m/s + 0 m/s)(1.0 × 10−6 s)
1
1
∆x = 3.9 × 10−3 m = 3.9 mm
3. m = 63 kg
vi = 7.0 m/s to the right
vf = 0 m/s
∆p mvf − mvi

F =  = 
∆t
∆t
(63 kg)(0 m/s) − (63 kg)(7.0 m/s)
−(63 kg)(7.0 m/s)
F =  =  = −32 N
14.0 s
14.0 s
∆t = 14.0 s
F = 32 N to the left
1
1
∆x = 2(vi + vf) ∆t = 2(7.0 m/s + 0 m/s)(14.0 s)
∆x = 49 m to the right
∆p = F∆t
4. m = 455 kg
∆t = 12.2 s
F = Fk = −mgmk
2
g = 9.81 m/s
vf = 0 m/s
∆p = −mgmk∆t = −(455 kg)(9.81 m/s2)(0.071)(12.2 s) = −3.9 × 103 kg•ms
∆p = 3.9 × 103 kg•ms opposite the polar bear’s motion
mvf − ∆p
(455 kg)(0 m/s) − (−3.9 × 103 kg•ms)
vi =  = 
m
455 kg
3.9 × 103 kg•ms
vi =  = 8.6 m/s
455 kg
1
1
∆x = 2(vi + vf)∆t = 2(8.6 m/s + 0 m/s)(12.2 s)
∆x = 52 m
5. m = 75.0 g
∆p = F∆t
∆t = 1.2 s
F = −mg
g = 9.81 m/s2
∆p = −mg∆t = −(75.0 × 10−3 kg)(9.81 m/s2)(1.2 s)
vf = 0 m/s
∆p = −0.88 kg•m/s
∆p = 0.88 kg•m/s downward
mvf − ∆p (75.0 × 10−3 kg)(0 m/s) − (−0.88 kg•m/s)
 = 
vi = 
m
75.0 × 10−3 kg
0.88 kg•m/s
vi = 
= 12 m/s upward
75.0 × 10−3 kg
1
V
V Ch. 6–4
1
∆x = 2(vi + vf)∆t = 2(12 m/s + 0 m/s)(1.2 s)
∆x = 7.2 m upward
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
mk = 0.071
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Givens
6. m = 4400 kg
F = 2200 N to the left
= −2200 N
∆t = 8.0 s
vi = 6.5 m/s to the right
= +6.5 m/s
Solutions
∆p = F∆t = (−2200 N)(8.0 s) = −1.8 × 104 kg•m/s
∆p = 1.8 × 104 kg•m/s to the left
∆p + mv
−1.8 × 104 kg•m/s + (4400 kg)(6.5 m/s)
vf = i = 
m
4400 kg
−1.8 × 104 kg•m/s + 2.9 × 104 kg•m/s
1.1 × 104 kg•m/s
vf =  =  = 2.5 m/s to the right
4400 kg
4400 kg
1
1
1
∆x = 2(vi + vf)∆t = 2(6.5 m/s + 2.5 m/s)(8.0 s) = 2(9.0 m/s)(8.0 s)
∆x = 36 m to the right
7. F = 25.0 N
∆t = 7.00 s
m = 14.0 kg
vi = 0 m/s
∆p = F∆t = (25.0 N)(7.00 s) = 175 kg•m/s
∆p + mv
175 kg•m/s + (14.0 kg)(0 m/s)
vf = i =  = 12.5 m/s
m
14.0 kg
1
1
∆x = 2(vi + vf)∆t = 2(0 m/s + 12.5 m/s)(7.00 s)
∆x = 43.8 m
8. m = 2.30 × 103 kg
vi = 22.2 m/s
vf = 0 m/s
F = −1.26 × 104 N
∆p mvf − mv
∆t =  = i
F
F
−5.11 × 104 kg•m/s
(2.30 × 103 kg)(0 m/s) − (2.30 × 103 kg)(22.2 m/s)

∆t = 
=
−1.26 × 104 N
−1.26 × 104 N
∆t = 4.06 s
1
1
∆x = 2(vi + vf)∆t = 2(22.2 m/s + 0 m/s)(4.06 s)
∆x = 45.1 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. m = 1.35 × 104 kg
vi = 66.1 m/s to the west
= −66.1 m/s
vf = 0 m/s
F = 4.00 × 105 N to the east
= +4.00 × 105 N
∆p mvf − mvi

∆t =  = 
F
F
(1.35 × 104 kg)(0 m/s) − (1.35 × 104 kg)(−66.1 m/s)
8.92 × 105 kg•m/s
∆t = 
= 
5
4.00 × 10 N
4.00 × 105 N
∆t = 2.23 s
1
1
∆x = 2(vi + vf)∆t = 2(−66.1 m/s + 0 m/s)(2.23 s) = −73.7 m
∆x = 73.7 m to the west
vf = 0 m/s
∆p mvf − mv
∆t =  = i
F
F
m = 1.50 × 103 kg
F = Fk = mamk
mk = 0.065
vf − v
∆t = i
amk
10. vi = 14.5 m/s
a = −1.305 m/s2
0 m/s − 14.5 m/s
∆t = 
(− 1.305 m/s2)(0.065)
∆t = 1.7 × 102 s
1
1
∆x = 2(vi + vf)∆t = 2(14.5 m/s + 0 m/s)(1.7 × 102 s)
V
∆x = 1.2 × 103 m = 1.2 km
Section Five—Problem Bank
V Ch. 6–5
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Additional Practice 6D
Givens
Solutions
1. m1 = 68 kg
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m2 = 68 kg
(68 kg)(−0.85 m/s) + (68 kg)(−0.85 m/s) − (68 kg)(0 m/s)
m1v1, f + m2v2, f − m2v2, i
v1, i = 
= 
68 kg
mi
v2, i = 0 m/s
v1, f = 0.85 m/s to the west
= −0.85 m/s
v1, i = −0.85 m/s + (−0.85 m/s) = −1.7 m/s
v2, f = 0.85 m/s to the west
= −0.85 m/s
v1, i = 1.7 m/s to the west
2. mi = 1.36 × 104 kg
3
m2 = 8.4 × 10 kg
v2, i = 0 m/s
v1, f = v2, f = 1.3 m/s
m1v1, i + m2v2, i = m1 v1, f + m2v2, f
m1v1, f + m2v2, f − m2v2, i
v1, i = 
m1
(1.36 × 104 kg)(1.3 m/s) + (8.4 × 103 kg)(1.3 m/s) − (8.4 × 103 kg)(0 m/s)
v1, i = 
1.36 × 104 kg
1.8 × 104 kg•m/s + 1.1 × 104 kg•m/s
2.9 × 104 kg•m/s
= 
v1, i = 
4
1.36 × 10 kg
1.36 × 104 kg
v1, i = 2.1 m/s
3. v1, f = 2.2 m/s backwards
= −2.2 m/s
v2, f = 5.5 m/s forward
= +5.5 m/s
m1 = 38 kg
m2 = 68 kg
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, i = v2, i, so
m1v1, f + m2v2, f (38 kg)(−2.2 m/s) + (68 kg)(5.5 m/s)
 = 
v1, i = 
m1 + m2
38 kg + 68 kg
290 kg•m/s
−84 kg•m/s + 370 kg•m/s
v1, i =  =  = 2.7 m/s
106 kg
106 kg
4. m1 = 38 kg
v1, i = 1.6 m/s to the north
m2 = 142 kg
v1, f = 0.32 m/s to the north
v2, f = 0.32 m/s to the north
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m1v1, f + m2v2, f − m1v1, i
v2, i = 
m2
(38 kg)(0.32 m/s) + (142 kg)(0.32 m/s) − (38 kg)(1.6 m/s)
v2, i = 
142 kg
12 kg•m/s + 45 kg•m/s − 61 kg•m/s
−4.0 kg•m/s
v2, i =  =  = −2.8 × 10−2 m/s
142 kg
142 kg
v2, i = 2.8 × 10−2 m/s to the south
5. m1 = 50.0 g
v1, i = 0 m/s
V
Because the initial velocities for both rifle and projectile are zero, the momentum conservation equation takes the following form:
v1, f = 400.0 m/s forward
m1v1, f + m2v2, f = 0
m2 = 3.00 kg
−m1v1, f
−(50.0 × 10−3 kg)(400.0 m/s)
v2, f =  =  = −6.67 m/s
m2
3.00 kg
v2, i = 0 m/s
v2, f = 6.67 m/s backward
V Ch. 6–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v1, i = 2.7 m/s forward
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Givens
Solutions
6. mi = 1292 kg
vi = 88.0 km/h to the east
mf = 1255 kg
mivi = mfvf
mv
(1292 kg)(88.0 km/h)
vf = ii = 
mf
1255 kg
vf = 90.6 km/h to the east
7. m = 5.0 × 1014 kg
vi = 74.0 km forward
mi = m2 =
1
m
2
v1, f = 105 km/s at 15.0°
above forward
v2, f is at an angle of −30.0°
to the forward direction
mvi = m1v1, f + m2v2, f
To solve for velocity in two dimensions, the momentum conservation equation must
be written as two equations, one for both the x and y directions.
In the x-direction:
mvi = m1v1, f (cos q1) + m2v2, f (cos q2)
1
1
vi = 2v1, f (cos q1) + 2v2, f (cos q2)
2vi − v1, f (cos q1)
v2, f = 
cos q2
(2)(74.0 km/s) − (105 km/s)(cos 15.0°)
v2, f = 
cos(−30.0°)
148 km/s − 101 km/s
47 km/s
v2, f =  = 
cos(−30.0°)
cos(−30.0°)
v2, f = 54 km/s
In the y-direction (check):
0 = m1v1, f (sin q1) + m2v2, f (sin q2)
v1, f (sin q1) = −v2, f (sin q2)
(105 km/s)(sin 15.0°) = −(54 km/s)[sin(−30.0°)]27.2 km/s = 27 km/s
The slight difference arises from differences in the number of significant figures and
from rounding.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v2,f = 54 km/s at 30.0° below the initial forward direction
8. v1, i = 0 cm/s
v1, f = 1.2 cm/s forward
= +1.2 cm/s
v2, i = 0 cm/s
v2, f = 0.40 cm/s backward
= −0.40 cm/s
m1v1, f + m2v2, f = 0
−m1v1, f
−(2.5 g)(1.2 cm/s)
m2 = 
= 
v2, f
−0.40 cm/s
m2 = 7.5 g
m1 = 2.5 g
V
Section Five—Problem Bank
V Ch. 6–7
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Givens
Solutions
9. vi = 0 cm/s
m1 = 25.0 g
m2 = 25.0 g
v1 = 7.0 cm/s to the south
= 7.0 cm/s at −90° from
east
v2 = 7.0 cm/s to the west
= 7.0 cm/s at 180° from
east
v3 = 3.3 m/s at 45° north of
east
m1v1, f + m2v2, f + m3v3, f = 0
In the x-direction:
m1v1, f (cos q1) + m2v2, f (cos q2) + m3v3, f (cos q3) = 0
m1v1, f (cos q1) + m2v2, f (cos q2)
m3 = 
−v3, f (cos q3)
− (25.0 g)(7.0 cm/s)(cos −90°) + (25.0 g)(7.0 cm/s)(cos 180°)
m3 = 
−(3.3 cm/s)(cos 45°)
(25.0 g)(7.0 cm/s)
m3 =  = 75 g
(3.3 cm/s)(cos 45°)
In the y-direction (check):
m1v1, f (sin q1) + m2v2, f (sin q2) + m3v3, f (sin q3) = 0
(25.0 g)(7.0 cm/s)[sin(−90°)] + (25.0 g)(7.0 cm/s)(sin 180°)
+ (75 g)(3.3 cm/s)(sin 45°) = 0
−180 g•cm/s + 0 g•cm/s + 180•g•cm/s = 0
10. v1, i = 0 m/s
v2, i = 5.4 m/s to the north
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, f = 1.5 m/s to the north
(63 kg)(1.5 m/s)
m1v1, f − m1v1, i
(63 kg)(1.5 m/s) − (63 kg)(0 m/s)
m2 =  =  = 
3.9 m/s
v2, i − v2, f
5.4 m/s − 1.5 m/s
v2, f = 1.5 m/s to the north
m2 = 24 kg
m1 = 63 kg
Additional Practice 6E
m2 = 770 kg
v2, i = 0 m/s
vf = 9.44 m/s forward
(m1 + m2)vf − m2v2, i
v1, i = 
m1
(1550 kg + 770 kg)(9.44 m/s) − (770 kg)(0 m/s)
(2320 kg)(9.44 m/s)
v1, i =  = 
1550 kg
1550 kg
v1, i = 14.0 m/s forward
m2 = 0.75 kg
(m1 + m2)vf − m2v2, i
v1, i = 
m1
v2, i = 0.50 m/s to the left
= −0.50 m/s
(0.17 kg + 0.75 kg)(4.2 m/s) − (0.75 kg)(−0.50 m/s)
v1, i = 
0.17 kg
2. m1 = 0.17 kg
vf = 4.2 m/s to the right
= +4.2 m/s
(0.92 kg)(4.2 m/s) + 0.38 kg•m/s
v1, i = 
0.17 kg
3.9 kg•m/s + 0.38 kg•m/s
4.3 kg•m/s
v1, i =  = 
0.17 kg
0.17 kg
v1, i = 25 m/s to the right
V
V Ch. 6–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. m1 = 1550 kg
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Givens
3. m1 = 45 g
m2 = 75 g
v2, i = 0 m/s
h = 8.0 cm
g = 9.81 m/s2
Solutions
Use the conservation of mechanical energy to calculate vf .
KE = PE
1
(m
1
2
+ m2)vf2 = (m1 + m2)gh
vf = 2g
m/s
h = (2
)(
9.
81
2)(8
.0
×10−2m
)
vf = 1.3 m/s
(m1 + m2)vf − m2v2, i
(45 g + 75 g)(1.3 m/s) − (75 g)(0 m/s)
v1, i =  = 
m1
45 g
(1.20 × 102 g)(1.3 m/s)
v1, i =  = 3.5 m/s
45 g
The height that the first ball must have can be determined by using the conservation
of mechanical energy.
PE = KE
1
m1gh1 = 2m1v1, i2
(3.5 m/s)
v i2
h1 = 1,
= 
2 = 0.62 m = 62 cm
2 g (2)(9.81 m/s )
4. m1 = m2 = m3 = 5.00 × 102 kg
vf = 3.67 m/s
v3, i = 3.00 m/s
v2, i = 3.50 m/s
m1v1, i + m2v2, i + m3v3, i = (m1 + m2 + m3)vf
(m1 + m2 + m3)vf − m2v2, i − m3v3, i
v1, i = 
m1
(5.00 × 102 kg + 5.00 × 102 kg + 5.00 × 102 kg)(3.67 m/s)
v1, i = 
5.00 × 102 kg
(5.00 × 102 kg)(3.50 m/s) − (5.00 × 102 kg)(3.00 m/s)
− 
5.00 × 102 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(15.00 × 102 kg)(3.67 m/s) − 1750 kg•m/s − 1.50 × 103 kg•m/s
v1, i = 
5.00 × 102 kg
5.50 × 103 kg•m/s − 3250 kg•m/s
v1, i = 
5.00 × 102 kg
2250 kg•m/s
v1, i = 
= 4.50 m/s
5.00 × 102 kg
5. m1 = 8500 kg
v1, i = 4.5 m/s to the right
= +4.5 m/s
m2 = 9800 kg
v2, i = 3.9 m/s to the left
= −3.9 m/s
m1v1, i + m2v2, i
vf = 
m1 + m2
(8500 kg)(4.5 m/s) + (9800 kg)(−3.9 m/s)
3.8 × 104 kg•m/s − 3.8 × 104 kg•m/s
vf =  = 
8500 kg + 9800 kg
1.83 × 104 kg
vf = 0 m/s
V
Section Five—Problem Bank
V Ch. 6–9
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Givens
Solutions
v1, i = 45 km/h to the north
m1v1, i + m2v2, i
vf = 
m1 + m2
m2 = 2500 kg
The component of vf in the x-direction is given by
v2, i = 33 km/h to the east
m2v2, i
(2500 kg)(33 km/h)
(2500 kg)(33 km/h)
vf, x =  =  = 
m1 + m2
3900 kg
1400 kg + 2500 kg
6. m1 = 1400 kg
vf, x = 21 km/h
The component of vf in the y-direction is given by
m1v1, i
(1400 kg)(45 km/h)
(1400 kg)(45 km/h)
vf, y =  =  = 
m1 + m2
3900 kg
1400 kg + 2500 kg
vf, y = 16 km/h
2
2
vf = vf
)2
+(16
)2
vf,
1km
/h
km
/h
, x +
y = (2
vf = 44
0km
2/h2+260
km
2/h2 = 7.
0×102km
2/h2
vf = 26 km/h
vf
16 km/h
q = tan−1 ,y = tan−1  = 37°
vf, x
21 km/h
vf = 26 km/h at 37° north of east
v1, i = 0.80 m/s to the west
= −0.80 m/s
m2 = 60.0 g
The component of final velocity in the x-direction is given by
m1v1, i + m3v3, i
(50.0 g)(−0.80 m/s) + (100.0 g)(0.20 m/s)
vf, x =  = 
m1 + m2 + m3 + m4
50.0 g + 60.0 g + 100.0 g + 40.0 g
v2, i = 2.50 m/s to the north
= +2.50 m/s
−4.0 × 101 g•m/s + 2.0 × 101 g•m/s −2.0 × 101 g•m/s
vf, x =  = 
250.0 g
250.0 g
m3 = 100.0 g
vf, x = −8.0 × 10−2 m/s
v3, i = 0.20 m/s to the east
= +0.20 m/s
The component of final velocity in the y-direction is given by
m4 = 40.0 g
v4, i = 0.50 m/s to the south
= −0.50 m/s
m2v2, i + m4v4, i
(60.0 g)(2.50 m/s) + (40.0 g)(−0.50 m/s)
vf, y =  = 
m1 + m2 + m3 + m4
50.0 g + 60.0 g + 100.0 g + 40.0 g
1.50 × 102 g•m/s − 2.0 × 101 g•m/s 1.30 × 102 g•m/s
vf, y =  = 
250.0 g
250.0 g
vf, y = 0.520 m/s
2
2
vf = vf
0−2m
)2
+(0.
)2
vf, 8.
0×1
/s
52
0m
/s
, x +
y = (−
vf = 6.
s2
+0.2
m2/
s2 = 0.
s2
4×10−3m
2/
70
27
6m
2/
vf = 0.525 m/s
vf
0.520 m/s
q = tan−1 ,y = tan−1 
= −81°
vf, x
−8.0 × 10−2 m/s
q = 81° north of west, or 9° west of north
vf = 0.53 m/s at 9° west of north
V
V Ch. 6–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. m1 = 50.0 g
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Givens
8. ms = 25.0 kg
Solutions
m1 = mass of child and sled = ms + mc = 25.0 kg + 42.0 kg = 67.0 kg
mc = 42.0 kg
m1v1, i + m2v2, i = (m1 + m2)vf
v1, i = 3.50 m/s
m1v1, i − m1vf
(67.0 kg)(3.50 m/s) − (67.0 kg)(2.90 m/s)
m2 =  = 
vf − v2, i
2.90 m/s − 0 m/s
v2, i = 0 m/s
vf = 2.90 m/s
40.0 kg•m/s
234 kg•m/s − 194 kg•m/s

m2 =  = 
2.90 m/s
2.90 m/s
m2 = 13.8 kg
9. v1, i = 5.0 m/s to the right
= +5.0 m/s
v2, i = 7.00 m/s to the left
= −7.00 m/s
vf = 6.25 m/s to the left
= −6.25 m/s
m2v2, i − m2vf
(150.0 kg)(−7.00 m/s) − (150.0 kg)(−6.25 m/s)
m1 =  = 
vf − v1, i
−6.25 m/s − 5.0 m/s
−1050 kg•m/s + 938 kg•m/s
−110 kg•m/s
m1 =  = 
−11.2 m/s
−11.2 m/s
m1 = 9.8 kg
m2 = 150.0 kg
10. v1, i = 8.0 × 103 m/s to the
right = +8.0 × 103 m/s
v2, i = 8.0 × 103 m/s to the
left = −8.0 × 103 m/s
vf = (0.900)v1, i
m2 = (52 000)(2.5 g)
= 1.3 × 105 g
m2vf − m2v2, i
(1.3 × 102 kg)(0.900)(8.0 × 103 m/s) − (1.3 × 102 kg)(−8.0 × 103 m/s)
m1 =  = 
v1, i − vf
(8.0 × 103 m/s)(1 − 0.900)
9.4 × 105 kg•m/s + 1.0 × 106 kg•m/s
1.9 × 106 kg•m/s
m1 = 
= 
3
(8.0 × 10 m/s)(0.100)
8.0 × 102 m/s
m1 = 2.4 × 103 kg
Additional Practice 6F
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. m1 = 55 g
v1, i = 1.5 m/s
m1v1, i + m2v2, i
(55 g)(1.5 m/s) + (55 g)(0 m/s)
(55 g)(1.5 m/s)
vf =  =  = 
m1 + m2
55 g + 55 g
1.10 × 102 g
m2 = 55 g
vf = 0.75 m/s
v2, i = 0 m/s
∆KE
KEf − KEi
KEf
percent decrease of KE =  × 100 =  × 100 =  − 1 × 100
KEi
KEi
KEi
KEi = 2m1v1, i2 + 2m2v2, i = 2(55 × 10−3 kg)(1.5 m/s)2 + 2(55 × 10−3 kg)(0 m/s)2
1
1
1
1
KEi = 6.2 × 10−2 J + 0 J = 6.2 × 10−2 J
KEf = 2(m1 + m2)vf2 = 2(55 g + 55 g)(10−3 kg/g)(0.75 m/s)3
1
1
KEf = 3.1 × 10−2 J
percent decrease of KE =
3.1 × 10−2 J
 − 1 × 100 = (0.50 − 1) × 100 = (−0.50) × 100
6.2 × 10−2 J
percent decrease of KE = −5.0 × 101 percent
V
Section Five—Problem Bank
V Ch. 6–11
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Givens
Solutions
2. m1 = 4.5 kg
v1, i = 0 m/s
m2 = 1.3 kg
vf = 0.83 m/s
(m1 + m2)vf − m1v1, i
(4.5 kg + 1.3 kg)(0.83 m/s) − (4.5 kg)(0 m/s)
v2, i =  = 
m2
1.3 kg
(5.8 kg)(0.83 m/s)
v2, i =  = 3.7 m/s
1.3 kg
1
1
1
1
KEi = 2m1v1, i2 + 2m2v2, i2 = 2(4.5 kg)(0 m/s)2 + 2(1.3 kg)(3.7 m/s)2
KEi = 0 J + 8.9 J = 8.9 J
1
1
1
KEf = 2(m1 + m2)vf2 = 2(4.5 kg + 1.3 kg)(0.83 m/s)2 = 2(5.8 kg)(0.83 m/s)2
KEf = 2.0 J
∆KE = KEf − KEi = 2.0 J − 8.9 J = −6.9 J
3. m1 = 1.50 × 1013 kg
v1, i = 250 m/s
m2 = 6.5 × 1012 kg
v2, i = 420 m/s
m1v1, i + m2v2, i
(1.50 × 1013 kg)(250 m/s) + (6.5 × 1012 kg)(420 m/s)
vf =  = 
m1 + m2
1.50 × 1013 kg + 6.5 × 1012 kg
3.8 × 1015 kg•m/s + 2.7 × 1015 kg•m/s
6.5 × 1015 kg•m/s
vf = 
= 
= 3.0 × 102 m/s
13
2.15 × 10 kg
2.15 × 1013 kg
1
1
1
1
KEi = 2m1v1, i2 + 2m2v2, i2 = 2(1.50 × 1013 kg)(250 m/s)2 + 2(6.5 × 1012 kg) (420 m/s)2
KEi = 4.7 × 1017 J + 5.7 × 1017 J = 10.4 × 1017 J = 1.04 × 1018
1
1
KEf = 2(m1 + m2)vf2 = 2(1.50 × 1013 kg + 6.5 × 1012 kg)(3.0 × 102 m/s)2
1
KEf = 2(2.15 × 1013 kg)(3.0 × 102 m/s)2
KEf = 9.7 × 1017 J
∆KE = KEf − KEi = 9.7 × 1017 J − 1.04 × 1018 J = −7.0 × 1016 J
v1, i = 15.0 m/s to the right
= +15.0 m/s
m2 = 0.950 kg
v2, i = 13.5 m/s to the left
= −13.5 m/s
m1v1, i + m2v2, i
(0.650 kg)(15.0 m/s) + (0.950 kg)(−13.5 m/s)
vf =  = 
m1 + m2
0.650 kg + 0.950 kg
9.75 kg•m/s − 12.8 kg•m/s
−3.0 kg•m/s
vf =  =  = −1.91 m/s
1.600 kg
1.600 kg
vf = 1.91 m/s to the left
1
1
1
1
KEi = 2m1v1, i2 + 2m2v2, i2 = 2(0.650 kg)(15.0 m/s)2 + 2(0.950 kg)(−13.5 m/s)2
KEi = 73.1 J + 86.6 J = 159.7 J
1
1
1
KEf = 2(m1 + m2)vf2 = 2(0.650 kg + 0.950 kg)(1.91 m/s)2 = 2(1.600 kg)(1.91 m/s)2
KEf = 2.92 J
∆KE = KEf − KEi = 2.92 J − 159.7 J = −1.57 × 102 J
V
V Ch. 6–12
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. m1 = 0.650 kg
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Givens
Solutions
5. m1 = 75.0 kg
v1, i = 1.80 m/s downstream
= 1.80 m/s
m2 = (8)(0.30 kg) = 2.4 kg
v2, i = 1.3 m/s upstream
= −1.3 m/s
m1v1, i + m2v2, i
(75.0 kg)(1.80 m/s) + (2.4 hg)(−1.3 m/s)
vf =  = 
m1 + m2
75.0 kg + 2.4 kg
135 kg•m/s − 3.1 kg•m/s
132 kg•m/s
vf =  =  = 1.71 m/s
77.4 kg
77.4 kg
1
1
1
1
KEi = 2m1v1, i2 + 2m2v2, i2 = 2(75.0 kg)(1.80 m/s)2 + 2(2.4 kg)(−1.3 m/s)2
KEi = 122 J + 2.0 J = 124 J
1
1
1
KEf = 2(m1 + m2)vf2 = 2(75.0 kg + 2.4 kg)(1.71 m/s)2 = 2(77.4 kg)(1.71 m/s)2 = 113 J
∆KE = KEf − KEi = 113 J − 124 J = −11 J
6. m1 = 8500 kg
1
1
1
1
KEi = 2m1v1, i2 + 2 m2v2, i2 = 2(8500 kg)(4.5 m/s)2 + 2(9800 kg)(−3.9 m/s)2
v1, i = 4.5 m/s
KEi = 8.6 × 104 J + 7.5 × 104 J = 16.1 × 104 J = 1.61 × 105 J
m2 = 9800 kg
1
1
KEf = 2(m1 + m2)vf2 = 2(8500 kg + 9800 kg)(0 m/s)2 = 0 J
v2, i = −3.9 m/s
∆KE = KEf − KEi = 0 J − 1.61 × 105 J = −1.61 × 105 J
vf = 0 m/s
7. m1 = 45 g
KEi = 2m1v1, i2 + 2m2v2, i2 = 2(45 × 10−3 kg)(3.5 m/s)2 + 2(75 × 10−3 kg)(0 m/s)2
1
v1, i = 3.5 m/s
1
1
1
KEi = 0.28 J + 0 J = 0.28 J
m2 = 75 g
KEf = 2(m1 + m2)vf2 = 2(45 g + 75 g)(10−3 kg/g)(1.3 m/s)2 = 2(1.20 × 10−3 kg)(1.3 m/s)2
1
v2, i = 0 m/s
1
1
KEf = 1.0 × 10−3 J
vf = 1.3 m/s
∆KE = KEf − KEi = 1.0 × 10−3 J − 0.28 J = −0.28 J
The decrease in kinetic energy is almost total.
8. m1 = 2.4 × 103 kg
1
v1, i = 8.0 × 103 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
KEi = 2m1v1, i2 + 2m2v2, i2
1
1
KEi = 2(2.4 × 103 kg)(8.0 × 103 m/s)2 + 2(1.3 × 102 kg)(−8.0 × 103 m/s)2
m2 = 1.3 × 102 kg
KEi = 7.7 × 1010 J + 4.2 × 109 J = 8.1 × 1010 J
3
v2, i = −8.0 × 10 m/s
1
3
vf = (0.900)(8.0 × 10 m/s)
1
KEf = 2(m1 + m2)vf 2 = 2(2.4 × 103 kg + 1.3 × 102 kg)[(0.900)(8.0 × 103 m/s)]2
1
KEf = 2(2.5 × 103 kg)(7.2 × 103 m/s)2
KEf = 6.5 × 1010 J
∆KE = KEf − KEi = 6.5 × 1010 J − 8.1 × 1010 J = −1.6 × 1010 J
9. m1 = m2 = m3 = 5.00 × 102 kg
v1, i = 4.50 m/s
v2, i = 3.50 m/s
v3, i = 3.00 m/s
vf = 3.67 m/s
1
1
1
KEi = 2m1v1, i2 + 2m2v2, i2 + 2m3v3, i2
1
1
KEi = 2(5.00 × 102 kg)(4.50 m/s)2 + 2(5.00 × 102 kg)(3.50 m/s)2
+
1
(5.00
2
× 102 kg)(3.00 m/s)2
KEi = 5.06 × 103 J + 3.06 × 103 J + 2.25 × 103 J = 10.37 × 103 J = 10.37 kJ
1
1
KEf = 2(m1 + m2 + m3)vf2 = 2(5.00 × 102 kg + 5.00 × 102 kg + 5.00 × 102 kg)
1
(3.67 m/s)2 = 2(1.500 × 103 kg)(3.67 m/s)2
KEf = 1.01 × 104 J = 10.1 kJ
V
∆KE = KEf − KEi = 10.1 kJ − 10.37 kJ = −300 J
Section Five—Problem Bank
V Ch. 6–13
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Givens
Solutions
10. m1 = 50.0 × 10−3 kg
v1, i = −0.80 m/s
−3
m2 = 60.0 × 10
1
1
1
1
KEi = 2m1v1, i2 + 2m2v2, i2 + 2m3v3, i2 + 2m4v4, i2
KEi = 2(50.0 × 10−3 kg)(−0.80 m/s)2 + 2(60.0 × 10−3 kg)(2.50 m/s)2
1
kg
v2, i = 2.50 m/s
m3 = 0.1000 kg
v3, i = 0.20 m/s
m4 = 40.0 × 10−3 kg
v4, i = −0.50 m/s
vf = 0.53 m/s
+
1
1
(0.1000
2
kg)(0.20 m/s)2 +
1
(40.0
2
× 10−3 kg)(−0.50 m/s)2
KEi = 1.6 × 10−2 J + 18.8 × 10−2 J + 0.20 × 10−2 J + 0.50 × 10−2 J
KEi = 0.211 J
1
KEf = 2(m1 +m2 + m3 + m4)vf 2
KEf = 2(50.0 × 10−3 kg + 60.0 × 10−3 kg + 0.1000 kg + 40.0 × 10−3 kg)(0.53 m/s)2
1
KEf = 2(0.2500 kg)(0.53 m/s)2 = 3.5 × 10−2 J
1
∆KE = KEf − KEi = 3.5 × 10−2 J − 0.211 J = −0.18 J
Additional Practice 6G
1. v1, i = 6.00 m/s to the right
= +6.00 m/s
v2, i = 0 m/s
v1, f = 4.90 m/s to the left
= −4.90 m/s
v2, f = 1.09 m/s to the right
= +1.09 m/s
m2 = 1.25 kg
Momentum conservation
m1v1, i + m2v2, i = m1v1, f + m2v2, f
1.36 kg•m/s
(1.25 kg)(1.09 m/s) − (1.25 kg)(0 m/s)
m2v2, f − m2v2, i
m1 = 
=  = 
10.90 m/s
6.00 m/s − (−4.90 m/s)
v1, i − v1, f
m1 = 0.125 kg
Conservation of kinetic energy (check)
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2m1v1, f2 + 2m2v2, f2
1
1
(0.125 kg)(6.00 m/s)2 + (1.25
2
2
1
+ 2(1.25 kg)(1.09 m/s)2
1
kg)(0 m/s)2 = 2(0.125 kg)(−4.90 m/s)2
2.25 J + 0 J = 1.50 J + 0.74 J
2.25 J = 2.24 J
2. m1 = 2.0 kg
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, i = 8.0 m/s
1
m v 2
2 1 1, i
v2, i = 0 m/s
m1v1, f − m1v1, i
m2 = 
v2, i − v2, f
v1, f = 2.0 m/s
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2m1v1, f2 + 2m2v2, f2
1 m1v1, f − m1v1, i
1
1 m1v1, f − m1v1, i
+ 2  v2, i2 = 2m1v1, f2 + 2  v2, f 2
v2, i − v2, f
v2, i − v2, f
v1, i2(v2, i - v2, f) + (v1, f - v1, i )v2, i2 = v1, f 2(v2, i - v2, f) + (v1, f - v1, i )v2, f 2
(v1, i2v2, i + v1, f v2, i2 - v1, iv2, i2 - v1, f 2v2, i + v2, f (v1, f 2 - v1, i2) = v2, f 2(v1, f - v1, i )
Because v2, i = 0, the above equation simplifies to
v1, f2 − v1, i2 = v2, f (v1, f − v1, i)
v2, f = v1, f + v1, i = 2.0 m/s + 8.0 m/s = 10.0 m/s
V
(2.0 kg)(2.0 m/s) − (2.0 m/s)(8.0 m/s)
4.0 kg•m/s − 16 kg•m/s
−12 kg•m/s
m2 =  =  = 
0 m/s − 10.0 m/s
−10.0 m/s
−10.0 m/s
m2 = 1.2 kg
V Ch. 6–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
The slight difference arises from rounding.
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Givens
3. m1 = m2 = 45 g
Solutions
Momentum conservation
v2, i = 0 m/s
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, f = 0 m/s
v1, i = v1, f + v2, f − v2, i = 0 m/s + 3.0 m/s − 0 m/s
v2, f = 3.0 m/s
v1, i = 3.0 m/s
Conservation of kinetic energy (check)
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2m1v1, f2 + 2m2v2, f2
v1, i2 + v2, i2 = v1, f2 + v2, f2
(3.0 m/s)2 + (0 m/s)2 = (0 m/s)2 + (3.0 m/s)2
9.0 m2/s2 = 9.0 m2/s2
4. m1 = 3.0 × 107 kg
Momentum conservation
m2 = 2.5 × 107 kg
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v2, i = 4.0 km/h to the north
= +4.0 km/h
m1v1, f + m2v2, f − m2v2, i
v1, i = 
m1
v1, f = 3.1 km/h to the north
= +3.1 km/h
v2, f = 6.9 km/h to the south
= −6.9 km/h
(3.0 × 107 kg)(3.1 km/h) + (2.5 × 107 kg)(−6.9 km/h) − (2.5 × 107 kg)(4.0 km/h)

3.0 × 107 kg
9.3 × 107 kg•km/h − 1.7 × 108 kg•km/h − 1.0 × 108 kg•km/h
v1, i = 
3.0 × 107 kg
−1.8 × 108 kg•km/h
v1, i = 
= −6.0 km/h
3.0 × 107 kg
v1, i = 6.0 km/h to the south
Conservation of kinetic energy (check)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2m1v1, f2 + 2m2v2, f2
1
(3.0
2
× 107 kg)[(−6.0 × 103 m/h)(1 h/3600 s)]2 + 2(2.5 × 107 kg)[(4.0 × 103 m/h)(1 h/3600 s)]2
1
1
1
= 2(3.0 × 107 kg)[(3.1 × 103 m/h)(1 h/3600 s)]2 + 2(2.7 × 107 kg)[(−6.9 × 103 m/h)(1 h/3600 s)]2
4.2 × 107 J + 1.5 × 107 J = 1.1 × 107 J + 4.6 × 107 J
5.7 × 107 J = 5.7 × 107 J
V
Section Five—Problem Bank
V Ch. 6–15
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Givens
Solutions
5. m1 = m2
Momentum conservation
v1, i = 3.0 m/s to the north
In the x-direction:
v1, f = 4.0 m/s to the west
m1 v1, i(cos q1, i) + m2 v2, i (cos q2, i) = m1v1, f (cos q1, f ) + m2 v2, f (cos q2, f )
v2, f = 3.0 m/s to the north
v2, i (cos q2, i) = v1, f (cos q1, f ) + v2, f (cos q2, f ) − v1, i (cos q1, i ) = (4.0 m/s)(cos 180°)
+ (3.0 m/s)(cos 90°) − (3.0 m/s)(cos 90°)
v1, i = 3.0 m/s
q1, i = 90° counterclockwise
from east
v2, i (cos q2, i) = −4.0 m/s + 0 m/s + 0 m/s = −4.0 m/s
v1, f = 4.0 m/s
In the y-direction:
q1, f = 180° counterclockwise from east
v2, f = 3.0 m/s
q2, f = 90° counterclockwise
from east
m1v1, i (sin q1, i) + m2v2, i(sin q2, i) = m1v1, f (sin q1, f ) + m2v2, f (sin q2, f )
v2, i (sin q2, i ) = v1, f (sin q1, f ) + v2, f (sin q2, f ) − v1, i (sin q1, i ) = (4.0 m/s)(sin 180°)
+ (3.0 m/s)(sin 90°) − (3.0 m/s)(sin 90°)
v2, i (sin q2, i) = 0 m/s + 3.0 m/s − 3.0 m/s = 0 m/s
This equation indicates that
sin q2, i = 0, or q2, i = 0° or 180°
v2, i (cos q2, i) = −4.0 m/s
v2, i = 4.0 m/s
q2, i = cos−1(−1.0) = 180°
v2, i = 4.0 m/s to the west
Conservation of kinetic energy (check)
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2m1v1, f2 + 2m2v2, f2
v1, i2 + v2, i2 = v1, f2 + v2, f2
(3.0 m/s)2 + (4.0 m/s)2 = (4.0 m/s)2 + (3.0 m/s)2
9.0 m2/s2 + 16 m2/s2 = 16 m2/s2 + 9.0 m2/s2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
25 m2/s2 = 25 m2/s2
V
V Ch. 6–16
Holt Physics Solution Manual
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Givens
6. m1 = 0.75 kg
Solutions
Momentum conservation
m2 = 0.50 kg
In the x-direction:
m3 = 0.50 kg
m1v1, i (cos q1, i) + m2v2, i (cos q2, i) + m3v3, i (cos q3, i)
= m1v1, f (cos q1, f) + m2v2, f (cos q2, f) + m3v3, f (cos q3, f)
v2, i = 0 m/s
v3, i = 0 m/s
v1, f = 0.80 m/s to the east
(at 0°)
v2, f = 3.4 m/s at 45° north
of east (at 45°)
v3, f = 3.4 m/s at 45° south
of east (at −45°)
m1v1, i (cos q1, i) = m1v1, f (cos q1, f ) + m2v2, f (cos q2, f ) + m3v3, f (cos q3, f )
− m2v2, i(cos q2, i) − m3v3, i(cos q3, i)
m1v1, i (cos q1, i) = (0.75 kg)(0.80 m/s)(cos 0°) + (0.50 kg)(3.4 m/s)(cos 45°)
+ (0.50 kg)(3.4 m/s)[cos (−45°)] − (0.50 kg)(0 m/s) − (0.50 kg)(0 m/s)
m1v1, i (cos q1, i) = 0.60 kg•m/s + 1.2 kg•m/s + 1.2 kg•m/s − 0 kg•m/s − 0 kg•m/s
m1v1, i (cos q1, i) = 3.0 kg•m/s
3.0 kg•m/s
v1, i (cos q1, i ) =  = 4.0 m/s
0.75 kg
In the y-direction:
m1v1, i (sin q1, i ) + m2v2, i (sin q2, i ) + m3v3, i (sin q3, i )
= m1v1, f (sin q1, f ) + m2v2, f (sin q2, f ) + m3v3, f (sin q3, f )
Because v2, i , v3, i , and sin q1, f equal 0, m1v1, i (sin q1, i ) = m2v2, f (sin q2, f )
+ m3v3, f (sin q3, f ) = (0.50 kg)(3.4 m/s)(sin 45°) + (0.50 kg)(3.4 m/s)[sin(−45°)]
= 1.2 kg•m/s − 1.2 kg•m/s = 0 kg•m/s
This result indicates that sin
q1, i = 0, or q1, i = 0° or 180°
v1, i (cos q1, i) = 4.0 m/s
v1, i = 4.0 m/s
q1, i = cos−1(1.0) = 0°
v1, i = 4.0 m/s to the east
Conservation of kinetic energy (check)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
m v 2
2 1 1, i
1
1
1
1
1
+ 2m2v2, i2 + 2m3v3, i2 = 2m1v1, f2 + 2m2v2, f2 + 2m3v3, f2
1
1
1
(0.75 kg)(4.0 m/s)2 + (0.50 kg)(0 m/s)2 + (0.50 kg)(0 m/s)2
2
2
2
1
1
1
= 2(0.75 kg)(0.80 m/s)2 + 2(0.50 kg)(3.4 m/s)2 + 2(0.50 kg)(3.4
m/s)2
6.0 J + 0 J + 0 J = 0.24 J + 2.9 J + 2.9 J
6.0 J = 6.0 J
V
Section Five—Problem Bank
V Ch. 6–17
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Givens
Solutions
7. v2, i = 2.000 m/s upward
Velocity of ball dropped from rest is
v2, f = 1.980 m/s upward
yi = ± (2
m/s
m) = ± 2 0 . 0 m / s
v1, i = ± 2a
∆
)(
−9.
81
2)(−
20
.4
∆yi = −20.4 m
v1, i = −20.0 m/s = 20.0 m/s downward
a = −g = −9.81 m/s2
Momentum conservation
m1 = 0.150 kg
m2 = 325.0 kg
hi = 20.4 m
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m1v1, i + m2v2, i − m2 v2, f
v1, f = 
m1
(0.150 kg)(−20.0 m/s) + (325.0 kg)(2.000 m/s)−(325.0 kg)(1.980 m/s)
v1, f = 
0.150 kg
−3.00 kg•m/s + 650.0 kg•m/s − 643.5 kg•m/s
3.50 kg•m/s
v1, f =  =   = 23.3 m/s
0.150 kg
0.150 kg
v1, f = 23.3 m/s upward
v1, f 2
(23.3 m/s)2
∆yf =  = 
= 26.9 m
2g
(2)(9.81 m/s2)
h = ∆yf − hi = 27 m − 20.4 m = 7 m above the shaft
Conservation of kinetic energy (check)
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2m1v1, f2 + 2m2v2, f2
1
1
(0.150 kg)(−20.0 m/s)2 + (325.0
2
2
1
+ 2(325.0 kg)(1.980 m/s)2
1
kg)(2.000 m/s)2 = 2(0.150 kg)(23 m/s)2
30.0 J + 650.0 J = 4.0 × 101 J + 637.1 J
680.0 J = 677 J
The slight difference arises from rounding.
v2, f = 2.017 m/s downward
= −2.017 m/s
v1, i = 20.0 m/s downward
= −20.0 m/s
2
g = 9.81 m/s
m1 = 0.150 kg
Momentum conservation
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m1v1, i + m2v2, i − m2v2, f
v1, f = 
m1
(0.150 kg)(−20.0 m/s) + (325.0 kg)(−2.000 m/s)−(325.0 kg)(−2.017 m/s)
v1, f = 
0.150 kg
m2 = 325.0 kg
−3.00 kg•m/s − 650.0 kg•m/s + 655.5 kg•m/s
2.5 kg•m/s
v1, f =  =  = 17 m/s
0.150 kg
0.150 kg
hi = 20.4 m
v1, f = 16.7 m/s upward
v1, f 2
(17 m/s)2
∆yf =  = 
= 14.2 m
2g
(2)(9.81 m/s2)
h = ∆yf − hi = 14.2 m − 20.4 m = −6.20 m
h = 6.20 m below the top of the shaft
V
V Ch. 6–18
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. v2, i = 2.000 m/s downward
= −2.000 m/s
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Givens
Solutions
Conservation of kinetic energy (check)
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2mi v1, f2 + 2m2v2, f2
1
1
(0.150 kg)(−20.0 m/s)2 + (325.0
2
2
1
+ 2(325.0 kg)(−2.017 m/s)2
1
kg)(−2.000 m/s)2 = 2(0.150)(17 m/s)2
30.0 J + 650.0 J = 22 J + 661.1 J
680.0 J = 683 J
The slight difference arises from rounding.
9. m1 = 0.500 kg
h = 40.0 cm
g = 9.81 m/s2
m2 = 2.50 kg
v2, i = 0 m/s
The velocity of steel ball at point of collision can be determined through conservation of mechanical energy.
PEi = KEf
1
m1gh = 2m1v1, i2
v1, i = 2g
m/s
m) = 2.80 m/s
h = (2
)(
9.
81
2)(0
.4
00
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m1v1, i + m2v2, i − m1v1, f
v2, f = 
m2
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2m1v1, f2 + 2m2v2, f2
m2 m1v1, i + m2v2, i − m1v1, f
m2v2, i2
vi, f2 = v1, i2 +  −  
m2
m1
m1
2
Because v2, i = 0 m/s, the equation simplifies to the following:
m1v1, i2 2m1v1, i v1, f m1v1, f2
v1, f2 = v1, i2 −  +  - 
m2
m2
m2
v +  − 1 (v
1 + m v − 
m m m1
2
2
1, f
2
m1
1, f
2
2
1, i) = 0
v
−  v +  − 1 (2.80 m/s) = 0
1 + 
2.50 kg 2.50 kg 2.50 kg 0.500 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2m1v1, i
(2)(0.500 kg)(2.80 m/s)
2
1, f
0.500 kg
1, f
2
(1.20)v1, f2 − (1.12 m/s)v1, f − 6.27 m2/s2 = 0
Solving for v1, f by using the quadratic equation,
v1, f
1.12 m/s ± (−
m/s
)2
−(4)
m2/
s2)
1.
12
(1
.2
0)
(−
6.
27

=
(2)(1.20)
v1, f
1.12 m/s ± 1.
m2/
s2
+30.
s2 1.12 m/s ± 31
m2/
s2
25
1m
2/
.3
=  = 
2.40
2.40
1.12 m/s ± 5.59 m/s
v1, f = 
2.40
V
Section Five—Problem Bank
V Ch. 6–19
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Givens
Solutions
The ball’s speed must have a smaller magnitude after collision than before. The positive
root gives a final, forward speed that is close to the ball’s initial speed. Therefore the negative root gives a more realistic result.
−4.47 m/s
v1, f =  = −1.86 m/s
2.40
m1v1, i + m2v2, i − m1v1, f
v2, f = 
m2
(0.500 kg)(2.80 m/s) + (2.50 kg)(0 m/s) − (0.500 kg)(−1.86 m/s)
v2, f = 
2.50 kg
1.40 kg•m/s + 0 kg•m/s + 0.930 kg•m/s
2.33 kg•m/s
v2, f =  =  = 0.932 m/s
2.50 kg
2.50 kg
v1, f = 1.86 m/s backwards
v2, f = 0.932 m/s forward
10. m1 = 7.00 kg
v1, i = 2.00 m/s to the east
(at 0°)
m2 = 7.00 kg
v1, i = 0 m/s
v1, f = 1.73 m/s at 30.0°
north of east
Momentum conservation
In the x-direction:
m1v1, i (cos q1, i) + m2v2, i (cos q2, i) = m1v1, f (cos q1, f ) + m2v2, f (cos q2, f )
v2, f (cos q2, f) = v1, i (cos q1, i) + v2, i (cos q2, i) − vi, f (cos q1, f)
v2, f (cos q2, f) = (2.00 m/s)(cos 0°) + 0 m/s − (1.73 m/s)(cos 30.0°)
v2, f = 2.00 m/s − 1.50 m/s = 0.50 m/s
In the y-direction:
m1v1, i (sin q1, i) + m2v2, i (sin q2, i) = m1v1, f (sin q1, i) m2v2, f (sin q2, f)
v2, f (sin q2, f ) = v1, i (sin q1, i) + v2, i (sin q2, i) − v2, f (sin q2, f)
v2, f (sin q2, f ) − 0.865 m/s

= 
v2, f (cos q2, f )
0.50 m/s
tan q2, f = −1.7
q2, f = tan−1(−1.7) = (−6.0 × 101)°
0.50 m/s
v2, f = 
= 1.0 m/s
cos(−6.0 × 101)°
v2, f = 1.0 m/s at (6.0 × 101)° south of east
Conservation of kinetic energy (check)
1
m v 2
2 1 1, i
1
1
1
+ 2m2v2, i2 = 2miv1, f2 + 2m2v2, f2
1
1
1
1
(7.00 kg)(2.00 m/s)2 + (7.00 kg)(0 m/s)2 = (7.00 kg)(1.73 m/s)2 + (7.00 kg)(1.0
2
2
2
2
m/s)2
14.0 J + 0 J = 10.5 J + 3.5 J
14.0 J = 14.0 J
V
V Ch. 6–20
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v2, f (sin q2, f ) = (2.00 m/s)(sin 0°) + 0 m/s − (1.73 m/s)(sin 30.0°) = −0.865 m/s
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Rotational Motion and the Law of Gravity
Chapter
7
Additional Practice 7A
Givens
1. ∆s = +24.0 m
r = 3.50 m
2. r = 5.55 m
∆s = +31.3 m
3. area = πr2 = 2730 km2
∆s = −545 km
Solutions
∆s 24.0 m
∆θ =  =  = 6.86 rad
r
3.50 m
∆s 31.3 m
∆θ =  =  = 5.64 rad
r
5.55 m
r=
area
 =
π
2730 km2
 = 29.5 km
π
∆s −545 km
∆θ =  =  = −18.5 rad
r
29.5 km
4. ∆s = 4.3 × 1011 m
∆θ = 0.39 rad
5. ∆s = 35.0 km
∆θ = 1.75 rad
6. ∆s = −36.6 µm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
−π
∆θ =  rad
6
7. r = 10.0 m
∆θ = +5.7 rad
8. r = 1.08 × 108 km
π
∆θ = +  rad
3
9. r = 4.48 × 109 km
π
∆θ = +  rad
3
10. r = 28.1 m
∆θ = –7.50 rad
∆s 4.3 × 1011 m
r =  =  = 1.1 × 1012 m
∆θ
0.39 rad
∆s 35.0 km
r =  =  = 20.0 km
∆θ 1.75 rad
∆s
−36.6 µm
r =  =  = 69.9 µm
∆θ
−π
 rad
6
∆s = r∆θ = (10.0 m)(5.7 rad) = 57 m
π
∆s = r∆θ = (1.08 × 108 km)  rad = 1.13 × 108 km
3
π
∆s = r∆θ = (4.48 × 109 km)  rad = 4.69 × 109 km
3
∆s = r∆θ = (28.1 m)(–7.50 rad)= 2.11 × 102 m
V
Section Five—Problem Bank
V Ch. 7–1
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Additional Practice 7B
Givens
Solutions
1. ∆θ = −106 rad
∆t = 7.5 s
2. ∆θ = +2π rad
∆t = 4.56 min
3. v = 280 m/s
∆x = 2.0 m
∆θ = +0.54 rad
∆θ −106 rad
ωavg =  =  = −14.1 rad/s
∆t
7.5 s
∆θ
2π rad
ωavg =  =  = 2.30 × 10−2 rad/s
∆t (4.56 min)(60 s/min)
∆x
∆t = 
v
∆θ ∆θ v∆θ (280 m/s)(0.54 rad)(1 rev/2π rad)
ωavg =  =  =  =  = 12 rev/s
2.0 m
∆t
∆x
∆x

v
4. ωavg = 0.75 rad/s
∆θ = 3.3 rad
5. ωavg = 8.6 × 10−3 rad/s
∆θ = (6)(2π rad)
6. ωavg = 2.75 rad/s
∆θ = (3)(2π rad)
7. ∆t = 4.2 h
ωavg = +2π rad/day
8. ωavg = 1 rev/212 × 106 year
3.3 rad
∆θ
∆t =  =  = 4.4 s
ωavg 0.75 rad/s
∆θ
(6)(2π rad)
∆t =  = 
= 4.4 × 103 s = 1.2 h
ωavg 8.6 × 10−3 rad/s
∆θ
(3)(2π rad)
∆t =  =  = 6.85 s
2.75 rad/s
ωavg
∆θ = ωavg∆t = (2π rad/day)(1 day/24 h)(4.2 h) = 1.1 rad
∆θ = ωavg∆t = (1 rev/212 × 106 year)(4.50 × 109 year)(2π rad/rev) = 133 rad
9. ωavg = −1 rev/243 day
∆tE = 365.25 day
∆tV = 224.7 day
10. ∆x = 6.0 m
v = 5.0 m/s
ωavg = 6.00 rad/s
ωavg = (−1 rev/243 day)(2π rad/rev) = −2.59 × 10−2 rad/day
∆θ = ωavg∆tE = (−2.59 × 10−2 rad/day)(365.25 day) = −9.46 rad
∆θ = ωavg∆tV = (−2.59 × 10−2 rad/day)(224.7 day) = −5.82 rad
∆x
∆t = 
v
∆x (6.00 rad/s)(6.0 m)
∆θ = ωavg∆t = ωavg =  = 7.2 rad
v
5.0 m/s
Additional Practice 7C
1. αavg = 1.5 rad/s2
V
ω1 = 3.0 rad/s
ω2 = ω1 + αavg∆t = 3.0 rad/s + (1.5 rad/s2)(4.0 s) = 3.0 rad/s + 6.0 rad/s
ω2 = 9.0 rad/s
∆t = 4.0 s
V Ch. 7–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 4.50 × 109 year
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Givens
2. ω1 = 9.5 rad/s
αavg = −5.4 × 10−3 rad/s2
Solutions
ω2 = ω1 + αavg∆t = 9.5 rad/s + (−5.4 × 10−3 rad/s2)(22 min)(60 s/min)
ω2 = 9.5 rad/s − 7.1 rad/s = 2.4 rad/s
∆t = 22 min
3. αavg = 32 rad/s2
ω2 = ω1 + αavg∆t = 0 rad/s + (32 rad/s2)(1.5 s) = 48 rad/s
∆t = 1.5 s
ω1 = 0 rad/s
4. ω2 = 76 rad/s
ω1 = 0 rad/s
ω2 − ω1 76 rad/s − 0 rad/s
 = 
∆t = 
= 8.0 s
αavg
9.5 rad/s2
αavg = 9.5 rad/s2
5. αavg = 3.91 rad/s2
ω2 = 7.70 rad/s
ω2 − ω1 7.70 rad/s − 2.50 rad/s
 = 
∆t = 
= 1.33 s
3.91 rad/s2
αavg
ω1 = 2.50 rad/s
6. ω1 = 5.14 × 10−2 rad/s
ω2 = 3.09 × 10−2 rad/s
ω2 − ω1 3.09 × 10−2 rad/s − 5.14 × 10−2 rad/s
−2.05 × 10−2 rad/s
 = 

∆t = 
=
αavg
−1.75 × 10−3 rad/s2
−1.75 × 10−3 rad/s2
αavg = −1.75 × 10−3 rad/s2
∆t = 11.7 s
7. ω1 = 4.0 rad/s
ω2 = 5.0 rad/s
ω2 − ω1 5.0 rad/s − 4.0 rad/s 1.0 rad/s
 =  =  = 0.13 rad/s2
αavg = 
7.5 s
7.5 s
∆t
∆t = 7.5 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. ω1 = 7.14 rad/s
ω2 = 2.38 rad/s
ω2 − ω1 2.38 rad/s − 7.14 rad/s −4.76
 =  =  = −0.529 rad/s2
αavg = 
∆t
9.00 s
9.0 s
∆t = 9.00 s
9. ω1 = 2.07 rad/s
ω2 = 1.30 rad/s
ω2 − ω1 1.30 rad/s − 2.07 rad/s −0.77 rad/s
 =  =  = −0.35 rad/s2
αavg = 
∆t
2.2 s
2.2 s
∆t = 2.2 s
10. ω1 = 2π rad/23.66 h
ω2 = 2π rad/24.00 h
∆t = 70.0 × 106 year
(2 π rad/24.00 h) − (2π rad/23.66 h)
ω2 − ω1
 = 
αavg = 
(70.0 × 106 year)(365.25 day/year)(24 h/day)
∆t
0.2618 rad/h − 0.2656 rad/h
αavg = 
(70.0 × 106 year)(365.25 day/year)(24 h/day)
−3.8 × 10−3 rad/h
αavg = 
= −6.2 × 10−15 rad/h2
(70.0 × 106 year)(365.25 day/year)(24 h/day)
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Section Five—Problem Bank
V Ch. 7–3
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Additional Practice 7D
Givens
Solutions
ωf − ω
3.33 rad/s − 0 rad/s
∆t = i = 
= 18.2 s
α
0.183 rad/s2
1. ωi = 0 rad/s
ωf = 3.33 rad/s
α = 0.183 rad/s2
Because ωi = 0 rad/s
2. ωi = 0 rad/s
α = 0.13 rad/s2
1
∆θ = 2α∆t2
∆θ = 1.6 rad
∆t =
3. ωi = 5.2 rad/s
2∆αθ = (
02.)1(31.r6
adra/sd) = 5.0 s
2
(2)(216 rad)
2∆θ
(2)(216 rad)
∆t =  =  =  = 16.6 s
26.1 rad/s
ωi + ωf 5.2 rad/s + 20.9 rad/s
ωf = 20.9 rad/s
∆θ = 216 rad
4. ωi = 0.111 rad/s
ωf2 − ωi2
(0.178 rad/s)2 − (0.111 rad/s)2
∆θ = 
= 
2a
(2)(1.1 × 10−2 rad/s2)
ωf = 0.178 rad/s
α = 1.1 × 10−2 rad/s2
5. ωi = 78.0 rev/min
∆t = 30.0 s
3.17 × 10−2 rad2/s2 − 1.23 × 10−2 rad2/s2
1.94 × 10−2 rad2/s2
∆θ = 
= 
= 0.87 rad
−2
2
2.2 × 10 rad/s
2.2 × 10−2 rad/s2
1
∆θ = ωi∆t + 2α∆t2
1
α = −0.272 rad/s
2
∆θ = (78.0 rev/min)(2π rad/rev)(1 min/60 s)(30.0 s) + 2(−0.272 rad/s2)(30.0 s)2
∆θ = 245 rad − 122 rad = 123 rad
∆θ = (123 rad)(1 rev/2π rad) = 19.6 rev = 123 rad = 19.6 rev
α = −44.0 rad/s2
∆θ = 276 rad
7. ωi = 0 rad/s
∆t = 13.0 s
ωf2 = ωi2 + 2α∆θ
2
ωf = ω
+2α
)2
+(2)
∆
θ = (2
98
rad
/s
(−
44
.0
rad
/s
2)(2
76
rad
)
i ωf = 8.
04rad
s2
−2.4
04rad
s2 = 6.
04rad
s2 f = 254 rad/s
88
×1
2/
3×1
2/
45
×1
2/
2∆θ
(2)(10.0 rev)(2π rad/rev)
ωf =  − ωi =  − 0 rad/s f = 9.67 rad/s
∆t
13.0 s
∆θ = 10.0 rev
8. ωi = 1200 rev/min
ωf = 3600 rev/min
∆t = 12 s
(3600 rev/min − 1200 rev/min)(2π rad/rev)(1 min/60 s)
ωf − ω
α = i = 
12 s
∆t
(2400 rev/min)(2π rad/rev)(1 min/60 s)
α =  = 21 rad/s2
12 s
V
V Ch. 7–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. ωi = 298 rad/s
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Givens
9. ∆θ = 158 rad
ωi = 0 rad/s
Solutions
ωf2 − ωi2
(70.0 rad/s)2 − (0 rad/s)2
α = 
=  = 15.5 rad/s2
2∆θ
(2)(158 rad)
ωf = 70.0 rad/s
10. ωi = 3.29 rad/s
∆t = 2.50 s
∆θ = 12.3 rad
(2)[(12.3 rad) − (3.29 rad/s)(2.5 s)]
2(∆θ − ωi∆t)
α = 
= 
(2.5 s)2
∆t2
(2)(12.3 rad − 8.2 rad) (2)(4.1 rad)
α = 
= 
(2.5 s)2
(2.5 s)2
α = 1.3 rad/s2
Additional Practice 7E
1. ω = 2.07 × 10−3 rad/s
vt = rω = (1.5 × 103 m)(2.07 × 10−3 rad/s) = 3.1 m/s
r = 1.5 km
2. ω = 188.5 rad/s
r = 3.73 cm
3. r = 15.2 m
ω = 6.28 rad/s
4. r = 0.30 m
vt = 4.5 m/s
5. r = 2.00 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vt = 94.2 m/s
6. vt = 0.63 m/s
r = 1.5 m
7. ω = 3.14 × 10−2 rad/s
vt = 0.45 m/s
8. ω = 10.0 rad/s
vt = 4.60 m/s
9. ω = 11 rad/s
vt = 4.0 cm/s
10. vt = 1.5 m/s
ω = 0.33 rad/s
vt = rω = (3.73 × 10−2 m)(188.5 rad/s) = 7.03 m/s
vt = rω = (15.2 m)(6.28 rad/s) = 95.5 m/s
v
4.5 m/s
ω = t =  = 15 rad/s
r
0.30 m
v
94.2 m/s
ω = t =  = 47.1 rad/s
r
2.00 m
v
0.63 m/s
ω = t =  = 0.42 rad/s
r
1.5 m
0.45 m/s
v
r = t = 
= 14 m
w 3.14 × 10−2 rad/s
v
4.60 m/s
r = t =  = 0.460 m = 46.0 cm
ω 10.0 rad/s
v
4.0 cm/s
r = t =  = 0.36 cm = 3.6 mm
ω
11 rad/s
1.5 m/s
v
r = t =  = 4.5 m
ω 0.33 rad/s
V
Section Five—Problem Bank
V Ch. 7–5
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Additional Practice 7F
Givens
Solutions
1. r = 6.0 cm
at = rα = (6.0 10−2 m)(35.2 rad/s2) = 2.1 m/s2
α = 35.2 rad/s
2
2. α = 105 rad/s2
at = rα = (1.75 cm)(105 rad/s2) = 184 cm/s2 = 1.84 m/s2
r = 1.75 cm
3. r = 5.87 m
α = 1.40 × 10
2
rad/s
4. ∆ω = 1.23 × 10−2 rad/s
∆t = 10.0 s
at = 7.50 × 10−2 m/s2
5. α = 42 rad/s2
at = 64 m/s2
6. α = 6.25 × 10−2 rad/s2
at = 0.75 m/s2
7. at = 0.157 m/s2
r = 0.90 m
8. r = 1.75 m
at = 0.83 m/s2
9. r = 0.50 m
∆v = 5.0 m/s
∆t = 8.5 s
10. r = 16 cm
at = 0.59 m/s2
at = rα = (5.87 m)(1.40 × 10−2 rad/s2) = 8.22 × 10−2 m/s2
∆ω 1.23 × 10−2 rad/s
α =  =  = 1.23 × 10−3 rad/s2
∆t
10.0 s
a
7.50 × 10−2 m/s2
r = t = 
= 61.0 m
α 1.23 × 10−3 rad/s2
64 m/s2
a
r = t = 2 = 1.5 m
42 rad/s
r
0.75 m/s2
a
r = t = 
= 12 m
6.25 × 10−2 rad/s2
r
a 0.157 m/s2
α = t =  = 0.17 rad/s2
r
0.90 m
a
0.83 m/s2
α = t =  = 0.47 rad/s2
r
1.75 m
∆v 5.0 m/s
at =  =  = 0.59 m/s2
∆t
8.5 s
a
0.59 m/s2
α = t =  = 1.2 rad/s2
r
0.50 m
a
0.59 m/s2
α = t = 
= 3.7 rad/s2
r
16 × 10−2 m
Additional Practice 7G
1. r = 3.81 m
vt = 124 m/s
v 2 (124 m/s)2
ac = t =  = 4.04 × 103 m/s2
r
3.81 m
V
V Ch. 7–6
Holt Physics Solution Manual
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−2
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Givens
Solutions
2. r = 6.50 cm
ω = 30.0 rad/s
3. r = 11 m
vt = 1.92 × 10−2 m/s
4. r = 8.9 m
ac = rω2 = (6.50 × 10−2 m)(30.0 rad/s)2 = 58.5 m/s2
v 2 (1.92 × 10−2 m/s)2
ac = t =  = 3.4 × 10−5 m/s2
r
11 m
vt = ra
m)(
m/s
c = (8
.9
20
.0
)(
9.
81
2) = 42 m/s
ac = (20.0) g
g = 9.81 m/s2
5. r = 4.2 m
ac = 2.13 m/s2
6. ac = g = 9.81 m/s2
vt = ra
m)(
m/s
c = (4
.2
2.
13
2) = 3.0 m/s
m)(
m/s
vt = ra
c = (1
50
9.
81
2) = 38 m/s
r = 150 m
7. vt = 75.0 m/s
2
ac = 22.0 m/s
8. ω = 3.5 rad/s
ac = 2.0 m/s2
9. vt = 0.35 m/s
ac = 0.29 m/s2
10. ac = 9.81 m/s2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vt = 15.7 m/s
v 2 (75.0 m/s)2
r = t = 
= 256 m
ac
22.0 m/s2
a
2.0 m/s2
r = c = 2 = 0.16 m = 16 cm
ω2 (3.5 rad/s)
v 2 (0.35 m/s)2
r = t = 
= 0.42 m = 42 cm
ac
0.29 m/s2
v 2 (15.7 m/s)2
r = t = 
= 25.1 m
ac
9.81 m/s2
Additional Practice 7H
1. m = 40.0 kg
ω = 0.50 rad/s
Fc = mrω2 = (40.0 kg)(18.0 m)(0.50 rad/s)2 = 180 N
r = 18.0 m
2. r = 0.25 m
vt = 5.6 m/s
v2
(5.6 m/s)2
Fc = mt = (0.20 kg) = 25 N
r
0.25 m
m = 0.20 kg
V
Section Five—Problem Bank
V Ch. 7–7
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Givens
Solutions
3. vt = 48.0 km/h
r = 35.0 m
[(48.0 km/h)(103 m/km)(1 h/3600 s)]2
v2
Fc = mt = (1250 kg) 
35.0 m
r
mk = 0.500
Fc = 6350 N
m = 1250 kg
2
g = 9.81 m/s
Ff = µkFn = µkmg = (0.500)(1250 kg)(9.81 m/s2)
Ff = 6130 N
The available frictional force is not large enough to maintain the automobile’s
circular motion.
4. m = 1250 kg
F = Ff + mg(sin θ) = µkFn + mg(sinθ) = µkmg(cos θ) + mg(sin θ)
r = 35.0 m
F = (0.500)(1250 kg)(9.81 m/s2)(cos 9.50°) + (1250 kg)(9.81 m/s2)(sin 9.50°)
θ = 9.50°
F = 6.05 × 103 N + 2.02 × 103 N
g = 9.81 m/s2
F = 8.07 × 103 N
µk = 0.500
Fc = F = 8.07 × 103 N
vt =
(8.07 × 103 N)(35.0 m)

1250 kg
Fr
c =
m
vt = 15.0 m/s = 54.0 km/h
5. m = 2.05 × 108 kg
r = 7378 km
Fc = 3.00 × 109 N
6. m = 55 kg
ω = 2.0 rad/s
vt =
Fcr
 =
m
(3.00 × 109 N)(7378 × 103 m)

2.05 × 108 kg
vt = 1.04 × 104 m/s = 10.4 km/s
Fc
135 N
=  = 0.61 m = 61 cm
r = 
mω2 (55 kg)(2.0 rad/s)2
7. m = 7.55 × 1013 kg
vt = 0.173 km/s
mvt2 (7.55 × 1013 kg)(0.173 × 103 m/s)2
r = 
=  = 4.47 × 1015 m
Fc
505 N
Fc = 505 N
8. ω = 36.7 rad/s
r = 0.10 m
Fc
670 N
= 5.0 kg
m = 
2 = 
rω
(0.10 m)(36.7 rad/s)2
Fc = 670 N
9. r = 35.0 cm
vt = 2.21 m/s
−2
F r (0.158 N)(35.0 × 10 m)
m = c2 = 
= 1.13 × 10−2 kg = 11.3 g
2
(2.21 m/s)
vt
Fc = 0.158 N
10. Fc = 8.00 × 102 N
V
r = 0.40 m
2
F r (8.00 × 10 N)(0.40 m)
m = c2 = 
= 8.9 kg
(6.0 m/s)2
vt
vt = 6.0 m/s
V Ch. 7–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fc = 135 N
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Additional Practice 7I
Givens
Solutions
1. m1 = 2.04 × 104 kg
5
m2 = 1.81 × 10 kg
N•m2 (2.04 × 104 kg)(1.81 × 105 kg)
m m2
−11 

Fg = G 1
= 0.11 N
2 = 6.673 × 10
kg2
r
(1.5 m)2
r = 1.5 m
N•m2
G = 6.673 × 10−11 
kg2
2. m1 = 1.4 × 1021 kg
m2 = 5.98 × 1024 kg
N•m2 (1.4 × 1021 kg)(5.98 × 1024 kg)
m m2
−11 

Fg = G 1
=
6.673
×
10
= 3.8 × 1018 N
kg2
(3.84 × 108 m)2
r2
r = 3.84 × 108 m
N•m2
G = 6.67 × 10−11 
kg2
3. m1 = 0.500 kg
m2 = 2.50 × 1012 kg
N•m2 (0.500 kg)(2.50 × 1012 kg)
m m2
−11 

Fg = G 1
= 8.34 × 10−7 N
2 = 6.673 × 10
kg2
(10.0 × 103 m)2
r
r = 10.0 km
N•m2
G = 6.673 × 10−11 
kg2
4. Fg = 2.77 × 10−3 N
r = 2.50 × 10−2 m
m1 = 157 kg
Fgr 2 (2.77 × 10−3 N)(2.50 × 10−2 m)2
m2 =  =  = 165 kg
Gm1
N•m2
6.673 × 10−11 
(157 kg)
kg2
N•m2
G = 6.673 × 10−11 
kg2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. Fg = 1.636 × 1022 N
m1 = 1.90 × 1027 kg
r = 1.071 × 106 km
Fgr 2
(1.636 × 1022 N)(1.071 × 109 m)2
m2 =  =  = 1.48 × 1023 kg
Gm1
N•m2
6.673 × 10−11 
(1.90 × 1027 kg)
kg2
N•m2
G = 6.673 × 10−11 
kg2
6. Fg = 1.17 × 1018 N
m1 = 1.99 × 1030 kg
r = 4.12 × 1011 m
Fgr 2
(1.17 × 1018 N)(4.12 × 1011 m)2
m2 =  =  = 1.50 × 1021 kg
Gm1
N•m2
6.673 × 10−11 
(1.99 × 1030 kg)
kg2
N•m2
G = 6.673 × 10−11 
kg2
7. m1 = m2 = 9.95 × 1041 kg
29
Fg = 1.83 × 10 N
N•m2
G = 6.673 × 10−11 
kg2
r=
G
mFm =
1 2
g
N • m2
6.673 × 10−11
(9.95 × 1041 kg)2
kg2

= 1.90 × 1022 m
1.83 × 1029 N
V
Section Five—Problem Bank
V Ch. 7–9
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Givens
Solutions
8. m1 = 1.00 kg
30
m2 = 1.99 × 10 kg
r=
Fg = 274 N
N•m2
G = 6.673 × 10−11 
kg2
Gm1m2
 =
Fg
9. m1 = 1.00 kg
31
m2 = 3.98 × 10 kg
−3
Fg = 2.19 × 10
r=
N
N•m2
G = 6.673 × 10−11 
kg2
Gm1m2
 =
Fg
10. Fg = 125 N
13
m1 = 4.5 × 10 kg
r=
14
m2 = 1.2 × 10 kg
2
N•m
N • m2
6.673 × 10−11 
(1.00 kg)(3.98 × 1031 kg)
kg2

= 1.10 × 1012 m
2.19 × 10−3 N
N • m2
6.673 × 10−11 
(4.5 × 1013 kg)(1.2 × 1014 kg)
kg2

= 5.4 × 107 m
125 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
G = 6.673 × 10−11 
kg2
Gm1m2
 =
Fg
N • m2
6.673 × 10−11 
(1.00 kg)(1.99 × 1030 kg)
kg2

= 6.96 × 108 m
274 N
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V Ch. 7–10
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Rotational Equilibrium and Dynamics
Chapter
8
Additional Practice 8A
Givens
1. d = 1.60 m
Solutions
t = 4.00 × 102 N • m
t
4.00 × 102 N • m
F =  = 
d(sin q) (1.60 m)(sin 80.0°)
q = 80.0°
F = 254 N
2. tnet = 14.0 N • m
tnet = t − t′
d′ = 0.200 m
t ′ = Fgd′(sin q ′) = t − tnet
q ′ = 80.0°
4.00 × 102 N • m − 14. N • m
t − tnet
Fg = 
= 
(0.200 m)(sin 80.0°)
d ′(sin q ′)
t = 4.00 × 102 N • m
386 N • m
Fg =  = 1.96 × 103 N
(0.200 m)(sin 80.0°)
3. d = 2.44 m
t = 50.0 N • m
t
50.0 N • m
F =  = 
d(sin q)
(2.44 m)(sin 90°)
q = 90°
F = 20.5 N
4. t = 1.4 N • m
d = 0.40 m
t
1.4 N • m
F =  = 
d(sin q)
(0.40 m)(sin 60.0°)
q = 60.0°
F = 4.0 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
tmax is produced when q = 90°, or
tmax = Fd = (4.0 N)(0.40 m) = 1.6 N • m
5. Fmax = 2.27 × 105 N • m
r = 0.660 m
1
d = 2r
6. m = 1.6 kg
Fmaxr

tmax = Fmaxd = 
2
(2.27 × 105 N • m)(0.660 m)
tmax =  = 7.49 × 104 N • m
2
t = F d(sin q) = mg (l − x)(sin q)
l = 43 cm
t = (1.6 kg)(9.81 m/s2)(0.43 m − 0.15 m)(sin 90°) = (1.6 kg)(9.81 m/s2)(0.28 m)
x = 15 cm
t = 4.4 N • m
q = 90°
g = 9.81 m/s2
7. t = 0.46 N • m
F = 0.53 N
0.46 N • m
t
d =  = 
F(sin q) (0.53 N)(sin 90°)
q = 90°
d = 0.87 m
V
Section Five—Problem Bank
V Ch. 8–1
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Givens
Solutions
8. t = 8.25 × 103 N • m
F = 587 N
t
8.25 × 103 N • m
d =  = 
F(sin q) (587 N)(sin 65.0°)
d = 15.5 m
q = 65.0°
9. m = 28 kg
2
g = 9.81 m/s
1.84 × 104 N • m
t
t
d =  =  = 
F(sin q) mg(sin q) (28 kg)(9.81 m/s2)(sin 89°)
d=
q = 89°
67 m
4
t = 1.84 × 10 N • m
10. Fb = 1.200 × 103 N
tnet = tg − tb = Fg d(sin qg) − Fbd(sin qb)
tnet
tnet
d = 
= 
Fg(sin qg) − Fb(sin qb)
mg(sin qg) − Fb(sin qb)
qb = 90.0°
m = 60.0 kg
g = 9.81 m/s2
qg = 87.7°
tnet = −2985 N • m
−2985 N • m
d = 
(60.0 kg)(9.81 m/s2)(sin 87.7°) − (1.200 × 103 N)(sin 90.0°)
−2985 N • m
d = 
588 − 1.200 × 103 N
−2985 N • m
d =  = 4.88 m
−612 N
Additional Practice 8B
1. m1 = 2.3 kg
Apply the second condition of equilibrium.
m2 = 0.40 kg
tnet = t1 − t2 = 0
l = 1.00 m
t1 = Fg,1d1 = m1gx
2
g = 9.81 m/s
t2 = Fg,2d2 = m2g(l − x)
m1gx = m2g(l − x)
ml
(0.40 kg)(1.00 m) (0.40 kg)(1.00 m)
x = 2 =  = 
m1 + m2
2.3 kg + 0.40 kg
2.7 kg
x = 0.15 m from the ostrich egg
2. mms = 139 g
Apply the second condition of equilibrium.
dms = 49.7 cm
tnet = tms − tw = 0
mw = 50.0 g
Let x be the distance from the fulcrum to the zero mark.
dw = 10.0 cm
2
g = 9.81 m/s
tms = mmsg(dms − x)
tw = mwg(x − dw)
mmsg(dms − x) = mwg(x − dw)
mmsdms + mwdw = (mw + mms)x
(139)(49.7 cm) + (50.0 g)(10.0 cm)
mmsdms + mwdw
 = 
x= 
139 g + 50.0 g
mms + mw
6.91 × 103 g • cm + 5.00 × 102 g • cm
7.41 × 103 g • cm
x =  = 
189 g
189 g
V
V Ch. 8–2
x = 39.2 cm from the zero mark
Holt Physics Solution Manual
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(m1 + m2)x = m2l
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Givens
Solutions
3. mA = 4.64 × 107 kg
Apply the first condition of equilibrium to solve for the weight of the cantilever arm, mcg.
dA = 3.00 × 102 m
Fs − mAg − mC g − Fcs = 0
Fcs = 3.22 × 107 N
mC g = Fs − mAg − Fcs = 7.55 × 108 N − (4.64 × 107 kg)(9.81 m/s2) − 3.22 × 107 N
Fs = 7.55 × 108 N
mC g = 7.55 × 108 N − 4.55 × 108 N − 3.22 × 107 N = 2.68 × 108 N
g = 9.81 m/s2
To solve for dC, apply the second condition of equilibrium using the support for the cantilever as the pivot point.
dA
dC
− mC g 
− FcsdC = 0
mAg 
2
2
2 + F d = 2
mC g
cs
C
mAgdA
(4.64 × 107 kg)(9.81 m/s2)(3.00 × 102 m)
mAgdA
 + 2Fcs = 
dC = 
2.68 × 108 N + (2)(3.22 × 107 N)
mC g
(4.64 × 107 kg)(9.81 m/s2)(3.00 × 102 m) (4.64 × 107 kg)(9.81 m/s2)(3.00 × 102 m)
dC = 
= 
2.68 × 108 N + 6.44 × 107 N
3.32 × 108 N
dC = 411 m
4. mf = 70.0 kg
Apply the first condition of equilibrium to find the weight of the ladder, ml g.
Fu − mf g − ml g = 0
2
g = 9.81 m/s
ml g = Fu − mf g
q = 10.0°
t = 7.08 × 10 N • m
ml g = 3.14 × 103 N − (70.0 kg)(9.81 m/s2) 3.14 × 103 N − 687 N = 2.45 × 103 N
Fu = 3.14 × 103 N
To solve for the length of the ladder (d), apply the second condition of an equilibrium, using the base of the ladder as the pivot point.
d
t − ml g  (sin q) − mf gd(sin q) = 0
2
ml g
 + mf g (sin q)d = t
2
2t
(2)(7.08 × 103 N • m)
d =  = 
3
(ml g + 2 mf g)(sin q)
[2.45 × 10 N + (2)(70.0 kg)(9.81 m/s2)](sin 10.0°)
3
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(2)(7.08 × 103 N • m)
(2)(7.08 × 103 N • m)
d = 
= 
= 21.3 m
3
3
(2.45 × 10 N + 1.37 × 10 N)(sin 10.0°)
(3.82 × 103 N)(sin 10.0°)
5. dc = 32.0 m
Apply the first condition of equilibrium in the x direction.
Fx = Rx,base − FI,x = 0
q = 60.0°
4
Rx,base = R(cos q) = FT,x
4
F ,x
R = T
cos q
FT,x = 1.233 × 10 N
FT,y = 1.233 × 10 N
g = 9.81 m/s2
Apply the first condition of equilibrium in the y direction; solve for the flagpole’s weight.
Fy = Ry,base − mg − FT,y = 0
Ry,base = R(sin q) − FT,y = mg
F ,x
T
(sin q) − FT,y = mg
cos q
mg = FT,x(tan q) − FT,y = (1.233 × 104 N)(tan 60.0°) − 1.233 × 104 N
mg = (1.233 × 104 N)[(tan 60.0°) − 1.00] = (1.233 × 104 N)(1.73 − 1.00)
mg = (1.233 × 104 N)(0.73) = 9.0 × 103 N
Section Five—Problem Bank
V
V Ch. 8–3
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To solve for the length of the flagpole (l ), apply the second condition of equilibrium,
using the base of the flagpole as the pivot point.
FT,y dc(sin q) − FT,ydc(cos q) − mg
6. mE = 5.98 × 1024 kg
2 (cos q) = 0
l
l
FT,xdc(sin q) − FT,ydc(cos q)
= 
1
mg(cos q)
2
l
(1.233 × 104 N)(32.0 m)(sin 60.0°) − (1.233 × 104 N)(32.0 m)(cos 60.0°)
1
= 
(9.0 × 103 N)(cos 60.0°)
2
l
3.42 × 105 N • m − 1.97 × 105 N • m
1.45 × 105 N • m

= 
=
1
1
(9.0 × 103 N)(cos 60.0°)
(9.0 × 103 N)(cos 60.0°)
2
2
l
= 64 m
Apply the first condition of equilibrium.
g = 9.81 m/s
Ffulcrum − Fapplied − mEg = c
dE = 1.00 m
Ffulcrum = Fapplied + mEg
dapplied = 3.8 × 1016 m
To solve for Fapplied, apply the second condition of equilibrium, using the fulcrum as
the pivot point.
2
Fapplieddapplied − mEgdE = 0
(5.98 × 1024 kg)(9.81 m/s2)(1.00 m)
m gdE
Fapplied = E 
= 
3.8 × 1016 m
dapplied
Fapplied = 1.5 × 109 N
Substitute the value for Fapplied into the first-condition equation and solve for
Ffulcrum.
Ffulcrum = 1.5 × 109 N + (5.98 × 1024 kg)(9.81 m/s2) = 1.5 × 109 N + 5.87 × 1025 N
7. d = 2.00 m
Apply the first condition of equilibrium in the x and y directions.
Fx = Frod − FT (cos q) = 0
q = 30.0°
3
t = 1.47 × 10 N • m
Frod = FT (cos q)
Fy = FT (sin θ) − Fg = 0
Fg = FT (sin q)
Apply the second condition of equilibrium, using the end of the rod anchored in the
wall as the pivot point.
t − Fgd = 0
t
1.47 × 103 N • m
Fg =  = 
d
2.00 m
Fg = 735 N
Substitute the value for Fy into the second first-condition equation to solve for FT.
Substitute the value of FT into the first first-condition equation to solve for Frod.
V
Fg
735 N
FT = 
=  = 1470 N
sin q sin 30.0°
Frod = FT (cos q) = (1470 N)(cos 30.0°)
Frod = 1270 N
V Ch. 8–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Ffulcrum = 5.87 × 1025 N
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8. Fg = 7.10 × 102 N
ms = 0.75
Apply the first condition of equilibrium to determine the largest force that will not
overcome static friction.
Fapplied − Fs = 0
dx = 1.22 m
Fapplied = Fs = msFn = msFg
dy = 1.00 m
Fapplied = (0.75)(7.10 × 102 N) = 532 N
Apply the second condition of equilibrium to determine the largest force that will not
lift the door from the rail. Choose one of the wheels for the axis of rotation.
d
Fapplieddy − Fg x = 0
2
(7.10 × 102 N)(1.22 m)
Fd
Fapplied = gx = 
(2)(1.00 m)
2dy
Fapplied = 433 N
The largest force that will not upset either equilibrium condition is the smaller of the
two forces.
Fapplied = 433 N
9. m = 307 kg
Apply the first condition of equilibrium in the x direction.
h = 2.44 m
Fx = Fapplied(sin q) − R(cos q) = 0
q = 70.0°
R = Fapplied(tan q)
d = 1.22 m
2
g = 9.81 m/s
To solve for Fapplied, either use the first condition of equilibrium in the y direction or
the second condition of equilibrium.
Fy = Fapplied(cos q) + R(sin q) − mg = 0
Fapplied(cos q) + [Fapplied(tan q)](sin q) = mg
(307 kg)(9.81 m/s2)
mg
Fapplied =  = 
(cos q) + (tan q)(sin q)
(cos 70.0°) + (tan 70.0°)(sin 70.0°0)
(307 kg)(9.81 m/s2)
(307 kg)(9.81 m/s2)
Fapplied =  = 
0.342 + 2.58
2.92
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fapplied = 1030 N
Alternatively,
h
Fappliedd − mg  (cos q) = 0
2
(307 kg)(9.81 m/s2)(2.44 m)(cos 70.0°)
mgh(cos q)
Fapplied =  = 
(2)(1.22 m)
2d
Fapplied = 1030 N
Substitute the value of Fapplied into the first-condition equation in the x direction to
solve for R.
R = Fapplied(tan q) = (1030 N)(tan 70.0°) = 2830 N
V
Section Five—Problem Bank
V Ch. 8–5
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Givens
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10. Fg = 1.96 × 103 N
Apply the first condition of equilibrium to the x and y directions.
d1 = 0.250 m
Fx = F − Rx = 0
d2 = 1.50 m
Rx = F
Fy = Ry − Fg = 0
Ry = Fg = 1.96 × 103 N
To solve for F, apply the second condition of equilibrium, using the hinge as the pivot
point.
Fgd1 − Fd2 = 0
(1.96 × 103 N)(0.250 m)
Fd
F = g1 = 
1.50 m
d2
F = 327 N
Substitute the value for F in the first-condition equation for the x direction, then
solve for R using the Pythagorean theorem.
Rx = 327 N
2
2
R= R
+
R
N)2
+(1.
N)2
27
96
×103
x y = (3
R = 1.
05
N2+
06
N2 = 3.
06
N2
07
×1
3.8
4×1
95
×1
R = 1.99 × 103 N
Additional Practice 8C
1. t = 2.98 N • m
t
2.98 N • m
t
I =  =  = 
a
wf − wi
55 rad/s − 0 rad/s


∆t
0.75 s
wi = 0 rad/s
wf = 55 rad/s
2
I = 0.041 kg • m
Dt = 0.75 s
For a hoop spinning on an axis along its diameter,
R = 20.0 cm
1
2I
(2)(0.041 kg • m2)
M = 2 = 
R
(0.200 m)2
M = 2.0 kg
2. t = 1.7 N • m
1.7 N • m
t
I =  = 2 = 0.3 kg • m2
a 5.5 rad/s
2
a = 5.5 rad/s
3. t = 0.750 N • m
2
a = 499 rad/s
4. M = 7.91 × 103 kg
R = 1.83 m
t 0.750 N • m
I =  = 
= 1.50 × 10−3 kg • m3
a 499 rad/s2
2
2
I = 5MR2 = 5(7.91 × 103 kg)(1.83 m)2 = 1.06 × 104 kg • m2
t = Ia = (1.06 × 104 kg • m2)(6.13 rad/s2) = 6.50 × 104 N • m
2
a = 6.13 rad/s
V
V Ch. 8–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
I = 2MR2
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Givens
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5. a = −6.53 × 10−27 rad/s2
I = 0.331 MR2 = (0.331)(5.98 × 1024 kg)(6.37 × 106 m)2 = 8.03 × 1037 kg • m2
t = Ia = (8.03 × 1037 kg • m2)(−6.53 × 10−22 rad/s2) = −5.24 × 1016 N • m
M = 5.98 × 1024 kg
R = 6.37 × 106 m
6. a = 1.05 rad/s2
4
t = Ia = (8.14 × 104 kg • m2)(1.05 rad/s2) = 8.55 × 104 N • m
2
I = 8.14 × 10 kg • m
7. t = 108 N • m
108 N • m
t
a =  = 2 = 20.0 rad/s2
5.40 kg • m
I
2
I = 5.40 kg • m
8. t = 1.01 N • m
−5
I = 3.85 × 10
2
kg • m
9. m = 0.15 kg
r = 0.35 m
t = 1.5 N • m
Dt = 0.26 s
10. M = 15 kg
R = 0.25 m
wi = 9.5 rad/s
wf = 0 rad/s
t = −0.80 N • m
t
1.01 N • m
a =  = 
= 2.62 × 104 rad/s2
I
3.85 × 10−5 kg • m2
t
t
a =  = 2
mr
I
1.5 N • m
a = 2 = 82 rad/s2
(0.15 kg)(0.35 m)
w = a∆t = (82 rad/s2)(0.26 s) = 21 rad/s
1
1
I = 2MR2 = 2(15 kg)(0.25 m)2 = 0.47 kg • m2
t −0.80 N • m
a =  = 2 =
I 0.47 kg • m
−1.7 rad/s2
−9.5 rad/s
wf − wi 0 rad/s − 9.5 rad/s
 = 
= 2
∆t = 
2
a
−1.7 rad/s
−1.7 rad/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆t = 5.6 s
Additional Practice 8D
1. r = 0.120 m
Li = Lf
m = 22.0 g
Iiwi = If wf
wi = 50.00 rad/s
(I + 25 mr 2)wi = Iwf
wf = 50.24 rad/s
I(wf − wi) = 25 mr 2 wf
(25)(22.0 × 10−3 kg)(0.120 m)2 (50.24 rad/s)
25 mr 2 w
I = f = 
50.24 rad/s − 50.00 rad/s
wf − wi
I = 1.7 kg • m2
2. M = 755 kg
li = 1.75 m × 2
= 3.50 m
wi = 1.25 rad/s
wf = 1.70 × 10−2 rad/s
Li = Lf
Iiwi = If wf
1
M 2w
i i
12
l
If =
= If wf
Mli 2wi

12wf
(755 kg)(3.50 m)2 (1.25 rad/s)
= 
(12)(1.70 × 10−2 rad/s)
4
V
2
If = 5.67 × 10 kg • m
Section Five—Problem Bank
V Ch. 8–7
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Givens
Solutions
Li = Lf
3. ri = 3.00 m
rf = 0.20 m
Iiwi = If wf
m = 55.0 kg
(I + 4 mri2)wi = (I + 4 mrf 2)wf
wi = 2.00 rad/s
I(wf − wi) = 4 m(ri2wi − rf 2wf)
wf = 2.35 rad/s
2
2
4 m(ri 2wi − rf 2wf) (4)(55.0 kg)[(3.00 m) (2.00 rad/s) − (0.20 m) (2.35 rad/s)]
 = 
I= 
2.35 rad/s − 2.00 rad/s
wf − wi
(4)(55.0 kg)(18.0 m2/s − 0.094 m2/s)
(4)(55.0 kg)(17.9 m2/s)
I =  = 
0.35 rad/s
0.35 rad/s
I = 1.1 × 104 kg • m2
Li = Lf
4. wi = 2.1 rad/s
ri = 1.2 m
Iiwi = If wf
rf = 0.50 m
mri 2wi = mrf 2wf
ri 2wi (1.2 m)2(2.1 rad/s)
 = 
wf = 
(0.50 m)2
rf 2
wf = 12 rad/s
L1 = L2
5. v1 = 43.5 km/s
r1 = 7.00 × 107 km
8
r2 = 1.49 × 10 km
I1w1 = I2w2
mr12w1 = mr22w2
r1v1 = r2v2
r1v1
(7.00 × 107 km)(43.5 km/s)
v2 =  = 
r2
1.49 × 108 km
v2 = 20.4 km/s
Mw = 3.81 kg
Iwwi = Itotwf
Rw = 0.350 m
MwRw2wi = Itotwf
Itot = 2.09 kg • m2
MwRw2wi
(3.81 kg)(0.350 m)2 (57.7 rad/s)
 = 
wf = 
Itot
2.09 kg • m2
wf = 12.9 rad/s
Li = Lf
7. wi = 1.50 rad/s
−2
wf = 2.04 × 10
M = 7.55 kg
li = 3.50 m
rad/s
Iiwi = If wf
1
  M 2w
i
i
12
l
lf =
lf =
1
= 12Mlf 2wf
l i2wi

wf
30.0 m
V
V Ch. 8–8
Holt Physics Solution Manual
=
(3.50 m)2 (1.50 rad/s)

2.04 × 10−2 rad/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Li = Lf
6. wi = 57.7 rad/s
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L1 = L2
8. v1 = 3.68 km/s
r1 = 7.35 × 10 km
I1w1 = I2w2
v2 = 6.14 km/s
mr12w1 = mr22w2
9
r1vi = r2v2
(7.35 × 109 km)(3.68 km/s)
rv
r2 = 1 1 = 
6.14 km/s
v2
r2 = 4.41 × 109 km
Li = Lf
9. v1 = 3403 m/s
I1w1 = I2w2
r1 = 3593 km
h1 = 2.00 × 10 km
mr12w1 = mr22w2
v2 = 3603 m/s
r1v1 = r2v2
2
(3593 km)(3403 m/s)
rv
r2 = 1 1 = 
3603 m/s
v2
r2 = 3394 km
r = r2 − h2 = r1 − h1
h2 = r2 − r1 + h1 = 3394 km − 3593 km + 2.00 × 102 km
h2 = 1 km
10. m1 = m2 = 55.0 kg
Li = Lf
5.00 m
ri =  = 2.50 m
2
Iiwi = If wf
vi = 5.00 m/s
ri 2wi = rf 2wf
vf = 15.0 m/s
(m1ri 2 + m2ri 2)wi = (mirf 2 + m2rf 2)wf
ri vi = rf vf
Copyright © by Holt, Rinehart and Winston. All rights reserved.
rv
(2.50 m)(5.00 m/s)
rf = i i = 
vf
15.0 m/s
rf = 0.833 m
distance between skaters = 2rf 2 = 1.67 m
Additional Practice 8E
1. k = 1.05 × 104 N/m
−2
x = 4.0 × 10
KErot = 2.8 J
m
MEi = MEf
PEelastic = KEtrans + KErot
1 2
kx
2
= KEtrans + KErot
KEtrans = 2kx 2 − KErot = 2(1.05 × 104 N/m)(4.0 × 10−2 m)2 − 2.8 J
1
1
KEtrans = 8.4 J − 2.8 J − 5.6 J
V
Section Five—Problem Bank
V Ch. 8–9
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Givens
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MEi = MEf
2. vi = 2.2 m/s
KEi = KEtrans − KErot
m = 55 g
2
1
mv 2
i
2
v
1
1
1
12
= 2mvf 2 + 2Iwf 2 = 2mvf2 + 23mr2 f
r
1
mv 2
i
2
= 2mvf2+3 = 2mvf23 = 3KEtrans
1
2
1
5
5
3
KEtrans = 10mvi2
3
1
2
1
KErot = KEi − KEtrans = 2mvi2 − 10mvi2 = 10mvi2 = 5mvi2
KErot = 5(55 × 10−3 kg)(2.2 m/s)2 = 5.3 × 10−2 J
1
3. h = 46.0 m
MEi = Ef
m = 25.0 kg
PEg = KEtrans + KErot
2
g = 9.81 m/s
2
v
1
1
1
1
1
mgh = 2mvf 2 + 2Iwf 2 = 2mvf 2 + 2(mr 2) f = 2mvf 2(1 + 1) = mvf 2
r
KEtrans =
1
mgh
2
=
1
(25.0
2
2
kg)(9.81 m/s )(46.0 m)
3
KEtrans = 5.64 × 10 J
4. k = 1.05 × 104 N/m
MEi = MEf
x = 4.0 cm
PEelastic = KEtrans + KErot
wf = 43.5 rad/s
1 2
kx
2
2
v
1
1
1
1
1
1
= 2mvf 2 + 2Iwf 2 = 2mr 2 f + 2Iwf 2 = 2Iwf 2 + 2Iwf 2 = Iwf 2
r
(1.05 × 10 N/m)(4.0 × 10−2 m)2
kx
I = 2 = 
(20943.5 rad/s)2
2wf
4
2
I = 4.4 × 10−3 kg • m2
1
1
wf = 27 rad/s
KEi = KEtrans + KErot = 2mvf2 + 2Iwf2
r = 0.11 m
v
1
1
15
1
1
5
7
KEi = 2mr2 f + 2Iwf2 = 22Iwf2 + 2Iwf2 = 2Iwf2(2 + 1) = 4Iwf2
r
2
4KE
(4)(45 J)
I = 2i = 2 = 3.5 × 10−2 kg • m2
7wf
(7)(27 rad/s)
m=
5
I
2
(5)(3.5 × 10−2 kg • m2)

=
=
(2)(0.11 m)2
r2
V
V Ch. 8–10
Holt Physics Solution Manual
7.2 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
MEi = MEf
5. KEi = 45 J
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Givens
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MEi = MEf
6. h = 0.60 m
2
g = 9.81 m/s
KEtrans + KErot = PEg
1
mv 2
i
2
1
+ 2Iwi2 = mgh
2
10gh
(10)(9.81 m/s +)(0.60 m)
 =

v = 7
7
1
mv 2
i
2
v
7
12
1
2
+ 25mr 2 i = 2mvi 1 + 5 = 10mvi2 = mgh
r
2
i
vi = 2.9 m/s
7. k = 150 N/m
MEi = Ef
PEelastic = KEtrans + KErot
x = 6.0 cm
2
(5)(150 N/m)(6.0 × 10 m)
5kx

v = 
7m = (7)(67 × 10 kg)
m = 67 g
1
kx 2
2
v
7
1
1
1
12
1 1
= 2mvf 2 + 2Iwf 2 = 2mvf 2 + 25mr 2 f = mvf22 + 5 = 10mvf2
r
−2
2
2
−3
f
vf = 2.4 m/s
MEi = MEf
8. vi = 2.4 m/s
KEtrans + KErot = PEg
q = 3.5°
2
g = 9.81 m/s
1
mv 2
i
2
+ 2Iwf2 = mgh
1
1
mv 2
i
2
v
12
+ 25mr2 i = mgd(sin q)
r
2
1
mvi2 2
+
=
1

5
7
mv 2
i
10
= mgd(sin q)
7vi2
(7)(2.4 m/s)2
d =  = 
10g(sin q) (10)(9.81 m/s2)(sin 3.5°)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 6.7 m
The ball has more than enough kinetic energy
to reach the back of the pinball machine.
9. vi = 4.6 m/s
g = 9.81 m/s2
MEi = MEf
KEtrans + KErot = PEg
1
mv 2
i
2
+ 2Iwi2 = mgh
1
1
mv 2
i
2
v
7
12
1 1
+ 25mr2 i = mvi22 + 5 = 10mvi2 = mgh
r
2
7vi2
(7)(4.6 m/s)2
h =   = 
10g
(10)(9.81 m/s2)
h = 1.5 m
V
Section Five—Problem Bank
V Ch. 8–11
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Givens
Solutions
10. r = 15 cm
MEi = MEf
wf = 102 rad/s
2
g = 9.81 m/s
PEg = KEtrans + KErot
1
1
mgh = 2mvf2 + 2Iwf 2
2
v
1
1
13
1
1
3
5
mgh = 2(mr 2) f + 2Iwf 2 = 22Iwf 2 + 2Iwf 2 = 2Iwf22 + 1 = 4Iwf2
r
3
I
5
2
mgh = 
gh = 4Iwf 2
r2
(5)(0.15 m)2(102 rad/s)2
5r2wf2
h = 
= 
(6)(9.81 m/s2)
6g
Copyright © by Holt, Rinehart and Winston. All rights reserved.
h = 2.0 × 101 m
V
V Ch. 8–12
Holt Physics Solution Manual
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Fluid Mechanics
Chapter
9
Additional Practice 9A
Givens
Solutions
1. rgasoline = 675 kg/m3
Vs = 1.00 m3
rgasoline msg
 =  rgasoline = Vsg rgasoline
FB = Fg 
rs
rs
FB = (1.00 m3)(9.81 m/s2)(675 kg/m3) = 6.62 × 103 N
g = 9.81 m/s2
2. rr = 2.053 × 104kg/m3
3
Vr = (10.0 cm)
FB = Fg − apparent weight
FB = mrg − apparent weight = rrVrg − apparent weight
g = 9.81 m/s
FB = (2.053 × 104 kg/m3)(10.0 cm)3(10−2 m/cm)3(9.81 m/s2) − 192 N = 201 N − 192 N
apparent weight = 192 N
FB = 9 N
2
3. mh = 1.47 × 106 kg
FB = Fg = mhg
Ah = 2.50 × 103 m2
FB = (1.47 × 106 kg)(9.81 m/s2) = 1.44 × 107 N
rsw = 1.025 × 103 kg/m3
m
msw
= h
volume of hull submerged = Vsw = 
rsw
rsw
2
g = 9.81 m/s
Vw
mh
h = s
= 
Ah Ahrsw
1.47 × 106 kg
h = 
= 0.574 m
3 2
(2.50 × 10 m )(1.025 × 103 kg/m3)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. msh = 1.47 × 106 kg
rsteel = 7.86 × 103 kg/m3
rgold = 1.93 × 104 kg/m3
Ah = 2.50 × 103 m2
rsw = 1.025 × 103 kg/m3
g = 9.81 m/s2
5. V = 166 cm3
msh
FB = Fg = mghg = rgoldVhg = rgold 
g
rsteel
(1.93 × 104 kg/m3)(1.47 × 106 kg)(9.81 m/s2)
= 3.54 × 107 N
FB = 
7.86 × 103 kg/m3
Vw
mgh
FB
= 
= 
h = s
Ah
Ahrsw Ahrswg
3.54 × 107 N
h = 
= 1.41 m
3 2
(2.50 × 10 m )(1.025 × 103 kg/m3)(9.81 m/s2)
Fg = FB + apparent weight
apparent weight = 35.0 N
rosmiumVg = rwVg + apparent weight
rw = 1.00 × 103 kg/m3
apparent weight
rosmium = rw + 
Vg
g = 9.81 m/s2
35.0 N
rosmium = 1.00 × 103 kg/m3+ 
3
−6 3
(166 cm )(10 m /cm3)(9.81 m/s2)
rosmium = 1.00 × 103 kg/m3+ 2.15 × 104 kg/m3
4
V
3
rosmium = 2.25 × 10 kg/m
Section Five—Problem Bank
V Ch. 9–1
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Givens
Solutions
6. V = 2.5 × 10−3 m3
Fg = FB + apparent weight
apparent weight = 7.4 N
rebonyVg = rwVg + apparent weight
rw = 1.0 × 103 kg/m3
apparent weight
rebony = rw + 
Vg
2
g = 9.81 m/s
7.4 N
rebony = 1.0 × 103 kg/m3+ 
(2.5 × 10−3 m3)(9.81 m/s2)
= 1.0 × 103 kg/m3+ 3.0 × 102 kg/m3
rebony = 1.3 × 103 kg/m3
msg = 45.0 g
msg
45.0 × 10−3 kg
rsg = 
= 
Vsg
(7.62 cm3)(10−6 m3/cm3)
Vlg = 7.38 × 10−6 m3
r of solid gallium = 5.91 × 103 kg/m3
g = 9.81 m/s2
Fg = FB
7. Vsg = 7.52 cm3
msgg = rlgVlgg
msg 45.0 × 10−3 kg
rlg = 
= 
Vlg 7.38 × 10−6 m3
r of liquid gallium = 6.10 × 103 kg/m3
8. rplatinum = 21.5 g/cm3
rw = 1.00 g/cm3
apparent weight = 40.2 N
g = 9.81 m/s2
Fg = FB + apparent weight
m
mg = rwVg + apparent weight = rw  g + apparent weight
rplatinum
r
 = apparent weight
mg 1 − w
rplat inum
apparent weight
40.2 N
m =  = 
r
1.00 g/cm3

g 1 − w
(9.81 m/s2) 1 − 3
21.5 g/cm
rplat inum
40.2 N
40.2 N
m = 
= 
(9.81 m/s2)(1 − 0.047)
(9.81 m/s2)(0.953)
m = 4.30 kg
9. rlithium = 534 kg/m3
rgasoline = 675 kg/m3
Vgasoline = 5.93 × 10−4 m3
Fg = FB
mlithiumg = rgasolineVgasolineg
mlithium = rgasolineVgasoline = (675 kg/m3)(5.93 × 10−4 m3)
mlithium = 0.400 kg
mlithium
0.400 kg
 = 3 = 7.49 × 10−4 m3
Vlithium = 
534 kg/m
rlithium
V
V Ch. 9–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Givens
Solutions
10. h1 = 5.00 cm
Before the mass is submerged,
h2 = 4.30 cm
Fg = FB
A = 3.4 m2
(mp + mb + mm)g = rwV1g = rwAh1g
3
3
rw = 1.0 × 10 kg/m
After the mass is submerged,
(mp + mb)g = rwV2g = rwAh2g
Substituting the second equation into the first,
rwAh2g + mmg = rwAh1g
mm = rwA(h1 − h2)
mm = (1.0 × 103 kg/m3)(3.4 m2)(5.00 cm − 4.3 cm)(10−2 m/cm)
mm = (1.0 × 103kg/m3)(3.4 m2)(0.70 × 10−2 m)
mm =
24 kg
Additional Practice 9B
1. P = 1.50 × 106 Pa
F = 1.22 × 104 N
2. P = 1.01 × 105 Pa
F = 2.86 × 108 N
F
1.22 × 104 N
= 8.13 × 10−3 m2
A =  = 
P 1.50 × 106 Pa
F
2.8 6 × 108 N
= 2.83 × 103 m2
A =  =  
P 1.01 × 105 Pa
A = 4pr 2
r=
3. F = 5.0 N
3
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P = 9.6 × 10 Pa
4. m = 1.40 × 103 kg
h = 0.076 m
rice = 917 kg/m3
0 m
4Ap = 2.
83×41p
=
3
2
15.0 m
F
5.0 N
= 5.2 × 10−4 m2
A =  = 
P 9.6 × 103 Pa
P1 = P2
F
F
1 = 2
A1 A2
mg miceg micehg
 =  =  = ricehg
A1
A2
Vice
1.40 × 103 kg
m
A1 =  = 
= 2.0 × 101 m2
3
riceh (917 kg/m )(0.076 m)
V
Section Five—Problem Bank
V Ch. 9–3
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Givens
Solutions
F = PnetA = (P1 − P2)A = (1 + 1.00 × 10−2 − 1)P2A = (1.00 × 10−2)P2A
5. A = 1.54 m2
P1 = (1 + 1.00 × 10−2)P2
P2 = 1.013 × 105 Pa
g = 9.81 m/s2
6. r = 6.0 km
F = (1.00 × 10−2)(1.013 × 105 Pa)(1.54 m2) = 1.56 × 103 N
F 1.56 × 103 N
m =   = 
= 159 kg
g
9.81 m/s2
F = PA = P(4pr2) = (1.2 × 1016 Pa)(4p)(6.0 × 103 m)2
16
P = 1.2 × 10 Pa
F = 5.4 × 1024 N
P1 = P2
7. F1 = 4.45 × 104 N
F
F
1 = 2
A1 A2
h1 = 448 m
h2 = 8.00 m
Fh
Fh
Fh
Fh
11 = 11 = 22 = 22
A1h1
V
A2h2
V
(4.45 × 104 N)(448 m)
Fh
F2 = 11 = 
8.00 m
h2
8. h = 760 mm
r = 13.6 × 10 kg/m
mg mgh mgh
F
P =  =  =  =  = rgh
A
Ah
V
A
g = 9.81 m/s2
P = (13.6 × 103 kg/m3)(9.81 m/s2)(760 × 10−3 m) = 1.0 × 105 Pa
3
3
9. m = 2.4 × 1013 kg
A = 3.14 km2
13
2
F
mg (2.4 × 10 kg)(9.81 m/s )
P =  =  = 
(3.14 km2)(106 m2/km2)
A
A
g = 9.81 m/s2
P = 7.5 × 107 Pa
10. F = 4.4 × 103 N
A = 2.9 × 10−2 m2
Po = 1.0 × 105 Pa
F
4.4 × 103 N
P =  = 
= 1.5 × 105 Pa
A 2.9 × 10−2 m2
Pgauge = P − Po = 1.5 × 105 Pa − 1.0 × 105 Pa = 5 × 104 Pa
Additional Practice 9C
1. P = 6.9 × 104 Pa
r = 0.55 kg/m3
g = 9.81 m/s2
Po = 1.01 × 105 Pa
2. P − Po = 1.47 × 106 Pa
r = 1.00 × 103 kg/m3
g = 9.81 m/s2
V
V Ch. 9–4
P = Po + pgh
P−P
6.9 × 104 Pa − 1.01 × 105 Pa
−3.2 × 104 Pa

h = o = 
=
rg
(0.55 kg/m3)(9.81 m/s2)
(0.55 kg/m3)(9.81 m/s2)
h = −5.9 × 103 m = 5.9 km above sea level
P = Po + rgh
1.47 × 106 Pa
P−P
h = o = 
(1.00 × 103 kg/m3)(9.81 m/s2)
rg
h = 1.50 × 102 m
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
F2 = 2.49 × 106 N
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Givens
Solutions
P = Po + rgh
3. Po = 9.0 × 104 Pa
r = 1.0 × 103 kg/m3
P−P
5.00 × 106 Pa − 9.0 × 108 Pa
4.91 × 106 Pa
h = o = 
3
3
2 = 
rg
(1.0 × 10 kg/m )(9.81 m/s )
(1.0 × 103 kg/m3)(9.81 m/s2)
g = 9.81 m/s2
h = 5.0 × 102 m
P = 5.00 × 106 Pa
P = Po − rgh
4. P = 4.03 × 105 Pa
r = 1.025 × 103 kg/m3
P−P
4.03 × 105 Pa − 1.01 × 105 Pa
3.02 × 105 Pa
h = o = 
3
3
2 = 
rg
(1.025 × 10 kg/m )(9.81 m/s )
(1.025 × 103 kg/m3 (9.81 m/s2)
g = 9.81 m/s2
h = 30.0 m
Po = 1.01 × 105 Pa
5. h = 9.1 m
Pgauge = P − Po = rgh
3
3
r = 1.0 × 10 kg/m
Pgauge = (1.0 × 103 kg/m3)(9.81 m/s2)(9.1 m) = 8.9 × 104 Pa
2
g = 9.81 m/s
6. h = 86 m
r = 1.29 kg/m3
P = Po + rgh = 1.01 × 105 Pa + (1.29 kg/m3)(9.81 m/s2)(86 m) = 1.01 × 105 Pa + 1.1 × 103 Pa
P = 1.02 × 105 Pa
Po = 1.01 × 105 Pa
g = 9.81 m/s2
7. r = 13.6 × 103 kg/m3
Po = 1.01 × 105 Pa
h = 1.50 m
P = Po + rgh = 1.01 × 105 Pa + (13.6 × 103 kg/m3)(9.81 m/s2)(1.50 m)
= 1.01 × 105 Pa + 2.00 × 105 Pa
P = 3.01 × 105 Pa
g = 9.81 m/s2
8. Po = 9.10 × 106 Pa
P = 8.60 × 106 Pa
8.60 × 106 Pa − 9.10 × 106 Pa
P−P
−5.0 × 105 Pa

r = o = 
=
(8.87 m/s2)(−1.00 × 103 m)
gVh
(8.87 m/s2)(−1.00 × 103 m)
gv = 8.87 m/s2
r = 56 kg/m3
h = −1.00 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P = Po + rgVh
9. Po = 1.01 × 105 Pa
P = Po + rgh
h = 3.99 m
2.23 × 105 Pa − 1.01 × 105 Pa
1.22 × 105 Pa
P−P

r = o = 
=
(9.81 m/s2)(3.99 m)
(9.81 m/s2)(3.99 m)
gh
g = 9.81 m/s2
r = 3.12 × 103 kg/m3
P = 2.23 × 105 Pa
10. Po = 1.01 × 105 Pa
P = Po + rgh
h = 3.99 m
P−P
1.29 × 105 Pa −1.01 × 105 Pa
2.8 × 104 Pa
r = o = 
= 
2
gh
(9.81 m/s )(3.99 m)
(9.81 m/s2)(3.99 m)
g = 9.81 m/s2
r = 7.2 × 102 kg/m3
P = 1.29 × 105 Pa
V
Section Five—Problem Bank
V Ch. 9–5
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Additional Practice 9D
Givens
Solutions
1. ∆P = P1 − P2 = 1.5 × 103 Pa
v2 = 17.0 m/s
3
1
1
P1 + 2rv12 + rgh1 = P2 + 2rv22 + rgh2
Assume v1 ≈ 0 m/s, so the above equation simplifies to
3
r = 1.00 × 10 kg/m
1
∆P = rg(h2 − h1) + 2rv22
1.5 × 103 Pa
(17.0 m/s)2
∆P v 2
h2 − h1 =  − 2 = 
3
3
2 − 
rg
2g (1.00 × 10 kg/m )(9.81 m/s ) (2)(9.81 m/s2)
h = h2 − h1 = 0.15 m − 14.7 m = −14.6 m
h = h2 − h1 = −14.6 m = 14.6 m below the surface
2. ∆x = 120 m
∆y = − 69 m
g = 9.81 m/s2
To calculate v2, use the equations for a horizontally-launched projectile.
∆x
v2 = vx = 
∆t
1
∆y = vy,i ∆t − 2g∆t2
vy,i = 0 m/s, so
1
∆y = − 2g∆t2
∆t =
−2g∆y
∆x
v2 = 
−2∆y

g
P1 +
1
rv 2
2 1
1
+ rgh1 = P2 + 2rv22 + rgh2
Because P1 = P2, and assuming v1 ≈ 0 m/s, Bernoulli’s equation simplifies to
2
∆x2
∆x2
h1 − h2 =  = 
−4∆y
−2∆t
2g 
g
(120 m)2
h = h1 − h2 =  = 52 m
(−4)(−69.0 m)
3. v1 = 2.00 m/s
v2 = 7.93 m/s
g = 9.81 m/s2
1
1
P1 + 2rv12 + rgh1 = P2 + 2rv22 +rgh2
P1 = P2, so
1
rg(h1 − h2) = 2r(v22 + v12)
v22 − v12
(7.93 m/s)2 − (2.00 m/s)2
 = 
h1 − h2 = 
2g
(2)(9.81 m/s2)
62.9 m2/s2 − 4.00 m2/s2
58.9 m2/s2
h1 − h2 = 
= 
2
(2)(9.81 m/s )
(2)(9.81 m/s2)
V
V Ch. 9–6
h = h1 − h2 = 3.00 m
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆x
1
1
rg(hi − h2) = 2rv22 = 2r 
−2∆y

g
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Givens
Solutions
4. v1 = 24.45 m/s
v2 = 0.55 m/s
g = 9.81 m/s2
1
1
P1 + 2rv12 + rgh1 = P2 + 2rv22 + pgh2
P1 = P2, so
1
rg(h2 − h1) = 2r(v12 − v22)
(24.45 m/s)2 − (0.55 m/s)2
v12 − v22
 = 
h2 − h1 = 
(2)(9.81 m/s2)
2g
597.8 m2/s2 − 0.30 m2/s2
597.5 m2/s2
h2 − h1 = 
= 
2
(2)(9.81 m/s )
(2)(9.81 m/s2)
h = h2 − h1 =
5. v1 = 2.50 m/s
h1 − h2 = 3.00 m
g = 9.81 m/s2
30.5 m
1
1
P1 + 2rv12 + rgh1 = r2 + 2rv22 + rgh2
P1 = P2, so
1
rv 2
2 1
1
+ rg(h1 − h2) = 2rv22
v2 = v12+
)2
+(2)
2g(
h1−h
.5
0m
/s
(9
.8
1m
/s
2)(3
.0
0m
)
2) = (2
v2 = 6.
m2/
s2
+58.
s2 = 65
m2/
s2
25
9m
2/
.2
v2 = 8.07 m/s
6. v1 = 0.90 m/s
∆P = P2 − P1 = 311 Pa
P = 1.025 × 103 kg/m3
1
1
P1 = 2rv12 + rgh1 = P2 + 2rv22 + rgh2
Because h1 = h2, so the above equation simplifies to
1
rv 2
2 1
v2 =
1
− ∆P = 2rv22
(2)(311 Pa)

v−2∆PP = (0.90m/s)−
1
.0
25×1
0 kg
/m
2
1
2
3
3
v2 = 0.
m2/s2
−0.6
m2s2 = 0.
m2/s2
81
07
20
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v2 = 0.45 m/s
7. rair = 1.3 kg/m3
For the static mercury columns, v1 = v2.
1
1
rmercury = 1.36 × 104 kg/m3
P1 + 2rmercuryv12 + rmercurygh1 = P2 + 2rmercuryv22 + rmercurygh2
h2 − h1 = 3.5 cm
P1 − P2 = rmercuryg(h2 − h1)
2
g = 9.81 m/s
For the flowing air, v1 = 0 m/s, because the tube is fixed to the wing, and h1 = h2.
1
1
P1 + 2rairv12 + rairgh1 = P2 + 2rairv22 + rairgh2
1
P1 − P2 = 2rairv22
Substituting the first equation into the second,
1
rmercury g (h2 − h1) = 2rairv22
v2 =
(2)(1.36 × 104 kg/m3)(9.81 m/s2)(3.5 × 10−2 m)

1.3 kg/m3
2rmercuryg(h2 − h1)
 =
rair
v2 = 85 m/s
V
Section Five—Problem Bank
V Ch. 9–7
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Givens
Solutions
1
8. r = 950 kg/m3
1
P1 + 2rv12 + rgh1 = P2 + 2rv22 + rgh2
v1 = 10.00 m/s
h1 = h2, so
v2 = 9.90 m/s
P2 − P1 = 2r(v12 − v22) = 2(950 kg/m3)[(10.00 m/s)2 − (9.90 m/s)2]
1
1
1
1
P2 − P1 = 2(950 kg/m3)(100.0 m2/s2 − 98.0 m2/s2) = 2(950 kg/m3)(2.0 m2/s2)
∆P = P2 − P1 = 950 Pa
9. A1 = 7.8 × 10−3 m2
From the continuity equation, an expression for v2 can be derived.
v1 = 2.0 m/s
A1v1 = A2v2
A2 = 3.1 × 10−4 m2
r = 1.00 × 103 kg/m3
A1
v2 = v
1
A2
h2 − h1 = 10.5 m
P1 + 2rv12 + rgh1 = P2 + 2rv22 + rgh2
1
1
A2
1
1
P1 − P2 = 2r(v22 − v12) + rg(h2 − h1) = 2rv12 12 − 1 + rg(h2 − h1)
A2
(7.8 × 10−3 m2)2
1
3
3
2
P1 − P2 = 2(1.00 × 10 kg/m )(2.0 m/s) 
−1
(3.1 × 10−4 m2)2
+ (1.00 × 103 kg/m3)(9.81 m/s2)(10.5 m)
g = 9.81 m/s2
1
P1 − P2 = 2(1.00 × 103 kg/m3)(2.0 m/s)2(6.3 × 102 − 1) + 1.03 × 105 Pa
1
P1 − P2 = 2(1.00 × 103 kg/m3)(2.0 m/s)2 (6.3 × 102) + 1.03 × 105 Pa
P1 − P2 = 1.3 × 106 Pa + 1.03 × 105 Pa
∆P = P1 − P2 = 1.4 × 106 Pa
10. A1 = 9.2 × 10−2 m2
From the continuity equation, an expression for v2 can be derived.
A1v1 = A2v2
v1 = 8.3 m/s
A2 = 4.6 × 10−2 m2
3
3
r = 1.0 × 10 kg/m
A1
v2 = v
1
A2
1
1
h1 = h2, so
(9.2 × 10 m )
P − P = (1.0 × 10 kg/m )(8.3 m/s)  − 1
(4.6 × 10 m ) A2
1
1
P1 − P2 = 2r(v22 − v12) = 2rv12 12 − 1
A2
1
2
1

2
3
3
1
−2
2 2
−2
2 2
1
P1 − P2 = 2(1.0 × 103 kg/m3)(8.3 m/s)2(4 − 1) = 2(1.0 × 103 kg/m3)(8.3 m/s)2(3)
∆P = P1 − P2 = 1.0 × 105 Pa
Additional Practice 9E
1. T1 = 184
At constant pressure:
T2 = 331 K
3
V1 = 3.70 m
V
V Ch. 9–8
V1
V
 = 2
T1 T2
VT
(3.70 m3)(331 K)
V2 = 12 =  =
T1
184 K
Holt Physics Solution Manual
6.66 m3
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P1 + 2rv12 + rgh1 = P2 + 2rv22 + rgh2
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Givens
Solutions
2. V1 = 0.455 m3
At constant temperature:
P1 = 9.1 × 106 Pa
P1V1 = P2V2
P2 = 5.0 × 104 Pa
PV
(9.1 × 106 Pa)(0.455 m3)
V2 = 1 1 = 
=
P2
5.0 × 104 Pa
83 m3
At constant temperature:
3. V1 = 65.4 m3
P1 = 1.03 × 105 Pa
P1V1 = P2V2
P2 = 5.84 × 104 Pa
PV
(1.03 × 105 Pa)(65.4 m3)
V2 = 1 1 = 
= 115 m3
P2
5.84 × 104 Pa
PV = NkBT
4. T1 = 295 K
T2 = 225 K
(5.55 × 1022)(1.38 × 10−23 J/K)(295 K)
Nk T
V1 = B1 = 
1.01 × 105 Pa
P1
P2 = 5.10 × 104 Pa
V1 = 2.24 × 10−3 m3
N = 5.55 × 1022 particles
(5.55 × 1022)(1.38 × 10−23 J/K)(225 K)
Nk T
V2 = B2 = 
5.10 × 104 Pa
P2
5
P1 = 1.01 × 10 Pa
kB = 1.38 × 10−23 J/K
V2 = 3.38 × 10−3 m3
5. P1 = 7.5 × 104 Pa
T1 = 250 K
6
P2 = 2.0 × 10 Pa
At constant volume:
P
P
1 = 2
T1 T2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(2.0 × 106 Pa)(250 K)
PT
T2 = 21 = 
= 6.7 × 103 K
7.5 × 104 Pa
P1
V
Section Five—Problem Bank
V Ch. 9–9
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Givens
Solutions
At constant pressure:
6. V1 = 1.00 m3
T1 = 295 K
V
V1
 = 2
T1 T2
V2 = 65.4 m3
VT
(65.4 m3)(295 K)
T2 = 21 = 
= 1.93 × 104 K
V1
1.00 m3
7. N = 2.1 × 1057 particles
PV = NkBI
PV
(2.1 × 1016 Pa)(2.1 × 1025 m3)
= 1.5 × 107 K
T =  = 
NkB
(2.1 × 1057)(1.38 × 10−23 J/K)
16
P = 2.1 × 10 Pa
V = 2.1 × 1025 m3
KB = 1.38 × 10−23 J/K
8. T1 = 295 K
At constant volume:
5
P1 = 2.50 × 10 Pa
P
P
1 = 2
T1 T2
T2 = 506 K
(2.50 × 105 Pa)(506 K)
PT
P2 = 12 =  = 4.29 × 105 Pa
295 K
T1
N = rV
9. T = 1.0 × 102 K
43
PV = NkBT = rVkBT
2
V = 3.3 × 10 m
3
r = 10.0 atoms/cm
−23
kB = 1.38 × 10
J/K
10. P1 = 1.42 × 105 Pa
P = rkBT = (10.0 atoms/cm3)(106 cm3/m3)(1.38 × 10−23 J/K)(1.0 × 102 K)
P = 1.38 × 10−14 Pa
At constant temperature:
P1V1 = P2V2
V2 = 1.83 m3
(1.42 × 105 Pa)(1.00 m3)
PV
P2 = 1 1 = 
=
1.83 m3
V2
7.76 × 104 Pa
Copyright © by Holt, Rinehart and Winston. All rights reserved.
V1 = 1.00 m3
V
V Ch. 9–10
Holt Physics Solution Manual
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Heat
Chapter 10
Additional Practice 10A
Givens
1. T1 = 463 K
T2 = 93 K
Solutions
TC,1 = (T − 273)°C = (463 − 273)°C = 1.90 × 102 °C
9
9
TF,1 =  TC,1 + 32 =  (1.90 × 102)°F + 32°F = 342°F + 32°F = 374°F
5
5
TC,2 = (T − 273)°C = (93 − 273)°C = −180 × 102 °C
9
9
TF,2 =  TC,2 + 32 =  (−1.80 × 102)°F + 32°F = −324°F + 32°F = −292°F
5
5
2. T = 330.0 K
TC = (T − 273)°C = (330.2 − 273.2)°C = 56.8°C
9
9
TF =  TC + 32 =  (56.8)°F + 32°F = 102°F + 32°F = 134°F
5
5
3. Ti = 237 K
Tf = 283 K
∆TC = (Tf − 273)°C − (Ti − 273)°C = Tf − Ti
∆TC = (283 − 237)°C = 46° C
9
9
9
∆TF = TC, f + 32 °F − TC, i + 32 °F =  ∆TC°F
5
5
5
9
∆TF = (46)°F = 83°F
5
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. TF,i = − 5 °F
TF,f = + 37°F
5
5
5
TC,i = (TF, i − 32)°C = (−5 − 32)°C = (− 37)°C = −21°C
9
9
9
5
5
5
TC,f = (TF, f − 32)°C = (37 − 32)°C = (5)°C = 3°C
9
9
9
∆T = (TC,f + 273 K) − (TC,i + 273 K) = TC,f − TC,i
∆T = [3 − (−21)] K = 24 K
5. TF = 78°F
5
T = TC + 273 = (TF − 32) + 273
9
5
5
T = (78 − 32) K + 273 K = (46) K + 273 K = 26 K + 273 K
9
9
T = 299 K
6. TC,1 = 47°C
TC, 2 = 42°C
T1 = (TC,1 + 273)K = (47 + 273) K = 3.20 × 102 K
T2 = (TC,2 + 273)K = (42 + 273) K = 315 K
V
Section Five—Problem Bank
V Ch. 10–1
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Givens
Solutions
5
5
T = (− 67 − 32) + 273 K = (− 97) + 273 K = (− 55 + 273)K
9
9
7. TF = −67°F
5
TC = (TF − 32) + 273 K
9
T = 218 K
8. TF = 2192°F
5
5
5
TC = (TF − 32)°C = (2192 − 32)°C = (2.160 × 103)°C
9
9
9
TC = 1.200 × 103°C
9. TF = 188.6°F
5
5
5
TC = (TF − 32.0)°C = (188.6 − 32.0)°C = (156.6)°C
9
9
9
TC = 87.00°C
10. T = 2.70 K
TC = (T − 273)°C = (2.70 − 273)°C = −270°C
Additional Practice 10B
∆PE + ∆KE + ∆U = 0
1. m = 5.25 g
vi = 3.27 m/s
∆Upenny =
1
∆U
2
k = 2.03 J/1.00° C
The change in potential energy from before to after impact is zero, as is KEf.
∆PE + ∆KE + ∆U = 0 + KEf − KEi + ∆V = − KEi + ∆U = 0
1
∆U = KEi = 2mvi2
mvi2
1
1 1
∆Upenny = 2 ∆U = 2 (2 mvi2) = 
4
∆U enny mvi2
∆T = p
=
k
4k
∆PE + ∆KE + ∆U = 0
2. h = 9.5 m
2
g = 9.81 m/s
∆Uacorn = (0.85)∆U
1200 J/kg
k/m = 
1.0°C
The change in kinetic energy from before the acorn is dropped to after it has landed
is zero, as is PEf.
∆PE + ∆KE + ∆U = PEf − PEi + 0 + ∆U = −PEi + ∆U = 0
∆U = PEi = mgh
∆Uacorn = (0.85) ∆U = (0.85)mgh
∆U
(0.85)mgh (0.85)gh
 =  = 
∆T = acorn
k
k
(k/m)
(0.85)(9.81 m/s2)(9.5 m)
∆T =  = 6.6 × 10−2 °C
1200 J/kg

1.0°C
V
V Ch. 10–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(5.25 × 10−3 kg)(3.27 m/s)2
∆T =  = 6.91 × 10−3 °C
(4)(2.03 J/1.00°C)
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Givens
Solutions
∆PE + ∆KE + ∆U = 0
3. PEi = 6.2 J
∆Ucone = (0.100)∆U
k = 180 J/1.0°C
The change in kinetic energy from before the ice cream cone is dropped to after it has
landed is zero, as is PEf.
∆PE + ∆KE + ∆U = PEf − PEi + 0 + ∆U = −PEi + ∆U = 0
∆U = PEi
∆Ucone = (0.100)∆U = (0.100)PEi
∆U
(0.100)PE
 =  i
∆T = cone
k
k
(0.100)(6.2 J)
∆T =  = 3.4 × 10−3 °C
180 J/1.0°C
∆PE + ∆KE + ∆U = 0
4. vi = 47.5 m/s
h = 151 m
∆U = −∆PE − ∆KE = PEi − PEf + KEi − KEf
m = 7.32 g
Both PEf and KEf equal zero.
∆Utwig = (0.100)∆U
∆U = PEi + KEi = mgh + 2mvi2
k = 8.5 J/1.0°C
∆Utwig = (0.100)(mgh + 2mvi2)
1
1
g = 9.81 m/s2
1
(0.100)(mgh + 2mvi2)
∆U
 =
∆T = twig
k
k
1
(0.100)[(7.32 × 10–3 kg)(9.81 m/s2)(151 m) + 2(7.32 × 10–3 kg)(47.5 m/s)2]
∆T = 
8.5 J/1.0°C
(0.100)(10.8 J + 8.26 J) (0.100)(19.1 J)
∆T =  =  = 0.22°C
8.5 J/°C
8.5 J/°C
∆PE = ∆KE + ∆U = 0
5. h = 561.7 m
∆U = 105 J
When the stone lands, its kinetic energy is transferred to the internal energy of the
stone and the ground. Therefore, overall, ∆KE = 0 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2
g = 9.81 m/s
∆PE = PEf − PEe = 0 − mgh = − ∆U
105 J
∆U
m =  = 
= 1.91 × 10−2 kg = 19.1 g
(9.81 m/s2)(561.7 m)
gh
6. h = 5.5 m
∆PE + ∆KE + ∆U = 0
2
g = 9.81 m/s
3
∆U = 2.77 × 10 J
When the prey lands, all of the kinetic energy is transferred to the ground. Therefore
∆KE = 0 J
∆PE = PEf − PEi = 0 − mgh = −∆U
2.77 × 103 J
∆U
m =  = 
=
gh (9.81 m/s2)(5.5 m)
7. vf = 0 m/s
51 kg
∆PE + ∆KE + ∆U = 0
vi = 13.4 m/s
The bicyclist remains on the bicycle, which does not change elevation, so ∆PE = 0 J.
∆U = 5836 J
∆KE = KEf − KEi = 0 − 2mvi2 = −∆U
1
2∆U (2)(5836 J)
m = 
= 2 = 65.0 kg
(13.4 m/s)
vi2
Section Five—Problem Bank
V
V Ch. 10–3
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Givens
Solutions
∆PE + ∆KE + ∆U = 0
8. ∆KE = KEf − KEi
4
= −2.15 × 10 J
There is no change in the height of the sticks, so ∆PE = 0 J.
∆U = −∆KE = −(−2.15 × 104 J) = 2.15 × 104 J
9. vi = 20.5 m/s
∆PE + ∆E + ∆U = 0
vf = 0 m/s
The height of the skater does not change, so ∆PE = 0 J.
m = 61.4 kg
∆KE = KEf − KEi = 0 − 2mvi2
1
1
1
∆U = −∆KF = −(− 2mvi2) = 2(61.4 kg)(20.5 m/s)2 = 1.29 × 104 J
10. ∆KE = KEf − KEi
= −7320 J
∆Uhands = (1 − 0.300)∆U
∆PE + ∆KE + ∆U = 0
The hands don’t change height, so ∆PE = 0 J.
∆U = −∆KE = −(−7320 J) = 7320 J
∆Uhands = (1− 0.300)(7320 J) = (0.700)(7320 J) = 5120 J
Additional Practice 10C
1. mw = 15 g = 0.015 kg
cp,hpmhp∆Thp = cp,wmw∆Tw
∆Tw = 1.0°C
cp,wm w ∆Tw (4186 J)(kg • °C)(0.015 kg)(1.0°C)
 = 
cp,hp = 
mhp ∆T hp
(0.015 kg)(1.6°C)
∆Thp = 1.6°C
cp,hp = 2.6 × 103 J/kg • °C
mhp = 15 g = 0.015 kg
cp,w = 4186 J/kg • °C
cp,vmv ∆Tv = cp,wmw ∆Tw
mw = 1.00 kg
∆Tw = Tw,i − Tf = 90.0°C − 73.7°C = 16.3°C
Tv,i = 21.0°C
∆Tv = Tf − Tv,i = 73.7°C − 21.0°C = 52.7°C
Tw,i = 90.0°C
(4186 J/kg • °C)(1.00 kg)(16.3°C)
cp,wm w ∆Tw
 = 
cp,v = 
(0.340 kg)(52.7°C)
mv ∆T v
Tf = 73.7°C
cp,w = 4186 J/kg • °C
3. ma = 0.250 kg
mw = 1.00 kg
∆Tw = 1.00° C
∆Ta = 17.5° C
cp,v = 3.81 × 103 J/kg •°C
cp,ama∆Ta = cp,wmw ∆Tw
(4186 J/kg • °C)(1.00 kg)(1.00°C)
cp,wmw ∆Tw
cpa = 
= 
(0.250 kg)(17.5° C)
ma∆Ta
cp,a = 957 J/kg • °C
cp,w = 4186 J/kg • °C
cp,imi ∆Ti = cp,wmw ∆Tw
4. mi = 3.0 kg
mw = 5.0 kg
∆Tw = 2.25° C
∆Ti = 29.6° C
V
(4186 J/kg • °C)(5.0 kg)(2.25°C)
cp,wmw ∆Tw
cp,i = 
= 
(3.0 kg)(29.6°C)
mi∆Ti
cp,i = 530 J/kg • °C
cp,w = 4186 J/kg • °C
V Ch. 10–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. mv = 0.340 kg
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Givens
Solutions
5. Q = −1.09 × 1010 J
cp,w = 4186 J/kg • °C
∆Tw = −5.0°C
6. cp,b = 121 J/kg • °C
Q = 25 J
∆Tb = 5.0°C
7. Q = 45 × 106 J
∆Ta = 55° C
cp,a = 1.0 × 103 J/kg • °C
8. mc = 0.190 kg
4
Q = 6.62 × 10 J
cp,c = 387 J/kg • °C
9. mt = 0.225 kg
3
cp,t = 2.2 × 10 J/kg • °C
Q = −3.9 × 104 J
10. cp,t = 140 J/kg • °C
mt = 0.23 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Q = −3.0 × 104 J
Q
cp,w = 
mw ∆Tw
Q
−1.09 × 1010 J
mw =   =  = 5.2 × 105 kg
cp,w ∆Tw
(4186 J/kg • °C)(−5.0°C)
Q
cp,b = 
mb ∆Tb
25 J
Q
mb =  =  = 4.1 × 10−2 kg
(121 J/kg • °C)(5.0°C)
cp,b∆Tb
Q
cp,a = 
ma ∆Ta
Q
45 × 106 J
ma =  = 
a = 818 kg
cp,a∆Ta
(1.0 × 103 J/kg • °C)(55°C)
Q
cp,c = 
mc∆Tc
6.62 × 104 J
Q
∆Tc =  =  =
mc cp,c (0.190 kg)(387 J/kg • °C)
9.00 × 102°C
Q
cp,t = 
mt∆Tt
−3.9 × 104 J
Q
∆Tt =  = 
= −79°C
mtcp,t (0.225 kg)(2.2 × 103 J/kg • °C)
Q
cp,t = 
mt∆Tt
−3.0 × 104 J
Q
∆Tt =  =  = −930°C
mtcp,t (0.23 kg)(140 J/kg • °C)
Additional Practice 10D
1. Q = 1.10 × 106 J
Q = mLf
Q 1.10 × 106 J
Lf =  =  = 2.06 × 105 J/kg
m
5.33 kg
m = 5.33 kg
2. Q = 9.6 × 105 J
m = 5.33 kg
3. Q = 3.72 × 105 J
m = 0.65 kg
4. Q = 8.5 × 104 J
6
Lv = 2.26 × 10 J/kg
Q 9.6 × 105 J
Lf =  =  = 1.8 × 105 J/kg
m
5.33 kg
Q 3.72 × 105 J
Lsubl =  =  = 5.7 × 105 J/kg
m
0.65 kg
Q = mLv
Q
8.5 × 104 J
m =  = 
= 3.8 × 10−2 kg
Lv 2.26 × 106 J/kg
Section Five—Problem Bank
V
V Ch. 10–5
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Givens
Solutions
5. Q = 2.11 × 106 J
Lv = 8.45 × 105 J/kg
Q
2.11 × 106 J
= 2.50 kg
m =  = 
Lv 8.45 × 105 J/kg
Qtot = mcp,i ∆T + mLf
6. m = 250 kg
Qtot = 1.380 × 10 J
mcp,i ∆T = Qtot − mLf
cp,i = 605 J/kg • °C
Q ot
L
∆T = t
− f
mcp,i cp,i
8
Lf = 2.47 × 105 J/kg
1.380 × 108 J
2.47 × 105 J/kg
∆T =  − 
(250 kg)(605 J/kg • °C)
605 J/kg • °C
∆T = (910 − 408)°C = 5.0 × 102°C
Qtot = mcp,i ∆T + mLf
7. m = 1.45 kg
Tf = 330°C
Qtot = 4.46 × 104 J
2.45 × 104 J/kg
4.46 × 104 J
Q ot
L
∆T = t
− f =  − 
120 J/kg • °C
mcp,i cp,i (1.45 kg)(120 J/kg • °C)
cp,i = 120 J/kg • °C
∆T = (2.6 × 102 − 2.0 × 102)°C = 60°C
Lf = 2.45 × 104 J/kg
∆T = Tf − Ti = 330°C − Ti = 60°C
Ti = 330°C − 60°C = 270°C
8. m = 0.75 g
Qtot = mcp,w ∆T + mLv
Ti = 100.0°C
Qtot = 2.0 × 10 J
2.26 × 106 J/kg
Q ot
Lv
2.0 × 103 J
∆T = t
− 
= 
− 
−3
4200 J/kg • °C
mcp,w cp,w (0.75 × 10 kg)(4200 J/kg • °C)
cp,w = 4200 J/kg • °C
∆T = (6.3 × 102 − 5.4 × 102)°C = 90°C
Lv = 2.26 × 106 J/kg
∆T = Ti − Tf = 100.0°C − Tf = 90°C
3
9. m = 0.35 kg
Q = mLf = (0.35 kg)(8.02 × 104 J/kg)
4
Lf = 8.02 × 10 J/kg
10. m = 55.0 g
Q = 2.8 × 104 J
Qtot = mcp,m ∆T + mLv = mcp,m (Tf − Ti) + mLv
Qtot = (55.0 × 10−3 kg)(130 J/kg • °C)(357°C − 20.0°C)
Ti = 20.0°C
+ (55.0 × 10−3 kg)(2.95 × 105 J/kg)
Tf = 357°C
cp,m = 130 J/kg • °C
Lv = 2.95 × 105 J/kg
Qtot = (55.0 × 10−3 kg)(130 J/kg • °C)(337°C)
+ (55.0 × 10−3 kg)(2.95 × 105 J/kg) = 2.4 × 103 J + 1.62 × 104 J
Qtot = 1.86 × 104 J
V
V Ch. 10–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Tf = 100.0°C − 90°C = 10°C
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Thermodynamics
Chapter 11
Additional Practice 11A
Givens
Solutions
W
3.29 × 106 J
P =  = 
= 1.50 × 103 Pa = 1.50 kPa
∆V
2190 m3
1. W = 3.29 × 106 J
∆V = 2190 m3
2. W = 1.06 × 106 J
∆V = Vf − Vi = 8.50 × 105 m3 − 2.80 × 103 m3 = 8.47 × 105 m3
Vi = 2.80 × 103 m3
W
1.06 × 106 J
P =  = 
= 1.25 Pa
∆V 8.47 × 105 m3
Vf = 8.50 × 105 m3
3. W = 1.3 J
−4
∆V = 5.4 × 10
W
1.3 J
P =  = 
= 2.4 × 103 Pa = 2.4 kPa
∆V 5.4 × 10−4 m3
3
m
4. W = 472.5 J
4
W
472.5 J
∆V =  = 
= 1.89 × 10−2 m3
P
2.50 × 104 Pa
5
W
393 J
∆V =  = 
= 6.00 × 10−4 m3
P
6.55 × 105 Pa
P = 25.0 kPa = 2.50 × 10 Pa
5. W = 393 J
P = 655 kPa = 6.55 × 10 Pa
6. W = 0.20 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P = 39 Pa
7. W = 3.2 × 102 J
P = 4.0 × 105 Pa
8. P = 2.07 × 107 Pa
W 0.20 J
∆V =  =  = 5.1 × 10−3 m3
P
39 Pa
W
3.2 × 102 J
∆V =  = 
= 8.0 × 10−4 m3
P
4.0 × 105 Pa
W = P ∆V = (2.07 × 107 Pa)(0.227 m3) = 4.70 × 106 J
∆V = 0.227 m3
9. Vi = 3.375 × 10−6 m3
Vf = 5.694 × 10−6 m3
∆V = Vf − Vi = 5.694 × 10−6 m3 − 3.375 × 10−6 m3 = 2.319 × 10−6 m3
W = P ∆V = (1.0133 × 105 Pa)(2.319 × 10−6 m3) = 0.2350 J
P = 101.33 kPa
= 1.0133 × 105 Pa
10. Vi = 2.0 × 10−3 m3
Vf = 5.0 × 10−3 m3
∆V = Vf − Vi = 5.0 × 10−3 m3 − 2.0 × 10−3 m3 = 3.0 × 10−3 m3
W = P ∆V = (900 Pa)(3.0 × 10−3 m3) = 2.7 J
P = 0.90 kPa = 900 Pa
V
Section Five—Problem Bank
V Ch. 11–1
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Additional Practice 11B
Givens
Solutions
1. mi = 2.0 kg
W = Fd = m2gd = (15 kg)(9.81 m/s2)(2.3 × 10−3 m) = 0.34 J
mf = 15 kg
Ti = 28°C
Tf = 52°C
d = 2.3 mm
∆U = Uf − Ui = Q − W
2. Ui = 39 J
Uf = 163 J
W = Q − ∆U = Q − (Uf − Ui) = Q − Uf + Ui
Q = 114 J
W = 114 J − 163 J + 39 J = −10 J
∆U = Uf − Ui = Q − W
3. Ui = 8093 J
4
Uf = 2.092 × 10 J
W = Q − ∆U = Q − (Uf − Ui) = Q − Uf + Ui
Q = 6932 J
W = 6932 J − 2.092 × 104 J + 8093 J = −5895 J
4. Q = 4.50 × 108 J
∆U = Q − W = 4.50 × 108 J − 3.21 × 108 J = 1.29 × 108 J
8
W = 3.21 × 10 J
5. Q = 632 kJ
∆U = Q − W = 632 kJ − 102 kJ = 5.30 × 102 kJ = 5.30 × 105 J
W = 102 kJ
6. Q = 867 J
∆U = Q − W = 867 J − 623 J = 244 J
W = 623 J
∆U = Q − W
7. W = 192 kJ
∆U = 786 kJ
∆U = Q − W
8. W = 602 kJ
5
∆U = 1.09 × 10 J
9. ∆U = 873 J
Q = ∆U + W = 602 kJ + 109 kJ = 711 kJ
∆V = 0, so W = 0 J
∆U = Q − W
Q = ∆U + W = 873 J + 0 J = 873 J
10. ∆U = 986 J
∆V = 0, so W = 0 J
∆U = Q − W
Q = ∆U + W = 986 J + 0 J = 986 J
V
V Ch. 11–2
Holt Physics Solutions Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Q = ∆U + W = 786 kJ + 192 kJ = 978 kJ
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Additional Practice 11C
Givens
Solutions
1. eff = 0.17
9
Qh = 5.5 × 10 J
2. eff = 0.35
Qh = 7.37 × 108 J
3. eff = 0.15
8
Qh = 9.36 × 10 J
4. eff = 0.29
Qh = 693 J
5. eff = 0.11
Wnet = 1150 J
6. eff = 0.19
Wnet = 998 J
7. Wnet = 544 J
eff = 0.2225
8. Qh = 365 J
Qc = 223 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. Qh = 571 J
Qc = 463 J
10. Wnet = 128 J
Qh = 581 J
Wnet = eff Qh = (0.17)(5.5 × 109 J) = 9.4 × 108 J
Wnet = eff Qh = (0.35)(7.37 × 108 J) = 2.6 × 108 J
Wnet = eff Qh = (0.15)(9.36 × 108 J) = 1.4 × 108 J
Wnet = eff Qh = (0.29)(693 J) = 2.0 × 102 J
W et 1150 J
=  = 1.0 × 104 J
Qh = n
eff
0.11
W et 998 J
=  = 5.3 × 103 J
Qh = n
eff
0.19
W et
544 J
=  = 2.44 × 103 J
Qh = n
eff
0.2225
Q
223 J
eff = 1 − c = 1 −  = 1 − 0.611 = 0.389
Qh
365 J
Q
463 J
eff = 1 − c = 1 −  = 1 − 0.811 = 0.189
Qh
571 J
W et 128 J
=  = 0.220
eff = n
Qh
581 J
V
Section Five—Problem Bank
V Ch. 11–3
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Vibrations and Waves
Chapter 12
Additional Practice 12A
Givens
Solutions
1. k = 420 N/m
−2
x = 4.3 × 10
Felastic = −kx = −(420 N/m)(4.3 × 10−2 m) = −18 N
m
2. k = 65 N/m
−1
x = –1.5 × 10
Felastic = −kx = −(65 N/m)(–1.5 × 10−1 m) =
3. k = 49 N/m
−1
x = –1.2 × 10
9.8 N
m
Felastic = −kx = −(49 N/m)(–1.2 × 10−1 m) = −5.9 N
m
4. k = 26 N/m
x = –5.0 × 10−2 m
Felastic = −kx = −(26 N/m)(–5.0 × 10−2 m) =
1.3 N
Fnet = 0 = Felastic + Fg = −kx + Fg
5. Fg = –669 N
x = –6.5 × 10−2 m
Fg = kx
–669 N
Fg
k =  = 
= 1.0 × 104 N/m
–6.5 × 10−2 m
x
Fnet = 0 = Felastic + Fg = −kx + Fg
6. Fg = –550 N
Fg = kx
Copyright © by Holt, Rinehart and Winston. All rights reserved.
x = –15 m
Fg –550 N
k =  =  = 37 N/m
x
–15 m
Fnet = 0 = Felastic + Fg = −kx + Fg
7. Fg = 620 N
−2
x = 7.2 × 10
m
Fg = kx
Fg
620 N
k =   = 
= 8.6 × 103 N/m
x
7.2 × 10−2 m
8. Felastic = 12 N
k = 180 N/m
9. Felastic = 52 N
k = 490 N/m
Felastic = −kx
12 N
Felastic
 = −  = –0.067 m =
x=− 
k
180 N/m
−6.7 cm
Felastic = −kx
52 N
Felastic
 = −  = − 0.11 m = −11 cm
x=− 
k
490 N/m
Section Five—Problem Bank
V
V Ch. 12–1
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Givens
Solutions
Fnet = 0 = Felastic + Fg = −kx −mg
10. m = 3.0 kg
g = 9.81 m/s
mg = −kx
k = 36 N/m
mg
(3.0 kg)(9.81 m/s2)
x = −  = −  = –0.82 m = −82 cm
k
36 N/m
2
Additional Practice 12B
1. L = 3.0 × 10−1 m
T = 1.16 s
T = 2p
Lg
4p 2 L
T2 = 
g
(4p 2)(3.0 × 10−1 m)
4p 2 L
g = 
= 8.8 m/s2
2 = 
(1.16 s)2
T
2. L = 0.650 m
T = 2p
T = 2.62 s
Lg
4p 2 L
T2 = 
g
4p 2 L (4p 2)(0.650 m)
g = 
= 
= 3.74 m/s2
T2
(2.62 s)2
3. L = 1.14 m
T = 2p
T = 3.55 s
Lg
4p 2 L (4p 2)(1.14 m)
g = 
= 
= 3.57 m/s2
T2
(3.55 s)2
4. L = 5.00 × 10−1 m
T = 2.99 s
T = 2p
Lg
4p 2 L
T2 = 
g
4p 2 L (4p 2)(5.00 × 10−1 m)
g = 
= 
= 2.21 m/s2
(2.99 s)2
T2
5. f = 1.0 Hz
g = 9.81 m/s2
T = 2p
Lg = 1f
1
4p 2 L
2 =  
g
f
V
V Ch. 12–2
g
9.81 m/s2
 = 0.25 m = 25 cm
L =  22 = 
4p f
(4p 2)(1.0 s−1)2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4p 2 L
T2 = 
g
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Givens
6. f = 0.50 Hz
g = 9.81 m/s2
Solutions
T = 2p
Lg = 1f
1
4p 2 L
2 =  
f
g
g
9.81 m/s2
 = 0.99 m =
L =  22 = 
4p f
(4p 2)(0.50 s−1)2
7. f = 2.5 Hz
g = 9.81 m/s2
T = 2p
99 cm
Lg = 1f
1
4p 2 L
2 =  
f
g
g
9.81 m/s2
 = 4.0 × 10−2 m
L =  22 = 
4p f
(4p 2)(2.5 s−1)2
8. L = 6.200 m
g = 9.819 m/s2
T = 2p
Lg = 2p
68.12900
m/s =
9.
m
2
1
1
f =  =  =
T 4.993 s
9. L = 2.500 m
g = 9.780 m/s2
T = 2p
4.993 s
0.2003 Hz
Lg = 2p
9.27.85000
m/s =
m
2
3.177 s
1
1
f =  =  = 0.3148 Hz
T 3.177 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
10. L = 3.120 m
g = 9.793 m/s2
T = 2p
Lg = 2p
9.37.91320
mm/s =
2
3.546 s
1
1
f =  =  = 0.2820 Hz
T 3.546 s
Additional Practice 12C
1. f = 3.00 × 102 Hz
k = 8.65 × 104 N/m
T = 2p
mk = 1f
1
4p 2 m
2 =  
f
k
k
8.65 × 104 N/m
 = 2.43 × 10−2 kg
m = 22 = 
4p f
(4p 2)(300 s−1)2
V
Section Five—Problem Bank
V Ch. 12–3
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Givens
Solutions
2. T = 0.079 s
T = 2p
k = 63 N/m
mk
4p 2 m
T2 = 
k
kT 2 (63 N/m) (0.079 s)2
m = 2 =  
= 9.96 × 10−3 kg
4p
4p 2
3. k = 2.03 × 103 N/m
f = 0.79 Hz
T = 2p
mk = 1f
1
4p 2 m
2 =  
f
k
k
2.03 × 103 N/m
 = 82 kg
m =  22 = 
4p f
(4p 2)(0.79 Hz)2
4. F = 32 N
F = mg
T = 0.42 s
2
g = 9.81 m/s
F
32 N
m =  = 2 = 3.3 kg
g 9.81 m/s
T = 2p
mk
4p 2 m
T2 = 
k
4p2 (32 N)
4p 2 m 4p 2 F
k = 
= 730 N/m
2 = 
2 = 
(9.81 m/s2)(0.42 s)2
T
gT
5. F = 66 N
T = 2.9 s
g = 9.81 m/s2
F
m = 
g
T = 2p
mk
4p 2 m
T2 = 
k
(4p 2)(66 N)
4p 2 m 4p 2 F



 = 32 N/m
k = 
=
=
(9.81 m/s2)(2.9 s)2
T2
gT 2
f = mg
6. f = 87 N
2
g = 9.81 m/s
T = 0.64 s
87 N
F
m =  = 2 = 8.9 kg
9.81 m/s
g
mk
T = 2p
V
4p2 m
T2 = 
k
4p 2 m 4p 2 F
(4p2) (87 N)



 = 850 N/m
k = 
=
=
(9.81 m/s2)(0.64 s)
T2
gT 2
V Ch. 12–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
f = mg
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Givens
7. k = 364 N/m
m = 24 kg
8. m = 55 kg
k = (36)(458 N/m)
9. m = 8.2 kg
k = 221 N/m
10. m = 3(24 g) = 72 g
= 7.2 × 10−2 kg
Solutions
kg
mk = 2p
36244N
/m =
T = 2p
1.6 s
mk = 2p
6)(5455
8kgN/m) =
(3
T = 2p
kg
mk = 2p
2281.2N
/m =
T = 2p
1.2 s
−2
0 kg
mk = 2p7.
29×91N
/m =
T = 2p
k = 99 N/m
0.36 s
0.17 s
Additional Practice 12D
1. l = 2.3 × 104 m
v = fl = (0.065 Hz)(2.3 × 104 m) = 1.5 × 103 m/s
f = 0.065 Hz
2. f = 2.8 × 105 Hz
l = 0.51 cm = 5.1 × 10−3 m
3. l = 6.0 m
T = 2.0 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. f = 288 Hz
l = 5.00 m
5. f = 1.6 Hz
l = 2.5 m
6. f = 2.5 × 104 Hz
v = 331 m/s
7. f = 20.0 Hz
v = 331 m/s
8. l = 14 m
v = 7.0 m/s
v = fl = (2.8 × 105 Hz)(5.1 × 10−3 m) = 1.4 × 103 m/s
l 6.0 m
v = fl =  =  = 3.0 m/s
T 2.0 s
v = fl = (288 Hz)(5.00 m) = 1.44 × 103 m/s
v = fl = (1.6 Hz)(2.5 m) =
4.0 m/s
v = fl
v
331 m/s
l =  = 
=
f
2.5 × 104 Hz
1.3 × 10−2 m
v = fl
v 331 m/s
l =  =  = 16.6 m
f
20.0 Hz
v = fl
v
7.0 m/s
f =  =  = 0.50 Hz
l
14 m
V
Section Five—Problem Bank
V Ch. 12–5
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Givens
Solutions
9. l = 10.6 m
v = fl
v = 331 m/s
v
331 m/s
f =  =  = 31.2 Hz
l
10.6 m
10. l = 1.1 m
v = fl
4
v 2.42 × 104 m/s
f =  =  = 2.2 × 104 Hz
l
1.1 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
v = 2.42 × 10 m/s
V
V Ch. 12–6
Holt Physics Solution Manual
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Sound
Chapter 13
Additional Practice 13A
Givens
Solutions
1. P = 5.88 × 10−5 W
P
Intensity =  2
4p r
Intensity = 3.9 × 10−6 W/m2
r=
5.88 × 10−5 W

(4p) (3.9 × 10−6 W/m2)
P
 =
(4p)(Intensity)
r = 1.1 m
2. P = 1.57 × 10−3 W
−3
Intensity = 5.20 × 10
2
W/m
P
Intensity =  2
4p r
r=
1.57 × 10−3 W

(4p)(5.20 × 10−3 W/m2)
P
 =
(4p)(Intensity)
r = 0.155 m
3. P = 4.80 W
−2
Intensity = 7.2 × 10
2
W/m
P
Intensity =  2
4p r
r=
4. P = 151 kW = 1.51 × 105 W
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Intensity = .025 W/m2
r = 32 m
6. P = 3.5 W
r = 0.50 m
7. P = 2.76 × 10−2 W
r = 5.0 × 10−2 m
8. Intensity = 9.3 × 10−8 W/m2
r = 0.21 m
4.80 W
−2
2
2.3 m
P
Intensity = 2
4p r
r=
5. P = 402 W
=  =

(4p)(InP
(4p)(7.2 × 10 W/m )
en
si
ty) t
1.51 × 10 W
 =
= 

(4p)(InP
/m)
t
en
si
ty) (4
p
)(
.025W
5
2
693 m
P
402 W
Intensity =  2 = 2 = 3.1 × 10−2 W/m2
4p r
(4p)(32 m)
P
3.5 W
Intensity =  2 = 2 = 1.11 W/m2
4p r
(4p)(0.50 m)
P
2.76 × 10−2 W
Intensity =  2 = 
= 0.88 W/m2
4p r
(4p)(5.0 × 10−2 m)2
P
Intensity = 2
4p r
P = 4pr2 Intensity = (4p)(0.21 m)2 (9.3 × 10−3 W/m2)
P = 5.2 × 10−3 W
V
Section Five—Problem Bank
V Ch. 13–1
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Givens
Solutions
9. Intensity = 4.5 × 10−4 W/m2
r = 1.5 m
P
Intensity =  2
4p r
P = 4pr2 Intensity = (4p) (1.5 m)2 (4.5 × 10−4 W/m2)
P=
10. Intensity = 1.0 × 104 W/m2
r = 1.0 m
1.3 × 10−2 W
P
Intensity =  2
4p r
P = 4pr2 Intensity = (4p) (1.0 m)2 (1.0 × 104 W/m2)
P=
1.3 × 105 W
Additional Practice 13B
1. n = 2
nv
fn = 
4L
f2 = 466.2 Hz
nv
(2)(331 m/s)
L =  =  = 0.355 m = 35.5 cm
4fn (4)(466.2 Hz)
v = 331 m/s
2. n = 3
nv
fn =  , n = 1, 3, 5, . . .
4L
f3 = 370 Hz
nv (3)(331 m/s)
L =  =  = 0.671 m = 67.1 cm
4fn
(4)(370 Hz)
v = 331 m/s
3. n = 1
nv
fn = 
4L
f1 = 392.0 Hz
4. n = 1
nv
fn = 
2L
f1 = 370.0 Hz
nv
(1)(331 m/s)
L =  =  = 0.447 m = 44.7 cm
2 fn (2)(370.0 Hz)
v = 331 m/s
5. n = 1
−1
L = 35.0 cm = 3.50 × 10
(1)(346 m/s)
f1 = 
= 494 Hz
(2)(3.50 × 10−1 m)
v = 346 m/s
6. n = 1
−1
L = 4.20 × 10
v = 329 m/s
m
nv
fn = 
2L
m
nv
fn = 
2L
(1)(329 m/s)
f1 = 
= 392 Hz
(2)(4.20 × 10−1 m)
V
V Ch. 13–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
nv
(1)(331 m/s)
L =  =  = 0.211 m = 21.1 cm
4fn (4)(392.0 Hz)
v = 331 m/s
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Givens
7. n = 1
v = 499 m/s
L = 0.85 m
8. n = 1
f1 = 277.2 Hz
L = 0.75 m
Solutions
nv
fn = 
2L
(1)(499 m/s)
f1 =  =
(2)(0.85 m)
294 Hz
nv
fn = 
2L
(2)(0.75 m)(277.2 Hz)
2 Lf
v = n = 
1
n
v = 420 m/s
9. n = 7
f1 = 466.2 Hz
L = 1.53 m
10. n = 1
f1 = 125 Hz
(4)(1.53 m)(466.2 Hz)
4 Lf
v = n =  = 408 m/s
7
n
nv
fn = 
2L
(2)(1.32 m)(125 Hz)
2Lf
v = n =  = 330 m/s
1
n
Copyright © by Holt, Rinehart and Winston. All rights reserved.
L = 1.32 m
nv
fn = 
4L
V
Section Five—Problem Bank
V Ch. 13–3
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Light and Reflection
Chapter 14
Additional Practice 14A
Givens
Solutions
1. f = 7.6270 × 108 Hz
l = 3.9296 × 10−1 m
2. f = 1.17306 × 1011 Hz
l = 2.5556 × 10−3 m
c = fl = (7.6270 × 108 s−1)(3.9296 × 10−1 m) = 2.9971 × 108 m/s
The radio wave travels through Earth’s atmosphere.
c = fl = (1.17306 × 1011 s−1)(2.5556 × 10−3 m) = 2.9979 × 108 m/s
The microwave travels through space.
3. l = 3.2 × 10−9 m
c
3.00 × 108 m/s
f =   = 
= 9.4 × 1016 Hz
l
3.2 × 10−9 m
4. l = 5.291 770 × 10−11 m
c
3.00 × 108 m/s
= 5.67 × 1018 Hz
a. f =  = 
l 5.291 770 × 10−11 m
b. You cannot see atoms because light in the visible part of the spectrum has wavelengths
that are much larger than atoms.
5. UVA: l1 = 3.2 × 10−7 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
l2 = 4.0 × 10−7 m
UVB: l1 = 2.8 × 10−7 m
l2 = 3.2 × 10−7 m
for UVA waves:
c
3.00 × 108 m/s
= 9.4 × 1014 Hz
f1 =  = 
l1
3.2 × 10−7 m
c
3.00 × 108 m/s
f2 =  = 
= 7.5 × 1014 Hz
l2
4.0 × 10−7 m
for UVB waves:
c
3.00 × 108 m/s
f1 =  = 
= 1.1 × 1015 Hz
l1
2.8 × 10−7 m
c
3.00 × 108 m/s
f2 =  = 
= 9.4 × 1014 Hz
l2
3.2 × 10−7 m
6. l = 1.67 × 10−10 m
c 3.00 × 108 m/s
f =  = 
= 1.80 × 1018 Hz
l 1.67 × 10−10 m
7. f = 9.5 × 1014 Hz
c 3.00 × 108 m/s
l =  = 
= 3.2 × 10−7 m = 320 nm
f
9.5 × 1014 s−1
8. f = 2.85 × 109 Hz
c 3.00 × 108 m/s
l =  = 
= 0.105 m = 10.5 cm
f 2.85 × 109 s−1
V
Section Five—Problem Bank
V Ch. 14–1
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Givens
Solutions
9. f1 = 1800 MHz = 1.8 × 109 Hz
f2 = 2000 MHz = 2.0 × 109 Hz
10. f = 2.5 × 1010 Hz
c 3.00 × 108 m/s
= 0.17 m = 17 cm
l1 =  = 
f1
1.8 × 109 s−1
c 3.00 × 108 m/s
l2 =  = 
= 0.15 m = 15 cm
f2
2.0 × 109 s−1
c 3.00 × 108 m/s
a. l =  = 
= 1.2 × 10−2 m = 1.2 cm
f
2.5 × 1010 s−1
b. 1.2 cm > 1. 2 mm
The microwave’s wavelength is larger than the holes, so it cannot pass through.
This is analogous to a ball being trapped by a net.
c. 400 nm < 1.2 mm
700 nm < 1.2 mm
Yes, visible light can pass through the holes in the microwave oven’s door. This is
why we can look through the holes and see the food as it cooks.
Additional Practice 14B
1. f = 32.0 cm
a. You want to appear to be shaking hands with yourself, so the image must appear
to be where your hand is. So
1 1 1 2
 =  +  = 
f
p q p
p=q
p = 2f = (2)(32.0 cm) = 64.0 cm
q = p = 64.0 cm
2. f = 9.5 cm
q = 15.5 cm
h′
q
b. M =  = − 
h
p
qh
(64.0 cm)(7.5 cm)
h′ = −  = −  = −7.5 cm
p
64.0 cm
1 1 1
a.  =  + 
f
p q
1 1 1
1
1
 =  −  =  − 
p f q 9.5 cm 15.5 cm
1 0.105 0.065 0.040
 =  −  = 
p
1 cm
1 cm
1 cm
p = 25 cm
h = 3.0 cm
3. f = 17 cm
q = 23 cm
h = 2.7 cm
qh
(15.5 cm)(3.0 cm)
b. h′ = −  = −  = −1.9 cm
p
25 cm
1 1 1
a.  =  + 
f
p q
1 0.059 0.0435 0.016
 =  −  = 
p
1 cm
1 cm
1 cm
p=
V
V Ch. 14–2
1 1 1
1
1
 =  −  =  − 
p f
q 17 cm
23 cm
62 cm
qh
(23 cm)(2.7 cm)
b. h′ = −  = −  = −1.0 cm
p
62 cm
Holt Physics Solutions Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
h = 75 cm
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Givens
4. p = 5.0 cm
Solutions
A car’s beam has rays that are parallel, so q = ∞.
1 1
 =  = 0
q ∞
1 1 1 1
1
 =  +  =  + 0 = 
f p q p
p
f = p = 5.0 cm
2 1 1 1
 =  +  = 
R p q f
R = 2 f = (2)(5.0 cm) = 1.0 × 101 cm
5. p = 19 cm
q = 14 cm
1
1
1 1 1
 =  +  =  + 
p q 19 cm 14 cm
f
0.12
1 0.053 0.071
 =  +  = 
1 cm
f
1 cm
1 cm
f = 8.3 cm
2
1 1
1
 =  +  = 
R p q
f
R = 2f = (2)(8.3 cm) = 17 cm
6. p = 35 cm
q = 42 cm
1 1 1
2
1
1
 =  +  =  =  + 
f
p q R 35 cm 42 cm
1 0.029 0.024 0.053
 =  +  = 
f
1 cm
1 cm
1 cm
f = 19 cm
2
1
 = 
R
f
Copyright © by Holt, Rinehart and Winston. All rights reserved.
R = 2f = (2)(19 cm) = 38 cm
7. p = 3.00 cm = 3.00 × 102 cm
f = 30.0 cm
1 1
1
1
1
a.  =  −  =  − 
q
f
p
30.0 cm 3.00 × 102 cm
1 0.0333 0.00333 0.0300
 =  −  = 
q
1 cm
1 cm
1 cm
q = 33.3 cm
h = 15 cm
8. f = 17.5 cm
p = 15.0 cm
h′
q
b. M =  = − 
h
p
qh
(33.3 cm)(15 cm)
h′ = −  = − 
= 1.7 cm
p
3.00 × 102 cm
1 1
1
1
1
a.  =  −  =  − 
q
f
p
17.5 cm 15.0 cm
1 0.0571 0.0667 0.00960
 =  −  = 
q
1 cm
1 cm
1 cm
q = −104 cm
V
q
−104 cm
b. M = −  = −  = 6.93
p
15.0 cm
Section Five—Problem Bank
V Ch. 14–3
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Givens
Solutions
9. f = 60.0 cm
p = 35.0 cm
1 1 1
1
1
0.0167 0.0286 −0.0119
a.  =  −  =  −  =  −  = 
q
f
p
60.0 cm 35.0 cm
1 cm
1 cm
1 cm
q = −84.0 cm
q
−84.0 cm
b. M = −  = −  = 2.4
p
35.0 cm
10. f = 23.0 cm
p = 3.00 cm
1 1 1
1
1
0.0435 0.333 −0.290
a.  =  −  =  −  =  −  = 
q
f p 23.0 cm 3.00 cm
1 cm
1 cm
1 cm
q = −3.45 cm
q
3.45 cm
M = −  = −  = 1.15
p
3.00 cm
The image is real, and upright, so you can read the writing.
p = 33.0 cm
1 1 1
1
1
0.0435 0.0303 0.0132
b.  =  −  =  −  =  −  = 
q
f p
1 cm
1 cm
1 cm
23.0 cm 33.0 cm
q = 75.8 cm
75.8 cm
q
M = −  = −  = −2.30
33.0 cm
p
The image is inverted and virtual, so you cannot read the writing (unless you can
read upside-down).
Additional Practice 14C
h = 1.5 m = 150 cm
p = 3 m = 300 cm
9.0 cm
h′
a. M =  =  = 0.060
150 cm
h
q
b. M = − 
p
q = −Mp = −(0.060)(300 cm) = −18 cm
1
1
1 1 1
0.33 −5.6 −5.3
 =  +  =  +  =  −  = 
p q 3 m −0.18 m
f
1m
1m
1m
f = −0.19 m = −19 cm
c. R = 2f = 2 (19 cm) = −38 cm
V
V Ch. 14–4
Holt Physics Solutions Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. h′ = 9.0 cm
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Givens
2. q = −5.2 cm
p = 17 cm
Solutions
1 1 1
0.059
−0.13
1
1
0.19
a.  =  +  =  +  =  −  = 
f
1 cm
1 cm
p q 17 cm −5.2 cm
1 cm
f = −7.7 cm
h = 3.2 cm
1
2
1 1
b.  =  =  + 
f
R p q
R = 2f = (2)(−7.7 cm) = −15 cm
3. f = −6.3 cm
q = −5.1 cm
1 1 1
1
1
0.159 0.196 0.037
 =  −  =  −  =  −  = 
p f q −6.3 cm −5.1 cm
1 cm
1 cm
1 cm
p = 27 cm
−q −(−5.1 cm)
M =  =  = 0.19
p
27 cm
4. f = −33 cm
q = −16.1 cm
1 1 1
1
1
−0.030 0.062 0.032
 =  −  =  −  =  −  = 
p f q −33 cm −16.1 cm
1 cm
1 cm
1 cm
p = 31 cm
q
(−16.1 cm)
M = −  = −  = 0.52
f
31 cm
5. f = −12 cm
q = −9.0 cm
1 1 1
1
1
0.083 0.111 0.028
 =  −  =  −  =  +  = 
p f q −12 cm −9.0 cm
1 cm
1 cm
1 cm
p = 36 cm
q
9.0 cm
M = −  = −  = 0.25
p
36 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. h = 1.75 m
M = 0.11
q = −42 cm = −0.42 m
q
h′
a. M =  = − 
p
h
h′ = Mh = (0.11)(1.75 m) = 0.19 m
q
0.42 m
b. p = −  = −  = 3.8 m
M
0.11
7. f = −27 cm
p = 43 cm
1 1 1
−0.037 0.023 −0.060
1
1
 =  −  =  −  =  −  = 
q
f p −27 cm 43 cm
1 cm
1 cm
1 cm
q = −17 cm
q
−17 cm
M = −  = −  = 0.40
43 cm
p
V
Section Five—Problem Bank
V Ch. 14–5
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Givens
Solutions
8. f = −8.2 cm
p = 18 cm
1 1
1
1
1
−0.122 0.056 −0.18
 =  −  =  −  =  −  = 
q
f
p −8.2 cm 18 cm
1 cm
1 cm
1 cm
q = −5.6 cm
q
−5.6 cm
M = −  = −  = 0.31
p
18 cm
9. f = −39 cm
p = 16 cm
1
1
1
1 1
−0.026 0.062 −0.088
a.  =  −  =  −  =  −  = 
p −39 cm 16 cm
q
f
1 cm
1 cm
1 cm
q = −11 cm
h = 6.0 cm
10. M = 0.24
q
M = − 
p
q = −Mp = −(0.24)(12 cm) = −2.9 cm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
p = 12 cm
q
h′
b. M =  = − 
p
h
qh
(−11 cm)(6.0 cm)
h′ = −  = −  = 4.1 cm
p
16 cm
V
V Ch. 14–6
Holt Physics Solutions Manual
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Refraction
Chapter 15
Additional Practice 15A
Givens
1. qr = 35°
nr = 1.553
Solutions
= sin 1.0
=
00
nr (sin qr)
(1.553)(sin 35°)
 = sin−1  = 63°
qi = sin−1 
ni
1.000
ni = 1.000
2. qr = 41°
nr = 1.486
nr (sin qr)

qi = sin−1 
ni
−1
(1.486)(sin 41°)
77°
ni = 1.00
3. qr = 33°
nr = 1.555
nr (sin qr)
(1.555)(sin 33°)
 = sin−1  = 58°
qi = sin−1 
ni
1.000
ni = 1.000
4. c = 3.00 × 108 m/s
v = 2.07 × 108 m/s
5. c = 3.00 × 108 m/s
v = 1.97 × 108 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. c = 3.00 × 108 m/s
v = 1.95 × 108 m/s
7. qi = 59.2°
nr = 1.61
c 3.00 × 108 m/s
= 1.45
n =  = 
v 2.07 × 108 m/s
c
3.00 × 108 m/s
n =  = 
= 1.52
v 1.97 × 108 m/s
c 3.00 × 108 m/s
= 1.54
n =  = 
v 1.95 × 108 m/s
ni(sin qi)
(1.00)(sin 59.2°)
 = sin−1  = 32.2°
qr = sin−1 
nr
1.61
ni = 1.00
8. qi = 35.2°
ni = 1.00
nr,1 = 1.91
nr,2 = 1.66
n (sin q )
(1.00)(sin 35.2°)
n  = sin 1.6
= 20.3°
6
ni(sin qi)
(1.00)(sin 35.2°)
 = sin−1  = 17.6°
qr,1 = sin−1 
nr,1
1.91
qr,2 = sin−1
i
i
−1
r,2
V
Section Five—Problem Bank
V Ch. 15–1
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Givens
Solutions
9. qi = 17°
1st surface:
ni = 1.00
ni(sin qi)
(1.00)(sin 17°)
 = sin−1  = 11°
qr = sin−1 
nr
1.54
nr = 1.54
qi = 11°
2nd surface:
(1.54)(sin 11°)
qr = sin−1  = 17°
1.00
ni = 1.54
nr = 1.00
10. qi = 22°
1st surface:
ni = 1.000
nr = 1.544
qi = 14°
ni = 1.544
nr = 1.000
ni(sin qi)
(1.000)(sin 22°)
 = sin−1  = 14°
qr = sin−1 
nr
1.544
2nd surface:
(1.544)(sin 14°)
qr = sin−1  = 22°
1.000
Additional Practice 15B
1. p = 13 cm
q = 19 cm
1
1
0.13
1 1 1
0.077 0.053
 =  +  =  +  =  +  = 
p q 13 cm 19 cm
1 cm
f
1 cm
1 cm
h′ = 3.0 cm
f = 7.7 cm
q
h′
M =  = − 
p
h
2. p = 44 cm
q = −14 cm
1 1 1
1
1
0.023 0.071 −0.048
 =  +  =  +  =  +  = 
f
p q 44 cm −14 cm
1 cm
1 cm
1cm
h = 15 cm
f = −21 cm
h′
q
M =  = − 
h
p
qh
(−14 cm)(15 cm)
h′ = −  = −  = 4.8 cm
p
44 cm
3. f = −13.0 cm
M = 5.00
q
M = − 
p
q = −Mp = −(5.00)p
1 1 1 1
1
−5.00 + 1.00
−4.00
4.00
 =  +  =  +  =  =  = 
f p q p
−(5.00)p
−(5.00)p
−(5.00)p (5.00)p
h′
q
M =  = − 
h
p
(4.00)f (4.00)(−13.0 cm)
p =  =  = −10.4 cm
5.00
5.00
V
V Ch. 15–2
Holt Physics Solutions Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ph′
(13 cm)(3.0 cm)
h = −  = −  = 2.1 cm
q
19 cm
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Givens
4. h = 18 cm
h′ = −9.0 cm
f = 6.0 cm
Solutions
h′ −9.0 cm
a. M =  =  = −0.50 cm
h
18 cm
q
b. M = − 
p
q = −Mp = −(−0.50)p = (0.50)p
1 1 1 1
1
0.50
1
1.50
3.0
 =  +  =  +  =  +  =  = 
f p q p (0.50)p (0.50)p (0.50)p (0.50)p
p
p = (3.0)f = (3.0)(6.0 cm) = 18 cm
c. q = −Mp = (0.50)(18 cm) = 9.0 cm
5. p = 4 m
f =4m
1 1 1
1
 =  + , but f = p, so  = 0. That means q = •
p q
f
q
q
•
M = −  = −  = •
p
4m
The rays are parallel, and the light can be seen from very far away.
6. p = 0.5 m
f = 0.5 m
1 1 1
 =  + 
f
p q
1
p = f, so  = 0, and q = •
q
−q −•
M =  =  m = −•
p 0.5
The rays are parallel, and the light can be seen from far away.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. f = 3.6 cm
q = 15.2 cm
1
0.066
1 1
1
1
0.28
0.21
 =  −  =  −  =  −  = 
f
q 3.6 cm 15.2 cm 1 cm
1 cm
p
1 cm
p = 4.8 cm
8. q = −12 cm
f = −44 cm
1 1 1
1
1
− 0.023
−0.083
0.060
 =  −  =  −  =  −  = 
p
f q −44 cm −12 cm
1 cm
1 cm
1 cm
p = 17 cm
9. f = 9.0 cm = 0.09 m
q = 18 m
1 1 1
1
1
11
0.056
11
 =  −  =  −  =  −  = 
p
f q 0.090 m 18 m 1 m
1m
1m
p = 0.091 m = 9.1 cm
10. f = 5.5 m
q = 5.5 cm = 0.055 m
0.18
1 1 1
1
1
−18
18
 =  −  =  −  =  −  = 
f q 5.5 m −0.055 m
1m
p
1m
1m
p = 5.5 × 10−2 m = 5.5 cm
V
Section Five—Problem Bank
V Ch. 15–3
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Additional Practice 15C
Givens
Solutions
1. qc = 37.8°
nr = 1.00
2. qc = 39.18°
nr = 1.000
3. qc,1 = 35.3°
qc,2 = 33.1°
nr = 1.00
n
sin qc = r
ni
nr
1.00
ni =  =  = 1.63
sin qc
sin 37.8°
n
sin qc = r
ni
nr
1.000
ni =  =  = 1.583
sin qc
sin 39.18
n
sin qc = r
ni
nr
1.00
ni,1 =  =  = 1.73
sin qc,1 sin 35.3°
nr
1.00
=  = 1.83
ni,2 = 
sin qc,2
sin 33.1°
4. ni = 1.670
qc = 62.85°
n
sin qc = r
ni
nr = ni(sin qc) = (1.670)(sin 62.85°) = 1.486
5. ni = 1.80
qc = 57.0°
n
sin qc = r
ni
nr = ni(sin qc) = (1.80)(sin 57.0°) = 1.51
qc = 69.9°
n
sin qc = r
ni
nr = ni(sin qc) = (1.64)(sin 69.9°) = 1.54
7. ni- = 1.766
nr = 1.000
8. ni = 1.774
nr = 1.000
9. ni = 1.61
n
sin qc = r
ni
n
1.000
qc = sin−1 r = sin−1  = 34.31°
ni
1.774
10. ni = 1.576
nr = 1.000
V Ch. 15–4
n
sin qc = r
ni
1.00
−1 nr
qc = sin  = sin−1  = 38.4°
ni
1.61
nr = 1.00
V
n
sin qc = r
ni
1.000
−1 nr
qc = sin  = sin−1  = 34.49°
ni
1.766
n
sin qc = r
ni
n
1.000
qc = sin−1 r = sin−1  = 39.38°
ni
1.576
Holt Physics Solutions Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. ni = 1.64
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Interference and Diffraction
Chapter 16
Additional Practice 16A
Givens
Solutions
1. 14 450 lines/cm
–7
l = 6.250 × 10 m
q < 90°
m = 1: q1 = sin–1(ml/d)
q1 = sin–1[(1)(6.250 × 10–7 m) ÷ (1 450 000 lines/m)–1]
q1 = 64.99°
m = 2: q2 = sin–1[(2)(6.250 × 10–7 m) ÷ (1 450 000 lines/m)–1]
q2 = ∞
Therefore, 1 is the highest-order number that can be observed.
2. 12 260 lines/cm
–7
l = 5.896 × 10 m
q < 90°
m = 1: q1 = sin–1(ml/d)
q1 = sin–1[(1)(5.896 × 10–7 m) ÷ (1 226 000 lines/m)–1]
q1 = 46.29°
m = 2: q2 = sin–1[(2)(5.896 × 10–7 m) ÷ (1 450 000 lines/m)–1]
q2 = ∞
Therefore, 1 is the highest-order number that can be observed.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. l= 5.461 × 10–7 m
m=1
(1)(5.461 × 10–7 m)
ml
d =  =  = 5.634 × 10–7 m
sin (75.76°)
sin q
q = 75.76°
d = 5.634 × 10–5 cm
# lines/cm = (5.634 × 10–5 cm)–1 = 1.775 × 104 lines/cm
4. l = 4.471 × 10–7 m
m=1
(1)(4.471 × 10–7 m)
ml
d =  =  = 6.920 × 10–7 m
sin (40.25°)
sin q
q = 40.25°
d = 6.920 × 10–7 cm
# lines/cm = (6.920 × 10–7 cm)–1 = 1.445 × 104 lines/cm
5. 1950 lines/cm
l = 4.973 × 10–7 m
m = 1: q1 = sin–1(ml/d)
q1 = sin–1[(1)(4.973 × 10–7 m) ÷ (195 000 lines/m)–1]
q1 = 5.565°
m = 2: q2 = sin–1[(2)(4.973 × 10–7 m) ÷ (195 000 lines/m)–1]
q2 = 11.18°
V
Section Five—Problem Bank
V Ch. 16–1
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Print
Givens
Solutions
6. 7500 lines/cm
–7
l1 = 5.770 × 10 m
–7
q1 = sin–1(ml1/d)
q1 = sin–1[(2)(5.770 × 10–7 m) ÷ (750 000 lines/m)–1]
l2 = 5.790 × 10 m
q1 = 59.94°
m=2
q2 = sin–1(ml2/d)
q2 = sin–1[(2)(5.790 × 10–7 m) ÷ (750 000 lines/m)–1]
q2 = 60.28°
q = q2 – q1 = 59.94° – 60.28° = 0.3400°
7. 3600 lines/cm
m=3
d(sin q) (360 000 m)–1sin (76.54°)
l =  =  = 9.000 × 10–7 m
3
m
q = 76.54°
8. 9550 lines/cm
m=2
d(sin q) (955 000 m)–1sin (54.58°)
l =  =  = 4.267 × 10–7 m
2
m
q = 54.58°
9. 12 510 lines/cm
m=1
d(sin q) (1 251 000 m)–1sin (38.77°)
l =  =  = 5.006 × 10–7 m
1
m
q = 38.77°
10. 2400 lines/cm
m=2
d(sin q) (240 000 m)–1sin (26.54°)
l =  =  = 9.296 × 10–7 m
2
m
Additional Practice 16B
1. l = 5.875 × 10–7 m
m=2
ml 2(5.875 × 10–7 m)
d =  =  = 5.18 × 10–4 m
sin (0.130°)
sinq
q = 0.130°
d = 0.518 mm
2. l = 6.563 × 10–7 m
m=4
–7
ml 4(6.563 × 10 m)
d =  =  = 2.40 × 10–4 m
sin (0.626°)
sinq
q = 0.626°
d = 0.240 mm
V
V Ch. 16–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 26.54 °
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Givens
Solutions
m=3
3(6.93 × 10–7 m)
ml
d =  =  = 2.06 × 10–4 m
sin (0.578°)
sin q
q = 0.578°
d = 0.206 mm
3. l = 6.93 × 10–7 m
4. d = 8.04 × 10–6 m
m=3
d(sin q) (8.04 × 10–6 m) sin (13.1°)
l =  =  = 6.07 × 10–7 m
3
m
q = 13.1 °
l = 607 nm
5. d = 4.43 × 10–6 m
m=3
d(sin q) (4.43 × 10–6 m) sin (21.7°)
l =  =  = 5.46 × 10–7 m
3
m
q = 21.7 °
l = 546 nm
6. d = 3.92 × 10–6 m
m=2
d(sin q) (3.92 × 10–6 m) sin (13.1°)
l =  =  = 4.44 × 10–7 m
2
m
q = 13.1 °
l = 444 nm
7. d = 2.20 × 10–4 m
l= 5.27 × 10–7 m
q = sin–1(ml/d)
q = sin–1[(1)(5.27 × 10–7 m) ÷ (2.20 × 10–4 m)] = 0.137°
m=1
8. d = 1.63 × 10–4 m
l = 4.308 × 10–7 m
q = sin–1(ml/d)
q = sin–1[(1)(4.308 × 10–7 m) ÷ (1.63 × 10–4 m)] = 0.151°
m=1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. d = 3.29 × 10–4 m
l = 5.83 × 10–7 m
q = sin–1(ml/d)
q = sin–1[(1)(5.83 × 10–7 m) ÷ (3.29 × 10–4 m)] = 0.102°
m=1
10. d = 2.67 × 10–4 m
l = 6.87 × 10–7 m
q = sin–1(ml/d)
q = sin–1[(1)(6.87 × 10–7 m) ÷ (2.67 × 10–4 m)] = 0.147°
m=1
V
Section Five—Problem Bank
V Ch. 16–3
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Electric Forces and Fields
Chapter 17
Additional Practice 17A
Givens
Solutions
1. q1 = − 1.30 × 10−5 C
q2 = −1.60 × 10−5 C
Felectric = 12.5 N
kC = 8.99 × 109 N • m2/C2
2. q1 = 9.99 × 10−5 C
−5
q2 = 3.33 × 10
C
Felectric = 87.3 N
3. q1 = −4.32 × 10−5 C
q2 = 2.24 × 10
C
Felectric = −6.5 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. q1 = −5.33 × 10−6 C
q2 = +5.3 × 1
−6
−2
r = 4.2 × 10
C
m
kC = 8.99 × 109 N • m2/C2
5. q1 = −1.40 × 10−8 C
q2 = +1.40 × 10−8 C
r = 7.1 × 10−2 m
kC = 8.99 × 109 N • m2/C2
6. q1 = −8.0 × 10−9 C
−9
q2 = +8.0 × 10
−2
r = 2.0 × 10
r = 0.387 m = 38.7 cm
kC q1q2

Felectric = 
r2
C
m
kC = 8.99 × 109 N • m2/C2
(8.99 × 109 N • m2/C2)(9.99 × 10−5 C)(3.33 × 10−5 C)1

87.3 N
kCq 1q2
 =
Fel ectric
r = 0.585 m = 58.5 cm
kC q1q2

Felectric = 
r2
r=
KC = 8.99 × 109 N • m2/C2
(8.99 × 109 N • m2/C2)(−1.30 × 10−5 C)(−1.60 × 10−5 C)

12.5 N
r=
kC = 8.99 × 109 N • m2/C2
−5
kC q1q2

Felectric = 
r2
kCq1q2
 =
r= 
Felectric
(8.99 × 109 N • m2/C2)(−4.32 × 10−5 C)(2.24 × 10−5 C)

−6.5 N
kCq1q2
 =
Felectric
r = 1.15 m
kC q1q2

Felectric = 
r2
(8.99 × 109 N • m2/C2)(−5.3 × 10−6 C)(5.3 × 10−6 C)
Felectric = 
(4.2 × 10−2 m)2
Felectric = −143 N
kC q1q2

Felectric = 
r2
(8.99 × 109 N • m2/C2)(−1.40 × 10−8 C)(1.4 × 10−8 C)
Felectric = 
(7.1 × 10−2 m)2
Felectric = 3.50 × 10−4 N
kC q1q2

Felectric = 
r2
(8.99 × 109 N • m2/C2)(−8.0 × 10−9 C)(8.0 × 10−9 C)
Felectric = 
(2.0 × 10−2 m)2
V
Felectric = 1.4 × 10−3 N
Section Five—Problem Bank
V Ch. 17–1
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Givens
Solutions
7. r = 8.3 × 10−10 m
Felectric = 3.34 × 10−10 N
kC = 8.99 × 109 N • m2/C2
kC q1q2 kCq2
 = 
Felectric = 
r2
r2
q=
−10
electric
2
9
C
−19
q = 1.60 × 10
8. r = 6.4 × 10−8 m
−14
Felectric = 5.62 × 10
9
N
2
kC = 8.99 × 10 N • m /C2
−10
(3.34 × 10 N)(8.3 × 10 N)

F
kr = 8.99 × 10 N m /C
•
2
2
2
C
kC q1q2 kCq2
 = 
Felectric = 
r2
r2
q=
−14
−8
(5.62 × 10 N)(6.4 × 10 m)

F
kr = 8.99 × 10 N m C
electric
2
9
C
•
2
2 2
q = 1.60 × 10−19 C
9. r = 9.30 × 10−11 m
kC q1q2 kCq2
 = 
Felectric = 
r2
r2
Felectric = 2.66 × 10−8 N
kC = 8.99 × 109 N • m2/C2
q=
(2.66 × 10−8 N)(9.30 × 10−11 m)2

8.99 × 109 N • m2/C2
Felectricr 2
 =
kC
q = 1.60 × 10−19 C
10. r = 6.5 × 10−11 m
−4
Felectric = 9.92 × 10
kC q1q2 kCq2
 = 
Felectric = 
r2
r2
N
9
2
kC = 8.99 × 10 N • m /C2
q=
−4
−11
(9.92 × 10 N)(6.5 × 10 m)

F
kr = 8.99 × 10 N m /C
electric
2
9
C
•
2
2
2
q = 2.2 × 10−17 C
qp = q1 = q2 = q3 = q4
1. qp = 1.60 × 10−19 C
−9
r4,1 = r2,1 = 1.52 × 10
m
kC = 8.99 × 109 N • m2/C2
r3,2 = (1
)2
+(1.
0−9m
)2 = 2.15 × 10−9 m
.5
2×10−9m
52
×1
kCqp2
(8.99 × 109 N • m2/C2)(1.60 × 10−19 C)2

F2,1 = 
= 9.96 × 10−11 N
2 = 
r2,1
(1.52 × 10−9 m)2
kCqp2
(8.99 × 109 N • m2/C2)(1.60 × 10−19 C)2

F3,1 = 
= 4.98 × 10−11 N
2 = 
r3,1
(2.15 × 10−9 m)2
kCq2p (8.99 × 109 N • m2/C2)(1.60 × 10−19 C)2
 = 
F4,1 = 
= 9.96 × 10−11 N
r4,12
(1.52 × 10−9 m)2
1.52 × 10−9 m
j = tan−1 
= 45°
1.52 × 10−9 m
F2,1: Fx = 0 N
Fy = 9.96 × 10−11 N
F3,1: Fx = F3,1 cos 45° = (4.98 × 10−11 N)(cos 45°) = 3.52 × 10−11 N
Fy = F3,1 sin 45° = (4.98 × 10−11 N)(sin 45°) = 3.52 × 10−11 N
V
F4,1: Fx = 9.96 × 10−11 N
Fy = 0 N
V Ch. 17–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Additional Practice 17B
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Givens
Solutions
Fx,tot = 0 N + 3.52 × 10−11 N + 9.96 × 10−11 N = 1.3 × 10−10 N
Fy,tot = 9.96 × 10−10 N + 3.52 × 10−11 N + 0 N = 1.3 × 10−10 N
2
)2 = (1
N
)2
+(1.
0−10
N
)2
Ftot = (F
x,to
(F
.3
×10−10
3×1
t)+
y,tot
Ftot = 1.8 × 10−10 N
1.3 × 10−10 N
= 45°
j = tan−1 
1.3 × 10−10 N
2. h = 3.50 m
Set the origin at the midpoint.
q1 = q2 = 4.50 C
r1,3 = r2,3 = (2
)2
+(3.
m
)2 = 4.30 m
.5
0m
50
q3 = 6.30 C
kC q1q3 (8.99 × 109 N • m2/C2)(4.50 C)(6.30 C)
 = 
F1,3 = 
= 1.38 × 1010 N
(4.30 m)2
r1,32
r1,2 = 5.00 m
kC = 8.99 × 109 N • m2/C2
kC q2q3 (8.99 × 109 N • m2/C2)(4.50 C)(6.30 C)
 = 
= 1.38 × 1010 N
F2,3 = 
(4.30 m)2
r2,32
3.50 m
q = tan−1  = 54.5°
2.50 m
F1,3: Fx = F1,3 cos q = (1.38 × 1010 N) cos (54.5°) = 8.01 × 109 N
Fy = F1,3 sin q = (1.38 × 1010 N) sin (54.5°) = 1.12 × 1010 N
F2,3: Fx = F2,3 cos q = (1.38 × 1010 N) cos (54.5°) = 8.01 × 109 N
The electrical force on q2 points in the −x direction, so Fx must be negative.
Fx = −8.01 × 109 N
Fy = F2,3 sin q = (1.38 × 1010 N) sin (54.5°) = 1.12 × 1010 N
Fx,tot = 8.01 × 109 N − 8.01 × 109 N = 0 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fy,tot = 1.12 × 1010 N + 1.12 × 1010 N = 2.24 × 1010 N
There is no x-component of the resultant force, so
Ftot = Fy,tot = 2.24 × 1010 N pointing upward along the y-axis
3. q1 = −9.00 × 10−9 C
q2 = −8.00 × 10−9 C
kC q1q2 (8.99 × 109 N • m2/C2)(−9.00 × 10−9 C)(−8.00 × 10−9 C)
 = 
F1,2 = 
= 1.62 × 10−7 N
r1,22
(2.00 m)2
r1,2 = 2.00 m
kC q1q3 (8.99 × 109 N • m2/C2)(−9.00 × 10−9 C)(−7.00 × 10−9 C)
 = 
F1,3 = 
= −6.29 × 10−8 N
r1,32
(3.00 m)2
r1,3 = 3.00 m
F1,2: Fx = 4.05 × 10−8 N
q3 = 7.00 × 10−9 C
kC = 8.99 × 109 N • m2/C2
Fy = 0 N
F1,3: Fx = 0 N
Fy = −6.29 × 10−8 N
Ftot = (1
)2
+(−
0−8N
)2 = 1.74 × 10−7 N
.6
2×10−7N
6.
29
×1
Ftot is negative because the larger, y-component of the force is negative.
−6.29 × 10−8 N
j = tan−1 
= −21.22°
1.62 × 10−7 N
Section Five—Problem Bank
V
V Ch. 17–3
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Givens
Solutions
4. q1 = −2.34 × 10−8 C
q2 = 4.65 × 10−9 C
kC q1q2 (8.99 × 109 N • m2/C2)(−2.34 × 10−8 C)(4.65 × 10−9 C)
 = 
F1,2 = 
r1,22
(0.500 m)2
q3 = 2.99 × 10−10 C
F1,2 = Fy = −3.91 × 10−6 N
r1,2 = 0.500 m
9
2 2
−8
−10
kC q1q3 (8.99 × 10 N • m /C )(−2.34 × 10 C)(2.99 × 10 C)


=
F1,3 = 
(1.00 m)2
r1,32
r1,3 = 1.00 m
kC = 8.99 × 109 N • m2/C2
F1,3 = Fy = − 6.29 × 10−8 N
Fy,tot = −3.91 × 10−6 N + −6.29 × 10−8 N = 3.97 × 10−6 N
There are
of the electrical force, so the magnitude of the electrical
no x-components
)2.
force is (F
y,t
ot
Ftot = 3.97 × 10−6 N directed along the −y-axis
qe = q1 = q2 = q3 = q4
5. qe = −1.60 × 10−19 C
r2,3 = r4,3 = 3.02 × 10−5 m
r1,3 = 2(
)2
3.
02
×10−5m
= 4.27 × 10−5 m
9
2
2
kC = 8.99 × 10 N • m /C
(8.99 × 109 N • m2/C2)(−1.60 × 10−19 C)2
kCq2e


F3,2 = Fx = 
= 2.52 × 10−19 N
2 =
(3.02 × 10−5 m)2
r3,2
kCqe2 (8.99 × 109 N • m2/C2)(−1.60 × 10−19 C)2
 = 
F3,4 = Fy = 
= 2.52 × 10−19 N
r3,42
(3.02 × 10−5 m)2
(8.99 × 109 N • m2/C2)(−1.60 × 10−19 C)2
kCqe2


F3,1 = 
= 1.26 × 10−19 N
2 =
(4.27 × 10−5 m)2
r3,1
3.02 × 10−5 m
= 45°
j = tan−1 
3.02 × 10−5 m
F3,1: Fx = F3,1 cos 45° = (1.26 × 10−19 N) cos 45° = 8.91 × 10−20 N
Fy = F3,1 sin 45° = (1.26 × 10−19 N) sin 45° = 8.91 × 10−20 N
Fx,tot = 8.91 × 10−20 N + 2.52 × 10−19 N + 0 N = 3.41 × 10−19 N
Fy,tot = 8.91 × 10−20 N + 0 N + 2.52 × 10−19 N = 3.41 × 10−19 N
Ftot = 4.82 × 10−19 N
Fy tot
3.41 × 10−19 N
j = tan−1 ,
= tan−1 
= 45°
Fx,tot
3.41 × 10−19 N
6. q1 = 2.22 × 10−10 C
9
2 2
−8
−9
kCq3q2 (8.99 × 10 N • m /C )(4.44 × 10 C)(3.33 × 10 C)


= 4.25 × 10−6 N
F3,2 = 
2
2 =
(0.559 m)
r3,2
q3 = 4.44 × 10−8 C
r1,2 = 1.00 m
h = 0.250 m
2
2
kC = 8.99 × 10 N • m /C
0.250 m
j = tan−1  = 26.6°
0.500 m
r3,1 = r3,2
F3,1: Fx = F3,1 cos q = (2.84 × 10−7 N) cos (−26.6°) = 2.54 × 10−7 N
= 0.559 m
F3,2: Fx = F3,2 cos q = (4.25 × 10−6 N) cos (−26.6°) = 3.80 × 10−6 N
= (0
m
)2
+(0.
)2
.5
00
25
0m
V
9
2 2
−10
−8
kCq1q3 (8.99 × 10 N • m /C )(2.22 × 10 C)(4.44 × 10 C)


F3,1 = 
= 2.84 × 10−7 N
2
2 =
(0.559 m)
r3,1
q2 = 3.33 × 10−9 C
9
V Ch. 17–4
Fy = F3,1 sin q = (2.84 × 10−7 N) sin (−26.6°) = 1.27 × 10−7 N
Fy = F3,2 sin q = (4.25 × 10−6 N) sin (−26.6°) = 1.90 × 10−6 N
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2
Ftot = (F
)2 = (3
0−19
N
)2(3.
N
)2
x,to
(F
.4
1×1
41
×10−19
t)+
y,t
ot
Menu
Print
Givens
Solutions
Fx,tot = 2.54 × 10−7 N + 3.80 × 10−6 N = 4.05 × 10−6 N
Fy,tot = 1.27 × 10−7 N + 1.90 × 10−6 N = 2.03 × 10−6 N
2
Ftot = (F
+(F
x,to
)2 = (4
.0
5×10−6N
)2+(2.
03
×10−6N
)
t y,tot
Ftot = 4.53 × 10−6 N
Fy tot
2.03 × 10−6 N
j = tan−1 ,
= tan−1 
= 26.6°
Fx,tot
4.05 × 10−6 N
7. q1 = q2 = q3 = 2.0 × 10−9 C
r1,2 = 1.0 m
(8.99 × 109 N • m2/C2)(2.0 × 10−9 C)2
kC q1q2


F1,2 = 
=
= 3.60 × 10−8 N
(1.0 m)2
r1,22
r1,3 = (1
m)2
+(2.
.0
0m
)2
= 2.24 m
(8.99 × 109 N • m2/C2)(2.0 × 10−9 C)2
kC q1q3


F1,3 = 
=
= 7.17 × 10−9 N
(−2.24 m)2
r1,32
kC = 8.99 × 109 N • m2/C2
F1,2 = Fx = 3.60 × 10−8 N
Fy = 0 N
2.0 m
j = tan−1  = 63.4°
1.0 m
F1,3: Fx = F1,3 cos q = (7.17 × 10−9 N) cos (63.4°) = 3.21 × 10-−9 N
Fy = F1,3 sin q = (7.17 × 10−9 N) sin (63.4°) = 6.41 × 10−9 N
Fx,tot = 3.60 × 10−8 N + 3.21 × 10−9 N = 3.92 × 10−8 N
Fy,tot = 0 N + 6.41 × 10−9 N = 6.41 × 10−9 N
2
Ftot = (F
)2 = (3
0−8N
)2
+(6.
)2
x,to
(F
.9
2×1
41
×10−9N
t)+
y,tot
Ftot = 3.97 × 10−8 N
F tot
6.41 × 10−9 N
j = tan−1 y,
= tan−1 
= 9.29°
Fx,tot
3.92 × 10−8 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. q1 = −4.0 × 10−3 C
q2 = −8.0 × 10−3 C
−3
3 = 2.0 × 10 C
k
9
2 2
C = 8.99 × 10 N • m /C
q
(8.99 × 109 N • m2/C2)(−4.0 × 10−3 C)(−8.0 × 10−3 C)
kC q1q2


F1,2 = 
2 =
(2.0 m)2
r1,2
F1,2 = Fx,tot = 7.2 × 104 N
r1,2 = 2.0 m
kCq1q3
(8.99 × 109 N • m2/C2)(−4.0 × 10−3 C)(2.0 × 10−3 C)

F1,3 = 
2 = 
r1,3
(2.0 m)2
r1,3 = 2.0 m
F1,2 = Fy,tot = −1.8 × 104 N
2
Ftot = (F
)2 = (7
04
N
)2
+(−
04
N
)2
x,to
(F
.2
×1
1.
8×1
t)+
y,tot
Ftot = 7.4 × 104 N
F tot
–1.8 × 104 N
j = tan−1 y,
= tan−1 
= –14°
Fx,tot
7.2 × 104 N
V
Section Five—Problem Bank
V Ch. 17–5
Menu
Print
Givens
Solutions
9. q1 = 9.00 × 10−3 C
2
2
x1,2
)2
+(1.
m
)2 = 1.41 m
r1,2 = r1,3 = 2+y1,
1,32+y1,
.0
0m
00
2 = x
3 = (1
q2 = 6.00 × 10−3 C
2
r2,3 = r1,2
)2
+(1.
m
)2 = 1.99 m
2+r1,
.4
1m
41
3 = (1
q3 = 3.00 × 10−3 C
(8.99 × 109 N • m2/C2)(9.00 × 10−3 C)(6.00 × 10−3 C)
kC q1q2


=
= 2.44 × 105 N
F1,2 = 
(1.41 m)2
r1,22
kC = 8.99 × 109 N • m2/C2
x1,2 = 1.00 m
y1,2 = 1.00 m
x1,3 = 1.00 m
y1,3 = 1.00 m
(8.99 × 109 N • m2/C2)(9.00 × 10−3 C)(3.00 × 10−3 C)
kCq1q3


=
= 1.22 × 105 N
F1,3 = 
(1.41 m)2
r1,32
1.00 m
j = tan−1  = 45°
1.00 m
F1,2: Fx = F1,2 cos q = (2.44 × 105 N) cos 45° = 1.73 × 105 N
Fy = F1,2 sin q = (2.44 × 105 N) sin 45° = 1.73 × 105 N
F1,3: Fx = F1,3 cos q = (1.22 × 105 N) cos 45° = 8.63 × 104 N
Fy = F1,3 sin q = (1.22 × 105 N) sin 45° = 8.63 × 104 N
Fx,tot = 1.73 × 105 N + 8.63 × 104 N = 2.59 × 105 N
Fy,tot = 1.73 × 105 N + 8.63 × 104 N = 2.59 × 105 N
2
Ftot = (F
)2 = (2
05
N
)2
+(2.
05
N
)2
x,to
(F
.5
9×1
59
×1
t)+
y,tot
Ftot = 3.66 × 105 N
2.59 × 105 N
= 45°
j = tan−1 
2.59 × 105 N
10. q1 = q2 = q3 = 4.00 × 10−9 C
kC = 8.99 × 109 N • m2/C2
r2,1 = r2,3 = 4.00 m
All forces are along the x-axis, so there are no y-components.
(8.99 × 109 N • m2/C2)(4.00 × 10−9 C)2
k q2
F2,1 = F2,3 = C2 = 
= 8.99 × 10−9 N
(4.00 m)2
r2,1
Fx,tot = 2(8.99 × 10−9 N) = 1.80 × 10−8 N
Fy,tot = 0 N
Additional Practice 17C
1. q1 = 9.0 mC
q2 = −19 mC
q3 = 9.0 mC
The charge, q3, cannot be in electrostatic equilibrium between q1 and q2, because the
forces point in the same direction. Because q2 is larger than q1, q3 will be close to q1,
and opposite q2.
kC = 8.99 × 109 N • m2/C2
kC q3q1
kC q3q2
 = −

F3,1 = −F3,2 = 
(y − 3.0 m)2
y2
r2,1 = 3.0 m
q1y 2 = −q2 (y − 3.0 m)2 = −q2y 2 + 6q2y − (9.0 m2)q2
(q1 + q2)y2 − 6q2y + (9.0 m2)q2 = 0
6q2 ± (6
q2
)2
−4(q
m2)(q
1+q
.0
2)(9
2)
y = 
2(q1 + q2)
6(−19 mC) ± (6
.0
m
)2
(−
19
m
C
)2
−4
(9
.0
m
C
−1
9
m
C
)(
9.0
m2)
(−
19
m
C)
y = 
2(9.0 mC − 19 mC)
y = 9.6 m = r2,3
V
9
2 2
−5
−6
kC q3q2 (8.99 × 10 N • m /C )(−1.9 × 10 C)(9.0 × 10 C)


F3,2 = 
2
2 =
(9.6 m)
r3,2
F3,2 =
V Ch. 17–6
−1.7 × 10−2 N
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2
Ftot = (F
)2 = 1.80 × 10−8 N along the x-axis
x,to
(F
t)+
y,tot
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Givens
Solutions
2. q1 = 25 mC
The charge, q3, cannot be in electrostatic equilibrium between q1 and q2, because the
forces point in the same direction. Because q1 is larger than q2, q3 will be closest to q2
on the side opposite of q1.
q2 = −5.0 mC
q3 = −35 mC
kC = 8.99 × 109 N • m2/C2
−kC q3q1
kC q3q2
 = 

F3,2 = −F3,1 = 
(x + 0.25 m)2
(−x)2
r1,2 = 0.25 m
(q2 + q1)x2 − (0.50 m)q2x + (0.625 m2)q2 = 0
)2
q22−
m2)
q2
(0.50 m)q2 ± (0
.5
0m
4(q
2+q
.6
25
1)(0
x = 
2(q2 + q1)
(0.50 m)(−5.0 mC) ± (0
)2(−
)2
−4(−
C)
.5
0m
5.
0mC
5.
0mC
+25mC
)(
0.
62
5m
2)(−
5.
0m
x = 
2(−5.0 mC + 25 mC)
x = −0.20 m = r3,2
9
2 2
−5
−6
kC q3q1 (8.99 × 10 N • m /C )(−3.5 × 10 C)(−5.0 × 10 C)
 = 
F3,2 = 
= 39 N
(−0.20 m)2
r3,2
3. q1 = 6.0 mC
−kCq1q2 −(8.99 × 109 N • m 2/C2)(6.0 × 10−6 C)(−12.0 × 10−6 C)
 = 
F2,3 = −F1,2 = 
r1,22
(5.0 × 10−2 m)2
q2 = −12.0 mC
q3 = 6.0 mC
F2,3 = 259 N
9
2
2
kC = 8.99 × 10 N • m /C
r1,0 = 5.0 × 10−2 m
4. q1 = 7.2 nC
q2 = 6.7 nC
q3 = −3.0 nC
kC q1q3
kC q2q3
 − 
 =0
F1,3 + F1,2 = 
x2
(x − 0.32 m)2
kC = 8.99 × 109 N • m2/C2
(q1 − q2)x2 − (0.64 m)q1x + (0.32 m)2q1x = 0
r1,2 = 3.2 × 10−1 m = 0.32 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
The charge, q3, must be between the charges to achieve electrostatic equilibrium.
(0.64 m)(7.2 nC) ± (0
)2(7.
)2
−4(7
m
)2(7
.6
4m
2nC
.2
nC
−6.7
nC
)(
0.
32
.2
nC
)
x = 
2(7.2 nC − 6.7 nC)
x = 16 cm
5. q1 = 5.5 nC
q2 = 11 nC
The charge, q3, must be between the charges to achieve electrostatic equilibrium.
q3 = −22 nC
kC q2q3
kC q1q3

 =0
F1,3 + F1,2 = 
− 
2
x
(x − 88 cm)2
kC = 8.99 × 109 N • m2/C2
(q1 − q2)x2 − (176 cm)q1x + (88 cm)2q1x = 0
r1,2 = 88 cm
(176 cm)(5.5 nC) ± (1
)2(5
)2
−4(5
)2(5
76
cm
.5
nC
.5
nC
−11nC
)(
88
cm
.5
nC
)
x = 
2(5.5 nC − 11 nC)
x = 36 cm
V
Section Five—Problem Bank
V Ch. 17–7
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Givens
Solutions
6. q1 = −2.5 nC
The charge, q3, must be between the charges to achieve electrostatic equilibrium.
q2 = − 7.5 nC
q3 = 5.0 nC
kC q1q3
kCq2q3

 =0
− 
F1,3 + F1,2 = 
2
x
(x − 20.0 cm)2
kC = 8.99 × 109 N • m2/C2
(q1 − q2)x2 − (40.0 cm)q1x + (20.0 cm)2q1x = 0
r1,2 = 20.0 cm
(40.0 cm)(−2.5 nC) ± (4
)2(−
)2
−4(−
)2(−
0.
0cm
2.
5nC
2.
5nC
+7.5
nC
)(
20
.0
cm
2.
5nC
)
x = 
2(−2.5 nC + 7.5 nC)
x = 7.3 cm
7. q1 = −2.3 C
−kC q3q1 kC q3q2
 −  = 0
F3,1 + F3,2 = 
r3,2
r3,12
q3 = −4.6 C
r1,2 = r3,1 = 2.0 m
r3,2 = 4.0 m
−q1r3,222 −(−2.3 C)(4.0 m)2
 = 
q2 = 
= 9.2 C
r3,1
(2.0 m)2
kC = 8.99 × 109 N • m2/C2
8. q1 = 8.0 C
−kC q3q1 kC q3q2
 − 
F3,1 + F3,2 = 
=0
r3,12
r3,22
q3 = −4.0 C
−q1r3,22 −(8.0 C)(2.0 m)2
 = 
q2 = 
= −32 C
r3,12
(1.0 m)2
r1,2 = 1.0 m
r3,1 = 1.0 m
r3,2 = 2.0 m
9. q1 = 49 C
− k q3q1 kCq3q2
F3,1 + F3,2 = C
− 
=0
r3,12
r3,22
q3 = −7.0 C
−q1r3,22 −(49 C)(25.0 m)2
 = 
= −94.5 C
q2 = 
r3,12
(−18.0 m)2
r1,2 = 7.0 m
r3,1 = −18.0 m
10. q1 = 72 C
− k q3q1 kCq3q2
F3,1 + F3,2 = C
− 
=0
r3,22
r3,12
q3 = −8.0 C
r1,2 = 15 mm = 1.5 × 10−2 m
r3,1 = −9.0 mm = −9.0 × 10−3 m
−(72 C)(2.4 × 10−2 m)2
−q1r3,22


q2 = 
=
= −512 C
(−9.0 × 10−3 m)2
r3,12
r3,2 = 2.4 × 10−2 m
Additional Practice 17D
1. Ex = 9.0 N/C
Felectric

Ex = 
q
q = −6.0 C
Felectric = Exq = (9.0 N/C)(−6.0 C)
Felectric = −54 N in the −x direction
2. Ey = 1500 N/C
V
−9
q = 5.0 × 10
C
Felectric

Ey = 
q
Felectric = Eyq = (1500 N/C)(5.0 × 10−9 C)
Felectric = 7.5 × 10−6 N in the +y direction
V Ch. 17–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
r3,2 = 25.0 m
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Givens
3. m = 3.35 × 10−15 kg
q = 1.60 × 10−19 C
Solutions
a. Felectric = −Fgravity = −mg = −(3.35 × 10−15 kg)(9.81 m/s2)
Felectric = −3.29 × 10−14 N upward
g = 9.81 m/s2
4. qq = 3.00 × 10−6 C
q2 = 3.00 × 10−6 C
r2 = (2
)2
+(2.
m
)2 = 2.02 m
.0
0m
00
kC = 8.99 × 109 N • m2/C2
9
2 2
−6
k q1 (8.99 × 10 N • m /C )(3.00 × 10 C)

E1 = Ey = Ey = C
=
(0.250 m)2
r12
r1 = 0.250 m
E1 = Ey,1 = 4.32 × 105 N/C
k q2 (8.99 × 109 N • m2/C2)(3.00 × 10−6 C)
E2 = C
= 
= 6.61 × 103 N/C
r22
(2.02 m)2
y
0.250 m
j = tan−1  = tan−1  = 7.12°
x
2.00 m
Ex,2 = E2 cos 7.12° = (6.61 × 103 N/C)(cos 7.12°) = 6.56 × 103 N/C
Ey,2 = E2 sin 7.12° = (6.61 × 103 N/C)(sin 7.12°) = 8.19 × 103 N/C
Ex,tot = 0 N/C + 6.56 × 103 N/C = 6.56 × 103 N/C
Ey,tot = 4.32 × 105 N/C + 8.19 × 103 N/C = 4.40 × 105 N/C
2
Etot = (E
+(E
)2
x,to
y,t
ot
t Etot = (6
03
N/C
)2
+(4.
05
N/C
)2
.5
6×1
40
×1
Etot = 4.40 × 105 N/C
E tot 4.40 × 105 N/C
= 
tan j = y,
Ex,tot 6.56 × 103 N/C
j = 89.1°
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. q1 = 1.50 × 10−5 C
q2 = 5.00 × 10−6 C
k q1 (8.99 × 109 N • m2/C2)(1.50 × 10−5 C)
= 
= 1.35 × 105 N/C
E1 = Ey,1 = C
(1.00 m)2
r12
r1 = 1.00 m
k q2 (8.99 × 109 N • m2/C2)(5.00 × 10−6 C)
= 
= 1.80 × 105 N/C
E2 = Ey,2 = C
(0.500 m)2
r22
r2 = 0.500 m
Ey,tot = Etot = 1.35 × 105 N/C + 1.80 × 105 N/C = 3.15 × 105 N/C
kC = 8.99 × 109 N • m2/C2
The electric field points along the y-axis.
V
Section Five—Problem Bank
V Ch. 17–9
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Givens
Solutions
6. E = 1663 N/C
−9
Felectric = 8.4 × 10
Felectric

E=
q
N
Felectric
8.42 × 10−9 N
q =  =  = 5.06 × 10−12 C
E
1663 N/C
7. E = 4.0 × 103 N/C
−9
Felectric = 6.43 × 10
Felectric

E= 
q
N
Felectric
6.43 × 10−9 N
q =  = 
= 1.61 × 10−12 C
E
4.0 × 103 N/C
8. q1 = −1.60 × 10−19 C
9
2
2
kC = 8.99 × 10 N • m /C
q2 = −1.60 × 10−19 C
−19
C
3 = 1.60 × 10
−10
r1 = 3.00 × 10 m
r2 = 2.00 × 10−10 m
q
k q k q2 kCq3
Fx,tot = C21 + C
+ 
=0
r
r22
x2
q1 = q2 = −q3
1
1
k q
k q  +  = 
r r x
1
1 1
kCq1 2 + 2 − 2 = 0
r1 r2 x
C 1
1
2
2
2
C 1
2
1
1
x2 =  = 
1
1
1
1
2 + 2

+ 
r1 r2
(3.00 × 10−10 m)2 (2.00 × 10−10 m)
x = 1.66 × 10−10 m
9. q1 = −9.00 C
q2 = 6.00 C
q3 = 3.00 C
The electric field of both q2 and q1 point toward q1. To balance the electric field, q3
must be placed opposite of q2. Also, because the electric field of q2 points in the −x
direction, E2,x is negative.
Ex,tot = Ex − E2,x + E3,x
r1 = 1.5 mm
kC = 8.99 × 109 N • m2/C2
r1 = r2
kC (q1 − q2) kCq3

+ 
=0
x2
r12
(q1 − q2) −q3

= 
r12
x2
−(3.00 C)(1.5 mm)2
−q r12
x2 = 3
=  = 0.45 mm2
(−9.00 C − 6.00 C)
(q1 − q2)
x = ±0.67 mm Since q3 must be opposite q2, and the position of q3 is positive, q2 must
be on the negative x-axis, so x = −0.67 mm
V
V Ch. 17–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
k q1 kCq2 kCq3
Ex,tot = C
− 
+ 
=0
x2
r12
r22
r2 = 1.5 mm
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Givens
Solutions
10. q1 = 5.50 × 10−8 C
q2 = 1.10 × 10−8 C
The electric field of both q2 and q1 point toward q1. To balance the electric field, q3
must be on the side opposite q2. Also, because q2 points in the −x direction, E2,x is
negative.
q3 = 5.00 × 10−9 C
Ex,tot = E1,x − E2,x + E3,x
−7
r1 = −5.00 × 10
−7
r2 = 5.00 × 10
m
m
kC = 8.99 × 109 N • m2/C2
k q1 kCq1 kCq3
Ex,tot = C
+ 
+ 
=0
x2
r12
r22
r1 = r2, so r12 = r22
kCq1 kCq2 −kCq3

− 
= 
r12
r22
x2
q1 − q2 −q3

= 
r12
x2
−q r12
−(5.00 × 10−9 C)(−5.00 × 10−7 m)2
x2 = 3
= 
= 1.89 × 10−14 m2
(q1q2)
(−5.50 × 10−8 C − 1.10 × 10−8 C)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
x = −1.37 × 10−7 m
V
Section Five—Problem Bank
V Ch. 17–11
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Electrical Energy and Capacitance
Chapter 18
Additional Practice 18A
Givens
1. q = 1.45 × 10−8 C
E = 105 N/C
d = 290 m
2. q1 = 6.4 × 10−8 C
q2 = −6.4 × 10−8 C
r = 0.95 m
Solutions
PEelectric = −qEd = −(1.45 × 10−8 C)(−105 N/C)(290 m)
PEelectric = 4.42 × 10−4 J
kc q1q2

PEelectric = 
r
(8.99 × 109 N•m2/C2)(6.4 × 10−8 C)(−6.4 × 10−8 C)
PEelectric = 
0.95 m
PEelectric = −3.9 × 10−5 J
3. q = 1.60 × 10−19 C
PEelectric = −qEd
E = 3.0 × 106 N/C
PEelectric = −(1.60 × 10−19 C)(3.0 × 106 N/C)(7.3 × 10−7 m)
d = 7.3 × 10−7 m
PEelectric = 3.5 × 10−19 J
4. q1 = −4.2 × 10−8 C
q2 = 6.3 × 10−8 C
PEelectric = −6.92 × 10−4 J
kc q1q2

r= 
PEelectric
(8.99 × 109 N•m2/C2)(−4.2 × 10−8 C)(6.3 × 10−8 C)
r = 
−6.92 × 10−4 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
r = 3.4 × 10−2 m = 0.34 cm
5. q1 = 1.6 × 10−8 C
q2 = 1.4 × 10−8 C
PEelectric = 2.1 × 10−6 J
kc q1q2

r= 
PEelectric
(8.99 × 109 N•m2/C2)(1.6 × 10−8 C)(1.4 × 10−8 C)
r = 
2.1 × 10−6 J
r = 9.5 × 10−1 m = 96 cm
6. q1 = −5.5 × 10−8 C
q2 = 7.7 × 10−8 C
PEelectric = −1.3 × 10−2 J
kc q1q2

r= 
PEelectric
(8.99 × 109N•m2/C2)(−5.5 × 10−8 C)(7.7 × 10−8 C)
r = 
−1.3 × 10−2 J
r = 2.9 × 10−3 m = 2.9 mm
V
Section Five—Problem Bank
V Ch. 18–1
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Print
Givens
Solutions
7. PEelectric = −1.39 × 1011 J
E = 3.4 × 105 N/C
−(−1.39 × 1011 J)
−PE lectric
q = e
= 
(3.4 × 105 N/C)(7300 m)
Ed
d = 7300 m
q = 56 C
q1 = q2
rPE ectric
q2 = el
kc
8. r = 1.25 × 10−3 m
PEelectric = 1.25 × 10−4 J
q=
(1.25 × 10−3 m)(1.25 × 10–4 J)

(8.99 × 109N•m2/C2
rPEelectric
 =
kc
q = 4.17 nC
a. q1 = 4.2 nC
b. q2 = −4.2 nC
c. It is unlikely that the hat would stay on a friend’s head only by charge because
the charge on the friend would most likely be different than the hat.
q1 = q2
9. r = 9.4 × 10−4 m
−10
PEelectric = 8.89 × 10
J
rPE ectric
q2 = el
kc
(9.4 × 10−4 m)(8.89 × 10–10 J)

(8.99 × 109 N•m2/C2)
q=
rPEelectric
 =
kc
q = 9.6 × 10−12 C
a. q1 = 9.6 × 10−12 C
b. q2 = −9.6 × 10−12 C
PEelectric = 6.3 × 10−6 J
q1 = q2
rPE ectric
q2 = el
kc
q=
(1.25 × 10−6 m)(6.3 × 10−6 J)

(8.99 × 109 N•m2/C2)
rPEelectric
 =
kc
q = 2.9 × 10−11 C
a. q1 = 2.9 × 10−11 C
b. q2 = −2.9 × 10−11 C
Additional Practice 18B
1. ∆V = 114.0 V
r = 6.695 × 106 m
2. ∆V = 18600 V
V
r = 1991 m
V Ch. 18–2
r∆V (6.695 × 106 m)(114.0 V)
= 8.49 × 10−2 C
q =  = 
8.99 × 109 N•m2/C2
kc
r∆V (1991 m)(18600 V)
q =  = 
= 4.12 × 10−3 C
kc
8.99 × 109 N•m2/C2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
10. r = 1.25 × 10−6 m
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Givens
Solutions
3. V = 1.0 V
–19
q = –1.60 × 10
C
k q (8.99 × 109 N•m2/C2)(−1.60 × 10−19 C)
r = c = 
1.0 V
V
r = 1.4 × 10−9 m
k q (8.99 × 109 N•m2/C2)(9.4 × 10−8 C)
r = c = 
9.0 V
V
4. V = 9.0 V
q = 9.4 × 10−8 C
r=
5. q = 1.28 × 10−18 C
93.4 m
k q (8.99 × 109 N•m2/C2)(1.28 × 10−18 C)
∆V = c = 
3.95 × 10−2 m
r
r = 3.95 × 10−2 m
∆V = 2.91 × 10−7 V
∆V = –E∆d = −(3.0 × 106 N/C)(6.25 × 10−4 m)
6. E = 3.0 × 106 N/C
∆d = 6.25 × 10−4 m
7. E = 95 N/C
∆V = −1.9 × 103 V
∆V = −E∆d = −(95 N/C)(3.0 × 102 m)
2
∆d = 3.0 × 10 m
∆V = −2.8 × 104 V
8. q1 = 8(1.60 × 10−19 C)
= 1.28 × 10−18 C
q 105°
f =  =  = 52.5°
2
2
r1 = dcosf = (9.58 × 10−11 m) cos (52.5°) = 5.83 × 10−11 m
r2 = dsinf = (9.58 × 10−11 m) sin (52.5°) = 7.60 × 10−11 m
q2 = 1.60 × 1−19 C
d = 9.58 × 10−11 m
kq
2k q
q 2q
V = c 1 + c2 = kc 1 + 2
r1
r2
r1 r2
q = 105°
1.28 × 10−18 C 2(1.60 × 10−19 C)
+ 
V = (8.99 × 109 N•m2/C2) 
5.83 × 10−11 m
7.60 × 10−11 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
V = 197 V + 37.9 V = 235 V
9. q1 = 3.04 × 10−18 C
q2 = 5.60 × 10−18 C
r1 = 1.89 × 10−10 m
r2 = −9.30 × 10−11 m
10. q1 = 1.60 × 10−19 C
q2 = −1.60 × 10−19 C
q3 = 1.60 × 10−19 C
−19
q4 = −1.60 × 10
−10
d = 2.82 × 10
m
C
kq kq
q q
V = c 1 + c2 = kc 1 + 2
r1
r2
r1 r2
3.04 × 10−18 C
5.60 × 10−18 C
V = (8.99 × 109 N•m2/C2) 
+ 
−10
1.89 × 10 m −9.30 × 10−11 m
V = 145 V − 541 V = −396 V
r = r1 + r2 + r3 + r4 = (0.707)d = (0.707)(2.82 × 10−10 m)
r = 1.99 × 10−10 m
k
k
V = c [q1 + q2 + q3 + q4] = c [2q1 + 2q2]
r
r
8.99 × 109 N•m2/C2
V = 
[2(1.60 × 10−19 C) + 2(−1.60 × 10−19 C)]
1.99 × 10−10 m
V= 0V
V
Section Five—Problem Bank
V Ch. 18–3
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Additional Practice 18C
Givens
Solutions
1. A = 4.8 × 10−3 m2
C = 1.8 × 10−5 F
−12 2
2
−3 2
ke A (1)(8.85 × 10 C /N•m )(4.8 × 10 m )
d = 0 = 
−5
1.8 × 10 F
C
k=1
d = 2.4 × 10−9 m
2. A = 6.4 × 10−3 m2
C = 4.55 × 10−9 F
e A (8.85 × 10−12 C2/N•m2)(6.4 × 10−3 m2)
d = 0 = 
4.55 × 10−9 F
C
d = 1.2 × 10−5 m
3. R = 6.4 × 106 m
R
6.4 × 106 m
Csphere =  = 9
= 7.1 × 10−4 F
kc 8.99 × 10 N•m2/C2
4. R = 0.10 m
R
0.10 m
= 1.1 × 10−11 F
Csphere =  = 9
kc 8.99 × 10 N•m2/C2
5. C = 1.4 × 10−5 F
Q = C∆V = (1.4 × 10−5 F)(1.5 ⫻ 104 V) = 0.21 C
∆V = 1.5 × 104 V
6. C = 1.0 × 10−9 F
∆V = 600 V
7. C = 5.0 × 10−13 F
Q2
(0.21 C)2
PEelectric =  = 
= 1.6 × 103 J
2C 2(1.4 × 10−5 F)
Q = C∆V = (1.0 × 10−9 F)(600 V) = 6.0 × 10−7 C
Q 2 (6.0 × 10−7 C)2
PEelectric =  = 
= 1.8 × 10−4 J
2C 2(1.0 × 10−9 F)
Q = C∆V = (5 × 10−13 F)(1.5 V) = 7.5 × 10−13 C
8. C = 1.0 × 10−6 F
Q = 3.0 × 10−2 C
9. C = 2.0 × 10−6 F
Q = 4.0 × 10−4 C
10. C = 5.0 × 10−5 F
Q = 6.0 × 10−4 C
Q 3.0 × 10−2 C
∆V =  = 
= 3.0 × 104 V = 30 kV
C
1.0 × 10−6 F
Q 4.0 × 10−4 C
∆V =  = 
= 2.0 × 102 V
C
2.0 × 10−6 F
Q 6.0 × 10−4 C
∆V =  = 
= 12 V
C
5.0 × 10−5 F
V
V Ch. 18–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆V = 1.5 V
Print
Current and Resistance
Chapter 19
Additional Practice 19A
Givens
Solutions
1. ∆Q = 76 C
∆Q 76 C
I =  =  = 4.0 A
∆t
19 s
∆t = 19 s
2. ∆Q = 1.14 × 10−4 C
∆t = 0.36 s
3. ∆Q = 2.9 × 10−2 C
∆t = 11 s
4. ∆Q = 98 C
∆Q 1.14 × 10−4 C
I =  =  = 0.32 mA
∆t
0.36 s
∆Q 2.9 × 10−2 C
I =  =  = 2.6 mA
∆t
11 s
∆Q
98 C
∆t =  =  = 70 s
I
1.4 A
I = 1.4 A
5. ∆Q = 30.9 C
∆Q 30.9 C
∆t =  =  = 3.20 s
I
9.65 A
I = 9.65 A
6. ∆Q = 56 C
∆Q
56 C
∆t =  =  = 7.2 s
I
7.8 A
I = 7.8 A
∆Q = I∆t = (9.3 A) (15 s) = 1.4 × 102 C
7. ∆t = 15 s
I = 9.3 A
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8. ∆t = 2.0 min = 120 s
∆Q = I∆t = (3.0 A) (120 s) = 3.6 × 102 C
I = 3.0 A
9. ∆t = 2.0 s
∆Q = I∆t = (0.70 A) (2.0 s) = 1.4 C
I = 0.70 A
10. ∆t = 4.3 s
∆Q = I∆t = (5.6 A) (4.3 s) = 24 C
I = 5.6 A
Additional Practice 19B
1. R = 1.0 × 105 Ω
−3
I1 = 1.0 × 10
I2 = 1.5 × 10
−2
A
∆V1 = I1R = (1.0 × 10−3 A) (1.0 × 105 Ω) = 1.0 × 102 V
∆V2 = I2R = (1.5 × 10−2 A) (1.0 × 105 Ω) = 1.5 × 103 V
A
V
Section Five—Problem Bank
V Ch. 19–1
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Givens
Solutions
2. I = 0.75 A
∆V = IR = (0.75 A) (6.4 Ω) = 4.8 V
R = 6.4 Ω
3. I = 4.66 A
∆V = IR = (4.66 A) (25.0 Ω) = 116 V
R = 25.0 Ω
4. ∆V = 120 V
120 V
∆V
I =  =  = 9.84 A
12.2 Ω
R
5. ∆V = 650 V
2
R = 1.0 × 10 Ω
6. R = 40.0 Ω
∆V1 = 120 V
V2 = 240 V
7. I = 0.75 A
∆V = 120 V
8. I = 0.89 A
∆V = 5.00 × 102 V
9. I = 0.545 A
∆V = 120 V
10. I = 0.65 A
∆V = 117 V
650 V
∆V
= 6.5 A
I =  = 
1.0 × 102 Ω
R
120 V
∆V
I1 = 1 =  = 3.00 A
40.0 Ω
R
240 V
∆V
I2 = 2 =  = 6.00 A
40.0 Ω
R
120 V
∆V
R =  =  = 1.6 × 102 Ω
0.75 A
I
∆V 5.00 × 102 V
R =  =  = 5.6 × 102 Ω
I
0.89 A
120 V
∆V
R =  =  = 220 Ω
0.545 A
I
117 V
∆V
R =  =  = 1.8 × 102 Ω
0.65 A
I
Additional Practice 19C
1. ∆V = 2.5 × 104 V
P = I∆V = (20.0 A) (2.5 × 104 V) = 5.0 × 105 W
I = 20.0 A
2. ∆V = 720 V
R = 0.30 Ω
3. ∆V = 120 V
R1 = 144 Ω
R2 = 240 Ω
(∆V )2 (720 V)2
P =  =  = 1.7 × 106 W
R
0.30 Ω
(∆V )2 (120 V)2
P1 =  =  = 100 W
R1
144 Ω
2
(∆V )
(120 V)2
P2 =  =  = 60.0 W
R2
240 Ω
The brighter bulb is the 60.0-W bulb, which has the 240−Ω resistance.
V
4. ∆V = 120 V
P = 1750 W
V Ch. 19–2
(∆V )2 (120 V)2
R =  =  = 8.22 Ω
P
1750 W
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
R = 12.2 Ω
Solutions
Print5. ∆V = 120 V
(∆V )2 (120 V)2
R =  =  = 22.2 Ω
P
650 W
P = 650 W
6. ∆V = 120 V
(∆V )2 (120 V)2
R =  =  = 38.9 Ω
P
370 W
P = 370 W
7. P = 350 W
R = 75 Ω
8. P = 230 W
R = 91 Ω
9. I = 8.0 × 106 A
P
I 2 = 
R
I=
=
R = 
75
Ω
2.16 A
P
I 2 = 
R
I=
=
R = 
91
Ω
1.59 A
P
P
350 W
230 W
P 6.0 × 1013 W
= 7.5 × 106 V
∆V =  = 
I
8.0 × 106 A
13
P = 6.0 × 10 W
10. I = 16.3 A
P 1.06 × 104 W
∆V =  =  = 6.50 × 102 V
I
16.3 A
4
P = 1.06 × 10 W
Additional Practice 19D
1. ∆t = 1.0 h
8
Energy = 2.7 × 10 J
2. ∆t = 3.0 h
Energy = 4.86 × 108 J
3. P = 1200 W = 1.200 kW
Energy = 1.512 × 108 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Givens
4. P = 600 W = 0.600 kW
9
Energy = 8.64 × 10 J
5. Cost of electricity
= $0.0650/kW • h
Energy = 200.0 kW • h
6. Cost of electricity
= $0.078/kW • h
Final Meter Reading
= 24422 kW • h
Previous Meter Reading
= 24204 kW • h
2.7 × 108 J
Energy
P =  = 
(1.0 h) (3.6 × 106 J/kW • h)
∆t
= 75 kW
4.86 × 108 J
Energy
P =  = 
(3.0 h) (3.6 × 106 J/kW • h)
∆t
= 45 kW
1.512 × 108 J
Energy
∆t =  = 
(1.200 kW) (3.6 × 106 J/kW • h)
P
= 35.00 h
8.64 × 109 J
Energy
∆t =  = 
(0.600 kW) (3.6 × 106 J/kW • h)
P
= 4.00 × 103 h
Cost = (Energy) ($0.065/kW • h)
Cost = (200.0 kW • h) ($0.0650/kW • h)
= $13.00
Energy = Final Meter Reading – Previous Meter Reading
Energy = 24422 kW • h – 24204 kW • h = 218 kW • h
Cost = (Energy) ($0.078/kW • h)
Cost = (218 kW • h) ($0.078/kW • h)
= $17.00
V
Section Five—Problem Bank
V Ch. 19–3
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Givens
Solutions
7. ∆t = 8.0 h
Energy = P∆t = (0.125 kW) (8.0 h) = 1.0 kW • h
P = 0.125 kW
8. ∆t = 24.0 h
Energy = P∆t = (0.75 kW) (24.0 h) (3.6 × 106 J/kW • h) = 6.5 × 107 J
P = 0.75 kW
9. ∆t = 10.0 min = 0.167 h
Energy = P∆t = (0.55 kW) (0.167 h) (3.6 × 106 J/kW • h) = 3.3 × 105 J
P = 0.55 kW
10. ∆t = 3.0 min
= 5.0 × 10–2 h
Energy = 1.5 × 105 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
P = 0.85 kW
Energy = P∆t = (0.85 kW) (5.0 × 10–2h) (3.6 × 106 J/kW • h)
V
V Ch. 19–4
Holt Physics Solution Manual
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Circuits and Circuit Elements
Chapter 20
Additional Practice 20A
Givens
1. ∆V = 12 V
R1 = 16 Ω
Solutions
∆V
12 V
R2 =  − R1 =  − 16 Ω = 29 Ω − 16 Ω = 13 Ω
I
0.42 A
I = 0.42 A
2. ∆V = 3.0 V
R1 = 24Ω
∆V
3.0 V
R2 =  − R1 =  − 24 Ω = 48 Ω − 24 Ω = 24 Ω
I
0.062 A
I = 0.062 A
3. ∆V = 9.0 V
R1 = 9.1 Ω
9.0 V
∆V
R2 =  − R1 =  − 9.1 Ω = 27 Ω − 9 Ω = 18 Ω
0.33 A
I
I = 0.33 A
4. 73 bulbs
Reach bulb = 3.0 Ω
5. 25 speakers
Reach speaker = 12.0 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. 57 lights
Reach light = 2.0 Ω
7. 4 speakers
Req = ΣReach bulb All bulbs have equal resistance.
Req = (73) (3.0 Ω) = 219 Ω
Req = ΣReach speaker All speakers have equal resistance.
Req = (25)(12.0 Ω) = 3.0 × 102 Ω
Req = ΣReach light All lights have equal resistance.
Req = (57)(2.0 Ω) = 114 Ω
Req = ΩReach speaker All speakers have equal resistance.
Reach speaker = 4.1 Ω
Req = (4)(4.1 Ω) = 16.4 Ω
∆V = 12 V
∆V
12 V
I =  =  = 7.3 × 10−1 A
Req 16.4 Ω
8. 10 bulbs
Req = ΣReach bulb All bulbs have equal resistance.
Reach bulb = 10 Ω
Req = (10)(10 Ω) = 100 Ω
∆V = 115 V
∆V 100 V
I =  =  = 1 A
Req 100 Ω
9. R1 = 96 Ω
R2 = 48 Ω
R3 = 29 Ω
Req = ΣR = R1 + R2 + R3 = 96 Ω + 48 Ω + 29 Ω = 173 Ω
∆V
115 V
I =  =  = 665 mA
Req 173 Ω
V
∆V = 115 V
Section Five—Problem Bank
V Ch. 20–1
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Givens
Solutions
10. R1 = 56 Ω
Req = ΣR = R1 + R2 + R3 = 56Ω + 82Ω + 24Ω = 162Ω
R2 = 82 Ω
R3 = 24 Ω
∆V
9.0 V
I =  =  = 55.6 mA
Req 162 Ω
∆V = 9.0 V
Additional Practice 20B
1. ∆V = 3.0 V
R1 = 3.3 Ω
I = 1.41 A
∆V = IReq
∆V ∆V ∆V
I =  =  + 
Req
R1
R2
∆V
∆V
 = I − 
R2
R1
3.0 V
∆V
3.0 V
R2 =  =  =  = 6.0 Ω
[1.41 A − 0.91 A]
∆V
3.0 V
I − 
1.41 A − 
R1
3.3 Ω
2. ∆V = 12 V
R1 = 56 Ω
I = 3.21 A
∆V = IReq
∆V ∆V ∆V
I =  =  + 
Req
R1
R2
∆V
∆V
 = I − 
R2
R1
12 V
∆V
12 V
R2 =  =  =  = 4.0 Ω
[3.21 A − 0.21 A]
∆V
12 V
I − 
3.21 A − 
R1
56 Ω
R1 = 18 Ω
I = 0.103 A
∆V = IReq
∆V ∆V ∆V
I =  =  + 
Req
R1
R2
∆V
∆V
 = I − 
R2
R1
1.5 V
∆V
1.5 V
R2 =  =  =  = 75 Ω
[0.103
A
− 0.083 A]
∆V
1.5 V
I − 
0.103 A − 
R1
18 Ω
4. R1 = 39 Ω
R2 = 82 Ω
R3 = 12 Ω
R4 = 22 Ω
∆V = 3.0 V
1
1
1
1
1
1
1
1
1
 =  +  +  +  =  +  +  + 
Req R1 R2 R3 R4 39 Ω 82 Ω 12 Ω 22 Ω
1 0.026 0.012 0.083 0.045 0.17
 =  +  +  +  = 
Req 1 Ω
1Ω
1Ω
1Ω
1Ω
Req = 6.0 Ω
V
V Ch. 20–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. ∆V = 1.5 V
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Givens
5. R1 = 10.0 Ω
R2 = 12 Ω
Solutions
1
1
1
1
1
1
1
1
1
 =  +  +  +  =  +  +  + 
Req R1 R2 R3 R4 10.0 Ω 12 Ω 15 Ω 18 Ω
R3 = 15 Ω
1 0.10 0.083 0.067 0.056
 =  +  +  + 
Req 1 Ω
1Ω
1Ω
1Ω
R4 = 18 Ω
Req = 3.3 Ω
∆V = 12 V
6. R1 = 33 Ω
R2 = 39 Ω
R3 = 47 Ω
R4 = 68 Ω
V = 1.5 V
7. ∆V = 120 V
R1 = 75 Ω
R2 = 91 Ω
1
1
1
1
1
1
1
1
1
 =  +  +  +  =  +  +  + 
Req R1 R2 R3 R4 33 Ω 39 Ω 47 Ω 68 Ω
1
0.030
0.026
0.021
0.015
 =  +  +  + 
Req
1Ω
1Ω
1Ω
1Ω
Req = 11 Ω
V
I1 = 
R1
V
I2 = 
R2
120 V
I1 =  = 1.6 A
75 Ω
120 V
I2 =  = 1.3 A
91 Ω
8. ∆V = 120 V
R1 = 82 Ω
R2 = 24 Ω
V
I1 = 
R1
V
I2 = 
R2
120 V
I1 =  = 1.5 A
82 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
120 V
I2 =  = 5.0 A
24 Ω
9. ∆V = 120 V
R1 = 11 Ω
R2 = 36 Ω
V
I1 = 
R1
V
I2 = 
R2
120 V
I1 =  = 11 A
11 Ω
120 V
I2 =  = 3.3 A
36 Ω
10. ∆V = 1.5 V
R1 = 3.3 Ω
R2 = 4.3 Ω
V
I1 = 
R1
V
I2 = 
R2
1.5 V
I1 =  = 0.45 A
3.3 Ω
1.5 V
I2 =  = 0.35 A
4.3 Ω
V
Section Five—Problem Bank
V Ch. 20–3
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Additional Practice 20C
Givens
1. I = 0.680 A
Solutions
Series:
V = 15.0 V
Group (a): Req,a = R1 + R2 = 15 Ω + 11 Ω = 26 Ω
R1 = 15 Ω
Group (b): Req,b = R3 + R4 = 6.0 Ω + 7.0 Ω = 13.0 Ω
R2 = 11 Ω
Parallel:
R3 = 6.0 Ω
R4 = 7.0 Ω
1
1
1
0.038 0.077 0.12
Groups (a) + (b):  =  +  =  +  = 
Req 26 Ω 13.0 Ω 1 Ω
1Ω
1Ω
Req = 8.3 Ω
R5 = 12.0 Ω
Series:
Req = 12.0 Ω + 8.3 Ω + R = 20.3 Ω + R
∆V = IReq = I(20.3 Ω + R)
15.0 V
∆V
R =  − 20.3 Ω =  − 20.3 Ω = 1.8 Ω
0.680 A
I
2. I = 0.375 A
Series:
V = 9.00 V
Group (a): Req,a = R2 + R3 = 3.0Ωeq + 4.0Ωeq = 7.0Ω
R1 = 18.0 Ω
Group (b): Req,b = R5 + R6 = 20.0 Ω + 22.0 Ω = 42.0 Ω
R2 = 3.0 Ω
Parallel:
R3 = 4.0 Ω
1
1
1
0.0556 0.143
Group (c):  =  +  =  + 
Req 18.0 Ω 7.0 Ω
1Ω
1Ω
R4 = 8.00 Ω
Req = 5.03 Ω
R5 = 20.0 Ω
R6 = 22.0 Ω
R7 = 6.0 Ω
1
1
1
0.0238 0.167
Group (d):  =  +  =  + 
Req 42.0 Ω 6.0 Ω
1Ω
1Ω
Req = 5.24 Ω
Group (e): Req,e = Req,c + R4 + Req,d + R
= 5.03 Ω + 8.00 Ω + 5.24 Ω + R = 18.27 Ω + R
∆V
9.00 V
R =  − 18.27 Ω =  − 18.27 Ω = 5.7 Ω
I
0.375 A
V
V Ch. 20–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Series:
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Givens
3. I = 0.185 A
Solutions
Series:
V = 12.0 V
Group (a): Req,a = R1 + R2 = 2.0 Ω + 3.0 Ω = 5.0 Ω
R1 = 2.0 Ω
Group (b): Req,b = R3 + R4 = 3.0 Ω + 4.0 Ω = 7.0 Ω
R2 = 3.0 Ω
Group (c): Req,c = R6 + R7 + R8 = 2.0 Ω + 3.0 Ω + 4.0 Ω = 9.0 Ω
R3 = 3.0 Ω
R4 = 4.0 Ω
R5 = 8.0 Ω
R6 = 2.0 Ω
R7 = 3.0 Ω
R8 = 4.0 Ω
R9 = 4.0 Ω
R10 = 5.0 Ω
R11 = 6.0 Ω
Group (d): Req,d = R9 + R10 + R11 = 4.0 Ω + 5.0 Ω + 6.0 Ω = 15.0 Ω
Parallel:
1
1
1
1
1
0.20 0.14
Group (e):  =  +  =  +  =  + 
Req,e Req,a Req,b 5.0 Ω 7.0 Ω
1Ω
1Ω
Req,e = 2.9 Ω
1
1
1
1
1
0.111 0.067
Group (f):  =  +  =  +  =  + 
Req,f Req,c Req,d 9.0 Ω 15.0 Ω
1Ω
1Ω
Req,f = 5.6 Ω
Series:
Req = 4R5 + 2Req,e + Req,f + R = 4(8.0 Ω) + 2(2.9 Ω) + 5.6 Ω + R
Req = 32.0 Ω + 5.8 Ω + 5.6 Ω + R = 43.4 Ω + R
12.0 V
∆V
R =  − 43.4 Ω =  − 43.4 Ω = 21.5 Ω
0.185 A
I
4. each resistor = 10.0 Ω
Parallel:
1
0.100 0.100
1
1
Group (a):  =  +  =  + 
1Ω
1Ω
Req,a 10.0 Ω 10.0 Ω
Req,a = 5.00 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1
1
1
1
1
Group (b):  =  +  +  + 
Req,b
10.0 Ω 10.0 Ω 10.0 Ω 10.0 Ω
1
0.100 0.100 0.100
0.100
 =  +  +  + 
Req,b
1Ω
1Ω
1Ω
1Ω
Req,b = 2.50 Ω
Group (c): Req,c = 10.0 Ω + 10.0 Ω + 5.00 Ω + 2.50 Ω + 5.00 Ω + 2.50 Ω + 10.0 Ω + 10.0 Ω
Req,c = 55.0 Ω
5. each resistor = 2.0 Ω
Series:
Group (a): Req,a = 2.0 Ω + 2.0 Ω = 4.0 Ω
Group (b): Req,b = 2.0 Ω + 2.0 Ω + 2.0 Ω = 6.0 Ω
Group (c): Req,c = 2.0 Ω + 2.0 Ω + 2.0 Ω = 2.0 Ω = 8.0 Ω
Parallel:
1
1
1
1
1
0.50 0.25 0.17 0.13
Group (d):  =  +  +  +  =  +  +  + 
Req,d 2.0 Ω 4.0 Ω 6.0 Ω 8.0 Ω
1Ω 1Ω 1Ω 1Ω
Req,d = 0.95 Ω
Series:
Req = 2.0 Ω + 2.0 Ω + 0.95 Ω = 5.0 Ω
Section Five—Problem Bank
V
V Ch. 20–5
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Givens
Solutions
6. each resistor = 10.0 Ω
Parallel:
1
1
1
1
0.100 0.100 0.100
Group (a):  =  +  +  =  +  + 
Req,a 10.0 Ω 10.0 Ω 10.0 Ω
1Ω
1Ω
1Ω
Req,a = 3.33 Ω
Series:
Req = 4(10.0 Ω) + 3.33 Ω + 2(10.0 Ω) + 3.33 Ω + 2(10.0 Ω) + 3.33 Ω + 3(10.0 Ω)
Req = 1.20 × 102 Ω
7. ∆Vtot = 12.0 V
Parallel:
1
1
1
1
1
0.20 0.50
Group (a):  =  +  =  +  =  + 
Req,a R2 R3 5.0 Ω 2.0 Ω
1Ω
1Ω
R1 = 3.0 Ω
R2 = 5.0 Ω
Req,a = 1.4 Ω
R3 = 2.0 Ω
1
1
1
1
1
0.20 0.17
Group (b):  =  +  =  +  =  + 
Req,b R5 R6 5.0 Ω 6.0 Ω
1Ω
1Ω
R4 = 4.0 Ω
R5 = 5.0 Ω
Req,b = 2.7 Ω
R6 = 6.0 Ω
R7 = 1.5 Ω
Series:
Group (c): Req,c = R1 + Req,a + R4 + Req,b + R7
Req,c = 3.0 Ω + 1.4 Ω + 4.0 Ω + 2.7 Ω + 1.5 Ω = 13 Ω
∆Vtot 12.0 V
I = 
=  = 0.92 A
Req,c
13 Ω
8. ∆Vtot = 15.0 V
R2 = 5.0 Ω
R3 = 5.0 Ω
R4 = 3.0 Ω
1
1
1
1
1
1
1
0.20
Group (a):  =  +  +  =  +  +  =  3
Req,a R1 R2 R3 5.0 Ω 5.0 Ω 5.0 Ω
1Ω
Req,a = 1.7 Ω
Series:
Req = 2R4 + 2Req,a = (2)(3.0 Ω) + (2)(1.7 Ω) = 9.4 Ω
∆Vtot 15.0 V
=  = 1.6 A
I = 
Req
9.4 Ω
V
V Ch. 20–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
R1 = 5.0 Ω
Parallel:
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Givens
9. ∆Vtot = 24.0 V
Solutions
Series:
R1 = 4.0 Ω
Group (a): Req,a = R6 + R7 + R8 = 2.0 Ω + 4.0 Ω + 2.0 Ω = 8.0 Ω
R2 = 4.0 Ω
Group (b): Req,b = R9 + R10 = 4.0 Ω + 2.0 Ω = 6.0 Ω
R3 = 4.0 Ω
Parrallel:
R4 = 3.0 Ω
1
1
1
1
1
0.25 0.25
Group (c):  =  +  =  +  =  + 
Req,c R1 R2 4.0 Ω 4.0 Ω
1Ω 1Ω
R5 = 1.0 Ω
R6 = 2.0 Ω
R7 = 4.0 Ω
R8 = 2.0 Ω
R9 = 4.0 Ω
R10 = 2.0 Ω
R11 = 2.0 Ω
Req,c = 2.0 Ω
1
1
1
1
1
0.13 0.17
Group (d):  =  +  =  +  =  + 
Req,d Req,a Req,b 8.0 Ω 6.0 Ω
1Ω
1Ω
Req,d = 3.4 Ω
Series:
Group (e): Req,e = Req,c + R3 + R4 + R5 + Req,d + R11
Req,e = 2.0 Ω + 4.0 Ω + 3.0 Ω + 1.0 Ω + 3.4 Ω + 2.0 Ω = 15 Ω
∆Vtot 24.0 V
=  = 1.6 A
I = 
Req,e
15 Ω
10. ∆V = 24.0 V
R1 = 4.0 Ω
R2 = 8.0 Ω
R3 = 2.0 Ω
R4 = 5.0 Ω
R5 = 3.0 Ω
R6 = 2.0 Ω
R7 = 3.0 Ω
Parallel:
1
0.25 0.13
1
1
1
1
Group (a):  =  +  =  +  =   +  
1Ω
Req,a R1 R2 4.0 Ω 8.0 Ω 1 Ω
Req,a = 2.6 Ω
1
0.50 0.33
1
1
1
1
Group (b):  =  +  =  +  =  + 
1Ω
1Ω
Req,b R6 R7 2.0 Ω 3.0 Ω
Req,b = 1.2 Ω
Series:
Req = 2Req,a + R3 + R4 + R5 + Req,b = (2)(2.7 Ω) + 2.0 Ω + 5.0 Ω + 1.2 Ω + 3.0 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Req = 17 Ω
∆Vtot 24.0 V
I = 
=  = 1.4 A
Req
17 Ω
Additional Practice 20D
1. ∆Vtot = 12.0 V
R1 = 3.0 Ω
R2 = 10.0 Ω
R3 = 10.0 Ω
R4 = 10.0 Ω
R5 = 4.0 Ω
Parallel:
1
1
1
1
1
1
1
0.100 0.100 0.100
 =  +  +  =  +  +  =  +  + 
Req,a R2 R3 R4 10.0 Ω 10.0 Ω 10.0 Ω
1Ω
1Ω
1Ω
Req,a = 3.3 Ω
Series:
Req,b = R1 + Req,a + R5 = 3.0 Ω + 3.3 Ω + 4.0 Ω = 10 Ω
∆Vtot 12.0 V
=  = 1.2 A
I = 
Req,b
10 Ω
∆V = IR5
∆V = (1.2 A)(4.0 Ω) = 4.8 V
V
Section Five—Problem Bank
V Ch. 20–7
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Givens
Solutions
2. ∆Vtot = 1.5 V
R1 = 6.0 Ω
Parallel:
R2 = 2.0 Ω
1
1
1
1
1
0.125 0.111
 =  +  =  +  =  + 
Req,a R4 R5 8.0 Ω 9.0 Ω
1Ω
Ω
R3 = 4.0 Ω
Req,a = 4.2 Ω
R4 = 8.0 Ω
Series:
R5 = 9.0 Ω
Req = R1 + R2 + R3 + Req,a = 6.0 Ω + 2.0 Ω + 4.0 Ω + 4.2 Ω = 16 Ω
∆Vtot 1.5 V
I = 
=  = 0.094 A
Req
16 Ω
∆V = IReq,a = (0.094 A)(4.2 Ω) = 0.39 V
∆V 0.39 V
I =  =  = 0.043 A
R5
9.0 Ω
3. ∆Vtot = 9.0 V
R1 = 5.0 Ω
Parallel:
R2 = 4.0 Ω
1
1
1
1
1
0.14 0.17
 =  +  =  +  =  + 
Req,a R3 R4 7.0 Ω 6.0 Ω
1Ω
1Ω
R3 = 7.0 Ω
Req,a = 3.2 Ω
R4 = 6.0 Ω
Series:
R5 = 3.0 Ω
Req = R1 + R2 + Req,a + R5 + R6 = 5.0 Ω + 4.0 Ω + 3.2 Ω + 3.0 Ω + 2.0 Ω = 17 Ω
R6 = 2.0 Ω
∆Vtot 9.0 V
I = 
=  = 0.52 A
Req
17 Ω
∆V = I Req,a = (0.52 A)(3.2 Ω) = 1.7 V
∆V
1.7 V
I =  =  = 0.28 A
R4
6.0 Ω
Series:
R1 = 7.0 Ω
Req,a = R2 + R3 + R4 = 5.0 Ω + 4.0 Ω + 3.0 Ω = 12.0 Ω
R2 = 5.0 Ω
Req,b = R5 + R6 + R7 = 2.0 Ω + 10.0 Ω + 6.0 Ω = 18.0 Ω
R3 = 4.0 Ω
Parallel:
R4 = 3.0 Ω
R5 = 2.0 Ω
1
1
1
1
1
0.083 0.056
 = 
  +  =  +  =  + 
Req,b Req,a Req,b 12.0 Ω 18.0 Ω
1Ω
1Ω
R6 = 10.0 Ω
Req,b = 7.2 Ω
R7 = 6.0 Ω
Series:
Req,d = R1 + Req,c = 7.0 Ω + 7.2 Ω = 14 Ω
∆Vtot 9.0 V
I = 
=  = 0.63 A
Req
14 Ω
∆V = IReq,c = (0.63 A)(7.2 Ω) = 4.5 V
4.5 V
∆V
I =  =  = 0.25 A
18.0 Ω
R6
V
V Ch. 20–8
∆V = IR6 = (0.25 A)(10.0 Ω) = 2.5 V
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. ∆Vtot = 9.0 V
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Givens
5. ∆Vtot = 3.0 V
Solutions
Series:
R1 = 2.0 Ω
Req,a = R1 + R2 + R3 = 2.0 Ω + 4.0 Ω + 6.0 Ω = 12 Ω
R2 = 4.0 Ω
Req,b = R4 + R5 = 1.0 Ω + 3.0 Ω = 4.0 Ω
R3 = 6.0 Ω
Parallel:
R4 = 1.0 Ω
R5 = 3.0 Ω
1
1
1
1
1
1
0.083 0.25 0.20
1
 =  +  +  =  +  +  =  +  + 
Req,c Req,a Req,b Req 12 Ω 4.0 Ω 5.0 Ω
1Ω
1Ω 1Ω
R6 = 5.0 Ω
Req,c = 1.9 Ω
3.0 V
∆Vtot
I = 
=  = 1.6 A
1.9 Ω
Req,c
∆V
3.0 V
I =  =  = 0.25 A
Req,a
12 Ω
∆V = IR2 = (0.25 A)(4.0 Ω) = 1.0 V
6. ∆Vtot = 3.0 V
Series:
R1 = 2.0 Ω
Req,a = R1 + R2 = 2.0 Ω + 3.0 Ω = 5.0 Ω
R2 = 3.0 Ω
Req,b = R3 + R4 = 2.0 Ω + 4.0 Ω = 6.0 Ω
R3 = 2.0 Ω
Parallel:
R4 = 4.0 Ω
R5 = 2.0 Ω
R6 = 2.0 Ω
1
1
1
1
1
0.20 0.17
 = 
 +  =  +  =  + 
Req,c Req,a Req,b 5.0 Ω 6.0 Ω
1Ω
1Ω
Req,c = 2.7 Ω
1
1
1
1
1
 =  +  =  + 
Req,d R5 R6 2.0 Ω 2.0 Ω
Req,d = 1.0 Ω
Series:
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Req = Req,c + Req,d = 2.7 Ω + 1.0 Ω = 3.7 Ω
3.0 V
∆Vtot
I = 
=  = 0.81 A
3.7 Ω
Req
∆V = IReq,c = (0.81 A)(2.7 Ω) = 2.2 V
∆V
2.2 V
I =  =  = 0.44 A
Req,a 5.0 Ω
∆V = IR2 = (0.44 A)(3.0 Ω) = 1.3 V
V
Section Five—Problem Bank
V Ch. 20–9
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Givens
Solutions
7. ∆Vtot = 12.0 V
R1 = 5.0 Ω
Parallel:
R2 = 5.0 Ω
1
0.20 0.20
1
1
1
1
 =  +  =  +  =  + 
1Ω
1Ω
Req,a R1 R2 5.0 Ω 5.0 Ω
R3 = 5.0 Ω
Req,a = 2.5 Ω
R4 = 2.0 Ω
Series:
R5 = 5.0 Ω
Req,b = R4 + R5 = 2.0 Ω + 5.0 Ω = 7.0 Ω
R6 = 5.0 Ω
Req,c = R6 + R7 = 5.0 Ω + 5.0 Ω = 10 Ω
R7 = 5.0 Ω
Parallel:
1
1
1
0.14 0.10
1
1
 =  +  =  +  =  + 
1Ω 1Ω
Req,d Req,b Req,c 7.0 Ω 10 Ω
Req,d = 4.2 Ω
Series:
Req = Req,a + R3 + Req,d = 2.5 Ω + 5.0 Ω + 4.2 Ω = 12 Ω
∆Vtot 12.0 V
I = 
=  = 1.0 A
Req
12 Ω
∆V = IReq,d = (1.0 A)(4.2 Ω) = 4.2 V
∆V
4.2 V
I =  =  = 0.6 A
Req,b 7.0 Ω
∆V = IR4 = (0.6 A)(2.0 Ω) = 1.2 V
R1 = 4.0 Ω
Parallel:
R2 = 5.0 Ω
1
1
1
1
1
1
1
0.50 0.33 0.14
 =  +  +  =  +  +  =  +  + 
Req,a R3 R4 R5 2.0 Ω 3.0 Ω 7.0 Ω
1Ω 1Ω 1Ω
R3 = 2.0 Ω
Req,a = 1.0 Ω
R4 = 3.0 Ω
R5 = 7.0 Ω
1
1
1
1
1
0.25 0.20
 =  +  =  +  =  + 
Req,b R1 R2 4.0 Ω 5.0 Ω 1 Ω 1 Ω
R6 = 3.0 Ω
Req,b = 2.2 Ω
Series:
Req,c = Req,a + Req,b + R4 = 1.0 Ω + 2.2 Ω + 3.0 Ω = 6.2 Ω
∆Vtot 12.0 V
I = 
=  = 1.9 A
Req,c
6.2 Ω
∆V = IReq,a = (1.9 A)(1.0 Ω) = 1.9 V
∆V 1.9 V
I =  =  = 0.27 A
7.0 Ω
R5
V
V Ch. 20–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. ∆Vtot = 12.0 V
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Givens
9. ∆Vtot = 1.5 V
Solutions
Series:
R1 = 4.0 Ω
Req,a = R3 + R4 = 6.0 Ω + 4.0 Ω = 10 Ω
R2 = 5.0 Ω
Req,b = R5 + R6 + R7 = 6.0 Ω + 12.0 Ω + 6.0 Ω = 24 Ω
R3 = 6.0 Ω
Req,c = R8 + R9 = 3.0 Ω + 3.0 Ω = 6.0 Ω
R4 = 4.0 Ω
Parallel:
R5 = 6.0 Ω
R6 = 12.0 Ω
1
1
1
1
0.10 0.042 0.17
 =  +  +  =  +  + 
Req,d Req,a Req,b Req,c
1Ω
1Ω
1Ω
R7 = 6.0 Ω
Req,d = 3.2 Ω
R8 = 3.0 Ω
Series:
R9 = 3.0 Ω
Req = R1 + R2 + Req,d + R10 = 4.0 Ω + 5.0 Ω + 3.2 Ω + 3.0 Ω = 15 Ω
R10 = 3.0 Ω
∆Vtot 1.5 V
I = 
=  = 0.10 A
Req
15 Ω
∆V = IReq,d = (0.10 A)(3.2 Ω) = 0.32 V
∆V
0.32 V
I =  =  = 0.013 A
Req,b
24 Ω
∆V = IR6 = (0.013 A)(12 Ω) = 0.16 V
10. ∆Vtot = 12.0 V
R1 = 5.0 Ω
Parallel:
R2 = 6.0 Ω
1
1
1
1
1
0.067 0.033
 =  +  =  +  =  + 
Req,a R3 R4 15.0 Ω 30.0 Ω
1Ω
1Ω
R3 = 15.0 Ω
Req,a = 10 Ω
R4 = 30.0 Ω
Series:
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Req = R1 + R2 + Req,a = 5.0 Ω + 6.0 Ω + 10.0 Ω = 21 Ω
∆Vtot 12.0 V
I = 
=  = 0.57 A
Req
21 Ω
∆V = IReq,a = (0.57 A)(10 Ω) = 5.7 V
5.7 V
∆V
I =  =  = 0.38 A
15.0 Ω
R3
V
Section Five—Problem Bank
V Ch. 20–11
Print
Magnetism
Chapter 21
Additional Practice 21A
Givens
Solutions
1.2 × 10−17 N
Fmagnetic
B =  = 
= 60.3 × 10−5 T
(1.60 × 10−19 C)(1.2 × 106 m/s)
qv
1. q = 1.60 × 10−19 C
v = 1.2 × 106 m/s
Fmagnetic = 1.2 × 10−17 N
1.9 × 10−22 N
Fmagnetic
= 3.0 × 10−10 T
B =  = 
(1.60 × 10−19 C)(3.9 × 106 m/s)
qv
2. q = 1.60 × 10−19 C
6
v = 3.9 × 10 m/s
Fmagnetic = 1.9 × 10−22 N
Fmagnetic
3.7 × 10−13 N
B =  = 
= 03.0 T
qv
(1.60 × 10−19 C)(7.8 × 106 m/s)
3. q = 1.60 × 10−19 C
v = 7.8 × 106 m/s
Fmagnetic = 3.7 × 10−13 N
4. q = 1.60 × 10−19 C
3
r = 1.0 km = 1.0 × 10 m
m = 1.67 × 10−27 kg
B = 3.3 T
5. q = 1.60 × 10−19 C
B = 5.0 × 10−5 T
mv 2
 = qv B
r
qr B (1.60 × 10−19 C)(1.0 × 103 m)(3.3 T)
v =  = 
= 3.2 × 1011 m/s
m
1.67 × 10−27 kg
6.1 × 10−17 N
Fmagnetic
= 7.6 × 106 m/s
v =  = 
(1.60 × 10−19 C)(5.0 × 10−5 T)
qB
Fmagnetic = 6.1 × 10−17 N
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6. B = 1 × 10−8 T
q = 1.60 × 10−19 C
3.2 × 10−22 N
Fmagnetic
v =  = 
= 2 × 105 m/s
(1.60 × 10−19 C)(1 × 10−8 T)
qB
Fmagnetic = 3.2 × 10−22 N
7. q = 1.60 × 10−19 C
v = 6 × 106 m/s to the right
q = 45°
Fmagnetic = qvBsin q = (1.60 × 10−19 C)(6 × 106 m/s)(3 × 10−4 T)sin 45°
Fmagnetic = 2 × 10−16 N
B = 3 × 10−4 T upward
8. q = 1.60 × 10−19 C
Fmagnetic = qvB = (1.60 × 10−19 C)(3.0 × 107 m/s)(0.8 T) = 4 × 10−12 N
B = 0.8 T
v = 3.0 × 107 m/s
9. q = 1.60 × 10−19 C
Fmagnetic = qvB = (1.60 × 10−19 C)(2.2 × 106 m/s)(1.1 × 10−2 T) = 3.9 × 10−15 N
6
v = 2.2 × 10 m/s
B = 1.1 × 10−2 T
V
Section Five—Problem Bank
V Ch. 21–1
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Givens
Solutions
10. q = 1.60 × 10−19 C
Fmagnetic = qvB = (1.60 × 10−19 C)(9.3 × 105 m/s)(4.1 × 10−10 T) = 6.1 × 10−23 N
v = 9.3 × 105 m/s
B = 4.1 × 10−10 T
Additional Practice 21B
1. B = 4.6 × 10−4 T
Fmagnetic = 2.9 × 10−3 N
l
Fmagnetic
2.9 × 10−3 N
=  = 
= 0.63 m
BI
(4.6 × 10−4 T)(10.0 A)
l
5.6 × 10−5 N
Fmagnetic
=  = −
= 2m
BI
(2.8 × 10 5 T)(1 A)
T = 10.0 A
2. I = 1 A
B = 2.8 × 10−5 T
Fmagnetic = 5.6 × 10−5 N
3.
l
Fmagnetic
7.3 × 10−2 N
B =  =  = 5.1 × 10−4 T
Il
(12 A)(12 m)
= 12 m
I = 12 A
Fmagnetic = 7.3 × 10−2 N
4. Fmagnetic = 7.8 × 105 N
l
= 12 m = 1.2 × 104 m
Fmagnetic
7.8 × 105 N
B =  = 
= 3.6 × 10−3 T
Il
(1.8 × 104 A)(1.2 × 104 m)
I = 1.8 × 104 A
5. I = 14.32 A
l
= 15.0 cm = 0.150 m
l
= 10 m
Fmagnetic
6.62 × 10−4 N
B =  =  = 3.08 × 10−4 T
Il
(14.32 A)(0.150 m)
Fmagnetic = 6.62 × 10−4 N
mg = BIl
m = 75 kg
B = 4.8 × 10−4 T
mg
(75 kg)(9.81 m/s2)
I =  = 
= 1.5 × 105 A
Bl (4.8 × 10−4 T)(10 m)
g = 9.81 m/s2
7.
= 1.0 m
Fmagnetic = 9.1 × 10−5 N
l
B = 1.3 × 10−4 T
8. I = 1.5 × 103 A
l
= 15 km = 1.4 × 104 m
q = 45°
9.1 × 10−5 N
Fmagnetic
I =  = 
= 0.7 A
(1.3 × 10−4 T)(1.0 m)
Bl
Fmagnetic = Bcos qI l = (2.3 × 10−5 T)cos 45°(1.5 × 103 A)(1.5 × 104 m)
Fmagnetic = 3.7 × 102 N
B = 2.3 × 10−5 T
Fmagnetic = BI l = (3.6 × 10−4 T)(14 A)(2 m) = 1 × 10−2 N
9. I = 14 A
l
=2m
B = 3.6 × 10−4 T
V
10. I = 0.5 A
l
= 5 cm = 5 × 10−2 m
Fmagnetic = BI l = (1.3 × 10−4 T)(0.5 A)(5 × 10−2 m) = 3 × 10−6 N
B = 1.3 × 10−4 T
V Ch. 21–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6.
Print
Induction and Alternating Current
Chapter 22
Additional Practice 22A
Givens
Solutions
emf ∆t
emf ∆t
B =  = 
−NA∆cosq
−NA[cosqf − cosqi ]
1. N = 540 turns
A = 0.016 m2
qi = 0°
(3.0 V)(0.05 s)
B = 
−(540)(0.016 m2)[cos 90.0° − cos 0°]
qf = 90.0°
∆t = 0.05 s
B = 1.7 × 10−2 T
emf = 3.0 V
emf ∆t
emf ∆t
B =  = 
−NA∆cosq
−NA(cosqf − cosqi )
2. N = 320 turns
2
A = 0.068 m
qi = 0°
(4.0 V)(0.25 s)
B = 
−(320)(0.068 m2)(cos 90.0° − cos 0°)
qf = 90°
∆t = 0.25 s
B = 4.6 × 10−2 T
emf = 4.0 V
emf ∆t
emf ∆t
B =  = 
−NA∆cosq
−NA(cosqf − cosqi )
3. N = 628 turns
2
A = 0.93 m
qi = 0°
(62 V)(0.30 s)
B = 
−(628)(0.93 m2)(cos 30.0° − cos 0°)
∆t = 0.30 s
qf = 30.0°
B = 0.24 T
emf = 62 V
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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4. N = 550 turns
A = 5.0 × 10−5 m2
∆B = 2.5 × 10−4 T
∆t = 2.1 × 10−5 s
q = 0°
−(550)(5.0 × 10−5 m2)(2.5 × 10−4 T)(cos 0°)
emf = 
2.1 × 10−5 s
emf = 0.33 V
5. N = 220 turns
−6
A = 6.0 × 10
2
m
∆B = 9.7 × 10−4 T
∆t = 1.7 × 10
q = 0°
−NA∆B cos q
emf = 
∆t
−6
s
−NA∆B cos q
emf = 
∆t
−(220)(6.0 × 10−6 m2)(9.7 × 10−4 T)(cos 0°)
emf = 
1.7 × 10−6 s
emf = 0.75 V
V
Section Five—Problem Bank
V Ch. 22–1
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Givens
Solutions
6. N = 148 turns
A = 1.25 × 10−8 m2
q = 0°
−4
∆B = 5.2 × 10
T
∆t = 8.5 × 10−9 s
emf
220 V
I =  =  = 1.8 A
R
120 Ω
R = 120 Ω
8. N = 180 turns
A = 5.0 × 10
−(148)(1.25 × 10−8 m2)(5.2 × 10−4 T)(cos 0°)
emf = 
8.5 × 10−9 s
emf = 0.11 V
7. emf = 220 V
−5
−NA∆B cos q
emf = 
∆t
2
m
∆B = 5.2 × 10−4 T
q = 0°
∆t = 1.9 × 10−5 s
R = 1.0 × 102Ω
−NA∆B cos q
emf = 
∆t
−(180)(5.0 × 10−5 m2)(5.2 × 10−4 T)(cos 0°)
emf = 
1.9 × 10−5 s
emf = 0.25 V
emf
0.25 V
= 2.5 × 10−3 A = 25 mA
I =  = 
R
1.0 × 102Ω
9. N = 246 turns
A = 0.40 m2
q = 0°
Bi = 0.237 T
Bf = 0.320 T
∆t = 0.9 s
−NA∆B cos q
−NA[Bf − Bi] cos q
∆t =  = 
emf
emf
−(246)(0.40 m2)[0.320 T − 0.237 T](cos 0°)
∆t = 
9.1 V
∆t = 0.90 s
10. N = 785 turns
A = 7.3 × 10−2 m2
∆B = 6.9 × 10−3 T
emf = 2.8 V
q = 0°
−(785)(7.3 × 10−2 m2)(6.9 × 10−3 T)(cos 0°)
−NA∆B cos q
∆t =  = 
2.8 V
emf
∆t = 0.14 s/oscillation
∆tearthquake = (0.14 s/oscillation)(120 oscillations)
∆t = 16.8 s
The earthquake lasted for 16.8 s.
Additional Practice 22B
1. N = 220 turns
A = 0.080 m2
B = 4.8 × 10−3 T
maximum emf = 12 V
2. N = 140 turns
V
D = 0.33 m
B = 9.3 × 10−2 T
maximum emf = 150 V
V Ch. 22–2
maximum emf
12 V
w =  = 
NAB
(220)(0.080 m2)(4.8 × 10−3 T)
w = 1.4 × 102 rad/s
maximum emf
w = 
NAB
D 2
0.33 m
A = π   = π 
2
2
Holt Physics Solutions Manual
= 8.6 × 10
2
−2
m2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
emf = 9.1 V
Solutions
Print
150 V
w = 
= 130 rad/s
−2 2
(140)(8.6 × 10 m )(9.3 × 10−2 T)
3. N = 195 turns
maximum emf = Imax R = (1.2 A)(16 Ω) = 19 V
2
A = 0.052 m
19 V
w = 
= 590 rad/s
(195)(0.052 m2)(3.2 × 10−3 T)
B = 3.2 × 10−3 T
Imax = 1.2 A
R = 16 Ω
4. N = 385 turns
w = 2πf
A = 0.38 m2
maximum emf = NABw = NAB2πf
B = 9.4 × 10−3 T
maximum emf = (385)(0.38 m2)(9.4 × 10−3 T)(2π)(45 Hz)
f = 45 Hz
maximum emf = 390 V
5. Imax = 14 A
maximum emf = Imax R = (14 A)(5 Ω) = 70 V
R=5Ω
6. N = 119 turns
maximum emf = NABw
A = 4.9 × 10−2m2
maximum emf = (119)(4.9 × 10−2 m2)(9.4 × 10−3 T)(345 rad/s)
B = 9.4 × 10−3 T
maximum emf = 19 V
w = 345 rad/s
7. maximum emf = 40 V
R=8Ω
8. N = 425 turns
2
A = 2.16 × 10 m
−2
B = 3.9 × 10
maximum emf 40 V
Imax =  =  = 5 A
R
8Ω
w = 2πf
−2
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Givens
T
f = 33 Hz
maximum emf = NABw = NAB2πf
maximum emf = (425)(2.16 × 10−2 m2)(3.9 × 10−2 T)(2π)(33 Hz)
maximum emf = 74 V
R = 25 Ω
74 V
maximum emf
Imax =  =  = 3.0 × 101 A
25 Ω
R
9. A = 1.20 × 10−2m2
−2
B = 6.0 × 10
T
w = 393 rad/s
maximum emf
213 V
N =  = 
−2 2
ABw
(1.2 × 10 m )(6.0 × 10−2 T)(393 rad/s)
N = 750 turns
maximum emf = 213 V
10. A = 0.60 m2
B = 0.012 T
f = 44 Hz
maximum emf = 320 V
maximum emf
maximum emf
N =  = 
ABw
AB 2πf
320 V
N = 
(0.60 m2)(0.012 T)(2π)(44 Hz)
N = 160 turns
V
Section Five—Problem Bank
V Ch. 22–3
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Print
Additional Practice 22C
Givens
Solutions
1. ∆Vrms = 320 V
∆Vrms 320 V
∆Vmax = 
=  = 450 V
0.707
0.707
R = 100 Ω
∆Vrms
320 V
=  = 3 A
Irms = 
R
100 Ω
I s
3A
=  = 4 A
Imax = rm
0.707 0.707
2. Irms = 1.3 A
I s
1.3 A
Imax = rm
=  = 1.8 A
0.707
0.707
3. Irms = 2.5 A
4
∆Vrms = 2.2 × 10 V
I s
2.5 A
=  = 3.5 A
Imax = rm
0.707 0.707
∆Vrms 2.2 × 104 V
=  = 8.8 × 104 Ω = 88 kΩ
R = 
Irms
2.5 A
4. ∆Vrms = 220 V
Irms = 1.7 A
5. Imax = 1.2 A
∆Vmax = 211 V
∆Vrms
∆Vmax = 
0.707
I s
Imax = rm
=
0.707
220 V
=  = 311 V
0.707
1.7 A
 = 2.4 A
0.707
Irms = 0.707 Imax = 0.707(1.2 A) = 0.85 A
∆Vrms = 0.707 Vmax = 0.707(211 V) = 149 V
6. Vmax = 170 V
∆Vrms = 0.707 Vmax = 0.707(170 V) = 120 V
7. ∆Vrms = 115 V
∆Vrms
115 V
Irms = 
=  = 2.30 A
R
50.0 Ω
Irms
2.30 A
Imax =  =  = 3.25 A
0.707
0.707
R = 50.0 Ω
8. Irms = 2.1 A
R = 16 k Ω = 1.6 × 104 Ω
I s
2.1 A
Imax = rm
=  = 3.0 A
0.707 0.707
P = (Irms)2R = (2.1 A)2(1.6 × 104 W) = 7 × 104 Ω P = 70 kW
9. ∆Vrms = 1.56 × 104 V
Irms = 1.3 A
R = 1.2 × 104 Ω
∆Vrms 1.56 × 104 V
∆Vmax =  =  = 2.2 × 104 V = 22kV
0.707
0.707
I s
1.3 A
Imax = rm
=  = 1.8 A
0.707 0.707
P = (Irms)2R = (1.3 A)2(1.2 × 104 Ω) = 2.0 × 104 W
V
V Ch. 22–4
Holt Physics Solutions Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
149 V
∆Vrms
R = 
=  = 175 Ω
0.85 A
Irms
Print10. I
rms
Solutions
= 2.2 × 1010 A
R = 6.1 × 10
−10
Ω
I s
2.2 × 1010 A
=  = 3.1 × 1010 A
Imax = rm
0.707
0.707
P = (Irms)2 R = (2.2 × 1010 A)2(6.1 × 10−10 Ω) = 2.9 × 1011 W
Additional Practice 22D
1. ∆V2 = 6.9 × 103 V
N1 = 1400 turns
N
1400
∆V1 = ∆V2 1 = (6.9 × 103 V)  = 6.9 × 104 V
N2
140
N2 = 140 turns
2. ∆V2 = 3.4 × 103 V
N1 = 90 turns
90
N
∆V1 = ∆V2 1 = (3.4 × 103 V)  = 1.4 × 102 V
2250
N2
N2 = 2250 turns
3. ∆V1 = 4.6 × 104 V
N1 = 1250 turns
250
N
∆V2 = ∆V1 2 = (4.6 × 104 V)  = 9.2 × 103 V
1250
N1
N2 = 250 turns
4. ∆V1 = 5600 V
N
840
∆V2 = ∆V1 2 = (5600 V)  = 3.36 × 104 V
N1
140
N1 = 140 turns
N2 = 840 turns
5. ∆V1 = 9200 V
N
1200
∆V2 = ∆V1 2 = (9200 V)  = 9.20 × 104 V
N1
120
N1 = 120 turns
N2 = 1200 turns
6. ∆V1 = 3.6 × 104 V
∆V2 = 7.2 × 103 V
∆V
7.2 × 103 V
= 11 turns
N2 = N1 2 = (55) 
∆V1
3.6 × 104 V
N1 = 55 turns
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Givens
7. ∆V1 = 240 V
N1 ∆V1 240 V
 =  =  = 48:1
N2 ∆V2 5.0 V
∆V2 = 5.0 V
∆V
3600 V
N2 = N1 2 = (58)  = 1.2 × 102 turns
∆V1
1800 V
8. ∆V1 = 1800 V
∆V2 = 3600 V
N1 = 58 turns
9. ∆V1 = 4900 V
∆V2 = 4.9 × 104 V
∆V
4900 V
N1 = N2 1 = (480) 
= 48 turns
∆V2
4.9 × 104 V
N2 = 480 turns
10. P = 1.38 × 106 W
3
∆V2 = 3.4 × 10 V
N1 = 340 turns
N2 = 17 turns
N
340
∆V1 = ∆V2 1 = (3.4 × 103 V)  = 6.8 × 104 V
N2
17
P = ∆V1I1
P
1.38 × 106 W
I1 =  = 
= 2.0 × 101 A
∆V1
6.8 × 104 V
Section Five—Problem Bank
V
V Ch. 22–5
Print
Atomic Physics
Chapter 23
Additional Practice 23A
Givens
Solutions
1. λ = 527 nm = 5.27 × 10−7 m
hc (6.63 × 10−34 J • s)(3.00 × 108 m/s)
E = hf =  = 
= 3.77 × 10−19 J
λ
5.27 × 10−7 m
2. λ = 430.8 nm
hc (6.63 × 10−34 J • s)(3.00 × 108 m/s)
= 4.62 × 10−22 J
E = hf =  = 
λ
4.308 × 10−7 m
λ = 4.308 × 10−7 m
3. E = 20.7 eV
E (20.7 eV)(1.60 × 10−19 J/eV)
f =  = 
= 5.00 × 1015 Hz
h
6.63 × 10−34 J • s
4. E = 1.24 × 10−3 eV
E (1.24 × 10−3 eV)(1.60 × 10−19 J/eV)
= 2.99 × 1011 Hz
f =  = 
h
6.63 × 10−34 J • s
5. E = 1.78 eV
E (1.78 eV)(1.60 × 10−19 J/eV)
f =  = 
= 4.30 × 1014 Hz
h
6.63 × 10−34 J • s
6. E = 12.4 MeV
7
E = 1.24 × 10 eV
7. E = 939.57 MeV
E = 9.3957 × 108 eV
hc (6.63 × 10−34 J • s)(3.00 × 108 m/s)
= 1.00 × 10−13 m
λ =  = 
E (1.24 × 107 eV)(1.60 × 10−19 J/eV)
hc
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
= 1.32 × 10−15 m
λ =  = 
E (9.3957 × 108 eV)(1.60 × 10 −19 J/eV)
1.32 × 10−15 m = 1.32 × 10−6 nm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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If a photon were to have this wavelength, it would not lie within the visible part of
the spectrum.
8. E = 3.1 × 10−6 eV
hc (6.63 × 10−34 J • s)(3.00 × 108 m/s)
= 0.401 m
λ =  = 
E (3.1 × 10−6 eV)(1.60 × 10−19 J/eV)
Additional Practice 23B
1. λ = 240 nm = 2.4 × 10−7 m
hft = 2.3 eV
hc
KEmax =  − hft
λ
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
KEmax = 
− 2.3 eV
2.4 × 10−7 m)(1.60 × 10−19 J/eV)
KEmax = 5.2 eV − 2.3 eV = 2.9 eV
V
Section Five—Problem Bank
V Ch. 23–1
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Givens
Solutions
2. λ = 519 nm = 5.19 × 10−7 m
hft = 2.16 eV
hc
KEmax =  − hft
λ
(6.63 × 10−34 J • s)(3.00 × 108 m/s)
KEmax = 
− 2.16 eV
(5.19 × 10−7 m)(1.60 × 10−19 J/eV)
KEmax = 2.40 eV − 2.16 eV = 0.24 eV
3. f = 6.5 × 1014 Hz
KEmax = 0.20 eV
hf − KEmax
ft = 
h
[(6.63 × 10−34 J • s)(6.5 × 1014 Hz) − (0.20 eV)(1.60 × 10−19 J/eV)]
ft = 
6.63 × 10−34 J• s
ft = 6.0 × 1014 Hz
4. f = 9.89 × 1014 Hz
KEmax = 0.90 eV
hf − KEmax
ft = 
h
(6.63 × 10−34 J • s)(9.89 × 1014 Hz) − (0.90 eV)(1.60 × 10−19 J/eV)
ft = 
6.63 × 10−34 J• s
5. ft = 1.36 × 1015 Hz
(6.63 × 10−34 J • s)(1.36 × 1015 Hz)
hft = 
= 5.64 eV
1.60 × 10−19 J/eV
6. ft = 1.1 × 1015 Hz
(6.63 × 10−34 J • s)(1.1 × 1015 Hz)
hft = 
= 4.6 eV
1.60 × 10−19 J/eV
7. hft = 4.1 eV
hc (6.63 × 10−34 J • s)(3.00 × 108 m/s)
= 3.0 × 10−7 m = 300 nm
λ =  = 
(4.1 eV)(1.60 × 10−19 eV)
E
8. hft = 5.0 eV
hc (6.63 × 10−34 J • s)(3.00 × 108 m/s)
= 2.5 × 10−7 m = 250 nm
λ =  = 
(5.0 eV)(1.60 × 10−19 eV)
E
9. KEmax = 0.62 V
KEmax = hf − hft = 2mev 2
1
me = 9.109 × 10−31 kg
v=
2(0.62 eV)(1.60 × 10−19 J/eV)


m
= 9.109 × 10 kg
2KEmax
−31
e
v = 4.7 × 105 m/s
10. KEmax = 1.2 eV
1
KEmax = hf − hft = 2 mev 2
me = 9.109 × 10−31 kg
v=
2KEmax
e
v = 6.5 × 105 m/s
V
V Ch. 23–2
2(1.2 eV)(1.60 × 10−19 eV)


m
= 9.109 × 10 kg
Holt Physics Solutions Manual
−31
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ft = 7.72 × 1014 Hz
Additional Practice 23C
PrintGivens
Solutions
1. v = 28 m/s
λ = 8.97 × 10−37 m
2. v = 7.1 × 102 m/s
λ = 5.8 × 10−42 m
3. v = 5.6 × 10−6 m/s
λ = 2.96 × 10−8 m
4. v = 12 m/s
−29
λ = 2.6 × 10
m
5. me = 9.109 × 10−31 kg
v = 2.19 × 106 m/s
6. m = 7.6 × 107 kg
v = 35 m/s
7. m = 5.94 × 1024 kg
v = 3.0 × 104 m/s
8. m = 4.0 × 1041 kg
v = 1.7 × 104 m/s
9. me = 9.109 × 10−31 kg
λ = 9.87 × 10−14 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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10. mn = 1.675 × 10−27 kg
λ = 5.6 × 10−14 m
h
6.63 × 10−34 J • s
m =  = 
= 26 kg
λv (8.97 × 10−37 m)(28 m/s)
h
6.63 × 10−34 J • s
= 1.6 × 105 kg
m =  = 
λv (5.8 × 10−42 m)(7.1 × 102 m/s)
h
6.63 × 10−34 J • s
= 4.0 × 10−21 kg
m =  = 
λv (2.96 × 10−8 m)(5.6 × 10−6 m/s)
h
6.63 × 10−34 J • s
m =  = 
= 2.1 × 10−6 kg
λv (2.6 × 10−29 m)(12 m/s)
h
6.63 × 10−34 J • s
= 3.3 × 10−10 m
λ =  = 
mv (9.109 × 10−31 kg)(2.19 × 106 m/s)
h
6.63 × 10−34 J • s
= 2.5 × 10−43 m
λ =  = 
mv (7.6 × 107 kg)(35 m/s)
h
6.63 × 10−34 J • s
= 3.7 × 10−63 m
λ =  = 
mv (5.94 × 1024 kg)(3.0 × 104 m/s)
h
6.63 × 10−34 J • s
= 9.7 × 10−80 m
λ =  = 
mv (4.0 × 1041 kg)(1.7 × 104 m/s)
h
6.63 × 10−34 J • s
= 7.37 × 109 m/s
v =  = 
mλ (9.109 × 10−31 kg)(9.87 × 10−14 m)
h
6.63 × 10−34 J • s
= 7.1 × 106 m/s
v =  = 
mλ (1.675 × 10−27 kg)(5.6 × 10−14 m)
V
Section Five—Problem Bank
V Ch. 23–3
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Subatomic Physics
Chapter 25
Problem Bank Solutions Chapter 25A
Givens
1. Z = 19
Solutions
N = A − Z = 39 − 19 = 20
A = 39
∆m = Z (atomic mass of H) + Nmn − atomic mass of K-39
atomic mass of K-39
= 38.963 708 u
∆m = 19(1.007 825) + 20 (1.008 665 u)–38.963 708 u
atomic mass of H = 1.007 825 u
mn = 1.008 665 u
∆m = 19.156 75 u + 20.1733 u – 38.963 708 u
∆m = 0.3653 u
Ebind = (0.3653 u)(931.50 MeV/u) = 340.3 MeV
2. Z = 50
A = 120
∆m = Z(atomic mass of H) + Nmn − atomic mass of Sn-120
atomic mass of Sn-120
= 119.902 197
∆m = 50(1.007 825 u) + 70(1.008 665 u) − 119.902 197 u
atomic mass of H = 1.007 825 u
mn = 1.008 665 u
Copyright © by Holt, Rinehart and Winston. All right reserved.
3.
N = A − Z = 120 − 50 = 70
∆m = 50.391 25 u + 70.606 55 u − 119.902 197 u
∆m = 1.095 60 u
Ebind = (1.095 60 u)(931.50 MeV/u) = 1020.55 MeV
For 107
47 Ag:
N = A − Z = 107 − 47 = 60
Z = 47
∆m = Z(atomic mass of H) + Nmn − atomic mass of Ag-107
A = 107
∆m = 47(1.007 825 u) + 60(1.008 665 u) − 106.905 091 u
atomic mass of Ag-107
= 106.905 091 u
∆m = 47.367 775 u + 60.5199 u − 106.905 091 u
atomic mass of H
= 1.007 825 u
Ebind = (0.9826 u)(931.50 MeV/u) = 915.29 MeV
∆m = 0.9826 u
mn = 1.008 665 u
For 63
29 Cu:
N = A − Z = 63 − 29 = 34
Z = 29
∆m = Z(atomic mass of H) + Nmn − atomic mass of Cu-63
A = 63
∆m = 29(1.007 825 u) + 34(1.008 665 u) − 62.929 599 u
atomic mass of Cu-63
= 62.929 599 u
∆m = 29.226 925 u + 34.2946 u − 62.929 599 u
∆m = 0.5919 u
Ebind = (0.5919 u)(931.50 MeV/u) = 551.4 MeV
The difference in binding energy is 915.29 MeV − 551.4 MeV = 363.9 MeV
V
Section Five—Solution Manual
V Ch. 25–1
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Givens
Solutions
N = A − Z = 12 − 6 = 6
4. For 12
6 C:
Z=6
∆m = Z (atomic mass of H) + Nmn − atomic mass of C-12
A = 12
∆m = 6(1.007 825 u) + 6(1.008 665u) − 12.000 000 u
atomic mass of C-12
= 12.000 000 u
∆m = 6.046 95 u + 6.051 99 u − 12.000 000 u
atomic mass of H
= 1.007 825 u
∆m = 0.098 94 u
Ebind = (0.098 94 u)(931.50 MeV/u) = 92.163 MeV
mn = 1.008 665 u
For 16
8 O:
N = A − Z = 16 − 8 = 8
Z=8
∆m = Z (atomic mass of H) + Nmn − atomic mass of O-16
A = 16
∆m = 8(1.007 825 u) + 8(1.008 665u) − 15.994 915 u
atomic mass of O-16
= 15.994 915
∆m = 8.0626 u + 8.06932 u − 15.994 915 u
∆m = 0.1370 u
Ebind = (0.1370 u)(931.50 MeV/u) = 127.62 MeV
The difference in binding energy is
127.62 MeV − 92.163 MeV = 35.46 MeV
5. Z = 17
N = A − Z = 35 − 17 = 18
A = 35
∆m = Z (atomic mass of H) + Nmn − atomic mass of CI-35
atomic mass of Cl-35
= 34.968 853 u
∆m = 17(1.007 825 u) + 18(1.008 665 u) − 34.968 853 u
atomic mass of H
= 1.007 825 u
∆m = 0.320 14 u
mn = 1.008 665 u
Ebind = (0.320 14 u)(931.50 MeV/u) = 298.21 MeV
∆m = 17.133 025 u + 18.155 97 u − 34.968 853 u
A=2
∆m = Z(atomic mass of H) + Nmn − atomic mass of H-2
atomic mass of H-2
= 2.014 102 u
∆m = 1(1.007 825 u) + 1(1.008 665 u) − 2.014 102 u
atomic mass of H
= 1.007 825 u
∆m = 2.388 × 10−3 u
Ebind = (2.388 × 10−3 u)(931.50 MeV/u) = 2.2244 MeV
mn = 1.008 665 u
N = A − Z = 58 − 28 = 30
7. A = 58
Z = 28
∆m = Z(atomic mass of H) + Nmn − atomic mass of Ni-58
atomic mass of Ni-58
= 57.935 345 u
∆m = 28(1.007 825 u) + 30(1.008 665 u) − 57.935 345 u
atomic mass of H
= 1.007 825 u
∆m = 0.5437 u
∆m = 28.2191 u + 30.259 95 u − 57.935 345 u
mn = 1.008 665 u
V
V Ch. 25–2
Holt Physics Solutions Manual
Copyright © by Holt, Rinehart and Winston. All right reserved.
N=A−Z=2−1=1
6. Z = 1
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Givens
8. A = 64
Solutions
N = A − Z = 64 − 30 = 34
Z = 30
∆m = Z(atomic mass of H) + Nmn − atomic mass of Zn-64
atomic mass of Zn-64
= 63.929 144 u
∆m = 30(1.007 825 u) + 34(1.008 665 u) − 63.929 144 u
atomic mass of H
= 1.007 825 u
∆m = 0.6002 u
∆m = 30.234 75 u + 34.2946 u − 63.929 144 u
mn = 1.008 665 u
9. A = 90
N = A − Z = 90 − 40 = 50
Z = 40
∆m = Z(atomic mass of H) + Nmn − atomic mass of Zr-90
atomic mass of Zr-90
= 89.904 702 u
∆m = 40(1.007 825 u) + 50(1.008 665 u) − 89.904 702 u
atomic mass of H
= 1.007 825 u
∆m = 0.842 u
mn = 1.008 665 u
10. A = 32
∆m = 40.313 u + 50.433 25 u − 89.940 702 u
Ebind = (0.842 u)(931.50 MeV/u) = 784 MeV
N = A − Z = 32 − 16 = 16
Z = 16
∆m = Z(atomic mass of H) + Nmn − atomic mass of S-32
atomic mass of S-32
= 31.972 071 u
∆m = 16(1.007 825 u) = 16(1.008 665 u) − 31.972 071 u
atomic mass of H
= 1.007 825 u
∆m = 0.2918 u
∆m = 16.1252 u + 16.1386 u − 31.972 071 u
mn = 1.008 665 u
Problem Bank Solutions Chapter 25B
Copyright © by Holt, Rinehart and Winston. All right reserved.
1.
210
4
84 Po → ? + 2He
A = 210 − 4 = 206
Z = 84 − 2 = 82
?=
2.
16
0
v
7 N → ? + −1e + 206
82
Pb
A = 16 − 0 = 16
Z = 7 − (−1) = 8
?=
3.
147
143
v
62 Sm → 60 Nd + ? + 16
8
O
A = 147 − 143 = 4
Z = 62 − 60 = 2
?=
4.
19
0
v
10 Ne → ? + 1e + 4
2 He
A = 19 − 0 = 19
Z = 10 − 1 = 9
?=
19
9 F
V
Section Five—Solution Manual
V Ch. 25–3
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Additional Practice 22C
Givens
5.
Solutions
0
? → 131
v
54 Xe + −1e + A = 131 + 0 = 131
Z = 54 + (−1) = 53
?=
0
6. ? → 90
v
39 Y + −1e + 131
53
I
A = 90 + 0 = 90
Z = 39 + (−1) = 38
?=
7.
160
156
74 W → 72 Hf
+?
90
38 Sr
A = 160 − 156 = 4
Z = 74 − 72 = 2
?=
4
8. ? → 107
52 Te + 2He
4
2 He
A = 107 + 4 = 111
Z = 52 + 2 = 54
?=
9.
157
72 Hf
→ 153
70 Yb + ?
111
54
Xe
A = 157 − 153 = 4
Z = 72 − 70 = 2
?=
10.
141
0
v
58 Ce → ? + −1e + 4
2 He
A = 141 − 0 = 141
Z = 58 − (−1) = 59
?=
141
59
Pr
1. mi = 5.25 × 10−3 g
mf = 3.28 × 10−4 g
mf 3.28 × 10−4 g 1
 = 
= 
mi 5.25 × 10−3 g 16
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1
If 16 of the sample remains after 12 h, then 8 of the sample must have remained after
1
1
6.0 h, 4 of the sample must have remained after 3.0 h, and 2 of the sample must have
remained after 1.5 h. So T1/2 = 1.5 h
∆t = 12 h
2. mi = 3.29 × 10−3 g
mf = 8.22 × 10−4 g
∆t = 30.0 s
3. mi = 4.14 × 10−4 g
mf 8.22 × 10−4 g 1
 = 
= 
mi 3.29 × 10−3 g 4
1
1
If 4 of the sample remains after 30.0 s, then 2 of the sample must have remained after
15.0 s, so T1/2 = 15.0 s .
mf = 2.07 × 10−4 g
mf 2.07 × 10−4 g 1
 = 
= 
mi 4.14 × 10−4 g 2
∆t = 1.25 days
If 2 of the sample remains after 1.25 days, then T1/2 = 1.25 days
1
V
V Ch. 25–4
Holt Physics Solutions Manual
Copyright © by Holt, Rinehart and Winston. All right reserved.
Problem Bank Solutions Chapter 25C
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Givens
Solutions
1
4. T1/2 = 10.64 h
For the sample to reach 2 its original strength, it takes 10.64 h. For the sample to
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1
reach 4 its original strength, it takes 2(10.64 h) = 21.28 h. For the sample to reach 8 its
original strength, it takes 3(10.64 h) = 31.92 h
5. T1/2 = 462 days
For the sample to reach 2 its original strength, it takes 462 days. For the sample to
1
reach 4 its original strength, it takes 2(462 days) = 924 days
6. T1/2 = 2.7 y
0.693
0.693
λ =  = 
=
T1/2 (2.7 y)(3.156 × 107 s/y)
9
N = 3.2 × 10
7. activity = 765.3 mCi
T1/2 = 22 h
8.1 × 10−9 s−1
0.693
0.693
λ =  =   = 8.75 × 10−6 s−1
T1/2 (22.0 h)(3600 s/h)
activity (0.7653 Ci)(3.7 × 1010 s−1/Ci)
N =  = 
= 3.2 × 1015 nuclei
8.75 × 10−6 s−1
λ
8. T1/2 = 21.6 h
6
N = 6.5 × 10
9. T1/2 = 12.33 y
8
N = 4.8 × 10
10. activity = 0.3600 Ci
0.693
0.693
λ =  =  = 8.90 × 10−6 s−1
(21.6 h)(3600 s/h)
T1/2
(8.9 × 10−6 s−1)(6.5 × 106)
activity = N λ = 
= 1.5 × 10−9 Ci
3.7 × 1010 s−1/Ci
0.693
0.693
λ =  = 
= 1.78 × 10–9 s–1
T1/2 (12.33 y)(3.56 × 107 s/y)
0.693 0.693
λ =  =  = 4.03 × 10−2 s−1
T1/2 17.2 s
activity (0.3600 Ci)(3.7 × 1010 s−1/Ci)
= 3.3 × 104 nuclei
N =   = 
λ
(4.03 × 10−2 s−1)
Copyright © by Holt, Rinehart and Winston. All right reserved.
T1/2 = 17.2 s
1
V
Section Five—Solution Manual
V Ch. 25–5