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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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Givens
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