퀴즈 및 연습문저l 하l 답 Answers to Quic뻐zzes a빼
Chapter 1
[퀴즈]
1. (a)
2. False
3. (b)
4. False. Your graph should look something like the one
shown in the next column. This v，、 -t graph shows that
the maximum speed is about 5.0 m / s , which is 18 km/ h
(= 11 m i/ h) , so the driver was not speeding.
Vx
(m/ s)
6
4
[연습문제]
2
1. (a) 5.52 X 10 3 kg/ m 3 (b) It is between the density of alu-
0
minum and that of iron and is greater than the densities
of typical surface rocks.
2. (a) 2.3 X 1017 kg/ m 3 (b) 1. 0 X 1013 times the density of
osmlum
3. 7.69 cm
-2
4π p(r:경 - rl)
3
5. The angle subtended by the Great Wall is less than the
visual acuity of the eye.
6. 70.0 m
7. 0 .1 41 nm
8. The value of k , a dimensionless constant cannot be
obtained by dimensional analysis
9. (b) only
10. (a) [A] = L/ T 3 and [B] = L/T (b) L/ T
11. 11 .4 X 10 3 kg/ m 3
4
12. The number of pages in Volume 1 is sufficient
13. 2.86 cm
14. r Fe (l.4 3)
15. 151 μm
16. (a) 3.39 X 10 5 ft3 (b) 2.54 X 10 4 1b
17. (a) ~102kg (b) ~10 3 kg
18. 100 tuners
19. The average distance between asteroids in the asteroid
belt is about 400 000 km.
20. (a) 3 (b) 4 (c) 3 (d) 2
2 1. 31 556 926.0 s
22. 1. 66 x 10 3 kg/ m 3
23. 19
24. 288 0 , 108 。
25. 63
27. :!::3 .4 6
28. (a) nine times smaller (b) ßt is inversely proportional to
the square of d (c) Plot ßt on the vertical axis and l/ d 2
on the horizontal axis (d) 4QL/ kπ (T" - 낀)
29. (a) 1014 bacteria (b) beneficial
30. 1O11 stars
-4
-6
5. (b)
6. (a)-(e) , (b)-(d) , (c)-(f)
7. (i) (e) (ii) (d)
[연습문제】
1. about 0.02 s
2. (a) 50.0 m/ s (b) 4 1. 0 m / s
3. (a) 2.30 m / s (b) 16.1 m / s (c) 11. 5 m/ s
4. (a) +L/ tl (b) -L/ t2 (c) 0 (d) 2L/ tl + t2
5. (a) -2 .4 m/s (b) -3.8 m / s (c) 4.0 s
6. (a) 20 m i/ h (b) 0 (c) 30 m i/ h
7. (a) 2.80 h (b) 218 km
8.
Position 잃 a function of time
100
를
ν
며i잉￡ 80
웅말 400
6
월훌 20
0
time
Velocity as a function of time
.........:;
옹용
.. =
‘
:
。 ‘
톨현 O
틱은
a흩
를
Chapter 2
I퀴즈】
g
μ
‘*
time
1. (b)
2. (c)
3. (b)
A-31
A-32
퀴즈 및 연습문제 해답
Acceleration as a function of time
{』여:
』a』 뼈-zμμ
·다잉
여』}
양야얘
g
nu
time
9. (a) 1. 3 m / s2 (b) t = 3 s, a = 2 m / s2 (c) t = 6 s, t > 10 s
(d) a = -1. 5 m /s2 , t = 8 s
10. (a)
x (m)
60
40
20
t (5)
2
4
(b) 23 m / s, 18 m / s, 14 m / s, and 9.0 m /s
(c) 4.6 m / s (d) zero
11. (a) 20 m / s, 5 m / s . (b) 262.5 m
12. (a) 4.98 X 10- 19 s (b) 1. 20 X 1015 m / s2
13. (a) 9.00 m / s (b) -3.00 m / s (c) 17.0 m / s (d) The graph
of velocity versus time is a straight line passing through
13 m / s at 10:05 a.m. and sloping downward , decreasing
by 4 m / s for each second thereafter. (e) Ifand only ifwe
know the object’ s velocity at one instant of time, knowing
its acceleration tells us its velocity at every other moment
as long as the acceleration is constan t.
14. (a) 20.0 s (b) 1 000 m (c) the plane cannot land
15. -16.0 cm / s2
16. (a) The idea is false unless the acceleration is zero. We
define constant acceleration to mean that the velocity
is changing steadily in time. So , the velocity cannot be
changing steadily in space. (b) This idea is true. Because
the velocity is changing steadily in time, the velocity halfway through an interval is equal to the average of its initial and final values .
17. The accelerations do not match, therefore the situation is
impossible.
18. (a) 19.7 cm/ s (b) 4.70 cm/ s2 (c) The length ofthe glider
is used to find the average velocity during a known time
interva l.
19. The equation represents that displacement is a quadratic
function of time.
1
<)
J
χj = V페 l- 걷 ax C-
20. (a) 3.75 s (b) 5.50 cm/ s (c) 0.604 s (d) 13.3 cm, 47.9 cm
(e) The cars are initially moving toward each other, so
they soon arrive at the same position x when their speeds
are quite different , giving one answer to (c) that is not an
answer to (a). The first car slows down in its motion to the
left , turns around , and starts to move toward the right ,
slowly at first and gaining speed steadily. At a particular
moment its speed will be equal to the constant rightward
speed of the second car, but at this time the accelerating
car is far behind the steadily moving car; thus , the answer
to (a) is not an answer to (c). Eventually the accelerating
car will catch up to the steadily coasting car, but passing
it at higher speed , and giving a nother answer to (c) that is
not an answer to (a) .
21. (a) -0.107 m / s2 (b) 4.53 m / s (c) pole #3 (stops at x =
96.3 m at t = 42 .4 s)
22. David will be unsuccessfu l. The average human reaction time is about 0.2 s (research on the Internet) and
a dollar bill is about 15.5 cm long , so David ’ s fingers are
about 8 cm from the end of the bill before it is dropped .
The bill will fall about 20 cm before h e can close his
fingers.
23. (a and b) The rock pass by the top of the wall with vJ
= 3.69 m / s (c) 2.39 m / s (d) does not agree
(e) The
average speed of the upward-moving rock is smaller
than the downward moving rock.
24. 7.96 s
25. 0.60 s
26. (a) 10.0 m / s up (b) 4.68 m / s down
h.gt
27. (a) -7+
c;.,2
t
",
h
gt
(b)
,-, -7--;'
t
2
28. (a) The box could reach the window according to the
data provided.
(b) Answers will vary.
29. (a) 4.00 m / s (b) 1. 00 ms (c) 0.816 m
30. (a) 3 .45 s (b) 10.0 ft
Chapter 3
【퀴 즈]
1. vectors: (b) , (c); scalars: (a) , (d) , (e)
2. (c)
3. (b) and (c)
4. (b)
5. (c)
[ 연습문제]
1. (a) 8.60 m (b) 4 .47 m , -63 .40 ; 4.24 m , 135。
2. (a) (2.17 ,1. 25) m , (-1. 90 , 3.29) m (b) 4.55m
3. (a) (-3.56 cm, -2 .4 0 cm) (b) (r = 4.30 cm , e = 3260 )
(c) (r = 8.60 cm , e = 34.0 0 ) (d) (r = 12.9 cm, e = 1460 )
4. (a) κ 180 0 - e (b) 180 0 + () (c)-()
5. This situation can ηeνer be true because the distance
is the length of an arc of a circle between two points ,
whereas the magnitude of the displacement vector is a
straight-line chord of the circle between the same points .
6. B is 43 units in the negative y direction
7. 9.5 N , 57 0 above the x axis
8. (a) The three diagrams are shown in the figure below.
~
퀴즈 및 연습문제 해답
Chapter 4
L..J
100m
”
/‘
-며 K
”
-•
A
•C/4
-•
B
A-33
!
판짧뀔
’
【퀴즈]
1. (a)
2. (i) (b) (ii) (a)
3. 150 , 30 0 , 45 0 , 60 0 , 75。
4. (i) (d) (ii) (b)
5. (i) (b) (ii) (d)
-
(b) The sum of a set ofvectors is not affected by the order
in which the vectors are added.
9. (a) 5.2 m at 60 0 (b) 3.0 m at 330 0 (c) 3.0 m at 150。
(d) 5.2 m at 300 。
10. approximately 420 ft at -2.63 。
11. (a) yes (b) The speed of the camper should be 28.3 m/s
or more to satisfy this requiremen t.
12. l. 31 km north and 2.81 km east
13. 9.4 8 m at 166。
14. (a)
【연습문저n
1. (a) (l. OOi + 0.750j) m/s
(b) (l. OOi + 0.500j) m/s , l. 12 m/s
2. (a) -5.00ωi m/s
(b) -5.00 때 m/s
(c) (4.00 m)J + (5.00 m)(- sinωti - cosω t])，
(5.00 m)ω[-cos ωi + sin ω 폐， (5.00 m)ω2 [sin ω ti + cos ωJ]
(d) a circle of radius 5.00 m centered at (0 , 4.00 m)
3. (여때
a띠) 강 = -12.0t셔j ， 때w‘~
seconds (b) 융 = -12.0j m/s2 (c) r = (3.00i - 6.0띠) m ,
강 = -12.0j m/s
4. (a) (lo.oi + 0.24파) mm (b) (1. 84 X 10 7 m/s)i + (8.78 X
10 5 m/s)j (c) 1. 85 X 10 7 m/s (d) 2.73。
5. (a) 칸= (3 .45 - 1. 79t)i + (2.89 - 0.650t)j
(b) 칸= (-25.3 + 3 .45t - 0.893t 2)i +
(28.9 + 2.89t - 0.325t 2)j
6. V xi
(b) 5.00i + 4.00j , -l. OOi + 8.00j
(c) 6 .4 0 at 38.7 0 , 8.06 at 97.2 。
15. (a) 185 N at 77.8 0 from the positive χ axis
(b) (-39.3i - 18 패 )N
16. (a) Its component parallel to the surface is (l. 50 m) cos
141 0 = -1. 17 m , or 1.1 7 m toward the top of the hill (b)
Its component perpendicular to the surface is (1. 50 m)
sin 141 0 = 0.944 m , or 0.944 m away from the snow.
17. (a) 5.66 m at 8 = 315 0 (b) 13 .4 m at 8 = 117 。
18. Cx = 7.30 cm , Cy = -7.20 cm
19. (a) 8.00 i + 12.0 j - 4.00k (b) 2.00 i + 3.00 j - 1. 00k
(c) -24.0 i - 36.0 j + 12.0k
20. (a) 5.00i - 1. 00j - 3.00k, 5.92 m
(b) (4.00i - 1l. 0j + 15.0k) m , 19.0 m
2 1. (a) - 3.00 i + 2.00 j (b) 3.61 at 1460 (c) 3.00 i - 6.00 j
22. (a) 49.5i + 27. 괴 (b) 56 .4, 28.7 。
23. (a) a = 5.00 and b = 7.00 (b) For vectors to be equal , all
their components must be equa l. A vector equation contains more information than a scalar equation.
24. 59.2 0 with the χ axis , 39.8 0 with the Y axis , 67.4 0 with the z
aXls
25. (a) (-20.5 i + 35.5 j) km/h (b) 25.0 j km/h
(c) (-6 1. 5i + 107j)km (d) 37.5jkm (e) 157km
26. (a) 10 .4 cm (b) 8 = 35.5。
27. l. 43 X 10 4 m at 32.2 0 above the horizontal
28. Impossible because 12 .4 m is greater than 5.00 m
29. 9 .4 60 west of north
30. 240 m at 237 。
얘
= dj
많옳 (여
b비)T파
he 社
d따
떼
m
ir때
rect
앉때
ec때
띠떼
Cα
t더ion
띠따
O야
off 야the ml
tan냐1끼(2h/d) below the horizonta l.
7. 12.0 m/s
8. 53.1 。
9.67.8 。
10. (a) 76.0 0
(b) ~ax = 2.13R (c) the same on every planet
11. dtan 8. --=---::: gd 。
t 2ut2 cos2 81
12. (a) 0.852 s (b) 3.29 m/s (c) 4.03 m/s
(d) 50.8 0 (e) t = 1. 12 s
13. (a) (0 , 0) (b) vxi = 18.0 m/s , vyi = 0
(c) Particle under constant acceleration
(d) Particle under constant velocity
(e) 왜 = νxi' vyJ = - gt (f) 장 = 값t， YJ= Yi - ~gt2
(g) 3.19 s (h) 36.1 m/s , -60 .1。
14. (a) 28.2 m/s (b) 4.07 s (c) the required initial velocity
will increase , the total time of flight will increase
(visin 8)2
15. (a) t = 낀 sin 8/g (b) hmax = h+ ' 'C)~ ,
2g
16. l. 21 s
17. 0.0337 m/s2 directed toward the center ofEarth
18. 7.58 X 10 3 m/s , 5.80 X 10 3 s
19. (a) 6.00 rev/s (b) 1. 52 X 103 m/s2 (c) l. 28 X 10 3 m/s2
20. 377 m/s2
2 1. (a) Yes. The particle can be either speeding up or slowing down , with a tangential component of acceleration of
magnitude~효"2 = 3.97 m/s2 (b) No. The magnitude of the acceleration cannot be less than v 2 /r = 4.5
m/s2
A-34
퀴즈 및 연습문제 해답
22. (a) ~
3. (a) (6.00 i + 15.0 j) N (b) 16.2 N
4. (a) -4 .47 x 1015 m / s2 (b) +2.09 X 10- 10 N
5. (a) (-45.oi + 15.0]) m/s (b) 162 0 from the + x axis
smaller
10. 2.38 kN
11. (a) .. =
|
“
·‘
/
1
cb
n
.
、
,
I mbg
FbC
(c) force: normal force of cushion on brick(rtcb)
• reaction force: force of brick on cushion (Fbc )
force: gra띠 tational force of Earth on brick(m b홍)
• reaction force: gra띠 tational force of brick on Earth
force: normal force of pavement on cushion(npc)
• reaction force: force of cushion on pavement
force: gra띠 tational force of Earth on cushion(mc륭)
• reaction force: gra띠 tational force of cushion on
Earth
25.0 、‘
: 25.0
: ~기1
-- - - - -...-
, '-
- ~- -- ---
9.80
/9.80
‘
/
、、
/
~‘(d) 26.8 m / s2 , 2 1. 4。
야터
νιι1
κ
-
12. (a) 3 .4 3 kN (b) 0.967 m / s horizontally forward
13. (a)
m5
i--
. ， - --
즈
퀴
이 이 이 비
ll‘ 、
/
1
------2
3
4
1l ‘ 、
/
/I，‘ 、
mn
m
·κ
me 4i
’
熾
」u
m
야
댔
ιm
ml m
m --‘ m
때
댔
/ t，‘、
나내
υ
j
내냐
/，，‘ 、
F
•
서띠
m
”
mgsin ()
빽
‘\
사U
/I，‘ 、
“v
U1·/
비 이삐
/l·
‘、
6
7
/I，‘ 、
5
κ
、、
INø끼
IIl
t'
、
//
/
\
‘
(b) ;1agnitude: 써(ν체 +
direction: 떠1 (뽕)
18
9. (a) 3.64 X 10- N (b) 8.93 X 10- 30 N is 408 billion times
fficg
(c)
-‘---
8. (a) 옹 vt
에
(b) 29.7 m / s2 (c) 6.67 m / s tangent to the circle
23. (a) 9.80 m / s2 down and 2.50 m / s2 south (b) 9.80 m / s2
down
(c) The bolt moves on a parabola with its axis
downward and tilting to the south. It lands south of the
point directly below its starting point.
(d) The bolt
moves on a parabola with a vertical axis.
24. 153 km/ h at 11. 3 0 north ofwest
25. 18.2 。
26. (a) 2.02 x 10 3 s (b) 1. 67 x 10 3 s (c) Swimming with the
current does not compensate for the time lost swimming
against the current.
27. 27.7 0 E ofN
28. 5i + 4t 3/ 2 j (b) 5ti + 1. 6t 5/ 2 j
29. The ball would not be high enough to have cleared the
24.0-m-high bleachers.
30. (a) 25.0 m / s2 (b) 9.80 m / s2
(c) (-225i + 75.0]) m (d) (-227i + 79.0]) m
6. 1. 59 m / s2 at 65.2 0 N ofE
7. (a) â is at 181 0 (b) 11. 2 kg (c) 37.5 m /s
(d) (-37.5i - 0.893j) m / s
]
m
rI
m m
[연습문제 l
1. 8.71 N
2. (a) force exerted by spring on hand , to the left; force
exerted by spring on wall , to the right (b) force exerted
by wagon on handle , downward to the left; force exerted
by wagon on planet , upward; force exerted by wagon
on ground , downward (c) force exerted by football on
player, downward to the right; force exerted by football
on planet , upward (d) force exerted by small-mass object
on large-mass object , to the left (e) force exerted by
negative charge on positive charge , to the left (f) force
exerted by iron on magnet , to the left
mg
(b) -2.54 m /s2 (c) 3 .1 9 m /s
15. (a) •
T1
(b) 613 N
퀴즈 및 연습문제 해답
A-35
(b) 9.80 N , 0.580 m /s2
29. (c) 3.56 N
N
30. (a)
4||
때
(b) 6.30 m / s2 (c) 3 1. 5 N
18. (a) 건 > 19.6 N (b) 건 드 -78 .4 N
(c)
|i------,
뻐
19. B =
3.37 X 10 3 N , A
=
3.83 X 10 3 N , B is in tension and A
20. The situation is impossible because maximum static friction cannot provide the acceleration necessary to keep
the book stationary on the seat.
2 1. (a) 14.7 m (b) neither mass is necessary
22. (a) 4.18 (b) Time would increase , as the wheels would
skid and only kinetic friction would act; or perhaps the
car would flip over.
23. 37.8 N
-•n
24. (a)
끓f
-•
T12 ~
n
480 N
2
(b) 0 .4 08 m /s
(c) 83.3 N
Chapter 6
I퀴즈1
1. (i) (a)
(ii) (b)
2. (i) Because the speed is constant, the only direction the
force can have is that of the centripetal acceleration.
The force is larger at @ than at @ because the radius at
@ is smaller. There is no force at @ because the wire is
straight. (ii) In addition to the forces in the centripetal
direction in part (a) , there are now tangential forces to
provide the tangential acceleration. The tangential force
is the same at all three points because the tangential
acceleration is constan t.
-•
를 T 23
-•
m2g
-•
T 12
-•
320N
IS III compress lO n.
N
-•
T 23
-•
F
-•
m3g
-•
Ft
@
삐
-•
mlg
(b) 2.31 m/s 2, down for m 1, left for m 2 , and up for m3
(c) 낀 2 = 30.0 N and 자 3 = 24.2 N
(d) 지 2 decreases and 자 3 increases
25. Yes, the coefficient of static friction is μs = 0.6 , which is
greater than regulated minimum value of 0.5.
26. (a) 48.6 N , 3 1.7 N
(b) IfP > 48.6 N , the block slides up the wal l.
IfP < 31. 7 N , the block slides down the wal l.
(c) 62.7 N , P> 62.7 N , the block cannot slide up the wal l.
If P < 62.7 N , the block slides down the wal l.
27. 834 N
28. (a) The free-body diagrams are shown in the figure below.
45.0N
____.
and η1 appear in both diagrams as action-reaction
palrs.
/1
m
3. (c)
4. (a)
[언습문제】
1. (a) 8.33 X 10 냉 N toward the nucleus
(b) 9.15 X 10 22 m / s2 inward
2. (a) 1. 65 X 10 3 m/s (b) 6.84 X 10 3 s
3. (a) (-0.233i + 0.163]) m / s2
(b) 6.53 m / s (c) (-0.181i + 0.181]) m / s2
4. 215 N , horizontally inward
5. 6.22 X 10- 12 N
6. The situation is impossible because the speed of the
0에 ect is too small , requiring that the lower string act like
a rod and push rather than like a string and pul l.
7. (a) no (b) yes
8. The radius of curvature is larger than 150 m , so the
driver is not justified in his claim as to faulty design of
the roadway.
9. (a) 1. 33 m / s2 . (b) 1. 79 m /s2 at 48.0 0 inward from the
direction of the velocity
10. (a) 4.81 m / s (b) 700 N
A-36
퀴즈 및 연습문제 해답
1_ f 2T
11. (a) v = 、/κ \\
ν
\
--=- - g J (b) 2Tup
ηl
~
I
12. (a) 20.6 N (b) 32.0 m/s2 inward , 3.35 m/s2 downward
tangent to the circle (c) 32.2 m/s2 inward and below the
cord at 5.98 0 (d) no change (e) acceleration is regardless of the direction of swing
13. (a) 8.62 m (b) Mg, downward (c) 8 .4 5 m/s2 (d) Calculation of the normal force shows it to be negative , which
is impossible. We interpret it to mean that the normal
force goes to zero at some point and the passengers wiU
fall out of their seats near the top of the ride if they are
not restrained in some way. We could arrive at this same
result without calculating the normal force by noting that
the acceleration in part (c) is smaller than that due to
gravity. The teardrop shape has the advantage of a larger
acceleration of the riders at the top of the arc for a path
having the same height as the circular path , so the passengers stay in the cars.
14. (a) 3.60 m/s2 (b) T = 0 (c) noninertial observer in the
car claims that the forces on the mass along χ are T and
a fictitious force (-Ma) (d) inertial observer outside the
car claims that T is the only force on M in the χ direction
15. (a) 491 N (b) 50 .1 kg (c) 2.00 m/s2
2(vt - L)
4. (a)
5. (b)
6. (c)
7. (i) (c)
8. (d)
(ii) (a)
I연습문제1
1. (a) l. 59 X 10 3 J (b) smaller (c) the same
2. (a) 472J (b) 2.76 kN
3. method one: -4.70 X 10 3 J , method two: -4.70 젠
5. 28.9
6. 5.33J
7. (a) 1l. 3 0 (b) 1560 (c) 82.3 。
8. (a) 7.50 J (b) 15.0 J (c) 7.50 J (d) 30.0 J
9. 7.37 N/m
10. (a) 0.938 cm (b) l. 25 J
11. Each spring should have a spring constant of 316 N/m.
12. (a) -l. 23 m/s2 , 0.616 m/s2 (b) -0.252 m/s2 if the force
of static friction is not too large , zero (c) 0
13. (b) mgR
14. (a) F(N)
25 120 •
15 i-10 •
5 1--
16. - - -
.......... .y'
_,.,,~
o ν-
(g+ α) t"-
17. 0.527 。
18. 0.212 m/s~ ， opposite the velocity vector
19. (a) 2.03 N down (b) 3.18 m/s~ down (c) 0.205 m/s down
20. (a) 32.7 çl (b) 9.80 m/s~ down (c) 4.90 m/s2 down
21. (a) l. 47 N . s/ m (b) 2.04 X 10- 3 s (c) 2.94 X 10- 2 N
23. 10 1 N
24. (a) At A , the velocity is eastward and the acceleration is
southward.
(b) At B , the velocity is southward and the acceleration is
westward.
25. 781 N
26. (a) 1.1 5 X 10 4 N up (b) 14 .1 m/s
ηw 2
27. (a) mg- 그~ (b) VgR
29. (a) v = V i e- bl/
1Il
(b)
v
Vi
0
(c) In this model , the 0비 ect keeps moving forever.
(d) It travels a finite distance in an infinite time interva l.
30. (a) 217 N (b) 283 N (c) 간 > 낀 always , so string 2 will
break first
o
0.05
(c)
1. (a)
2. (c) , (a) , (d) , (b)
3. (d)
0.15
0.2
L(m)
(b) The slope of the line is 116 N/m. (c) We use all the
points listed and also the origin. There is no visible evidence for a bend in the graph or nonlinearity near either
end. (d) 116 N/m (e) 12.7 N
15. (a) 0.600 J (b) -0.600 J (c) l. 50 J
16. (a) l. 94 m/s (b) 3.35 m/s (c) 3.87 m/s
17. 878 kN up
18. The dart does not reach the ceiling.
19. (a) 2.5J (b) -9.8J (c) -12J
20. (a) 애 = 0 , 2.59 X 10 5 J
(b) 돼 = 0 , -2.59 X 10 5 J , -2.59 X 10 5 J
21. (a) -196 J (b) -196 J (c) -196 J (d) The gravitational
force is conservative.
23. (a) 125 J (b) 125 J (c) 66.7 J (d) nonconservative
(e) The work done on the particle depends on the path
followed by the particle.
24. The book hits the ground with 20.0 J of kinetic energy.
The book-Earth now has zero gravitational potential
energy, for a total energy of 20.0 J , which is the energy
put into the system by the librarian.
25. (a) 40.0 J (b) -40.0 J (c) 62.5 J
26. (17 - 9χ상)i - 3χ3i
27. (a) Fx is zero at points A , C , and E; 뀐 is positive at point B
and negative at point D (b) A and E are unstable , and C
is stable
Fχ
Chapter 7
[퀴즈】
0.1
B
퀴즈 및 연습문제 해답
28.
Stable
Unstable
Neutral
29. (a) U(x) = 1 + 4e- 2x
(b) The force must be conservative because the work the
force does on the particle on which it acts depends only
on the original and final positions of the particle , not on
the path between them.
30. 0.131 m
Chapter 8
[쿠|즈]
1. (i) (b)
(ii) (b)
(iii) (a)
2. (a)
3. V 1 = v 2 = V 3
4. (c)
I 연습문제】
1. (a) Â. K + ð. U= 0 , V = 냉항 (b) Â. K = W, V = 얘학
2. (a) l. 85 x 10 4 m , 5.10 x 10 4 m (b) l. 00 X 10 7 J
3. (a) 5.94 m / s , 7.67 m / s (b) 147 J
4. (a) 1.1 1 X 10 9 J (b) 0.2
5. 5.49 m/ s
6.1빨
near terminal speed.
15. Both trails result in the same speed.
16. (a) 8.01 W (b) Some of the energy transferring into the
system of the train goes into internal energy in warmer
track and moving parts and some leaves the system by
sound. To account for this as well as the stated increase
in kinetic energy, energy must be transferred at a rate
higher than 8.01 W.
17. $145
18. The power of the sports car is four times that of the older一
model car.
19. ~10 4 W
20. Your grandmother can accompany you.
2 1. (a) 423 m i/gal (b) 776 m i/gal
22. (a) 854 (b) 0 .1 82 hp (c) This method is impractical
compared to limiting food intake.
23. (a) 0.225 J (b) -0.363 J (c) no (d) It is possible to find
an effective coefficient of friction but not the actual value
of μ since η and fvary with position.
24. ~102W
25. (a) 2 .49 m / s (b) 5 .45 m / s (c) l. 23 m (d) no (e) Some
of the kinetic energy of m2 is transferred away as sound
and to internal energy in m 1 and the floo r.
26. (a) x = -4.0 mm (b) -l. 0 cm
27. We find that her arms would need to be l. 36 m long to
perform this task. This is significantly longer than the
human arm .
28. (a) -1 490 J (b) -7 570 J (c) 4590 J
1
9
1
9
" .
/ 1
9
1
9\
29. (a) 2m따 -2 mvt (b) -mgh-~2mv;-2mvt)
1
') 1
')
(c) 2m따 - 5 muf + mgh
7. (a) -168] (b) 184J (c) 500 J (d) 148 J (e) 5.65 m/ s
8. (a) 650 J (b) 588 J (c) 0 (d) 0 (e) 62.0 J (f) l. 76 m / s
9. (a) 5.60] (b) 2.29 rev
10. (a) vB = 1. 65 m/ s2 (b) green bead
11. (a) 22.0] , 40.0 J
(b) Yes
(c) The total mechanical
energy has decreased , so a nonconservative force must
have acted.
12. (a) 0.381 m (b) 0.371 m (c) 0.143 m
13. (a) Isolated. The only external influence on the system
is the normal force from the slide. but this force is
always perpendicular to its displacement so it performs
no work on the system. (b) No , the slide is frictionless.
(c) Es)'stem = mgh
(e) Esystem
A-37
(d) Es)'stem = 한ηgh + !mvl
mgymax + !mvxl
j8gh , ,
1h
4
')
(f) 까 = 、/ 좋 (g) Ym 값 = h(l - gCOS 2 fJ) (h) If friction is
present, mechanical energy of the system would ηotbe conserved , so the child ’s kinetic energy at all points after leaving the top of the waterslide would be reduced when compared with the frictionless case. Consequently, her launch
speed and maximum height would be reduced as well.
14. (a) 24.5 m / s (b) Yes. This is too fast for safety (c) 206 m
(d) The air drag is proportional to the square of the skydiver's speed , so it will change quite a bi t. It will be larger
than her 784-N weight only after the chute is opened. It
will be nearly equal to 784 N before she opens the chute
and again before she touches down whenever she moves
30. (a) l. 53 J at χ = 6.00 cm , 0 J at x = 0 (b) l. 75 m/ s
(c) l. 51 m/ s (d) The answer to part (c) is not half the
answer to part (b) , because the equation for the speed of
an oscillator is not linear in position
Chapter 9
【퀴즈l
1. (d)
2. (b) , (c) , (a)
3. (i) (c) , (e) (ii) (b) , (d)
4. (a) All three are the same. (b) dashboard , seat belt, air bag
5. (a)
6. (b)
7. (b)
8. (i) (a) (ii) (b)
【연습문제】
1. (b) P = V감E
2. (a) Px = 9.00 kg' m / s, Py = -12.0 kg . m / s
(b) 15.0 kg . m/ s
3. Fon ba l = (326 i - 399i) kN, where positive x is from the
pitcher toward home plate and positive y is upward.
4. (a) 4.71 m/ s East (b) 717 J
5. (a) -6.00 i m / s (b) 8 .4 0 J (c) The original energy is
in the spring. (d) A force had to be exerted over a dis-
A-38
퀴즈 및 연습문제 해답
placement to compress the spring , transferring energy
into it by work. The cord exerts force , but over no displacemen t.
(e) System momentum is conserved with
the value zero.
(f) The forces on the two blocks are
internal forces , which cannot change the momentum
of the system; the system is isolated. (g) Even though
there is motion afterward , the final momenta are of
equal magnitude in opposite directions, so the final
momentum of the system is sti ll zero.
6. 10- 23 m / s
7. (c) no difference
8. 3.2 x 1O :~ N
9. (a) 12.0i N . s (b) 4.80i m / s (c) 2.80i m / s (d) 2 .4 0 i N
(c) Most of the
10. (a) 20.9 m / s East (b) -8.68 x 10 3 J
e nergy was transformed to internal energy with some
being carried away by sound.
11. (a) 2.50 m / s (b) 37.5 년
12. (a) 2.50 m / s (b) 37.5 kJ (c) The event considered in
this problem is the time reversal of the perfectly inelastic
collision in previous problem. The same mome ntum conservation equation describes both processes.
ηI
l3· (a) uf = J(U
+. 2U ) (b) AK = --(U12
+ ug - 2U U2)
3
3
\~ ]
-~21
\~I
-~~
14. (a) 4.85 m / s (b) 8.41 m
15. The defendant was traveling at 4 1. 5 m i/ h.
16. V o = 3.99 m / s and V y = 3.01 m / s
17. (a) The opponent grabs the fullback and does not let go ,
so the two players move together at the end of their interaction (b) e = 32 .3 0 , 2.88 m /s (c) 786 J into internal
e nergy
U
18. v = -느 45.0 0 • -45.0。
、
2
’
19. 11.7 cm , 13.3 cm
20. 3.57 X 10 8 J
2 1. (a) 15.9 g (b) 0 .1 53 m
22. (a)
y
ι
•
.....
(c) Yes , we could
the floor or from the Earth through the floor
(d) No.
The kinetic energy came from the original gravitational
potential energy of the Earth-elevated load system , in the
a mount 27.0 J
(e) Yes. The acceleration is caused by the
static friction force exerted by the floor that prevents the
wheels from slipping backward.
25. (a) yes (b) no (c) 103 kg . m/s , up (d) yes (e) 88.2 J
(f) No , the energy came from potential energy stored in
the person from previous meals.
26. 15.0 N in the direction of the initial velocity of the exiting
water stream.
27. (a) 442 metric tons (b) 19.2 metric tons (c) It is much
less than the suggested value of 442/ 2.50. Mathematically, the logarithm in the rocket propulsion equation is
not a linear function. Physically, a higher exhaust speed
has an extra-large cumulative effect on the rocket body ’s
fina l speed by counting again and again in the speed the
body attains second after second during its burn.
28. (b)
t (s)
v (m/ s)
0
20
40
60
80
100
120
132
0
224
488
808
1 220
1780
2690
3730
v (m/s)
4α)0
3500
3000 •
2500
2000
1500
10(JO
500
~----
O O 20
40
6O
SO lOO 120 l40
40
60
80 100 120 140
60
80 100 120 180
t(5)
0
20
40
60
80
100
120
132
a (m/ s2)
10 .4
12.1
14 .4
17.9
23 .4
34.1
62.5
125
이 (m/s 2)
1 쩌}
120
1ω
잉〕
60
40
20
OO
20
t (s)
(f)
•%-
(b) (-2.00i - 1. 00j) m
(c) (3.00i - 1. 00j) m /s
(d) (1 5.0i - 5.00j) kg' m / s
(e) (-220i + 140J) μN
over no distance , so it does zero work
say that the final momentum of the card came from
t (s)
v]
x
(c) (- 4.94i + 1. 39J) cm/ s
(b) No. The friction force exerted
by the floor on each stationary bit of caterpillar tread acts
(d)
m,
23. (a) (-2.89i - 1. 39J) cm
24. (a) Yes. 18.0i kg . m /s
(b) (-44.5i + 12.5J) g' cm/ s
(d) (- 2 .4좌 + 1. 56J) cm/ s 2
x(km)
t (s)
x(km)
160
o
20
40
60
80
100
0
2.19
9.23
22.1
42.2
71. 7
140
120 ’
100
80
60
40
120
132
115
153
20 |
00
i~
20
40
t (s)
퀴즈 및 연습문제 해답
29. In order for his motion to reverse under these conditions ,
the final mass of the astronaut and space suit is 30 kg ,
much less than is reasonable.
30.
(빨)壞
Chapter 10
【쿠|즈]
1. (i) (c) (ii) (b)
2. (b)
3. (i) (b) (ii) (a)
4. (b)
5. (b)
6. (a)
7. (b)
[연습문제]
1. (a) 7.27 X 10- 5 rad/ s
(b) Because of its angular speed ,
the Earth bulges at the equator.
2. 144 rad
3. (a) 4.00 rad/ s2 (b) 18.0 rad
4. (a) 3.5 rad (b) increase by a factor of 4
5. (a) 8.21 X 10 2 rad/ s2 (b) 4.21 X 10 3 rad
6. Because the disk ’s average angular speed does not match
the average angular speed expressed as (ω i + ω1) / 2 in
the model of a rigid object under constant angular acceleration , the angular acceleration of the disk cannot be
constan t.
7. ~ 10 7 rev/ yr
8. (a) 25.0 rad/ s (b) 39.8 rad/ s2 (c) 0.628 s
9. (a) 5.77 cm
10. (a) 54.3 rev (b) 12.1 rev/ s
11. (a) 3 .47 rad/ s (b) 1.74 m / s (c) 2.78 s (d) 1. 02 rotations
12. -3.55 N . m
13. (a) 1. 03 s (b) 10.3 rev
14. (a)
A-39
18. η= -0.0398 N . m
19. 10 0 kg . m 2 = 1 kg . m 2
{
{L
20. L = 1
Mχ31 L
CJ M
r Z dm = 1 x~~ dx = ~ --:-1
1
__ _ n
= 윷 MD
J0 L
L 3 I0 ι
2
22. (a) 92.0 kg . m (b) 184J (c) 6.00 m / s , 4.00 m / s ,
8.00 m / s (d) 184J (e) The kinetic energies computed
in parts (b) and (d) are the same.
23. (a) 24.5 m / s (b) no (c) no (d) no (e) no (f) yes
24. 1. 03 X 10- 3 J
Jall mass
a)
/ 2(~~)gh
,.
/
~ '~T
(b) A
쩌+쩌+효
2(m1 -
~)gh
V m R 2 + ~R2+1
1
26. (a) 11. 4 N (b) 7.57 m / s 2 (c) 9.53 m / s (d) 9.53 m / s
27. (a) 2.38 m / s (b) The centripetal acceleration at the top
얘
(2.38 m / s) 2
。
s -=- =
= 12.6 m / sz > g. Therefore , the ball
r
0 .450 m
must be in contact with the track , with the track pushing
downward on it. (c) 4.31 m / s (d) The speed of the ball
turns out to be imaginary. (e) When the ball is projected
with the same speed as before , but with only translational
kinetic energy, there is insufficient kinetic energy for the
ball to arrive at the top of the track.
28. (a) the cylinder (b) v2 / 4gsin () (c) The cylinder does
not lose mechanical energy because static friction does
not work on i t. Its rotation means that it has 50 % more
kinetic energy than the cube at the start , and so it travels
50 % farther up the incline.
29. The demolition company should not have guaranteed an
undamaged smokestack. Strong shear forces act on the
stack as it falls and , without performing an analysis of the
shear strength of the stack , such a guarantee should not
have been made.
30. (a) 12.5 rad/ s (b) 128 rad
A
H'A
'
Chapter 11
【쿠|즈l
1. (d)
2. (i) (a)
3. (b)
(ii) (c)
4. (a)
I연습문제l
1. i + 8.00 j + 22.0 k
2. (a) 740 cm 2 (b) 59.5 cm
3.45.0 。
(b) 0.309 m / s2 (c) 낀 = 7.67 N , 간 = 9.22 N
15. (a) 24.0 N . m (b) 0.0356 rad/ s 2 (c) 1. 07 m / s2
16. (a) For F = 25.1 N , R = 1. 00 m. For F= 10.0 N , R = 25.1 m
(b) No. Infinitely many pairs of values that satisfy this
requirementmayexist: forany F 드 50.0N , R= 25 .1 N. m/ F,
as long as R 드 3.00m.
17. (a) 0.312 (b) 117 N
5. (a) 찍 =F1 + 작 (b) no
6. (a) No (b) No , the cross product could not work out that
way.
7. (a) (-10.0 N . m)k (b) yes
(c) yes
(d) yes
(f) 5.00 j m
8. (-22 .0 kg' m 2/ s)k
9. m(x낌 - yV)k
10. (a) (-9.03 x 10 9 kg' m 2/ s)J (b) No (c) Zero
(e) no
A-40
퀴즈 및 연습문제 해답
11. (a) zero (b) (-mv/ sin 2 8 cos 8/2g)k
(c) (-2mv/sin 2 8 cos 8/g)k
(d) The downward gravitational force exerts a torque on
the pr이 ectile in the negative z direction.
13. mvR[cos (vν R) + lJ k
14. (a) 2[~ i + [2)
(b) The particle starts from rest at the
origin , starts moving into the first quadrant , and gains
speed faster while turning to move more nearly parallel
(c) (12 퍼 + 2)) m l 강
(e) -40[3k N. m
(d) (60ti + 10)) N
2
(f) -10t k kg' m Is
(g) (90 염 + 10 견)]
(h) (360t 3 + 20t) W
to the x axis
15. (a) - mCgt cos 8 k
torque on the bal l.
16.
L>
29. (a)
= TR - TR = 0
(b) monkey and bananas move
upward with the same speed at any instan t. (c) The monkey will not reach the bananas.
30. (a) 3 750 kg . m 2 /s (b) l. 88 년 (c) 3 750 kg . m 2 /s
(d) 10.0 m /s (e) 7.50 띤 (f) 5.62 새
Chapter 12
4
(b) The Earth exerts a gravitational
(c) - mgC cos 8 k
E= (450 kg · m 2/s)k
(a) 0.360 kg . m 2/s
energy is transformed to internal energy in the collision.
(f) 0.0193 m
28. 5 .4 6 X 10 22 N . m
(b) 0.540 kg . m 2 /s
1. (a)
2. (b)
3. (b)
4. (i) (b)
(ii) (a)
(iii) (c)
【언습문제l
1. Safe arrangements: 2-3-1 , 3-1-2 , 3-2-1; dangerous
arrangements: 1-2-3 , 1-3-2 , 2-1-3
2. The situation is impossible because χ is larger than the
remaining portion of the beam , which is 0.200 m long.
3. (3.85 cm , 6.85 cm)
5. x = 0.750 m
6. Sam: 176 N , ]oe: 274 N
7. 177 kg
8. (a)
(b) τcot8
ηUY
(c) T = μs째 (d) μs
웅 cot8
(e) The ladder slips
9. (a) 29.9 N
10. (a)
H
(b) 22.2 N
+
에
17.
19. l. 20 kg . m 2 /s
20. (a) 7.06 X 10 33 kg . m 2 / s, toward the north celestial p이e
(b) 2.66 X 10 40 kg . m 2 /s , toward the north ecliptic p이e
(c) The periods differ only by a factor of 365 (365 days for
orbital motion to 1 day for rotation). Because of the huge
distance from the Earth to the Sun , however, the moment
of inertia of the Earth around the Sun is six orders of
magnitude larger than that of the Earth about its axis.
2 1. 8.63 m /s2
22. 9 .4 km
23. (a) The mechanical energy of the system is not constan t. Some potential energy in the woman ’s body from
previous meals is converted into mechanical energy.
(b) The momentum of the system is not constan t. The
turntable bearing exerts an external northward force on
the axle. (c) The angular momentum of the system is
constant. (d) 0.360 rad /s counterclockwise (e) 99.9]
24. (a) 2.91 s (b) Yes , because there is no net external torque
acting on the puck-rod-putty system (c) No , because the
pivot pin is always pulling on the rod to change the direction of the momentum (d) No. Some mechanical energy
is converted into internal energy. The collision is perfectly
inelastic.
25. (a) 11 .l rad /s counterclockwise (b) No , 507 ] is transformed into internal energy in the system. (c) No , the
turntable bearing promptly imparts impulse 44.9 kg' m/s
north into the turntable-clay system and thereafter keeps
changing the system momentum as the velocity vector of
the clay continuously changes direction.
26. When the people move to the center, the angular speed
of the station increases. This increases the effective gravity by 26 %. Therefore , the ball will not take the same
amount of time to drop
27. (a) yes (b) 4.50 kg . m 2/ s (c) No. In the perfectly inelastic collision , kinetic energy is transformed to internal
energy.
(d) 0.749 rad/ s
(e) The total energy of the
system must be the same before and after the collision ,
assuming we ignore the energy leaving by mechanical
waves (sound) and heat (from the newly-warmer door to
the cooler air) . The kinetic energies are as follows: KEi =
2.50 X 10 3 J; K타 = l. 68 J. Most of the initial kinetic
【퀴즈]
T 、‘
30.0 0~
4
|
_1196N
(b) 392 N (c) 339 N to the right (d) 0
(e) V = 0 (f) 392 N (g) 339 N to the right
(h) The two solutions agree precisely. They are equally
accurate.
11. (a) 27.7 kN (b) 1l. 5 kN (c) 4 .l 9 kN
12. (a) No time interva l. The horse ’s feet lose contact with
the drawbridge as soon as it begins to move
(b) l. 73 rad/s (c) 2.22 rad/s
(d) 6.62 kN. The force at the hinge is (4.72i + 6.6건)kN
(e) 59.1 앤
13. (a) l. 04 kN at 60.0 0 upward and to the right
(b) (370 i + 910 j) N
퀴즈 및 연습문제 해답
14. (a)
A-41
Chapter 13
I쿠|즈l
1. (e)
2. (c)
3. (a)
4. (a) perihelion (b) aphelion (c) perihelion (d) all points
【연습문제】
1. 7.41 X 10- 10 N
2. (a) 4.39 X 10 20 N
(b) 218 N
15. (a) 859 N
(c) 72 .4 N (d) 2 .4 1 m
(b) 1. 04 kN at 36.9 to the left and upward
16. (a)
?nrr';2 Rh ~
0
(R - h) cos
(b)
h2
e - ';2 Rh - h2 sin e
ηl，g낸굶:과 cose
~
(R - h)cos e - ';2 Rh - h2 sin e
L+
and mgll
.;2Rh - h2 cos
e~
~- ~ ~
’
-“ ‘
’.
II
l (R - h) cos e - ';2Rh - h2 sin e J
17. (a) -0.053 8 m 3 (b) 1. 09 X 10 3 kg/ m 3 (c) With only a
5 % change in volume in this extreme case , liquid water
can be modeled as incompressible in biological and student laboratory situations.
18. ~ 1 cm
19. 23.8 μm
20. 1. 0 X 1011 N / m 2
2 1. (a) 3.1 4 X 10 4 N (b) 6.28 X 10 4 N
22. 1. 65 X 10 8 N / m 2
23. 9.85 X 10- 5
24. (a) Rigid 0에 ect in sta tÏ c equilibrium
(b)
Y
Img
n,
:→---4.00 m--커
(c) The woman is at x = 0 when η1 is greatest (d) η1=0
(e) l. 42 X 10 3 N (f) 5.64 m (g) same as answer (f)
25. ηA = 5.98 x 10 5 N , ηB = 4.80 X 10 5 N
26. 0.896 m
27. (a)
0
- -표딴뤘
잃oN
(b) T = 343 N , Rx = 171 N to the right , Ry = 683 N up
(c) 5 .1 4 m
28. (a) T = 1. 46 kN (b) H = l. 33 kN , V = 2.58 kN
29. (a) T = 갤 (L + d)/[sin e (2L + d)]
(b) Rx = 갤 (L + d) cot e/ (2L + d); Ry = 갤L/ (2L + d)
30. (a) 5.08 kN (b) 4.77 kN (c) 8.26 kN
(b) 1. 99 X 10 20 N (c) 3.55 X 10 22 N
(d) The force exerted by the Sun on the Moon is much
stronger than the force of the Earth on the Moon.
3. ~1O -7 N
4. The situation is impossible because no known element
could compose the spheres.
5. (a) 7.61 cm/ s2 (b) 363 s (c) 3.08 km
(d) 28.9 m / s at 72.9 0 below the horizontal
2MGr
6. (a)
- ~ . ~--::.~ .~ toward the center ofmass
(? + a 2)
(b) At r = 0 , the fields of the two 0에 ects are equal in
magnitude and opposite in direction , to add to zero
(c) As r • η ， 21vIGr(? + a 2f 3/2 approaches 2 1V1G(0) /α3 =0
(d) When r is much greater than a, the angles the field
vectors make with the x axis become smaller. At very
great distances , the field vectors are almost parallel to
the axis; therefore they begin to look like the field vector
from a single obj ect of mass 2M
(e) As r becomes much larger than a , the expression
approaches 2MGr(션 + 02r 3/2 = 2MGr/? = 2MG/ ?
as required.
7. (a) 1. 31 X 1017 N (b) 2.62 X 1012 N/ kg
8. l. 50 h or 90.0 min
9. (a) 0.708 yr (b) 0.399 yr
10. (a) The particle does posses angular momentum because
it is not headed straight for the origin.
(b) Its angular momentum is constan t. There are no
identified outside influences acting on the objec t.
11. 4.99 days
12. l. 27
13. (a) yes (b) 3.93 yr
14. (a) 6.02 X 1024 kg (b) The Earth wobbles a bit as the
Moon orbits it , so both objects move nearly in circles
about their center of mass , staying on opposite sides of i t.
15. 4 .1 7 X 1010 J
16. (b) 340 s
17. (a) -l. 67 X 10- 14J (b) The particles collide at the center
of the triangle.
19. 1. 58 X 1010 J
GM,cm
20. ---:-;::;:二12RE
21. l. 78 X 10 3 m
22. (a) 42 .1 km/ s (b) 2.20 X 1O11 m
23. (a) same size force (b) 15.6 km/ s
24. (a) 3.07 X 10 6 m (b) the rocket would travel farther from
Earth
25. 492 m/ s
A-42
퀴즈 및 연습문제 해답
26. For the typical data provided for a neutron star, the gravitational acceleration is an order of magnitude larger
than the centripetal acceleration.
27. l. 30 X 10 3 m /s
/ GM
28. If one uses the result v = ~ 당二 and the relation v
(2πγ T) , one finds the radius of the orbit to be smaller
than the radius of the Earth , so the spacecraft would
need to be in orbit underground.
29. (a) l. 00 X 10 7 m (b) l. 00 X 104 m / s
30. (c) l. 85 X 10- 5 m / s2
Chapter 14
【쿠|즈]
1. (a)
2. (a)
3. (c)
4. (b) or (c)
5. (a)
[연습문제】
1. 2.96 X 10 6 Pa
2. (a) ~4 X 10 17 kg/ m 3
3. 5.27 X 1018 kg
4. The situation is impossible because the longest straw
Superman can use and still get a drink is less than 12.0 m.
5. 7.74 X 10- 3 m 2
6. 2.71 X 10 5 N
7. 0.0721 mm
8. (a) l. 57 kPa , 0.015 5 atm , 11.8 m (b) Blockage of the
fluid within the spinal column or between the skull and
the spinal column would prevent the fluid level from
nsmg.
9. (a) 10.5 m (b) No. The vacuum is not as good because
some alcohol and water in the wine will evaporate. The
equilibrium vapor pressures of alcohol and water are
higher than the vapor pressure of mercury.
10. (a) P = Po + pgh (b) Mg/ A
11. 333 × 103 kg/m3
12. (a) 1 250 kg/ m 3 (b) 500 kg/ m 3
13. (a) B = 25.0 N (b) horizontally inward (c) The string
tension increases. The water under the block pushes
up on the block more strongly than before because the
water is under higher pressure due to the weight of the
oil above it (d) 62.5 %
14. (a) 408 kg/ m 3 (b) When m is less than 0.310 kg , the
wooden block will be only partially submerged in the
water. (c) When m is greater than 0.310 kg , the wooden
block and steel object will sink.
16. (a) 3.7 kN (b) l. 9 kN (c) Atmospheric pressure at this
high altitude is much lower than at the Earth ’s surface.
17. 20.0 g
18. (a) 0 .471 m / s (b) 4.24 m / s
19. (b) 616 MW
20. (a) 2.28 N toward Holland (b) l. 74 X 10 6 s
2 1. (a) (3.93 x 10-6 m 3/ s) lð:P where ðPis in pascal
(b) 0.305 L/ s (c) 0 .4 31 L/ s
22. (a) 28.0 m / s
(b) 28 .0 m / s (c) The answers agree
precisely. The models are consistent with each other.
(d) 2.11 MPa
23. (a) Answers will vary, but will depend on the Bernoulli
effec t. (b) 452 N (c) l. 81 X 10 3 N
24. 0.120 N
25. 1. 51 X 10 8 Pa
26.0.200 mm
27. (a) 6.80 X 10 4 Pa
(b) Higher. With the inclusion of
another upward force due to def1 ection of air downward ,
the pressure difference does not need to be as great to
keep the airplane in fligh t.
28. (a) 4 .4 3 m / s (b) 9 .1 0 m
29. (a) particle in equilibrium
tb1 ZFy = B - Fb - FHe - Fs = 。
(이 m s = 원Pair - P He ) πr 3 -
mb
(d) 0.0237 kg (e) 0.948 m
Chapter 15
[쿠|즈]
1. (d)
2. (f)
3. (a)
4. (b)
5. (c)
6. (i) (a)
(ii) (a)
I연습문제】
1. (a) 17 N to the left (b) 28 m /s2 to the left
2. (a) 18.8 m / s (b) 7.11 km/ s2
3. (a) 1. 50 Hz (b) 0.667 s (c) 4.00 m (d) π rad (e) 2.83 m
4.40.9 N/ m
5. (a) -2.34 m (b) -l. 30 m / s (c) -0.0763 m (d) 0.315 m / s
6. (a) motion is periodic (b) l. 81 s (c) The motion is not
simple harmonic. The net force acting on the ball is a
constant given by F = -mg (except when it is in contact
with the ground) , which is not in the form of Hooke ’s law.
7. (a) χ = 2.00 cos (3.00 '11" t - 90 0 ) or x = 2.00 sin (3.00 '11" t)
where xis in centimeters and lis in seconds (b) 18.8 cm/ s
(c) 0.333 s (d) 178 cm/ s2 (e) 0.500 s (f) 12.0 cm
9. (a) yes (b) The value of k in Equation 15.13 is proportional to the mass m , so the mass cancels in the equation ,
leaving only the extension of the spring and the acceleration due to gravity in the equation: T = 0.859 s.
10. 2.23 m / s
11. 2.60 cm or -2.60 cm
12. (a) E increases by a factor of 4 (b) V m ax is doubled
(c) αmax also doubles (d) the period is unchanged.
13. (a) 앓 (b) 랜 (c) x = :!::않A
(d) No; the maximum potential energy is equal to the
total energy of the system. Because the total energy must
remain constant , the kinetic energy can never be greater
than the maximum potential energy.
14. (a) Particle under constant acceleration (b) 1. 50 s
(c) isolated (d) 73 .4 N/ m (e) 19.7 m below the bridge
(f) 1. 06 rad/ s (g) +2.01 s (h) 3.50 s
퀴즈 및 연습문제 해답
15. (a) 4.58 N
(b) 0.125 J (c) 18.3 m/s 2 (d) 1. 00 m/s
(e) smaller (f) the coefficient ofkinetic friction between
the block and surface (g) 0.934
16. (a) The motion is simple harmonic. (b) 0.628 s
(c) 1. 00 kg (d) 0.800 m/s
17. (a) 1. 50 s (b) 0.559 m
19. 0.944 kg . m 2
ηu!a
20. I=~
4πζfζ
21. (a) 0.820 m/s (b) 2.57 rad/s 2 (c) 0.641 N
(d) V max = 0.817 m/s , a max = 2.54 rad/s 2, Fmax = 0.634 N
(e) The answers are close but not exactly the same. The
angular amplitude of 150 is not a small angle , so the
simple harmonic oscillation model is not accurate. The
answers computed from conservation of energy and from
Newton ’s second law are more accurate.
23. (a) 5.00 X 10- 7 kg . m 2 (b) 3.16 x 10- 4 N . m/rad
26. (a) 3.16 çl (b) 6.28 çl (c) 5.09 cm
72E
28. k = ~ ~- ~
V2 a- ζ
29.
I
1
kh 2
-=--=、 JgL +---;-:;2'TTL \1
30. If the damping constant is doubled , b/2m = 120 çl. In
this case , however, b/2m > ω。 and the system is overdamped. Your design objective is not met because the system does not oscillate.
Chapter 16
【퀴즈]
1. (i) (b) (ii) (a)
2. (i) (c) (ii) (b)
3. (c)
4. (f) and (h)
5. (d)
(iii) (d)
6. (c)
7. (b)
8. (b)
9. (e)
10. (e)
11. (b)
[연습문저|】
1. 184 km
2. (a) longitudinal P wave (b) 666 s
3. (a) L = (380 m/s)~t (b) 48.2 m (c) 48 cm
4. 2 .4 0 m/s
5. (a)
6. :::!::6.67 cm
7. 185 m/s
8. 13.5 N
9. (a) 0.051 0 kg/m (b) 19.6 m/s
10. (a) 1 (b) 1 (c) 1 (d) increased by a factor of 4
11. (a) As for a string wave , the rate of energy transfer is proportional to the square of the amplitude to the speed.
The rate of energy transfer stays constant because each
wavefront carries constant energy, and the frequency
stays constan t. As the speed drops , the amplitude must
increase (b) The amplitude increases by 5.00 times
12. (a) y = 0.075 sin (4 .l9x - 314t) , where χ and y are in
meters and t is in seconds (b) 625 W
15. (b) J(x + vt) = 웅(χ + νt)2 and g(x - vt) = 융(x- vd
(c) J(χ + vt) = 웅 sin(x+ νt) and g(χ - vt) = 융 sin(χ - vt)
16. (a) 2.00 μm (b) 40.0 cm (c) 54.6 m/s (d) -0 .433 μm
(e) 1. 72 mm/s
17. 1 X 1O11 Pa
18. 5.81 m
19. (a) The speed gradually changes from v = (331 m/s) (1 +
27 oC/273 OC)1/ 2 = 347 m/s to (331 m/s)(1 + 0/273 OC)1/ 2
=3진 앉//s ， a 4.5 % 낀ecrease. The cooler air at the same
pressure is more dense (b) The frequency is unchanged
because every wave crest in the hot air becomes one
crest without delay in the cold air (c) The wavelength
decreases by 4.6 %, from νσ = (347 m/s) (4 OOO/s) = 86.7
mm to (331 m/s)(4 OOO/s)
82.8 mm. The crests are
more crowded together when they move more slowly.
20. 335 m/s
2 1. (a) 153 m/s (b) 614 m
22. (a) 3.75 W/m 2 (b) 0.600 W/m 2
23. (a) 0.691 m (b) 691 km
24. We assume that both lawn mowers are equally loud and
approximately the same distance away. We found in
Example 16.8 that a sound of twice the intensity results in
an increase in sound level of 3 dB. We also see from the
What If? section of that example that a doubling of loudness requires a lO -dB increase in sound leve l. Therefore ,
the sound of two lawn mowers will not be twice the loudness , but only a little louder than one!
25. (a) B (b) positive (c) negative
(d) 1 533 m/s (e) 5.30 X 10 3 Hz
26. 2.82 X 10 8 m/s
27. This is much faster than a human athlete can run.
28. (a) 441 Hz (b) 439 Hz (c) 54.0 dB
29. 14.7 kg
30. 0.883 cm
y (cm)
Chapter 17
m
w
【퀴즈l
0
t (s)
-10
(b) 0.125 s (c) This agrees with the period found in the
example in the tex t.
A-43
1. (c)
2. (i) (a)
3. (d)
4. (b)
5. (c)
(ii) (d)
A-44
퀴즈 및 연습문제 해답
y( m)
【 연습문제】
1. (a) -1. 65 cm
(b) -6.02 cm
(c) 1. 15 cm
2. (a) I Y (111)
3.
y (cm)
y (cm)
6
t= 1s
4
4
2
2
0
0
x
4
8
0
12
14
16
x
4
8
12
4
8
12
y (cm)
y (cm)
10
0
12
18 x (m)
4
2
y (m)
--
I
’
'\1\
0|
0
4
l
l
8
12
χ
0
χ
0
y (cm)
6
t=
3s
4
2
01
o
4
-
8
-
12
• x
4. The man walks only through two minima; a third minimum is impossible.
5. (a) Yl: positive χ direction; Y2: negative χ direction
(b) 0.750 s (c) 1. 00 m
7. (a) The separation of adjacent nodes is ðχ = 깐 = 죠. The
k
2
nodes are still separated by half a wavelength.
(b) &s The nodes are located aI K +
$ ηπ
=
so th따
x= 깐 - 쁘‘ which means that each node is shifted 쁘
k
2k'
2k
to the left by the phase difference between the traveling
waves in comparison to the case in which φ = o.
퀴즈및연습문제해답
8. (a)
8
F/\
6
4
1
2
0
\
i
J
까
\
4
\
.*
4
I
/
/\
\
。、
\
\
\../
I
I
/\
、
78
\
10
/
\.._/
。
4
I=OS
8
6
4
3
0
1
J
4
4
6
JO
S
I
4
4
t= 5 ms
r
|
:J
‘|
l1\
-~ I\
/
/、
、
/"
/
/
Þ
J
\、
/
아
끼s
\
J
、
씨
\
1m
,
1‘
J
A-45
10. (a) 5.20 m (b) No. We do not know the speed ofwaves on
the string.
11. (a) 78.6 Hz (b) 157 Hz, 236 Hz, 314 Hz
12. 1. 86 g
Mgcos8
13. m=-~-4I(.L
14. 291 Hz
15. The resonance frequency of the bay calculated from the
data provided is 12 h , 24 min. The natural frequency of
the water sloshing in the bay agrees precisely with that
of lunar excitation , so we identify the extra-high tides as
amplified by resonance.
16. 57.9 Hz
17. (a) 0.656 m (b) 1. 64 m
18. (a) 349 m/s (b) 1. 14 m
19. η(206 Hz) and η(84.5 Hz)
20. η(0.252 m) with η = 1, 2 , 3 , ...
2 1. (a) 0.085 8η Hz , with η = 1, 2 , 3 ..
(b) It is a good rule. A car horn would produce several or
many of the closely-spaced resonance frequencies of the
air in the tunnel , so it would be great amplified.
22. 158 s
___ 2
23. 二，
v
2R!
t= 10 ms
8
6
4
3
0
](’
J
‘
4
24. -1O .0 o C
25. It is impossible because a single column could not produce both frequencies.
26. (a) 1. 99 beats/s (b) 3.38 m/s
27. (a) 521 Hz or 525 Hz (b) 526 Hz (c) reduced by 1. 14 %
28. The coefficients beyond η = 1 are approximate: Al = 100 ,
A 2 = 156 , A3 = 62 , A4 = 104 , As = 52 , A6 = 29 , A7 = 25.
ι1얘
n
y (nm)
400
t
= l5 ms
oa
6
‘‘
3
0
1
--
•
4
46
r뜻}、 ß
\../
t= 20 ms
(b) In any one picture , the wavelength is the smallest
distance along the x axis that contains a nonrepeating
shape. The wavelength is A = 4 m.
(c) The frequency is the inverse of the period. The
period is the time the wave takes to go from a full amplitude starting shape to the inversion of that shape and
then back to the original shape. The period is the time
interval between the top and bottom graphs: 20 ms. The
frequency is 1/0.020 s = 50 Hz.
(d) 4 m. By comparison with the wave function. y
(2A sin kχ) cos ω t， we identify k = π/2 and then compute
A=2π/k.
(e) 50 Hz. By comparison with the wave function y
(2A sin kχ:) cos ω t， we identify ω =2πj= 100 71'.
9. (a) 0.600 m
(b) 30.0 Hz
0
-400
29. 146 Hz
mg(L- d)
• (a)
-;:=~-=--2\" L" - 2dL
(b) 호
2d
Chapter 18
[퀴즈】
1. (c)
2. (c)
3. (c)
4. (c)
5. (a)
6. (b)
【연습문제]
1. (a) -738 0 N
0
2. (a) 106.7 F
(b) -105 0 N
(c) 270 0 N (d) 153 0 N
A-46
퀴즈및연습문제해답
(b) Yes. The normal body temperature is 98.6 oF, so the
patient has a high fever and needs immediate attention.
3. (a) -109 oF, 195 K (b) 98.6 oF, 310 K
4. (a) -320 oF (b) 77.3 K
5. ßr = 0.663 mm to the right at 78.2 below the horizontal
6. 3.27 cm
7. 55.0 oC
8. 1. 54 km. The pipeline can be supported on ro11ers. In
addition , !1 -shaped loops can be built between straight
sections; these loops bend as the steel changes length.
9. (a) 0.109 cm 2 (b) increase
10. 2.74 m
0
+Li ,haß
([)
1'+ a 2 (ß 1')2
12. (a) 437 0 C (b) 2.1 X 10 3 0 C (c) No; aluminum melts at
660 0 C (Table 19.2). Also , although it is not in Table 19.2 ,
Internet research shows that brass (an alloy of copper
and zinc) melts at about 900 oC.
13. Required l' = -376 oC is below absolute zero.
14. (a) 99.8 mL
(b) It lies below the mark. The acetone has reduced in
volume , and the flask has increased in volume.
15. (a) 2.52 X 10 6 N/m2 (b) the concrete will not fracture
16. (a) 396 N (b) -101 oC (c) The originallength divides
out of the equations in the calculation , so the answers
would not change.
17. In each pump-up-and-discharge cycle , the volume of air
in the tank doubles. Thus 1. 00 L of water is driven out by
the air injected at the first pumping, 2.00 L by the second , and only the remaining 1. 00 L by the third. Each
person could more efficiently use his device by starting
with the tank half full of water, instead of 80 % fu11.
18. 1. 50 X 10 29 molecules
19. (a) 1. 17 X 10- 3 kg (b) 11. 5 mN (c) 1. 01 kN
(d) molecules must be moving very fast
20. (a) 4 1. 6 mol (b) 1. 20 kg
(c) This value is in agreement with the tabulated density.
2 1. 6.64 X 10- 27 kg
22. 2 .4 2 X 1011 molecules
23. 6.58 X 10 6 Pa
24. 473 K
m1 - 앤=료쁨(운-높)
Water +
steam
Ic e
815
Energy added (J)
3. Situation
System
(a) Rapidly pumping
Air in the pump
up a bicycle tire
(b) Pan of roomWater in the pan
temperature water
sitting on a hot stove
(c) Air quickly leaking Air originally in
out of a balloon
the balloon
Q
W AEint
O +
+
+ 0
+
0
4. Path A is iso、Tolumetric ， path B is adiabatic , path C is isothermal , and path D is isobari c.
5. (b)
[ 연습문제】
1. (a) 2.26 X 10 6 ] (b) 2.80 X 10 4 steps (c) 6.99 X 10 3 steps
2. 16.9 C
3. 23.6 0 C
4. 0.234 댐/kg'OC
5. 0.918 kg
0
26. ~102 kg
27. Scenario (ii) is better for a dive of this depth.
11 +a “ T~\
29. (a) ø = 2 sin-l~--2 이 ~) (b) yes
Steam
r
〕딩믹
11. Y =
of energy added. Notice that this graph looks quite different from Figure 19.3 in that it doesn ’ t have the flat
portions during the phase changes. Regardless of how
the temperature is varying in Figure 19.3 , the internal
energy of the system simply increases linearly with energy
input; the line in the graph below has a slope of 1
(c) yes
! 1 +a I \
(d) ø = 2 sin - 11
∞ ι
I (e) 6 1. 0 。 (f) 59.6。
\2(1 + a in잉 r 간 )J
“
30. (b) It assumes a ß 1' is much less than 1.
Chapter 19
[퀴즈】
1. (i) iron , glass , water (ii) water, glass , iron
2. The figure on the next page shows a graphical representation of the internal energy of the system as a function
6. im씨 C꾀 + mccw) J; + η211 Cψ 낀
mAl CAl + m clll + m" clll
,.
7. (a) 1 822 ]/kg . oC (b) We cannot make a definite identification. It might be bery11ium. (c) The material might
be an unknown a110y or a material not listed in the table.
8. (a) 16.1 oC
(b) 16.1 oC
(c) It makes no difference
whether the drill bit is dull or sharp , or how far into the
block it cuts. The answers to (a) and (b) are the same
because all of the work done by the bit on the block constitutes energy being transferred into the internal energy
of the stee l.
9. (a) 25.8 0 C (b) The symbolic result from part (a) shows
no dependence on mass. Both the change in gravitational
potential energy and the change in internal energy of the
퀴즈 및 언습문제 해답
system depend on the mass , so the mass cancels.
10. l. 22 X 10 5 J
11. 2.27 km
12. 0.294 g
13. (a) OOC (b) 114 g
14. (a) 7 (b) As the car stops , it transforms part of its kinetic
energy into internal energy due to air resistance. As soon
as the brakes rise above the air temperature , they transfer energy by heat into the air and transfer it very fast if
they attain a high temperature.
15. (a) -4Pi~ (b) According to T = (P/ηR~)V2 ， it is proportional to the square of the volume.
16. (a) -12.0 MJ (b) +12.0 MJ
17. 720 J
18. From the first law of thermodynamics , ð. E int = Q + W =
10.0 J + 12.0 J = +22.0 J. The change in internal energy
is a positive number, which would be consistent with an
increase in temperature of the gas , but the problem statement indicates a decrease in temperature.
19. (a) 0 .041 0 m 3 (b) +5 .48 년 (c) -5 .4 8 댐
20. (a) -3.10 젠 (b) 37.6 띤
2 1. (a) -0.048 6 J (b) 16.2 앤 (c) 16.2 년
22. (a) 1 300 J (b) 100 J (c) -900 J (d) -1 400 J
23. 74.8 앤
24.667 W
25. (a) 1. 19 (b) l. 19
26. 30.3 kca l/ h
27. (a) 1. 85 ft 2 • oF . h / Btu (b) 2.08
28. (a) 0.964 kg or more (b) The test samples and the inner
surface of the insulation can be pre-warmed to 37.0 oc
as the box is assembled. Then , nothing changes in temperature during the test period and the masses of the test
samples and insulation make no difference.
29. l. 90 x 10 3 J / kg . oc
30. (a) 9.31 X 1010 J (b) -8 .47 X 1012 J (c) 8.38 X 1012 J
12. (a) 0.719 kJ/ kg . K (b) 0.811 kg
13. between 1O- 30 C and 1O-20 C
15. (a) l. 08 (b) no
A-47
(c) 233 세 (d) 327 앤
16. The maximum possible value of ì' = 1 + 푸 = 1.6
야70c띠
cu
υv
for the lowest possible value for ι =
3R Therefore [he
claim of ì' = l. 75 for the newly discovered gas cannot be
true.
17. 5.74 x 10 6 Pa
18. (a) 0.118 (b) 2.35 (c) Q = 0 (d) 135 J (e) +135 J
19. 227 K
20. This is equivalent to 668 0 C , which is higher than the
melting point of aluminum which is 660 oC. Also , the
temperature will rise much more when ignition occurs.
The engine will melt when put into operation! Therefore ,
the claim of improved efficiency using an engine fabricated out of aluminum cannot be true.
2 1. (a) 2 .45 X 10- 4 m 3 (b) 9.97 X 10- 3 mol (c) 9.01 X 10 5 Pa
(d) 5 .l 5 X 10- 5 m 3 (e) 560 K (f) 53.9 J (g) 6.79 X 10- 6 m 3
(h) 53.3 g (i) 2.24 K
22. (a) l. 03 (b) 35Cl
23. (a) 2.37 X 10 4 K (b) l. 06 X 10 3 K
25. (b) 0.278
26. (b) 8.31 km
27. (a) 3.90 km /s (b) 4.18 km /s
28. (b) 447 J / kg . K. This agrees with the tabulated value of
448 J / kg . K within 0.3 %.
(c) 127 J / kg . K. This agrees with the tabulated value of
129 J / kg . K within 2 %.
29. (a) pressure increases as volume decreases
(d) 0.500 atm- 1 (e) 0.300 atm- 1
Chapter 21
[쿠|즈]
Chapter 20
[퀴즈]
1. (i) (b)
2. (i) (a)
(ii) (a)
(ii) (c)
3. (d)
4. (c)
[언습문제]
1. 3.32 mol
--2. ~ -TTP~
2 }Ç.v동 NA
3. 5.05 X 10- 21 J
4. 2.78 X 10- 23 kg . m /s
5. (a) 2.28 띤 (b) 6.21 X 10-21 J
6. (a) 4.00 u = 6.64 X 10- 27 kg (b) 55.9 u = 9.28 X 10- 26 kg
(c) 207 u = 3 .4 4 X 10- 25 kg
7. 17.4 kPa
8. (a) 385 K (b) 7.97 X 10- 21 J
(c) the molecular mass of the gas
9. 74.8J
10. True
11. (a) W= 0 (b) ð. E int = 209J (c) 317 K
1. (i) (c)
2. (d)
(ii) (b)
3. C , B, A
4. (a) one (b) six
5. (a)
6. false (The adiabatic process must be reversible for the
entropy change to be equal to zero.)
【연습문제]
1. (a) 10.7 띤
(b) 0.533 s
2. (a) 0.25 or 25 % (b) IQc I/I ß!1 = 3/ 4
3. 55 .4 %
4. (a) 75.0 띤 (b) 7.33
5. (a) 4.51 X 10 6 J (b) 2.84 X 10 7 J (c) 68 .l kg
6. (a) 2.65 X 10 7J (b) 3.20
7. (a) 67.2 % (b) 58.8 kW
8. The efficiency of a Carnot engine operating between
these temperatures is 6.83 %. Therefore , there is no way
that the inventor ’s engine can have an efficiency of 0.110
= 1l. 0 %.
9. l. 86
11. (a) 564 C (b) No , a real engine wiU always have an efficiency less than the Carnot efficiency because it operates
0
A-48
퀴즈 및 연습문제 해답
in an irreversible manner.
12. 0.330 or 33.0 %
13. (a) 5 .1 2 % (b) 5.27 X 10 J 2 J / h (c) 5.68 X 10 4
(d) 4.50 X 10 6 m 2 (e) yes (f) numerically, yes; feasibly,
probably not
14. (a) 0.300 (b) 1. 40 X 10- 3 K- J (c) -2.00 X 10- 3 K- J
(d) No. The derivative in part (c) depends only on Th .
Qc _. ~ (0 .5T" + 383\
15. (a) .-~ = 1. 40 I ~"~~~ ), where Qr/D. l is in megaD. t
\ T" - 383 )
watts and T h is in kelvins (b) The exhaust power decreases
as the firebox tempe rature increases.
(c) 1. 87 MW
(d) 3.84 X 10 3 K
(e) No answer exists. The energy
exhaust cannot be that smal l.
r
16. (b) 1 - 칸
(띠
이) 돼
C
T h e com빼 a
없t디ionωO
아
f reversib
비le en
맨
1땅g
밍
멘ines iss
itself a reversible engine so it has the Carnot efficiency.
No improvement in net efficiency has resulted.
(d) τ = 웅(낀 + 1~)
(e)
τ = 따F
All H
HHHH
3H , lT
THHH , HTHH , HHTH ,
2H , 2T
TTHH , THTH , THHT,
4
6
HTTT, THTT, TTHT,
4
TTTT
(b) 2 heads and 2 tails
21. (a)
Microstates
Number ofways to draw
All R
RRR
1
2 R, 1 G
GRR , RGR , RRG
3
1 R, 2 G
GGR , GRG , RGG
3
All G
GGG
1
Number of ways to draw
비
Macrostate
‘
l
、
Macrostate
Microstates
All R
RRRR
4R , lG
GRRRR , RGRRR ,
RRGRR , RRRGR ,
RRRRG
3R , 2G
2R , 3G
GGRRR , GRGRR ,
GRRGR , GRRRG ,
RGGRR , RGRGR ,
RGRRG , RRGGR ,
RRGRG , RRRGG
RRGGG , RGRGG ,
RGGRG , RGGGR ,
GRRGG , GRGRG ,
22. 143J/ K
23. 1. 02 kJ/ K
24. 0.507 J / K
25. 195J/ K
26. 244J/ K
27. (a) -3 .45J/ K (b) +8.06J/ K (c) +4.62J/ K
28. 1 W/ K
29. 8.36 X 10 6 J / K . s
30. (a) 39 .4 J (b) 65 .4 rad/ s = 625 rev/ min
(c) 293 rad/ s = 2.79 X 10 3 rev/ min
2. (e)
3. (b)
4. (a)
5. A.B. C
TTTH
/
GGGGG
1. (a) , (c) , (e)
Total
HTTH , HTHT, HHTT
t
5
【퀴즈]
HHHT
All T
All G
10
RGGGG , GRGGG ,
GGRGG , GGGRG ,
GGGGR
Chapter 22
17. 1.1 7
18. (a) 51. 2 % (b) 36.2 %
20. (a)
Result
Possible Combinations
lH , 3T
lR , 4G
GRGGR , GGRRG ,
GGRGR , GGGRR
1
5
[연습문제]
1. (a) +1. 60 X 10- 19 C , 1. 67 X 10- 2 7 kg
(b) +1. 60 X 10- 19 C , 3.82 X 10- 26 kg
(c) -1. 60 X 10- 19 C , 5.89 X 10- 26 kg
(d) +3.20 X 10- 19 C , 6.65 X 10- 26 kg
(e) -4.80 X 10- 19 C , 2.33 X 10- 26 kg
(f) +6 .4 0 X 10- 19 C , 2.33 X 10- 26 kg
(g) +1. 12 X 10- 18 C , 2.33 X 10- 26 kg
(h) - 1. 60 X 10- 19 C , 2.99 X 10- 26 kg
2. (a) 9.21 X 10- 10 N
(b) No. The electric force depends only on the magnitudes of the two charges and the distance b etwe en them .
3. 3.60 X 10 6 N downward
4. - 10 26 N
5. (a) 8.74 X 10- 8 N (b) repulsive
6. The electric force is 18 orders of magnitude smaller than
the gravitational force .
7. (a) 0.951 m (b) yes , if the third bead has positive charge
8. (a)
써
써+/좌
d
(b) Yes , if the third bead has a positive charge.
9. (a) 8.24 X 10- 8 N (b) 2.19 X 10 6 m / s
10. (a) 46.7 N to the left (b) 157 N to the right (c) 111 N to
the left
υ2 1 1 ^ I
11. k -즌 I -----;= i + I 2 -
10
e
d L.
I 2V2
\
1 \^I
Ii I
2V2 / ‘ |
-----;=
12. (a) 0 (b) 30.0 N (c) 2 1. 6 N (d) 17.3 N (e) -13.0 N
(f) 17.3 N (g) 17.0 N (h) 24.3 N at 44.50 above the +x
direction
퀴즈 및 연습문제 해답
π
써~d3
Ik,qQ
13.(b) ~ '\ I~ (c) 4a" I~끔
2 、 keqQ
、 η~d3
Chapter 23
14. The unknown charge on each dust particle is about
half of the smallest possible free charge , the charge of
the electron. No such free charge exists. Therefore , the
forces cannot balance.
15. (a) -(5.58 x 10- 11 N/C)j (b) (1. 02 X 10- 7 N/C)J
16.
A-49
I퀴즈l
2. (b) and (d)
1. (e)
【연습문제1
k _ Oχi
ι ‘
(a 2 + χ2)3/ 2
17. (a)
뚫[(1 녕) i + 깜]
에 -ke 욕
[(1 + 4냉)i + 4얘 J]
앞 α18. The field at the origin can be to the right , if the unknown
charge is -9Q, or the field can be to the left , if and only
if the unknown charge is +27Q.
19. (a) 1. 80 X 10 4 N/C to the right
(b) 8.98 X 10-5 N to the left
20. (a) 1. 29 x 10 4J N/C (b) -3.86 x 10-2J N
2 1. (a) (-0.5991 - 2.70J) kN/C (b) (-3.001 -
13.5J) μN
QL1_
3. (a) 윌
(b)
κ τrd
(b) at the center (c) 1.7ke옳
(d) upward in the plane of the page
24. (a) 6.13 X 1010 m/s2 (b) 1. 96 X 10 25 s
(c) 11. 7 m (d) 1. 20 X 10- 15 J
25. (a) 111 ns (b) 5.68 mm
(c) (4501 + 102J) km/s
26. (a) Particle under constant velocity
(b) Particle under constant acceleration
(c) the proton moves in a parab이 ic path just like a projectile in a gravitational field
??ιv;sin 28
'eE
(e) 36.9 0 or 53 .1。
(f) 166 ns or 221 ns
27. 4.52 X 10- 14 C
28. (a) -3 .4 3 0 < 8 < 3 .43 0
(b) 34.0 N/C
π 2k， q ~
29. -~i
6a z
30. (a) 1. 09 X 10- 8 C (b) 5 .4 4 X 10• N
1~
1
. Q
1
- (d2 + L2) 2J' Ey = ke d(값 + L2)l / 2
l/
말 = 0, E~ = κeA
k， 욕
때h때 i따
S앞the
뼈
le 社
fie리l뼈d 아뼈a때po이intcha때n
d ;w
Y
<
L.
Q at a distanced along the y axis above the change.
λ、
4. (a) ko •
ι
23. (a)
(d)
2. (a) 6.64 X 10 6 N/C away from the center of the ring
(b) 2 .41 X 10 7 N/C away from the center of the ring
(c) 6.39 X 10 6 N/C away from the center of the ring
(d) 6.64 X 105 N/C away from the center of the ring
(b) to the left
χo
6. 355 kN . m 2/C
7. (a) 1. 98 X 10 6 N . m 2/C (b) 0
8. chψ2/2
9. 28.2 N . m 2/C
10. -226 N . m 2/C
11. -Q/EO for SI; 0 for S2; -2Q/EO for S3; 0 for S4
12. (a) 3.20 X 10 6 N . m 2/C (b) 1. 92 X 10 7 N . m 2/C
(c) The answer to part (a) would change because the
charge could now be at different distances from each face
of the cube. The answer to part (b) would be unchanged
because the flux through the entire closed surface
depends only on the total charge inside the surface.
13. (a) 339 N . m 2/C (b) No. The electric field is not uniform on this surface , so the integral in Equation 23.7
cannot be evaluated.
14.
Q-61ql
，~
,
OEO
q
15. (a) ~
2Eo
q
(b) ~
\-, 2E。
(c) The fluxes are the same. The plane
and the square look the same to the charge.
16. (a) The net flux is zero through the sphere because the
number of field lines entering the sphere equals the
number of lines leaving the sphere (b) The net flux is
2'TTR2E through the cylinder (c) The net charge inside
the cylinder is positive and is distributed on a plane parallel to the ends of the cylinder.
17. (a) EA cos 8 (b) -EA sin 8 (c) -EA cos 8 (d) EA sin 8
(e) 0 for both faces (f) 0 (g) 0
18. 2.33 X 10 21 N/C
19. 3.50 kN
20. 508 kN/C up
2 1. (a) 4.86 X 10 9 N/C away from the wall
(b) So long as the distance from the wall is small compared to the width and height of the wall , the distance
does not affect the field.
A-50
퀴즈 및 연습문제 해답
22. (a) 51. 4 kN/C outward (b) 645 N . m 2/C
23. (a) - Q (b) yes
2:L
24. Ê = pr/2Eo = 21T ke pr away from the axis
25. (a) 15.0 N . m 2 /C (b) 1. 33 x 10- 10 C
(c) No , fields on the faces would not be uniform.
26. (a) 0 (b) 3.65 x 10 5 N/C (c) 1. 46 x 10 6 N/C
(d) 6 .49 x 10 5 N/C
27. (a) +913 nC (b) 0
I x (cm)
234
-20 뉴-
1- 까 1 / 3
28. (a) r = 커챙j
3;
(b) Yes , it is possible for aηy value of r > a.
x (cm)
Chapter 24
2
345
I퀴즈l
1. (i) (b)
(ii) (a)
2. @ to @, @ to @, @ to @, @ to @
19. (a) C/m 2 (b)
3. (i) (c) (ii) (a)
4. (i) (a) (ii) (a)
5. (a)
햄 [L - dIn(1 +~)]
[연습문제]
1. 1. 35 M]
2. (a) -2.31 kV (b) Because a proton is more massive than
an electron , a proton traveling at the same speed as an
electron has more initial kinetic energy and requires a
greater magnitude stopping potential
(c) Ll κ/ ð. ~ = -mp/me
3. (a) 1.1 3 x 10 5 N/C (b) 1. 80 x 10-14 N (c) 4.37 X 10- 17 ]
5. (a) 0 .4 00 m/s (b) It is the same. Because the electric
field is uniform , each bit of the rod feels a force of the
same size as before.
6.
6.93ke 옴
7. (a) 103 V (b) -3.85 X 10- 7 ] , positive work must be done
8. -k 갇
eR
qQ
9. (a) 4ν2ko ~ (b) 4V2 ko
a
a
ι
ι
10. (a) 5 .4 3 kV (b) 6.08 kV (c) 658 V
21. keÀ(π + 2 ln 3)
kF Q
22.
~"_-_
2VR2+ x2
+ 2k_, Q:. ln! R +, 값급즙]
,
I
, C l 1.
1._ (
.LL
'1í.
\
y""
.LL
x
/
23. No. A conductor of any shape forms an equipotential surface. However, if the surface varies in shape , there is no
clear way to relate electric field at a point on the surface
to the potential of the surface.
24. The electric field just outside the surface occurs at 16.0
kN/C. The peak in the figure occurs at about 6.5 kN/C.
Therefore , it is not possible that this figure represents the
electric field for the given situation.
σ
25. -
eo
26 쉰쏟앓
)숭훌활
2k~q
12. (a) no point (b) 김」
13. (a) 10.8 m/s and 1. 55 m/s
(b) They would be greate r.
The conducting spheres will polarize each other, with
most of the positive charge of one and the negative
charge of the other on their inside faces. Immediately
before the spheres collide , their centers of charge will be
closer than their geometric centers , so they will have less
electric potential energy and more kinetic energy.
ν
πu
mm E
S
사댐-때
ι
폐때
냈-
_2
14. 22.8k~
27. Eglass = EA1
28. (a) 0 , 1. 67 MV (b) 5.84 MN/C away, 1. 17 MV
(c) 11. 9 MN/C away, 1. 67 MV
29. Even if the charge were to accelerate to infinity, it would
only achieve a maximum speed of 30.0 m/s , so it cannot
strike the wall of your laboratory at 40.0 m/s.
la +L+ 、/(a + 디 2 + b2 I
~
I
I
a + Va 2 + b 2
I
30. k_Àlnl
Chapter 25
[퀴즈l
1. (d)
2. (a)
퀴즈 및 연습문제 해답
3. (a)
4. (b)
5. (a)
A-51
Chapter 26
[퀴즈】
1. (a) > (b) = (c) > (d)
【연습문제]
2. (b)
3. (b)
4. (a)
1. (a) 9.00 V (b) 12.0 V
2. (a) 1. 00 μF (b)100V
3. 4 .4 3 μm
4. (a) 1. 36 pF (b) 16.3 pC (c) 8.00 X 10 3 V/m
(2N - 1)E Il (π - fJ )R 2
5.--d
【연습문제】
1. 27.0 yr
2. -!!:ω
2π
6. m:gdtan8
q
7. (a) 2.81 μF (b) 12.7 μF
8. None of the possible combinations of the extra capacitors
is 좋 C , so the desired capacitance cannot be achieved
9. ten
10. (a) 5.96 μF (b) 89.5 μC on 20 μF， 63.2 μC on 6 μF， and
26.3 μC on 15 μFand 3 μF
11. 12.9 μF
12. 0.672 mF < Cextra < 1. 74 mF and 0.6 μF < Cextra < 1. 6 μF
13. 6.00 pF and 3.OOpF
1 ~<) ~ ~ ~
1 ~
/ 1 ~<)
1 ~
14. C1 =*C
CC
2 '-'pb +, μ~Cζ
'v 4 '-'p -CC.c)=~c'-'P'-'S ’ 2 - 2 '-'P μ
'v ~C:4 '-'p '-'p
,
/
,
15. (a) 216 μJ (b) 54.0 μJ
16. (a) 12.0 μF (b) 8.64 X 10- 4 J
(c) U1 = 5.76 X 10- 4 J and 다 = 2.88 X 10- 4 J
(d) UEl + U，ι'2 = 5.76 X 1O-4 J + 2.88 X 10- 4 J = 8.64 X
1O-4 J = UE , 씩， which is one reason why the 12.0 μF capacitor is considered to be equivalent to the two capacitors.
(e) The total energy of the equivalent capacitance will
always equal the sum of the energies stored in the individual capacitors.
(f) 5.66 V
(g) The larger capacitor C2 stores more energy.
17. (a) 2.50 x 10- 2J (b) 66.7 V (c) 3.33 X 10- 2J (d) Positive work is done by the agent pulling the plates apart.
kÆ2 ζ (Q-ql)2 "
R1Q
'"
R 2Q
19. (b) ~ε:- + . ': ~ -,
(c) ~ '~
(d) ~ ~-_
2Rl
2R 2
' , Rl + R 2 " R 1 + R 2
kpQ
_ __
kpQ
(e) 11, = ~ ~ -~ and V: = -:::-느=::- (f) 0
1
Rl + R 2
:!
~ + 뀔
20. (a) Consider two sheets of aluminum foil , each 40 cm by
100 cm , with one sheet of plastic between them
(b) 10- 6 F (c) 10 2 V
21. (a) 81. 3 pF (b) 2.4 0 kV
22. (a) K = 3 .4 0 (b) nylon (c) The voltage would lie somewhere between 25.0 V and 85.0 V.
23. 1. 04 m
24. 22.5 V
25. (a) 40.0 패 (b) 500 V
26. -9 .4 3 X lO-2i N
29. (a) 100 pF (b) 0.22 μC (c) 2.2 kV
30. (a) 11.2 pF (b) 134 pC (c) 16.7 pF (d) 66.9 pF
3. 1. 05 mA
4. (a) 5.57 X 10- 5 m/s (b) The drift speed is smaller because
more electrons are being conducted.
5. (a) 0.63210 T (b) 0.999 9510T (c) 10T
6. (a) 99.5 kA/m2 (b) The current is the same.
(c) The current density is smaller. (d) 0.800 cm
(e) 1 = 5.00 A (f) 2 .49 X 10 4 A/ m2
7. (a) 17.0 A (b) 85.0 kA/ m2
8. 0.256 C
9. Silver
10. 8.89 n
11. 2.71 MD
12. (a) 1. 82 m (b) 280 μm
다표
13. (a) 'v
~ 눴~
PPm
파 ( pm \1/ 4
(b) '" 쇠-::----n I
\~"~ π \PmRJ
14. 6.00 X 10- 15 (D . m)-l
15. (a) 5.58 X 10- 2 kg/mol (b) 1. 41 X 10 5 moljm 3
(c) 8 .49 X 10 28 atoms/ m 3 (d) 1. 70 X 10 29 electrons/ m 3
(e) 2.21 X 10- 4 m/s
16. 0 .1 2
17. T = 1.44 X 10 3 0 C
18. (a) 3 1. 5 nD . m (b) 6.35 MA/ m2 (c) 49.9 mA
(d) 658 μm /s (e) 0 .4 00 V
19. 227 C
20. 1. 63 mV < ~ V < 3.53 mV
2 1. (a) 3.00 X 108 W (b) 1.75 X 1017 W
22. 36.1 %
23. 15.0 J..LW
24. 449 원
25. 290원
26. (a) 4.75 m (b) 340 W
28. ~$10
0
Q
29. (a) .:
4C
",
Q
~ 3Q
(b) -..:; on C, -τ"::on 3C
"4
' 4
Q2. ~ 3Q2. ~ ~
(c)n':-""inC, nn-""in3C
32C
’ 32C
,., 3Q2
(d) ,,:
' , 8C
ε、f
ε、t' vLl V
30. (a) ~(t' + 2x + Kt' - 2KX) (b) 二-，-(K - 1)
2d'
Chapter 27
【퀴즈]
1. (a)
2. (b)
,
"
d
clockwise
A-52
퀴즈 및 연습문제 해답
3. (a)
4. (i) (b)
5. (i) (c)
16. 어L)
(ii) (a)
(ii) (d)
(iii) (a)
(iv) (b)
1. (a) 4.59 D (b) 8.16 %
2. (a) 50 % (b) 0 (c) High efficiency (d) High power transfer
3. (a) 75 W (b) 100 \1\1 (c) 175 W (d) Two: switch positions
3 and 4. 1n both cases , the power is 100 W.
4. (a) The 120-V potential difference is applied across the
series combination of the two conductors in the extension cord and the light bulb. The potential difference
across the light bulb is less than 120 V, and its power is
less than 75 W
비
(c) 73.8 W
5. (a) 1A = 8/ R, 1B = 1c = 8 / 2R
(b) B and C have the same brightness because they carry
the same current.
(c) A is brighter than B or C because it carries twice as
much curren t.
6. 0 6 Q < ftX[l3 < 1 6 Q and 0.672 kQ < R lla < 1.74 kQ
7. Nichrome: 0.393 m , carbon: 8.98 X 10- 3 m
8. (a) 1. 00 k !1 (b) 2.00 k!1 (c) 3.00 k!1
~
~~/1
'"
1\
ÎJ (b)
~ = 28\과 -τ)
10. (a) The single hot dog and the two in parallel will all cook
first. (b) single hot dog and the two in parallel: 57.3 s; two
hot dogs in series: 229 s
4
11. None of these is -;:- R , so the desired resistance cannot be
3
achieved.
12. 14.2 W to 2.00 D , 28 .4 W to 4.00 !1, 1. 33 W to 3.00 !1,
4.00 W to 1. 00 D
13. (b) The current never exceeds 50 μA.
-;,
28
48
28
14. (a) ß~ = 효’ ß~= τ， ß~ =
ß~ = τ
and 13 decrease
,
U
I
370 μ
、 1
•
;) .. Õ V
-+
J.
’‘
(b) 11. 0 mA in the 220 D resistor and out of the positive
pole of the 5.80-V battery; The current is 1. 87 mA in the
150 D resistor and out of the negative pole of the 3.10-V
battery; 9.13 mA in the 370 !1 resistor
17. 50.0 mA from a to e
18. (a) 12 is directed from b toward a and has a magnitude of
2.00 A (b) ι = 1. 00 A (c) No. Neither of the equations
used to find ι and 객 contained 8 and R. The third equation that we could generate from Kirchhoff ’ s rules contains both the unknowns. Therefore , we have only one
equation with two unknowns.
19. (a) No. The circuit cannot be simplified further , and
Kirchhoff ’ s rules must be used to analyze i t.
(b) 1] = 3.50 A (c) 12 = 2.50 A (d) 13 = 1. 00 A
20. (a) 13.0/1 + 18.012 = 30.0 (b) 18.012 - 5.0013 = -24.0
(이 1] - /2 - 13 = 0 (c이 13 = 11 - 12
(e) 5.00/1 - 23.012 = 24.0
([) 12 = -0 .4 16 A and 긴 = 2.88 A (g) 13 = 3.30 A
(h) The negative sign in the answer for 12 means that this
current flows in the opposite direction to that shown in
the circuit diagram and assumed during the solution .
That is , the actual current in the middle branch of the
circuit flows from right to left and has a magnitude of
0 .4 16 A.
2 1. (a) 2.00 ms (b) 1. 80 X 10- 4 C (c) 1. 14 X 10- 4 C
23. (a) 1. 50 s (b) 1. 00 s
(c) i = 200 + 100e- t, where i is in microamperes and t is
in seconds
24. (a) 빠 R 2 ) C (b) R 2 C (c) 8(숭+숲 e νU
!l
1\
(c) & = E(I-τ)
(b) 긴 =[, E2,= ζ
-3
‘
3 정’ i
I 연습문제1
I 2
2
1\
9. (a) R] = 8\- ~ +τ+
A11. ;. I、、
1
2/
3q'’ 144 = -;;3
25. 587 kD
27. No.
28. (a) For the heater, 12.5 A; For the toaster, 6.25 A; For the
grill , 8.33 A (b) The current draw is greater than 25.0
amps , so this circuit will trip the circuit breaker.
29. 7.4 9 D
Chapter 28
[퀴즈l
1. (e)
(c)
14 increases and 11, /2'
'-'-4
3/
31
(d) 긴 = J , 4 = %= 0 , 강 = 도
15. (a) 0.846 A down in the 8.00-D resistor, 0 .4 62 A down
in the middle branch , 1. 31 A up in the right-hand
branch (b) -222 J by the 4.00-V battery, 1. 88 kJ by the
12.0-V battery (c) 687 J to 8.00 !1, 128 J to 5.00 !1, 25.6 J
to the 1. 00-α resistor in the center branch , 616 J to
3.00 !1, 205 J to the 1. 00-D resistor in the right branch
(d) Chemical potential energy in the 12.0-V battery is
transformed into internal energy in the resistors. The
4.00-V battery is being charged , so its chemical potential
energy is increasing at the expense of some of the chemical potential energy in the 12.0-V battery. (e) 1. 66 젠
2. (i) (b) (ii) (a)
3. (c)
4. (i) (c) , (b) , (a) (ii) (a) = (b) = (c)
[연습문제l
1. Gravitational force: 8.93 X 10- 30 N down , electric force:
1. 60 X 10- 17 N Up , and magnetic force: 4.80 X 10- 17 N
down.
2. (a) west (b) zero deflection (c) up (d) down
3. (a) into the page (b) toward the right (c) toward the
bottom of the page
4. 48.9 0 or 131 。
5. (a) 1. 25 X 10- 13 N (b) 7.50 X 1013 m/ s2
6. (a) 1. 44 X 10- 12 N (b) 8.62 X 1014 m/s2 (c) A force would
퀴즈 및 연습문제 해답
be exerted on the electron that had the same magnitude
as the force on a proton but in the opposite direction
because of its negative charge. (d) The acceleration of
the electron would be much greater than that of the proton because the mass of the electron is much smaller.
A-53
[연습문제l
1. 1. 60 X 10- 6 T
7. -20.9 j mT
2. (a) 4.06 X 10 얘 T (b) The error was caused by the operator of the compass by taking a reading under a power
line , and is not caused bya defect in the compass.
3. 12.5 T
8. (a) V2 rp (b) 녕우
4. 감; into the paper
μ。 I
9. 115 keV
10
5. 칸i(조+괴
2r \ π
4/
魔h2 + 념)
1 1. (a) 5.00 cm (b) 8.79 X 10 6 m / s
12. (a) 4.31 X 10 7 rad/ s (b) 5.17 X 10 7 m / s
13. 1. 56 X 10 5
14. (a) 7.66 X 10 7 S-l (b) 2.68 X 10 7 m / s (c) 3.75 MeV
(d) 3 .1 3 X 10 3 revolutions (e) 2.57 X 10- 4 s
15. (c) 682 m / s (d) 55.9 μm
16. (a) 0 .1 18 N
(b) Neither the direction of the magnetic
field nor that of the current is given. Both must be known
in order to determine the direction of the magnetic force.
17. -2.88 j N
19. 1. 07 m/ s
20. (a) east (b) 0.245 T
2 1. (a) Thε magnêtic forcε and thε gravitational force both
act on the wire. (b) When the magnetic force is upward
and balances the downward gravitational force , the net
force on the wire is zero , and the wire can move can move
upward at constant velocity. (c) 0.196 T , out of the page
(d) Ifthe field exceeds 0.20 T , the upward magnetic force
exceeds the downward gravitational force , so the wire
accelerates upward.
22. (a) 21Tr1B sin
(b) up , away from magnet
e
23. (a) 0 (b) -40.0i mN
(c) -40.0k mN
(d) (40. Oi + 40.0k) mN
(e) The forces on the four segments must add to zero , so
the force on the fourth segment must be the negative of
the resultant of the forces on the other three.
24. (a) north at 48.0 0 below the horizontal
(b) south at 48.0 0 above the horizontal (c) 1. 07 씨
25. 4.91 X 10- 3 N . m
26. (a) 0.713 A (b) Current is independent of angle.
27. (a) 118 μN . m (b) -118 패 드 UB 드 +118 μJ
28. (a) 37.7 mT (b) 4.29 X 10 25 m - 3
29. 2.75 Mrad/ s
30. (a) 12.5 km (b) It will not arrive at the center. Because
the radius of curvature of the proton ’s path is much
smaller than the radius of the cylinder, the proton enters
the magnetic field only for a short distance before turning around and exiting the field.
Chapter 29
[퀴즈】
I. B > C>A
2. (a)
3. c > a> d> b
4. a = c = d > b = 0
5. (c)
B Along Axis of Circular Loop
6.
x/R
0.8L
64-----←←→」二
E/fb00o20
0
3
2
4
x/R
5
0.00
1. 00
2.00
3.00
4.00
5.00
B/B。
1. 00
0.354
0.0894
0.0316
0.0143
0.00754
7. (a) 53.3 μ.T toward the bottom of the page
(b) 20.0 μ.T toward the bottom of the page (c) zero
8. (a) aty = -0 .4 20 m (b) 3 .47 x 10- 2 N(-))
(c) -1. 73 X 10 4 ) N / C
f..L
J
9. ~- u- .cVd릅감 - d) into the page
2παd
4.50μ0 1
10. (a) -까τ::_
(b) stronger
11. (a) 8.00 A (b) opposite directions (c) force of interaction would be attractive and the magnitude of the force
would double
12. (a) 3.00 X 10- 5 N/ m (b) attractive
13. (a) The situation is possible injust one way. (b) 12.0 cm
to the left ofwire 1 (c) 2 .4 0 A down
μ(l1 2 L
14. k= --_~
4π d(d + f)
15. This is the required center-to-center separation distance
of the wires , but the wires cannot be this close together.
Their minimum possible center-to-center separation distance occurs if the wires are touching , but this value is 2r
= 2(250 μm) = 500 μm ， which is ìnuch larger than the
required value above. We could try to obtain this force
between wires of smaller diameter, but these wires would
have higher resistance and less surface area for radia디 ng
energy. It is likely that the wires would melt very shortly
after the current begins.
16. 500 A
17. (a) 3.60 T (b) 1. 94 T
18. (a) 6.34 X 10- 3 N/ m (b) inward toward the center ofthe
bundle (c) greatest at the outer surface
19. (a) 4.00 m (b) 7.50 nT (c) 1. 26 m (d) zero
μ Obr]2
20. (a) 창수 (for r 1 < R or inside the cylinder)
μ obR3
(b) 렇J- (kr r 2 > R or outside the cylinder)
21. 31. 8 mA
22. 4.77 X 10 4 turns
A-54
퀴즈및 연습문제해답
23. 5.96 x 1O - ~ T
24. (a) Make the wire as long and thin as possible without
melting whe n it carries the 5-A current. (b) As small
in radius as possible with your experiment fitting inside.
Then with a smaller circumference , the wire can form a
solenoid with more turns.
25. (a) πBR~ cos8 (b) πBR 2 cos8
27. (a) 7.4 0 μWb (b) 2.27 μWb
28. (a) 8.63 X 10 4 :; e - (b) 4.01 X 10~o kg
29. 143 pT
30. (a) 2.74 X 10- 4 T (b) 2.74 X 10- 4 T(-j) (c) Under the
assumption that the rails are infinitely long , the length of
rail to the left of the bar does not depend on the location
of the ba r. (d) 1. 15 X 10- 3 N (e) +χ direction (f) Yes ,
length of the bar, current , and field are constant , so force
is constant (g) (0.999 m / s)i
14. The speed of the car is equivalent to about 640 km/ h or
400 m i/ h , much faster than the car could drive on the
curvy road and much faster than any standard automobile could drive in genera l.
15. (a) 0.729 m / s (b) counterclockwise (c) 0.650 mW
(d) Work must be done by an external force if the bar
is to move with constant speed. This input of energy by
work appears as internal energy in the resisto r.
16. (a) 6.00 μT
(b) Yes . The magnitude and direction of the Earth ’ s field
varies from one location to the other, so the induced voltage in the wire changes. Furthermore , the voltage will
change if the tether cord or its velocity changes their orientation relative to the Earth ’s field.
(c) Either the long dimension of the tether or the velocity
vector could be parallel to the magnetic field at some instan t.
17. 3.32 x 10 3 rev/ min
Chapter 30
18. _.
[퀴즈】
19. 1. 00 T
μ~ 1 V r. 1_
e \1 2
-Ilnll +二 11
2
_"
\
4π 'L Rl
1. (c)
2. (c)
a lJ
ngRsin8
?
20.
V
3. (b)
4. (a)
1. 2.26 mV
2. (a)
예편)
oL찍
뀔수=t
\~V
ClTl
2R
-
-;'.
ßl
(b)
,-,
μ 해π 냥 ßl
A
Tl
4r]R
ßt
e ~ cos 28
and t is in seconds (b) B = 3.02 sin 120 1T t, where B is in
volts and t is in seconds (c) 1 = 3.02 sin 120 1T t, where Ii s
in amperes and t is in seconds (d) P = 9.10 sin 2 120π t，
where P is in watts and t is in seconds
(e) T = 0.024 1 sin 2 120 1T t, where T is in newton meters
and t is in seconds
26. (a) 1. 60 V (b) zero (c) no change in either answer
이
(b) 625 m / s
3. 1. 89 X 10 카 lV
4. B = 68.2e- I. GO / , where t is in seconds and B is in mv.
5. (a) l. 60 A counte rclockwise when viewed from the left of
the figure (b) 20 .1 μT (c) left
μ 。 ηπ 년 ßl
m ax
2 1. (a) 8.00 X 10- 21 N
(b) tangent to a circle of radius η in a clockwise direction
(c) t = 0 or 1 = 1. 33 s
22. (a) E = 9.87 cos 100 1T t where E is in millivolts/ meter and t
is in seconds. (b) clockwise
23. 13.3 V
24. (a) amplitude doubles and period is unchanged
(b) doubles the amplitude and cuts the period in half
(c) amplitude unchanged and period is cut in half
25. (a) φ B
8.00 X 10- 3 cos 120 1T t, where φ B is in T . m 2
[연습문제]
6. (a)
_v
E
(c) left
7. 272 m
8. (b) The emf induced in the coil is proportional to the
line integral of the magnetic field around the circular
axis of the toroid. Amp 응 re ’s law says that this line integral
depends only on the amount of current the coil encloses.
9. B = 0 .4 22 cos 120πt， where B is in volts and t is in seconds
10. (a) 1l. 8 mV (b) The wingtip on the pilot’ s left is positive.
(c) no change (d) No. If you try to connect the wings to
a circuit containing the light bulb , you must run an extra
insulated wire along the wing. In a uniform field the total
emf generated in the one-turn coil is zero.
11. 2.83 mV
12. (a) 39.9 mV (b) The west end is positive.
,·
8 11l ax
/
i
I
-
I
I
’
’’
’’I
-
-
I
l
-
’
-’
’
’’
(e)
E
/
Emax
t
l
I
--
l
I
l
l
I
l
4
l
13. 펀E
B" -{;Z
l
I
l
/
/
l
퀴즈 및 연습문제 해답
28. ~1O -4V
29.3.79 mV
30. (a) Cπa 2 K (b) upper plate (c) The changing magnetic
field through the enclosed area of the loop induces a
clockwise electric field within the loop , and this causes
electric force to push on charges in the wire.
Chapter 31
A-55
ence across the register is 0 , and the emf across the coil is
24.0 V.
(b) After several seconds , the potential difference across
the resistor is 24.0 Vand that across the coil is O.
(c) The two voltages are equal to each other, both being
12.0 V,just once , at 0.578 ms after the circuit is connected.
(d) As the current decays , the potential difference across
the resistor is always equal to the emf across the coi l.
30. (a) 선 = i2 + i (b) 8 - i1 Rl - ~ ~ = 0
【쿠|즈l
(c) 8- 자1 R， -L띄
dt = 0
~ ~l
~
~
(d)
8' - iR' - L따
\
dt =0
~，
-
V~ ~
~
1. (c) , (f)
2. (i) (b) (ii) (a)
3. (a) , (d)
4. (a)
5. (i) (b) (ii) (c)
Chapter 32
【퀴즈]
[언습문제】
1. 100 V
2. 1. 36μH
3. 19.2 μT.m 2
4. (a) 188 μT (b) 3.33 x 10- 8 T . m 2 (c) 0.375 mH
(d) B and φ B are proportional to current: L is independent of curren t.
【연습문제]
5. 8 0
2
Lk
7. 8
-18.8 cos 12017t, where 8 is in volts and t is in
seconds
8. (a) 5.90 mH (b) 23.6 mV
9. (a) 0 .4 69 mH (b) 0.188 ms
10. (a) 1. 00 kil (b) 3.00 ms
12. (a) 20.0 % (b) 4.00 %
13. (a) i L = 0.500(1 - e- lO .Ot ) , where i L is in amperes and t
is in seconds (b) is = 1. 50 - 0.250e- 1O .Ot , where is is in
amperes and t is in seconds
14. (a) 흐(1
- e-5Rt/ 2L)
5R ，~~
I
(b)
, ~n (6 - e-5Rt/ 2L)
\-, 10R
15. (a) 6.67 A/s (b) 0.332 A/s
16. 64.8 mJ
17. 2 .4 4μJ
19. (a) 18.0 J (b) 7.20 J
20. (a) M 12 = μ。πR 22N1 N2 /e (b) ~l = μ。πR22N1 샌 /e
(c) Theyare the same.
21. (b) 3.95 nH
22.281 mH
23. If the energy is split equally between the capacitor and
inductor at some instant , the energy would be half this
value , or 200 때. Therefore , there would be no time when
each component stores 250 μJ.
24. (a) 6.03 J (b) 0.529 J (c) 6.56 J
25. (a) 2.51 kHz (b) 69.9 n
{ 2L \
__ . __ . ~ ( 2L \
2"
(c) B 二 μ ols ， zero
n
μ oJ;
(d) .
2"
(e) The energy
density found in part (d) agrees with the magnetic pressure found in part (b).
29. (a) Just after the circuit is connected , the potential differ-
n
1. (a) 193
(b) 144
2. (a) 170 V (b) 2 .4 0 x 10 2 n
(c) Because 건vg
(ßV; ms)2
국--
•
~
(ß까mi
R=~ ， a 100-Wbulb
- avg
has less resistance than a 60.0-W bulb.
3. 14.6 Hz
4. (a) The rms current in each 150-W bulb is 1. 25 A. The
rms current in the 100-W bulb is 0.833 A.
(b) Rl = 96.0 n , ~ = 96.0 n , and R3 = 144 n
(c) 36.0 n
5. (a) 25.3 rad/s (b) 0.114 s
6. (a) 0.0424 H (b) 942 rad/s
7. 5.60 A
8. 3.80J
9. (a) 12.6 n (b) 6.21 A (c) 8.78 A
10. 0 .450 Wb
11. 32.0 A
12. (a) 221 α (b) 0.163 A (c) 0.230 A (d) no
13. (a) 141 mA (b) 235 mA
14. !2C(따ms)
15. (a) 146 V (b) 212 V (c) 179 V (d) 33 .4 V
16. 11.1 A
17. (a) 25.0 sin wt at ω t = 90.0 。
(b) 30.0 sin ω t at ω t = 60.0。
(c) 18.0 sin ω t at ω t = 300 。
(a)
27. (a) 0.693 1 굽 I (b) 0.3471 효)
28. (b) .
1. (i) (c) (ii) (b)
2. (b)
3. (a)
4. (b)
5. (a) X L < Xc (b) XL = Xc (c) XL > Xc
6. (c)
7. (c)
A-56
퀴즈 및 연습문제 해답
18. (a) 17.40 (b) the voltage
19. (a) 88.4
(b) 107
(c) 1. 12 A (d) the v이 tage lags
behind the current by 55.8 0 (e) Adding an inductor will
change the impedence , and hence the current in the
circui t. The current could be larger or smaller, depending on the inductance added. The largest current would
result when the inductive reactance equals the capacitive
reactance , the impedance has its minimum value , equal
to 60.0 n , and the current in the circuit is
ßv"、ax
ß v", ax
1.20 X 10 2 V = 2.00 A
, .....'\.→:二
Z
R
60.0 n
n
n
20. 353 W
22. 88.0W
23. (a) 66.8 n (b) 0.953 n
24. (a) 156 pH (b) 8 .84 n
25. 1. 41 X 10" rad/ s
26. 242 mJ
27. 687 V
28. 1. 88 V
29. 2.6 cm
22. (a)
/ 2μ。 cP
,,
1-'
v
,,-
_v
7T r-
(b) 50.0 여 (c) 1. 67 X 10- 19 kg . m / s
,. , pe
(b) _:_;ν
"
pe
(c) τ
c
23. (a) 1. 60 X lO - lO i kg . m / s each second
(b) 1. 60 X lO - lo i N
(c) The answers are the same. Force is the time rate of
momentum transfer (Eq. 9.3).
24. (a) 1. 00 X 10 3 km or 621 mi
(b) While the pr이 ect may be theoretically possible , it is
not very practical , due to the required size of the antenna.
25. 56.2 m
(c) 45.4 W
26. 걷깐깐E
qB
27. (a) ~ 10 8 Hz radio wave (b) ~ 1013 Hz infrared
28. Listeners 100 km away will receive the news before the
people in the newsroom bya total time difference of 8 .4 1
X 10 - 3 s.
29. (a) 3.85 x 10 26 W (b) 1. 02 kV/ m (c) 3.39 μT
30. (a) 388 K (b) 363 K
Chapter 33
Chapter 34
【퀴즈]
1. (i) (b)
2. (c)
3. (c)
2 1. (a) 1. 90 kN/ C
【퀴즈]
(ii) (c)
1. (d)
2. Beams @ and @ are reflected; beams G) and @ are
refracted.
3. (c)
4. (c)
5. (i) (b) (ii) (b)
4. (b)
5. (a)
6. (c)
7. (a)
[언습문제l
I 언습문제]
1. (a) 7.19 X 10 11 V/ m . s
(b) 2.00 X 10- 7 T
2. (a) 3.1 5 x 10 ~J N / C (b) 5.25k x 10- 7 T
(c) 4.83(-J) X 10 - 16 N
3. (-2 .87j + 5.75k) X 109 m/s2
4. 1 1. 0 m
5. (a) 681 yr (b) 8 .32 min (c) 2.56 s
6. 60.0 km
7. 2.25 X 10 8 m / s
8. 2.9 X 10 8 m / s ::!::5 %
10. The ratio of ω to k is higher than the speed of light in a
vacuum , so the wave as described is impossible.
1 1. 8.64 X 10 10 m
12. 3.34 μJ/m ~
13. (a) 2.33 mT (b) 650 MW/ m 2 (c) 511 W
14. 33 .4 oC. 2 1. 7 oC
15. ~1 X 10 4 m 2
16. (a) 5 .1 6 X 10- 10 T
1. 114 rad/ s
2. (a) 3.00 X 10 8 m / s
(b) The sizes ofthe objects need to be taken into accoun t.
Otherwise the answer would be too large by 2 %.
3. 2.27 X 10 8 m/s
4. (c) φ = 0.0557 。
5. ß = 28
7. (a) 1. 94 m (b) 50.0 0 above the horizontal
8. (a) 1. 81 X 10 8 m / s (b) 2.25 X 10 8 m / s
(c) 1. 36 X 10 8 m / s
9. 8 2 = 19.50 , 83 = 19.50 , 84 = 30.0 。
10. (a) 29.0 0 (b) 25.8 0 (c) 32.0。
11. (a) 78.3 0 (b) 2.56 m (c) 9.72 0 (d) 442 nm
(e) The light wave slows down as it moves from air to
water‘, but the sound wave speeds up by a larger facto r.
The light wave bends toward the normal and its wavelength shortens , but the sound wave bends away from the
normal and its wavelength increases.
12. (a) 1. 52 (b) 417 nm (c) 4.74 X 1014 Hz (d) 198 Mm/s
13. (a) 30.0 0 , 19 0 , 41 0 , 77 0 (b) 30 ， 41 。
14. ~ 10- 11 s , ~ 10 3 wavelengths
15. (a) Yes , if the angle of incidence is 58 . 9 。
(b) No. Both the reduction in speed and the bending
toward the normal reduce the component of velocity
parallel to the interface. This component cannot remain
constant for a nonzero angle of incidence.
0
(b) Since the magnetic field of the Earth is approximately 5 X 10- 5 T , the Earth ’s field is some 100 000 times
stronge r.
17. 5.16 m
18. (a) 540 V/ m (b) 2.58 μJ/ m ~ (c) 773 W/ m 2
19. 5.31 X 10 - 5 N/ m2
20. (a) 5.82 X 10 8 N (b) 6.10 X 101 ~ times stronger
퀴즈및연습문제해답
16. η = 1. 55
17. (b) 37.2 0 (c) 37.3 0 (d) 37. 3。
18. (b) 4.73 cm (c) For η = 1, h = O. For η = 2 , h = ∞. For η
> 2, h has no real solution.
19. The index of refraction of the atmosphere decreases
with increasing altitude because of the decrease in density of the atmosphere with increasing altitude , just like
the index of refraction of the slabs as you move upward
from the bottom in Figure P34.19. Imagine that the Sun
is the source of light at the upper left of the diagram.
Imagine yourself to be at the point where the light strikes
the lower surface of the bottom slab. The direction from
which the refracted light from the Sun comes to you is
higher in angle relative to the horizontal than the actual
geometric position of the Sun.
띠
、
% ,/
. ‘
A-57
28. 81 reflections
h In + 1. 00\
29. (a) 낀-τ~)
)
/η+ 1. 00\
(b) ~ '" . 2~~ times larger
Chapter 35
【퀴즈】
1. false
2. (b)
3. (b)
4. (d)
5. (a)
6. (b)
7. (c)
4
|
|
|
h
[연습문제]
|
|
|
4
1. (a) younger
water
(d) 26.9。
3.
a
N
n =1. 333
]·g×응
:• ---2.10xl0 m
2
[φ - sin- l (搬)]}
24. (a)
(b) ~10 → s younger
?‘f /
12 -•
,.
O~
‘ .‘ 2
(e) 107 m
Sin- I {ηv sin [φ - sin- l (빨)]} sin- 1{ nRsin
2.
(c)
Mj
3. (a) PI + h, behind the lower mirror (b) virtual (c) upright
(d) 1. 00 (e) no
4. (a) 1. 00 m behind the nearest mirror
(b) the palm
(c) 5.00 m behind the nearest mirror
(d) the back of her hand
(e) 7.00 m behind the nearest mirror
(f) the palm
(g) all are virtual images
5. (a) 33.3 cm in front of the mirror (b) -0.666 (c) real
(d) inverted
6. (a)
25. (a) 27.0 0 (b) 37.1 0 (c) 49.8。
26. (a) 10.7 0 (b) air (c) Looking at Table 16.1 , we see that
the speeds of sound for solids are an order of magnitude
larger than the speed of sound in ai r. Therefore , we can
estimate the critical angle for the air-concrete interface
by using Equation 34.9 and letting the ratio of indices of
refraction be ~0. 1. This gives a critical angle ofabout 6 0 •
Therefore , all sound striking the wall at angles greater
than 6 0 is completely reflected.
27. (a) 과맏
η-
1
(b) R min •
O. Yes; for very small d , the light
strikes the interface at very large angles of incidence.
(c) R min decreases. Yes; as η increases , the critical angle
becomes smaller. (d) R min • ∞. Yes; as η • 1, the critical
angle becomes close to 90 0 and any bend will allow the
light to escape. (e) 350 μm
(b) q = -40.0 cm , so the image is behind the mirror
(c) M = +2.00 , so the image is enlarged and upright
7. (a) 7.50 cm behind the mirror
(b) upright (c) 0.500 cm
8. A convex mirror diverges light rays incident upon it , so the
mirror in this problem cannot focus the Sun ’ s rays to a
pomt.
A-58
퀴즈 및 연습문제 해답
28. The image is inverted , real , and diminished in size.
29. -40.0 cm
30. 8.00 cm
9. (a) 8.00 cm
(b)
Chapter 36
[퀴즈]
1. (c)
2. The graph is shown in the figure below. The width of
the primary maxima is slightly narrower than the N = 5
primary width but wider than the N = 10 primary width.
Because N = 6 , the seco
아
on떠
darηyμmax
처ima
없s the pr
a
디imarηymaxima.
I
- -
i
a
7 il-----‘
~ ‘
lllll
L ’k
2‘A
i
』
I a
O
}!、A
I””내，---’
”
2A
- -
----
냉
/
I
!
、
U-
-
-
ζ
n
i
g
4
18. (a) The image is in back of the lens at a distance of 1. 251
from the lens. (b) -0.250 (c) real
19. 20.0 cm
1 삐川川삐
2(η -1)
a
I
mx
|||
-
ι
，，
「------，」
r
_, R
기 lll--
(2 - η)
rface of':-:
개 ”””렉
(c) virtual
10. -0.790 cm
1 1. (b) 0.639 s , 0.782 s
12. (a) 25.6 m (b) 0.0587 rad (c) 2.51 m
(d) 0.0239 rad (e) 62.8 m
13. (a) 16.0 cm from the mirror
(b) +0.333 (c) upright
14. (a) 45 .1 cm (b) -89.6 cm (c) -6.00 cm
15. 3.75 mm
17. The track must be placed a radial distance from the outer
Jue
S
·m ao
3. (a)
‘
【연습문제]
1. 641
2. The sine of the angle for m = 1 fringe is greater than 1,
(a) 20.0 cm in back of the lens (b) real
(c) inverted (d) M = -1. 00
(e) Algebraic answers agree , and we can express values to
three significant figures: q = 20.0 cm , M = -1. 00
(ii)
which is impossible.
3.632 nm
4. (a) 1. 77 μm (b) 1. 47 μm
5. 2 .4 0μm
6. 36.2 cm
7. 0.318 m /s
8. v tan [sin - 1 (뽕)]
F
(a) 10 cm front of the lens (b) virtual
(c) upright (d) M = +2.00
(e) Algebraic answers agree , and we can express values to
three significant figures: q = -10.0 cm , M = +2.00
(f) Small variations from the correct directions of rays
can lead to significant errors in the intersection point of
the rays. These variations may lead to the three principal
rays not intersecting at a single poin t.
2 1. (a) 20.0 cm from the lens on the front side
(b) 12.5 cm from the lens on the front side
(c) 6.67 cm from the lens on the front side
(d) 8.33 cm from the lens on the front side
22. (b) M10ng = -1\(2
23. 2 1. 3 cm
24. (a) -4.00 diopters (b) diverging lens
25. 2.18 mm away from the CCD
26. (a) -800 (b) inverted
27. -575
10. (a) 13.2 rad (b) 6.28 rad
(c) 1. 27 X 10- 2 deg (d) 5.97 x 10- 2 deg
11. 506 nm
12. 6.59 m
13. (a) 1. 93 μm (b) 3.00À
(c) It corresponds to a maximum. The path difference is
an integer multiple of the wavelength.
14. (a) 22.6 cm (b) 2.51 x 10- 3 (c) 6.03 X 10- 7 m
(d) 7.21 0 (e) 2.28 cm
(f) The two answers are close but do not agree exactly.
The fringes are not laid out linearly on the screen as
assumed in part (a) , and this nonlinearity is evident for
relatively large angles such as 7.21 0 •
15. ER = 10.0 and φ = 53 . 1 。
17.
I/ Imax
(} (0)
-0.2
0
0.2
퀴즈 및 언습문제 해답
18. (c) 9:1
19. 96.2 nm
20. (a) 638 nm (b) A thicker film would require a higher
order of reflection , so use a larger value of m
(c) 360 nm , 600 nm
21. (a) 276 nm , 138 nm , 92.0 nm (b) No visible wavelengths
are intensified.
22. (a) green (b) violet
23. 1. 31
24. 11. 3 m 3
25. (a) 238 nm (b) The wavelength of the transmitted light
increases. (c) 328 nm
26. (a) 97.8 nm
(b) Yes.
Destructive interference occurs when
2nt=(m+ 웅)A (Eq. 36.12)
잉)μ
， M빠
"\、 ‘
i잉s
a phase change at both faces of the film in Figure
P36.26.) Hence , for m = 1, 2, ... we obtain thicknesses of
293 nm , 489 nm ,
27. 39.6μm
28. 1. 25 m
29. 1. 62 cm
30. 20.0 X 1O- 60 C - l
A-59
15. (a) blue (b) 186 m to 271 m
16. (a) three (b) 0 0 , +45.2 0 , -45.2 。
17. (a) 479 nm , 647 nm , 698 nm (b) 20.5 ， 28.3 ， 30.7。
18. (a) five (b) ten
19. fJ 21. > fJ 3v and these orders must overlap.
20. 514 nm
2 1. (a) 0.738 mm
22. fJ = 3 1. 9 。
23. (a) 0.109 nm (b) four
24. (a) four (b) two
25. (a) 93.3 % (b) 50.0 % (c) 0.00 %
26. In Equation 37.10 , tan fJÞ = n2/ηl' the index of refraction
η2 of the solid material must be larger than that of air
(η1 = 1. 00). Therefore , we must have tan 야 > 1. For this
to be true , we must have 야 > 45 0 , s。 이
4 1. 0 0 is not
possible.
27. 60.5。
0
0
28. fJþ = tan- 1 (삶) or 야 = tan- 1 (csc κ) or fJ þ = c헤sin fJc)
29. (a) 0.045 0 (b) 0.0162
30. (a) 25.6 (b) 18.9 。
0
Chapter 38
Chapter 37
[퀴즈]
[퀴즈】
1. (c)
2. (d)
3. (d)
4. (a)
1. (a)
2. (b)
3. (a)
4. (c)
5. (b)
6. (c)
5. (c)
6. (d)
7. (i) (c) (ii) (a)
8. (a) m 3 > m2 = m1 (b) K3 = K 2 > K 1 (c) μ2 > μ3 - μ1
[연습문제】
【연습문제]
1. 4.22 mm
2. The equation generated is identical to Equation 37.1
3. (a) 1. 50 m (b) 4.05 mm
4. 11
5.
1
3. (a) 0 .4 36 m (b) less than 0 .436 m
4. (a) 2.18 μs (b) 649 m
5. 5.00 s
6. (a) 65.0 beats/ min (b) 10.5 beats/ min
7. 0.140c
8. The driver ’ s own testimony shows him blatantly vi이 ating
any Earth-based speed limit; look for another defense.
9. 0.800c
cLι
10. v=
φ
7T
7. 1. 62 X 10- 2
8.0.284 m
9. (a) 79.8 μrad (b) violet , 54.2 μrad
(c) The resolving power is improved , with the minimum
resolvable angle becoming 60.0 μrad.
10. 30.5 m
11. 3.09 m
12. 0 .4 0 μrad
13. none of them
14. 16 .4 m
‘
/c 1:1? + L~
2
11. (b) 0.0504c
12. (a) 39.2 μs (b) accurate to one digit
13. (c) 2.00 kHz (d) 0.075 m / s = 0.168 m i/ h (0.250 %)
14. (a) v = 0.943c (b) 2.55 X 103 m
15. (a) 17.4 m (b) 3.30。
16. (a) 2.50 X 108 m/s = 0.834c (b) 4.98 m (c) -1. 33 X 10- 8 s
17. The driver was traveling at 0.763c relative to the Earth.
He was speeding.
18. 0.960c
19. 0.893c, 16.8 0 above the x' axis
20. (a) 2.73 X 10- 24 kg . m / s (b) 1. 58 X 10- 22 kg . m / s
(c) 5.64 X 10- 22 kg . m / s
A-60
퀴즈 및 연습문제 해답
22. (a) 929 MeV/ c (b) 6.58 x 10 3 MeV/ c (c) No
23. (a) $800 (b) $2.12 x 10 9
24. (a) 4.38 x 10 11 J (b) 4.38 x 10 11 J
25. (a) 0.582 MeV (b) 2 .4 5 MeV
26. (a) smaller (b) 3.18 x 10- 12 kg
(c) It is too small a fraction of9.00 g to be measured
28. (a) 2.66 X 10 7 m (b) 3.87 km/ s (c) -8.35 X 10- 11
(d) 5.29 X 10- 10 (e) +4 .4 6 X 10- 10
29. When Speedo arrives back on Earth , 118 years have
passed , and Goslo would be 158 years old. That is impossible at the present time.
30. (c) μ
o'i'.,ct
r
/
,
、
7.χ = _(,r. (써2 강 + q2E 2 t 2 - mc)
/m2c2+ q2 쉰 p ’
qE \ V •• ~
'1 ~
V~
v
v
..
/
Chapter 39
【퀴즈1
1. (b)
2. Sodium light , microwaves , FM radio , AM radio.
3. (c)
4. The classical expectation (which did not match the
experiment) yields a graph like the following drawing:
Kmax
High intensity
Low intensity (delayed)
f
5. (d)
6. (c)
7. (b)
8. (a)
[ 연습문제]
14. (a) 43.0 0 (b) E = 0.601 MeV; p = 0.601 MeV/c = 3.21 X
10- 22 kg . m/s (c) E = 0.279 MeV; p = 0.601 MeV/ c =
3.21 X 10- 22 kg . m / s
I 1?ιc2 + E.:.. \
15. (a) (j = cos - 1 1 ν ~ I
\2mo cι +Eo/
(b) E' =
E。 (2mr 강 +1',ô) . ,
~ :
'""
_ ,V', P
2(m c2 + Eo) ’ r
r
(c)κ
2c(따 c2 + Eo)
Eo (2me c2 + Eo)
EÕ
r\./
1;。 ( 2me C2 + Eo)
Pe= 2c(m 연 + Eo)
2(m" c + Eo) ’ e
9
2
I
T"""
\.'
e
16. (a) 0.101 nm (b) 80.8 。
17. (a) It is because Compton ’s equation and the conservation of vector momentum give three independent equations in the unknowns À' , Ào , and μ
(b) 3.82 pm
18. To have photon energy 10 eV or greater, according to
this definition , ionizing radiation is the ultraviolet light ,
x-rays , and l' rays with wavelength shorter than 124 nm;
that is , with frequency higher than 2 .4 2 X 10 15 Hz.
19. (a) 14.0 kV/ m (b) 46.8 μT (c) 4.19 nN (d) 10.2 g
20. (a) 1. 66 X 10- 27 kg . m / s (b) 1. 82 km/ s
2 1. (a) 14.8 keV, 15.1 keV (b) 124 keV
22. (a) ~ 10 8 eV (b) ~ -10 6 eV
(c) The electron could not be confined to the nucleus.
23. The speed with which the student must pass through
the door to experience diffraction is extremely low. It
is impossible for the student to walk this slowly. At this
speed , if the thickness of the wall in which the door is
built is 15 cm , the time interval required for the student to pass through the door is 1. 4 X 10 33 s , which
is 10 15 times the age of the Universe.
24. (a) 융
26. 105 V
27. 3 X 10- 29 J = 2 X 10 - 10 eV
Chapter 40
1. (a) lightning: ~ 10- 7 m; explosion: ~ 10- 10 m
【퀴즈]
1. (d)
2. (i) (a)
(ii) (d)
3. (c)
4. (a) , (c) , (f)
【연습문제l
(b) 5.27 X 10- 24 kg . m / s (c) 95.3 eV
1. (a) 126 pm
2. (a)
3
‘
J
o
o
m
-i
끼
3
ι
(b) lightning: ultraviolet; explosion: x-ray and gamma ray
2. (a) 999 nm (b) The wavelength emitted at the greatest
intensity is in the infrared (greater than 700 nm) , and
according to the graph in Active Figure 39.3 , much more
energy is radiated at wavelengths longer than Àmax than at
shorter wavelengths.
3. 2.27 X 10 30 photon/ s
4. (a) 5200 K (b) This is not blackbody radiation.
5. (a) 5.78 X 10 3 K (b) 501 nm
6. (i) (a) 2.57 eV (b) 1. 28 X 10- 5 eV (c) 1. 91 X 10- 7 eV
(ii) (a) 484 nm (b) 9.68 cm (c) 6.52 m
(iii) (a) visible light(blue) (b) radio wave (c) radio wave
8. (a) 295 nm , 1. 02 PHz (b) 2.69 V
9. (a) 148 days (b) The result for part (a) does not agree at
all with the experimental observations.
10. (a) 288 nm (b) 1. 04 X 10 15 Hz (c) 1.1 9 eV
1 1. (a) 8.27 eV (b) The photon energy is larger than the
work function (c) 1. 92 eV (d) 1. 92 V
12. 4 .85 X 10- 12 m
13. (a) 64 .40 (b) 0.120 nm
4
”‘
)
퀴즈 및 연습문제 해답
(b) 속
va
3. ~
A-61
~(χ) = ψ1 (x) 12 = 중 cos 2 (뿔)
(c) 0.865
1
4. The photon does not have the smallest possible energy to
cause the transition between states η = 1 to η = 2.
5. (a) 0.511 MeV, 2.05 MeV, 4.60 MeV
(b) They do; the MeV is the natural unit for energy radiated by an atomic nucleus.
6. (a) 5.13 x 10- 3 eV (b) 9 .4 1 eV
(c) The electron has a much higher energy because it is
much less massive.
η = 2: 야2 (x) = 끊 sin(빨);
뮤 (x) = ψ2 (x)12 = 중 sin 2 (빨)
1
η = 3: ψ3 (x) = 끊 cos(빨) ;
쩍 (x) = 1 ψ3 (x) 12 = 중 c헥빨)
,‘
/
t
m
、
7. (a)
η
4
없
〉끼여며 며
)
603 eV
-
n
j
A í\
-_\1- ~_、1 __ ~
η
~
‘----------------
nJ
Z
151 eV
1
37.7 eV
J
씨
2
(b) 2.20 nm , 2.75 nm , 4.12 nm , 4.71 nm , 6.59 nm , 11. 0 nm
8. (a) 2.00 X 10- 9 J (b) 1. 66 X 10- 28 m
(c) No. The length of the box would have to be much
smaller than the size of a nucleus (_10 - 14 m) to confine
the particle.
9. (a) 호
(b) h 2/ 8mL2
2L
(c) This estimate is too low by 47T 2 = 40 times , but it correctly displays the pattern of dependence of the energy
on the mass and on the length of the wel l.
11. (a)
i
L
0
x
x
18. (a)
0
EI
에
·‘
、
0
_\
h 2 (2 x2
16. (a) u= ~τ I=':τ - 31
mLζ
\
LZ
m
/
비
19. (a)
U(x)
~= -j흥L
L
x
/
1
f= 펴L
χ
(b) 2L
20. (a) 1. 03 X 10- 3
---- U= 펀5
η = 1: ψ1 (x) = 찮 cos(환) ;
(b) 1. 91 nm
2 1. 3.92 %
22. 600 nm
23. 2πc많
ε
2
(b) 5.26 X 10- 5 (c) 3.99 X 10- 2
(d) In the η
2 graph in the text ’s Figure 40 .4b , it is
more probable to find the particle either near x = L / 40r
x = 3L/ 4 than at the center, where the probability density
is zero. Nevertheless , the symmetry of the distribution
means that the average position is χ = L/2.
12. (a) x = L / 4 , L/ 2 , and 3L/ 4
(b) We look for sin (3 7T X/ L) taking on its extreme values
1 and - 1 so that the squared wave function is as large as
it can be. The result can also be found by studying Figure
40 .4b.
13. (a) 0.196 (b) 0.609
14. (a) 0.332 nm
(b) 45.6 nm , quite different from 102.6 nm
17. (a)
0
[
2
4
L
χ
u
퀴즈및언습문제해답
A-62
(b) 54 .4 eV
10. (a) l. 31 μm
30. (a)
-
m nm
At
11
-2
-r
e
= 꿇J꽁
%-짧
시μη
(c) first excited state
28. f
(m
11 (a) ￡
25. (b) b=~띤
2ñ and
_.. - 효hω
2
에
빠(뚫r (b) 0 (뚫t
f싫 싫e r
4
(d) 냐삼 = ao
Ze κe eη
(e) -13.6 eV (f) We find our results are in agreement
with the Bohr theory.
|ψ1 2
12.
2
ψl s ( X 10 15 m- 3/ 2 )
Q
P1 s ( X 10
10
m - 1)
1. 2
1. 5
1. 0
a
,‘ N n
U
/
1
、
1. 0
0.8
0.5
0.4
x
0.6
(d) 0.865
0.2
r
O
Chapter 41
0.5
O
me
ms
-----
0.5
1. 5 a"
2e
14. (b) 0 .497
3. (b)
4. (a) five
5. (c)
6. true
‘
1. (c)
2. (a)
O
O
v
-”애
E=
ικ -。/i
에
않
【퀴 즈l
r
1. 5 a"
15. (a)
(b) nine
n
e
3
2
2
3
2
3
2
1
3
2
0
3
2
0
3
2
-1
3
2
3
2
3
2
-2
-2
n
e
me
3
1
1
3
2
2
---9---(2
1la4
[연습문제】
1lR4
1. (a) 12 l. 5 nm , 102.5 nm , 97.20 nm
(b) ultraviolet
2. l. 94 μm
3. (a) Àm. = Il/À
m1
~ 1μ
E(eV)
0
4
3
-3 .40
6.05
2
-13.6
/
‘
、
η
c。
l
l
”
-(Z
4
I--n4
이
4. (b) 0.846 ns
5. (a) 2.86 eV (b) 0 .472 eV
7. (a) l. 89 eV (b) 656 nm
(c) 3.02 eV
(d) 410 nm
(e) 365 nm
8. (a) 0 .476 nm (b) 0.997 nm
9. (a) En = -54 .4 eV/η2 for η = 1, 2, 3 , ...
--9
----9----9----
ms
---2
1-2
3
1
3
0
3
0
3
-1
3
-1
1-2
1-2
1-2
--2
16. (a) ν닮 (b) -짧， -lí , 0 , Ií and 짧.
(c) 1450 , 1140 , 90.0 0 , 65.9 0 , and 35.3。
17. (a) 2 (b) 8 (c) 18 (d) 32 (e) 50
며이
rψ
~
Z
{
며
-54 .4
v
퀴즈및연습문제해답
18. (a) 3.99 x 1017 kg/m 3 (b) 8.17 am (c) 1. 77 Tm/s
e
me
0
o
0
0
2
1
1
2
1
1
1
0
1
2
2
2
1
」 」
(d) It is 5.91
speed of light-and impossible.
(b) η
」
X 10 3 c, which is huge
2
1
i
2
2
0
0
0
0
compared with the
19. 0.05 T
me-2
q4
。J
‘
94
2
<-‘
。J
「
「
i
」
1li
「
-
〕
--i
「
。J
q4
1li
---1li
。3
。4
。3
。/‘
2
2
1li
1li
----li
o
-
2 1. The electron energy is not enough to excite the hydrogen
atom from its ground state to even the first excited state.
22. (a) the 4s subshell (b) We would expect [Ar]3 없 4s 2 to
have lower energy, but [Ar]3d 5 4s1 has more unpaired
spins and lower energy according to Hund ’s rule.
3s
3p
4s
만JD [그 口
[ls2 2S2 2 p 6 3s 2
J
[ls22s차6J3s 3pl
Si
만꾀맏그 [[] 口 口 [1i2s22 p 6J3i3p2
E굉 마] 텐그 따] 口 [ls22s앨6J3s 3 p 3
S16
맏펴 마피 마그 마그 口 [ls22s차6J3s￦
口
며
口
2
14
l
D E
Cl
17
2
때 뼈 꽤 마그 口 [ls 22s 22 p 6J3s 23 p 5
Ar 18 때 뼈
K 19
텐꾀 마꾀 口
[ls 22s 22 p 6J3s 23 p 6
R펴 마꾀 마꾀 맘꾀 텐그 [ls 22s 22pB 3s 23 p 6J4s 1
24. (a) ls22s 22p3
1-2
--2
15
A
1-2
1-2
1-2
[퀴즈l
(ii) (a)
(iii) (c)
3. (b)
12
Mg 때 D 口 口 口
외 13 때마]
--2
4. (c)
[li2s쐐6J3s 1
ll
1-2
2. (e)
口
Na
1-2
Chapter 42
1. (i) (b)
(c) chromium
23.
---2
28. (a) 시 = 632.809 14 nm , 시 = 632.80857 nm , 선 = 632.809
71 nm , three (b) 679 m/s
29. (b) 2.53 X 10 74 (c) 1. 18 X 10- 63 m
(d) This number is mμch smaller than the radius of an
atomic nucleus( ~ 10- 15 m) , so the distance between quantized orbits of the Earth is too small to observe.
30. 301
--「 。
9{
1ll
0
ms
25. ls , 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s , 4f, 5d, 6p, 7s
27. (a) 14 keV (b) 8.8 X 10- 11 m
i
디/‘
。3
n/.
--i
」
。3
oJ
o
1
1
1
1
디/·o4
o3
1li
ms-
•
--「 。
O/-。/·oz
i
/
-
。J
〈
。J
‘
o3
,
1
i
1
1
1
0
0
0
o3
Oz
/i
1li
‘-‘2
n/·O/i
9J
s
1lA
。
oj
-」‘
e
20. n
A-63
[언습문제]
1. ~ 10 28 protons (b) ~ 10 28 neutrons (c) ~ 10 28 electrons
2. (a) 68 (b) 짧 Zn (c) Isotopes of other elements to the left
and right of zinc in the periodic table (from manganese
to bromine) may have the same mass number.
3. (a) 0.360 MeV (b) Figure P42.3 shows the highest point
in the curve at about 4 MeV, a factor of ten higher than
the value in (a).
4. (a) 5.18 fm (b) A is much less than the distance of closest
approach
5. (a) 2.82 X 10- 5 (b) 1. 38 X 10- 14
6. 3.54 MeV
7. (a) 0.210 MeV (b) There is less proton repulsion in 염Na;
it is a more stable nucleus.
E.
8. 56Fe has a greater 강 than its neighbors
9. (a) 84.2 MeV (b) 342 MeV (c) The nuclear force is so
strong that the binding energy greatly exceeds the minimum energy needed to overcome electrostatic repulsion.
10. ~200 MeV
A-64
퀴즈 및 연습문제 해답
11. (a) Nucleons on the surface have fewer neighbors with
which to interact. Th e surface term is negative to reduce
the estimate from the volume term , which assumes that
all nucleons have the same number of neighbors.
"61 L. The sphere has a larger ratio
(b) sphere, 효 r cube,
to its characteristic length , so it would represent a larger
binding energy and be more plausible for a nuclear
sh ape.
13. 9 .47 X 10 9 nuclei
R，T. /。
In2
15. (a) cannot occur (b) cannot occur (c) can occur
16. (a) 잃Ni ' (b) 2싫Pb (c) 협Co
17. (a) e- + p • n + v (b) 2.75 MeV
18. (a) l. 05 X 10 2 1 (b) l. 37 X 10 9 (c) 3.83 X 10- 12 S-1
(d) 3.17 X 10 3 decays/week (e) 951 decays/week
(f) 9.95 X 10 3 yr
19. (a) 0.281 (b) l. 65 X 10• 9
(c) Radon is continuously created.
N
댔 mχ
145
U
2혔 Th
~V
140 ~
‘
2맥 Bi
。"~
28. (a)
(b)
,-‘
,
- -2
I_
1= 4
5
It
4fl
3fl
5 ‘
τn
-’ 2·r
2fl
-- iI
-2
--2 i
lfl
0
- lfl
-2 f
5
-2 i
-2 3 1n ~
91 ~ μ
,
-3 fl
'
\
"
29. The proposed reaction can b e written as
19B + ~He •
130
)1Al pha decay
、'- Beta (-) decay
125
80
27. (a) ~ 10 6 atoms (b) ~ 10- 15 g
\
2짧 Fr 、
2맨 At
(b) 1 쏠 Xe (c) e + + v
-흥 h
~
2잃Ac，
135 ~
22. (a) Iõ Ne
23. (a) 2.5 mrem/x-ray (b) The technician’s occupational
exposure is high: 38 times the local background radiation of 0.13 rem/yr.
24. It would take over 24 days to raise the temperature of the
water to 100 oC and even longer to boil it , so this technique will not work for a 20-minute coffee break!
25. 3.96 X 10- 4 J / kg
26. (b) 감
14. ~二 (2 - t ， llì /~ - 2- tdη /~)
20.
2 1. (a) l~~Au + 6n • 댐Au. • l 없 Hg + _?e + v (b) 7.89 MeV
85
90
95
Z
lH + I ~ C
While electric charge is conserved (5 + 2 = 1 + 6) , the
number of nucleons is not (1 0 + 4 =1= 1 + 12). Therefore ,
this reaction cannot occu r.
30. The Q value of this hypothetical decay is calculated to
be -7.62 MeV, which m eans you would have to a dd this
much energy to the 238U nucleus to make it emit a proton.
찾아보기 Index
‘’
lï l
가시 스펙트럼
가시광선
가역
845
822
구역
계의 경계
134
구역벽
고리 법칙
669
구조모형
가우스 법칙
고유 길이
각배율
883
골
각변위
225
공간양자화
각속력
77
공명
각운동량
261
각운동량보존
각위치
간섭
267
224
각진동수
354 , 383
412
간섭 무늬
간섭계
1046
공명 각진동수
371
공명 진동수
425 , 797
268
837
굴절 망원경
885
굴절각
837
굴절률
838
굴절파
840
궤도 양자수
1040 , 1045
그래프표현
그레이
11
1099
470
과감쇠
370 , 776
그림 표현
10
과냉각
470
극성 분자
629
근시
882
90
근점
880
235
근축광선
858
금지 전이
1057
91
95
갈릴레이 속도 변환식
938
광년
5
기멋값
갈릴레이 시공 변환식
937
광선
811
기본진동수
422
832
기하학적 모형
9
832
기화열
갈릴레이의 상대성 원리
감마붕괴
1094
936
광선광학
광선 근사
감마선
822
광선 추적도
감속재
1097
광섬유
848
광자
963 , 983
감쇠 진동자
강자성
광전 효과
31
광전자
23
거시 상태
528
거울방정식
겉보기 힘
결론
860
120
33
결합 에너지
경로차
계
광전 측광
728
개념화
거리
370
1075
894
계기 압력
광축
328
861
길이
껍질
468
4
949
1041
986
껍질 모형
981
꼭지각
10σ7
841
끌림 계수
981
598
1008
길이 수축
126
928
교류발전기
749
교류 변압기
799
‘’
IL I
교환 법칙
138
나노 기술
구면 거울
858
난류
구면 수차
858 , 878
난반사
구면파
134
1040
과가열
관성 질량
81
10
997
그레이디언트 연산자
관성 모멘트
갈릴레이 변환식
728
궤도 자기 양자수
371 , 796
관성 기준틀
893
946
379
관성
892
903
간섭성
949
고유시간간격
895
굴절
168
고립계 모형의 각운동량
579
728
군속력
167
고립계 모형
690
각 위치
1031
고립계
519
가우스
계열 한계
구심 가속도
833
내부 에너지
398
77 , 114
1019
332
내부 저항
154, 460
660
1-1
1-2
찾아보기
디
93
뉴턴·미터
뉴턴의 운동 제 1 법칙
뉴턴의 원무늬
901
뉴턴의 제 2 법칙
92
뉴턴의 제 3 법칙
96
90
딸핵
단색
단순화모형
단순화한그림 표현
단열 과정
단열 자유 팽창
446
단위 벡터
대류
821
1037
흰트겐
대체 표현
10
뤼드베리 상수
53
덧셈의 교환법칙
53
데이비슨-거머의 실험
548
돋보기
돌림힘
7
137
402
1079
955
809
반감기
1082
반도체
548
반사
1098
1031
반사망원경
886
반사의 법칙
834
182
반중성미자
1091
마루
379
발머 계열
1031
발산거울
860
발열 반응
1096
마이스너 효과
마이크로파
730
방시능
821
231
296
방출률
말뤼스의 법칙
925
배
991
417
1099
430
배기 행정
526
966
맥스웰 방정식
663
맥스웰-볼츠만 속력 분포 함수
75
483
배경 복사
등가원리
등속원운동
1030
885
맥놀이
618 , 619
1080
망원경
1012
등가 전기용량
96
방출분광학
105
들몬상태
등가저항
1096
729
마력
만유인력 법칙
드브로이 파장
834
반작용력
마찰력
1070
1020
반응 에너지
1079
맴돌이 전류
모멘트팔
배진동
808
752
232
506
1012
419
반사 계수
830
883
동위 원소
502 , 1012
417
마법수
11
1062
바닥상태 에너지
992
도제
독립 입자모형
밀도
반자성
마디
도플러 효과
4
바닥상태
871
‘| 口 |’
882
도표표현
미터
반사파
도수
도트곱
528
‘| 닙 |’
898
대응 원라
덧셈의 결합법칙
미시 상태
976
744
뢰머의 측정법
482
856
915
로이드 거울
324
165
물체 거리
1062
로렌츠 힘의 법칙
353
물질 전달
밀도 반전
923
로렌츠 변환식
57
단조화운동
대기압
224
렌츠의 법칙
519
367
물의 삼중점
렌즈 제작자의 식
476 , 503
1096
1085
레일리-진스의 법칙
11
209
1077
레일리 기준
9
95
물방울모형
라우에 무늬
레이저
무게
물리 진자
라디오파
893
452
882
라디안
Ic l
몰수
문턱 에너지
‘12 1’
‘’
496
무게 중심
592
697
디옵터
136
μr
뉴턴
μr
등전위변
‘
8
몰비열
5
여근
680
880
모형
낀 낌
눈
낀
누전 차단기
비
1020
‘
4
정
네모 장벽
정
1017
5
여근
네모 우물
I0
비
516
정
846
과 몰 과 과 몰
냉동기
아H 아님오」 처「 처「
드〔드
〕。
드 〈U 드〈〕
내부전반사
422
배타 원리
1052
벼금 껍질
1041
베르누이 방정식
336
찾아보기
베인브리지 질량 분석계
베크렐
벡터 모형
붕괴 상수
1089
붕괴율
1046
브래그의 법칙
1082
924
258
브루스터 각
벡터양
23 , 51
브루스터의 법칙
23
변위 마디
426
879
선 스펙트럼
1085
벡터곱
변위
색 수차
1081
붕괴 에너지
1082
베타 붕괴
696
1030
선 위치
895
선운동량
191
선전하밀도
927
927
571
선택 규칙
1057
924
블랙홀
314 , 968
선형 편광
비가역
519
선형 편광파
813
426
비간섭성
893
섭씨 온도눈금
변위 벡터
64
비고립계
166
성능 계수
변위 전류
806
비관성 기준틀
변위 진폭
393
비극성 분자
변위 배
변형
284
변형력
284
비보존력
155
비압축성
332
세차운동
보강 간섭
412 , 894
비오-사바르 법칙
보존력
645
비저항의 온도 계수
594
보편 기체상수
453
볼트
부력
수직항력
11
96
수차
975
순간가속도
34, 66
순간각가속도
225
‘’
I .A. I
878
순간각속도
225
순간속도
26 , 65
사건 지평선
314
순간속력
27
328
사이클로이드
247
순간 일률
182
사이클로트론
697
순간전류
640
287
부피 선속
286
570
사진기
285 , 287
산란
부피 전하밀도
부피 탄성률
부하저항
660
669
32
분배 법칙
138
381
692
순수 굴림 운동의 조건
순환과정
숨은열
879
467
슈테판의 법칙
1096
상 거리
856
상변화
467
상보성의 원리
992
스칼라곱
137
스칼라양
23 , 51
스핀
844
상상표현
10
스핀 각운동량
분석
32
상쇄 간섭
412 , 895
스핀 다운
10
상자성
불변 질량
963
상태 변수
불안정 평형
158
스핀 업
729
1051
1049
1049
스핀 자기 양자수
471
상호유도계수
768
시간
839
727
분산
분석 모형
314
483 , 974
스넬의 굴절 법칙
856
248
474
슈바르츠실트 반지름
929
산란사건
상
665
분기점 법칙
사이클로트론 진동수
사인형 파동
333
857
332
860
부피 변형력
분류
505
589
부피 변형
분기점
722
수 밀도
369
빈의 변위 법칙
454
솔레노이드
수직 (방향) 배율
비회전성 흐름
볼츠만상수
695
200
352
505
속도선택기
비탄성 충돌
복원력
볼츠만분포 법칙
399
수식 표현
비틀림 진자
볼록 거울
650
소리 준위
272
332
928
884
271
비점성 유체
복굴절
복합현미경
712
643
비저항
1035
155
보존력장
463
비옴
397
세차진동수
비열
보어 반지름
951
세기
617 , 664
728
517
56
세계선
629
병렬 연결
보어 마그네톤
성분
90
444
6
1049
1-3
1-4
찾아보기
시간간격
6
역기전력
시간 상수
125 , 674 , 763
역반사
시간 팽창
946
835
역제곱 법칙
시간에 무관한 슈뢰딩거 방정식
1016
운동마찰계수
751
296
역학적 에너지
155
시공 그래프
951
역학적 에너지 보존
시공간의 휩
967
역학적 파동
시버트
실상
열
1099
시험 전하
쌍둥이 역설
열량계
949
아르키메데스의 원리
아보가드로수
328
452
안정 평형
158
안정도 선
1074
알파 붕괴
1022 , 1084
암페어
압력
열역학 제 1 법칙
473
위상
354
열역학 제 2 법칙에 대한 클라우지우스
열접촉
앙페르의 법칙
앞뒤반전
720
857
앓은 렌즈 방정식
양력
871
442
1097
641
641
442
유전 강도
626
열효율
515
유전상수
625
유전체
285
977
온도
양자점
1019
온도 기울기
양자화
421 , 977
옴
양전자
1081
옴의 법칙
어미핵
1085
와트
525
음파
479
643
643
완전 비탄성 충돌
에너지 막대그래프
153
용량 리액턴스
우주선
200
789
142
468
392
이상광선
928
이상 기체
452
이상유체
332
이상흡수체
이심률
303
695
이옹화
1036
740
453
이상 기체의 상태 방정식
1017
운동 기전력
333
15
이상 기체 법칙
182
494
우물
625
유효숫자
융해열
443
용수철 상수
333
유체의 연속 방정식
858
에너지 등분배 정리
528
유동속도
열평형
양자수
엔트로피
1061
유동속력
오토순환과정
502 , 977
유도 흡수
유선
오목 거울
에너지 준위 도표
1061
447
994
166
유도 방출
516
양자 입자
166
787
열펌프
977
에너지 보존
548
유도 리액턴스
479
양자상태
에너지 보존식
64
열팽창
영률
342
유도
479
열중성자
807
22
위치 벡터
478
열전도의 법칙
527
783
위치
525
앙페르-맥스웰 법칙
791
463
압축 행정
압축비
997
위상각
열용량
열전도도
287
위상속력
위상자
열전도
555
354, 384
517
393
1071
위상 상수
표현
462
1037 , 1070
880
725
압력 진폭
압축률
75
웨버
열의 일당량
426
881
원천 전하
515
39
원자 질량단위
원점
454
197
490
443
표현
287 , 322
압력 배
660
열역학 저12 법칙에 대한 켈빈-플랑크
315
운동량보존식
열역학 제 O 법칙
640 , 718
암흑물질
197
원자 변호
974
열역학 변수
146
운동량보존
원운동
465
열복사
운동 에너지
원시
465
열린 회로 전압
‘10 1’
106
운동학식
514
열량 측정법
856
운동마찰력
운동론
165
165 , 460
열기관
555
168
106
이온화 에너지
483
1036
453
찾아보기
이중선
인공방사능
인덕터
저항
1095
285
인장 변형력
285
일반상대성 이론
일-운동 에너지 정리
일함수
146
984
928
정상상태
1033
정상류
332
저항력
123
정상파
416
적외선
821
정적 평형
279
정전기적 평형 상태
880
전기 수송
966
정상광선
644
적응
135 , 164
982
644
저항기
762
인장 변형
일
저지 전압
1049
정지 마찰계수
165
106
557
정지 마찰력
105
전기 쌍딱자 모벤트
정지 에너지
962
전기 쌍딱자
627
602
제곱-평균-제곱근
임계 감쇠
370 , 776
전기력
임계 온도
651
전기력선
559
제곱-평균-제곱근 속력
전기선속
574
제만효과
임계각
846
임피던스
전기장
792
입자모형
546
전기장 벡터
8
조화열
555
전기적 위치 에너지
‘’
전달 변수
I^ I
전도
588
종파
422
주기
478
124
378
좌우반전
472
857
77 , 355 , 382
자기 공명 영상법
1104
전도도
643
주기율표
자기 궤도 양자수
1046
전동기
751
주사 터널링 현미경
자기 생L극자 모멘트
자기병
전력 인자
702
전류
694
자기에 대한 가우스 법칙
자기장
686
726
주축
640
준경험적 결합 에너지 공식
전압전폭
782
준안정 상태
전위
자발붕괴
1085
전위차
준정적
588
줄
588
자연 방사능
1095
전자 스핀
자연 진동수
370
전자 스핀 공명
1061
472
136
중력
1049
전자포획
822
858
643
1061
자외선파탄
95
중력 가속도
1104
중력 질량
1091
42
95
전자 현미경
993
중력 퍼텐설 에너지
자유 공간의 유전율
551
전자기 복사
165 , 482
중력장
자유 공간의 투자율
712
전자기파의 스펙트럼
976
자유낙하가속도
자유물체 도형
자체 유도
자체 유도 기전력
760
365
820
전체 순간 에너지 밀도
962
전하결합소자
전하운반자
986
641
816
중립 평형
중성미자
1090
중성자수
1070
중성자포획
절대 압력
328
중첩의 원리
절대 영도
445
직각성분
56
직렬 연결
618 , 662
445
절연체
548
1020
점성도
332 , 340
직류발전기
장벽 투과
1020
점전하
550
진동수
저감쇠
88
370
1097
중성자활성화분석
장벽 높이
장힘
158
678
직류
98
정규모드
정반사
421
833
149
300
중립선
589
절대 온도눈금
299
장력
전자볼트
전체 에너지
760
96
작은각도근사
장
97
760
자체 유도계수
작용력
42
1022
1040
전류 밀도
자발 방출
자외선
1055
주양자수
795
495
1047
종단속력
555
784
412
660
748
355 , 382
진폭
354 , 382
진행파
382
1101
1078
1-5
1-6
찾아보기
질량
5 , 91
큐 인자
질량분석계
질량중심
크기의 정도
696
크로스곱
207
질량중심의 가속도
질량중심의 속도
질량수
211
210
849
키르히호프의 법칙
차단파장
183
탄성 퍼텐셜 에너지
985
탄성 한계
12
탄성률
초전도체
651
탄소 연대 측정법
초점 거리
추적
284
탈출속력
860 , 871
1100
충돌
338
하이젠베르크의 불확정성 원리
398
합선
679
파수
383
항력
342
322
파스칼 법칙
핵
파장
382
핵력
380
핵반응
305
패릿
퍼텐설 에너지
콤프턴 효과
988
퍼댄설 에너지 함수
퀴리
1082
퀴리 온도
729
149
156
1096
핵스핀 궤도 효과
핵융합
1022
핵자수
1070
핵종
1072
허물체
편광각
927
허상
편향각
841
34, 65
평균각가속도
225
헨리
870
856
355 , 382
761
홀 전압
705
1080
1103
1070
헤르츠
평균가속도
1103
1074
핵스핀 양자수
526
989
페르미
737
612
팽창행정
콤프턴 파장
550
1032
핵 마그네톤
325
패러데이의 유도 법칙
550
842
파선
304
쿨롱상수
하위헌스의 원리
341
398
케플러의 제 2 법칙
쿨롱의 법칙
831
하겐-푸아죄유 식
998
파형
550 , 718
977
‘| 를 |’
303
쿨롬
911
파면
파스칼
989
341
1I I
케플러의 제 1 법칙
케플러의 제 3 법칙
432
피조의 측정법
380 , 1006
콤프턴 이동식
570
푸리에 급수
플랑크상수
파동-입자 이중성
520
표면 전하밀도
1058
파동 함수
카르노 정리
815
특성 X선
285 , 286
520
포인팅 벡터
프레넬 렌즈 873
층밀림 탄성률
520
69
프라운호퍼 회절 무늬
995
카르노순환과정
포물체 운동
1020
파동묶음
카르노 기관
352
푸아죄유의 법칙
420
286
‘| 격 | ’
평형 위치
240
투과 계수
1
332
811
433
층밀림 변형력
층흐름
640
푸리에 합성
‘’
286
평균 전류
690
406
199
795
태슬라
투과
층밀림 변형
평균 전력
432
토리첼리의 법칙
충격파
182
푸리에 정리
195
196
평균 일률
1020
612
충격량-운동량 정리
24
터널링
충격량
197
1092
312
축전기
충격량근사
151
285
차원분석
303 , 860
24 , 64
평균속력
평행축 정리
199
449
448
평균속도
평면파
탄성 충돌
983
894
초점
669
IEI
차단진동수
차수
평균선팽창계수
258
‘’
I
225
평균 부피 팽창 계수
14
킬로와트-시
1070
‘’
l ;t
클래딩
평균각속도
798
1000
찾아보기
홀로그래피
921
횡가속도
화씨 온도계
446
횡속도
385
‘1 71 타|’
385
확률 밀도
1007
횡파
확률진폭
1006
혹의 법칙
142 , 352
LC 회로
770
1054
PV 도표
472
378
Btu
461
환경
134
훈트의 규칙
활선
678
흐름률
333
P파
379
흑제
483 , 974
RBE
1099
회로 기호
회로도
616
흑체 복사
616
회전운동에너지
회전 운동에 대한 일-운동 에너지 정리
회전 평형
회절
278
894
974
흡수분광학
242
245
흡수율
1030
483
RC 회로
rem
672
1099
RL 회로
762
784
흡열 반응
1096
rms 전류
흡입 행정
525
SI
4
회절 격자
919
힘 도형
97
S파
379
회절 무늬
910
힘상수
142
X선
822
1-7