Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1

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ANSWERS 
Chapter 1 
1 . (c) 
3. (c) 
5. (n) 
7. (e) 
9. (n) True 
(b) False 
(c) True 
11 . 1 . 1 9 X 1057 
13. (n) 10-8 m 
(b) 20 atoms 
15. (a) 3 X 1010 diapers 
(b) 1 .5 x 107 m3 
(c) 0.6 mi2 
17. $177 M 
19. (n) 0.000040 W 
(b) 0.000000004 s 
(c) 3,000,000 W 
(d) 25,000 m 
21. (n) C1 is in m; C2 is in 111/S 
(b) CI is in m/s2 
(c) C1 is in m/s2 
(d) CI is in m; C2 is ill S-I 
(e) C1 is in m/s; C2 is in S-l 
23. (a) 4 x 107 m 
(b) 6.37 X 106111 
(c) 2.48 X 104 mi; 3.96 X 103 mi 
25. 210 em 
27. 1.28 km 
29. (n) 36.0 km/h·s 
(b) 10.0 m/s2 
(e) 88.0 ft/s 
(d) 26.8 m/s 
31. 4050 m2 
33. (n) m/s2 
(b) s 
(c) m 
35. T- I 
37. IIIv2/r 
41 . M/L3 
43. (n) 30,000 
45. 
47. 
49. 
51 . 
53. 
55. 
57. 
(b) 0.0062 
(c) 0.000004 
(d) 217,000 
(a) 1 .44 X 105 
(b) 255 X 10-8 
( c) 8.27 X 103 
(d) 6.27 X 102 
4 X 106 
(a) 1 .69 X 103 
(b) 4.8 
(c) 5.6 
(d) 10 
31 . 7 Y 
2.0 x 1023 
(n) 1.41 X 1017 kg/m3 
(b) 216 m 
(a) 4.85 X 10-6 parsec 
(b) 3.08 X 106 m 
(c) 9.47 x 1015 m 
(d) 6.33 X 104 AU 
(e) 3.25 c ·y 
59. The claim is conservative as the actual weight of water used is 
closer to 55,000 tons. 
61 . (n) 1 1 = 3/2; C = 17.0 y/(Gm)3/2 
(b) 0.510 Gm 
63. 1 . 1 6 X 1019 lb 
Chapter 2 
--------------------
1 . 0 
3. It is safer to land against the wind. 
5. (a) Negative 
(b) During the last five steps, gradually slow the speed of 
walking, until the wall is reached . 
(c) 1,-----------, 
O�--�-�--� 
t (5) 
A-l 
A-2 Answers 
7. (a) True 
(b) True in one dimension 
9. False 
11 . (a) 
13. (b) 
15. Yes. In any round-trip, A to B, and back to A, the average veloc­
ity is zero. 
17. No. If the velocity is constant, a graph of position as a function 
of time is linear with a constant slope equal to the velocity. 
19 . (b) 
21 . (a) False 
(b) True 
23. Vtop of fHght = 0; atop of fHght = -g 
25. (b) 
27. (e) 
29. (e) 
31 . (c) 
33. (a) 
35. (d) 
37. (d) 
39. Velocity: (a) negative at 10 and 11; (b) positive at 13, 14, 16, and 17; 
(e) zero at 12 and 15 
Acceleration: (a) negative at 14; (b) positive at 12 and 16; (e) zero 
at 1o, I" 13, 15, and 17 
41 . (a) Graphs (a), (f), and (i) 
(b) Graphs (e) and (d) 
(e) Graphs (a), (d), (e), ( f), (11), and (i) 
(d) Graphs (b), (e), and (g) 
(e) Graphs (a) and (i) are mutually consistent. Graphs (d) and 
(h) are mutually consistent. Graphs (f) and (i) are also 
mutually consistent. 
43. (a) 54.2 m/s; (b) -123g 
45. 
47. 
49. 
51 . 
53. 
55. 
4.02 m/s2 
14.2 ms 
(a) 0.278 km /min 
(b) -0.0833 km/ min 
(e) 0 
(d) 0.128 km/min 
(a) 2.25 h 
(b) 4.99 h 
(e) 880 km/h 
(d) 611 km/h 
(n) 4.33 y 
(b) 4.33 x 106 y; No 
35.8 m 
57. (a) 0 
(b) 0.333 m/s 
(e) -2.00 m/s 
(d) 1 .00 m/s 
59. 1 22 km/h; 1 .04v,v 
61 . (a) 400 ,-�������--�� 
350 
300 
250 :§: 200 
(b) 15 s 
(e) 300 m 
(d) 100 m 
63. 6 h 
65. -2.00 m/s2 
67. (a) 2 m 
" 150 
100 
50 
O ¥-��L+�--������ o 2 4 6 8 10 12 14 16 18 20 
I (s) 
(b) �x = (21 - 5)M + (Mj2, where x is in meters if I is in 
seconds. 
(e) v = 21 - 5, where v is in meters per second if I is in seconds. 
69. (a) a,v,All = 3.33 m/s2; a,v,BC = 0; a,v,CE = -7.50 m/s2 
(b) 75.0 m 
(c) 90 
80 
70 
60 
:§: 50 " 40 
30 
20 
10 
0 
0 1 2 3 4 5 6 7 8 9 
I (s) 
A B C 0 
10 
E 
(d) At point 0, I = 8 s, the graph crosses the time axis and so 
v = 0, 
71 . (a) 80.0 m/s 
(b) 400 11l 
(e) 40,0 m/s 
73. 15.6 Ill / S2 
75. (a) 4,68 s 
(b) 20,4 11l 
(e) 0.991 s and 3.09 s 
77. (a) 
(b) 7.27 m 
(c) 1 ,73 s 
S 
>< 
(d) 11 .9 m/s 
8 
7 
6 
5 
4 
3 
2 
0.2 0.4 0.6 0.8 1.0 1 ,2 1 .4 1 .6 1 .8 
t (s) 
79. 43.6 m 
81 . 68 .0 m/s 
83. (n) 666 m 
(b) 13.6 m/s 
85. (n) 10.4 5 
(b) 27.4 5 
(C) 12.8 5 
87. (n) 19.0 km 
(b) 2 min 1 8 5 
(c) 610 m/s 
89. 40.0 cm/s; -6.88 cm/s2 
91 . (n) 4.76 m/s or 10.7 mi/h 
(b) 0.595 
93. 10.9 m 
95. 27.6 m 
97. 4.59 km 
99. 2.40 m; 1 .40 5 
103. (n) -25.7 m/s2 
(b) 2.33 5 
105. (n) 1 .03 coy /y2 = 1 .03 ely 
(b) = 2 d 
107. 4.80 m/ 5 60 
50 
40 
5 30 
'"' 
20 
10 
2 4 6 8 10 
1 (5) 
109. ! " 
111 . (n) 34.7 5 
(b) 1 .20 km 
(c) 1400 
1200 
1000 
S 800 
'"' 600 
400 
200 
0 
0 5 10 15 20 
1 (5) 
113. (n) L l = � L 
(b) I = � If;n 
12 14 16 
25 30 35 
Chapter 2 
115. (n) A = 90 m 35 
30 
25 
(j) 20 " 
r 15 S 
;::> 10 
5 
0 
0 0.5 1 1 .5 2 2.5 
1 (5) 
(b) x(t) = (3 m/s2) t2 + (3 111/5)1; nx = 90.0 m 
117. x(t) = ( � 111/S3 ) 13 - (5 m/s)1 
119. (n) 0.250 m/s per box 
3 
A-3 
(b) v(l 5) = 0.930 m/s; v(2 5) = 3.20 m/s; v (3 5) = 6.20 m/s 
121 . 
(c) x(3 s) = 7.00 m 
123. (n) v(l) = (0.1 m/s3)12 
(b) 2.23 m/s 
125. 12 .8 m/s2; 30.5% 
7 
6 
5 
4 
3 
2 
o o 
V / 
v vs. I 
/ / V 
/ 
2 3 
1 (5) 
4 
127. (n) a = wvmaxcos(w l); Because a varies sinusoidally with time, 
it is not constant. 
VOla>: 
(b) X = Xo + -' [1 - cos(wI)] 
w 
A-4 
129. (a) 
(b) 
Answers 
(b = 1 s) 
Because the numerical value of b, expressed in 51 units, is 
one, the numerical values of a, v, and x are the same at 
each instant in time. 
Chapter 3 
1 . The magnitude o f the displacement o f a particle is less than or 
equal to the distance it travels along its path. 
3. The displacement for any trip around the track is ZERO. Thus 
we see that no matter how fast the race car travels, the average 
velocity is always ZERO at the end of each complete circuit. 
5. No. The magnitude of a component of a vector must be less 
than or equal to the magnitude of the vector. If the angle () 
shown in the figure is equal to 0° or multiples of 90°, then the 
magnitude of the vector and its component are equal. 
7. No 
9. (e) 
11 . (c) 
13. (a) The velocity vector, as a consequence of always being in 
the direction of motion, is tangent to the path. 
(b) Ij 
v 
v 
x 
15. (a) A car moving along a straight road while braking 
17. 
(b) A car moving along a straight road while speeding up 
(c) A particle moving around a circular track at constant 
speed 
(a) 
(b) 
(c) 
t - U V? V2 1 -", A� = ,,- v, 
V1 t a = /:'vl/1t 
� -v2 1 t� = v2 - v1 V1 
1 V2 t a = Mil/1t 
�Vl 
- - - ""V2 . V2 t;.V = V2- Vl ,� 
-V1 
t a = /:,vl/1t 
19. 
21. True 
23. (d) 
25. (a) False 
(b) True 
27. 
-VA VB -VA 
VB 
29. (a) 
Direction of velocity 
Path vector 
AB north 
BC northeast 
CD east 
DE southeast 
EF south 
(b) Direction of acceleration 
Path vector 
AB north 
BC southeast 
CD 0 
DE southwest 
EF north 
(c) The magnitudes are approximately equal. 
31 . The droplet leaving the bottle has the same horizontal velocity 
as the ship. During the time the droplet is in the air, it is also 
moving horizontally with the same velocity as the rest of the 
ship. Because of this, it falls into the vessel, whjch has the samehorizontal velocity. Because you have the same horizontal 
velocity as the ship, you see the same thing as if the ship were 
standing still. 
33. True 
35. 
37. 
39. 
The principal reason is aerodynamic drag; when moving 
through a fluid such as the atmosphere, the ball's acceleration 
will depend strongly on its velocity. 
14.8 m/s 
R = 22.2 m; Cl' = 22.5° 
N 
Chapter 3 A-S 
41 . (11) Y 5l. D = (3 m){ + (3 m)] + (3 m)k; 0 = 5.20 m 
53. v,v = (14.1 km/h)£ + ( -4.1 km/h)] 
� 
55. (b) 
A 
57. (a) v". = (33.3 m/s)i + (26.7 m/s)] 
A + B (b) a,v = (-3.00 m/s2)£ + (-1 .77 m/s2)J 
59. v = (30 m/s)£ + [40 m/s - (10 m/s2)11J; a = ( - 10 m/s2)] x 
61 . (a) vov = (20 m/s)( - [ + J) 
(b) y (b) a", = (-2 m/s2)i 
(c) M = (600 m)(-£ + J) 
63. (a) 13.1° west of north 
(b) 300 km/h 
A - B 65. 8.47°; 2.57 h 
67. You should fly your plane acrOS5 the wind. 
x 69. (a) rAil (6 5) = (120 m)i + (4 m)J 
(b) V AB (6 5) = (-20 m/5)£ - (12 m/5)J 
(c) y (c) aAB = (-2 m/s2)J 
71. 1 .52 X 10-6 m/52; 1 .55 X 1O-7g 
73. 3.44 X 1O-3g; 6.07 X 1O-4g 
75. 33.4 min-1 
77. 11 = 
(va sin Ba)2 
x 2g 
(d) Y 79. 33.8 m/s 
81 . 20.3 m / s; 36.2° 
x 83. 69.3° 
85. (a) 18.0 111/5 
B - A (b) 14.0° 
87. (11) 8.14 m/s 
(b) 23.2 m/5 
89. -63.4° 
91. 209 m 
(e) y 93. (a) 0.452 5 
(b) 22.6 m x 
95. (a) 485 km 
2B 
(b) 1.70 km/5 
101. L = 2v� tan B 
g cos B 
103. 10.8 m/s; v = (6.50 m/s)l + (-21.6 m/5)J 
105. 40.5 m/5; 0.994 5 
43. (b) 107. 7.41 111/S; 0.756 5; 15.9 m /5; 17.5 m/5; 25.0° 
45. 109. 0.785 m 
A () Ax Ay lll . 4.91 m/s2; 8.50 m/52 
(a) 10 m 30° 8.66 m 5 m 113. (a) y (m) 
(b) 5 m 45° 3.54 m 3.54 m 25 
(c) 7 km 60° 3.50 km 6.06 km 
(d) 5 km 90° 0 5 km 20 
(e) 15 km/s 150° - 13.0 km/5 7.50 km/s 15 (j) 10 m/5 240° -5.00 m/5 -8.66 m/5 
(g) 8 m/s2 270° 0 -8.00 m/52 10 
5 
47. (a) 5.83; 31 .0° 
(b) 122; -35.0° 0 5 10 15 
(c) 5.39; B = 42.1°; <p = 236° x (m) 
49. (a) v = (5 m/s)i + (8.66 m/5)J (b) v = (5 m/5)[ + (10 m/5)J; 11 .2 m/s 
(b) A = (-3.54 m)[ + (-3.54 m)J 
(c) r = (14 m)l - (6 m)J 
A-6 
115. 
117. 
119. 
Answers 
31 .3°; S.06 m 
Fourth step 
x I g (a) Vmin = cos 0 \j 2(x tan 0 - /1) 
(b) vmin > 26.0 m/s = 5S.0 mi/h 
(e) hmax < x tan 0 
Chapter 4 
1 . I f an object with no net force acting on i t i s a t rest o r i s moving 
with a constant speed in a straight line (i.e., with constant ve­
locity) relative to the reference frame, then the reference frame 
is an inertial reference frame. 
3. No. If the net force acting on an object is zero, its acceleration is 
zero. The only conclusion one can draw is that the net force act­
ing on the object is zero. 
5. No. Correctly predicting the direction of the subsequent 
motion requires knowledge of the initial velocity as well as the 
acceleration. 
7 . The mass of an object is an intrinsic property of the object 
whereas the weight of an object depends directly on the local 
gravitational field. Therefore, the mass of the object would not 
change and Wgrav = I11g]oca]· Note that if the gravitational field is 
zero then the gravitational force is also zero. 
9 . Your apparent weight would be greater than your true weight 
when observed from a reference frame that is accelerating up­
ward. That is, when the surface on which you are standing has 
an acceleration a such that ay is positive. 
1 1 . (a) Fn21 = 1111g 
(b) F,112 = 1111g 
(e) FnT2 = (1111 + 1112) g 
(d) Fn2T = (1111 + 1112) g 
13. (b) 
15. (e) 
17. (a) 
y' 1 
� 
T1 
w 
1;' 
19. (a) True 
(b) False 
(e) False 
(d) False 
21 . (d) 
(b) Tf 
F 
T' 1 
23. The velocity of the elevator has no effect on the person's 
apparent weight. 
25. (a) 7S2 N; 62.6 N 
(b) Because there is no acceleration, the forces are the same 
going up and going down the incline. 
27. (a) 6.00 m/ S2 
(b) 1 /3 
(e) 2.25 m/S2 
29. 
31. 
33. 
-3.75 kN 
(a) 4.24 m/ S2 @ 45.0° from each force 
(b) S.40 m/s2 @ 14.6° from iFo 
12.0 kg 
35. (a) 4.00 m/ S2 
(b) 2.40 m/ S2 
37. (a) a = ( 1 .50 m/s2)i + (-3.50 m/s2)j 
(b) v = (4.50 m/s)i + (- 10.5 m/s)j 
(e) r = (6.75 m)i + (- 15.S m)j 
39. (a) 530 N 
(b) 119 1b 
41 . (a) 60.0 N 
(b) 57.7 N 
43. 
45. 
47. 
49. 
51. 
53. 
T2 > T1 
(a) 36.9° 
(b) 4.0S N 
(e) 3.43 N; 2.40 N; 3.43 N 
(a) a = (0.500 m/ s2)i + (2.60 m/ S2)j 
(b) 1\ = (-5.00 N)i + (-26.0 N)j 
(a) 
(b) 
(a) 
(b) 
(e) 
(a) 
(b) 
T = _,_v_ . 0 = 90° · T -7 T as 0 -7 0° 2 sin Of I ITIi1X 
19 .6 N 
11 .S kN 
9.S1 kN 
7.S1 kN 
3.S2 kN 
4.30 kN 
55. 56.0 N 
57. (d) 
59. (a) 50S N; 50S N 
(b) 111g; 0 
61 . 552 N 
63. (a) 
65. (e) 
67. (a) 19.6 N 
(b) 19.6 N 
(e) 25.6 N 
(d) 14.6 N 
69. (a) 1 .31 m/ S2 
71 . 
75. 
77. 
79. 
S1 . 
S3. 
(b) 16.7 N; 21 .3 N 
(a) 
(b) 
(a) 
(a) 
(b) 
(a) 
(b) 
(a) 
(b) 
(a) 
(b) 
a = __ F __ . F Fm1 
1111 + 1112' 2
,1 111] + 1112 
0.400 m/s2; O.SOO N 
g(m2 - 1111 sin 0) gI1111112( 1 + sin 0) a = ; T = --'--=-------1111 + 1112 1111 + 1112 
2.45 m/ S2; 36.S N 
1 .37 m/s2; 61 .4 N 
1 .19; The answer is the ratio of two quantities with the 
same uni ts and so has no units. 
39S N 
36S N 
5.00 cm 
aSkg = 4.91 m/s2; a20 kg = 2.45 m/s2; T = 24.5 N 
85. 
87. 
91 . 
93. 
95. 
97. 
1 .36 kg or 1 .06 kg 
4ml1Ti2g 
F = 2T = ---Ii'll + 1112 
(n) -100 m/52 
(b) 6.13 cm 
(c) 35.0 ms 
305 N; 1 .55 kN 
(n) 
(b) 
(c) 
(d) 
(a) 
(b) 
(c) 
(d) 
1 .50 m/s 
1 .50 m 
0.500 m/s 
12.0 N 
F a = ---1111 + 1'/'12 
Fm2 
F = ---net IH 1 + m2 Frll 1 T = ---m'1 + 1112 
, � �F 
Yes . . . correct answers appear above. 
99. (a) 55.0 g 
(b) 2.45 m/s2; 2.03 N 
101. (a) ! (F2 + 2F1 ) 
(b) 
3To 
4C 
Chapter 5 
1 . The force of friction between the object and the floor of the 
truck must be the force that causes the object to accelerate. 
3. (d) 
5. (b) 
7. As the spring is extended, the force exerted by the spring on the 
block increases. Once that force is greater than the maximum 
value of the force of static friction on the block, the block will be­
gin to move. However, as it accelerates, it will shorten the length 
of the spring, decreasing the force tha t the spring exerts on the 
block. As this happens, the force of kinetic friction can then slow 
the block to a stop, which starts the cycle over again. One inter­
esting application of this to the real world is the bowing of a vio­
lin string: the string under tension acts like the spring, while the 
bow acts as the block, so as the bow is dragged across the string, 
the string periodically sticks and frees itself from the bow. 
9. (e) 
11 . Block 1 will hit the pulley before block 2 hits the wall. 
13. (d) 
15. (d) 
17. For a rock, which has a relatively small surface area compared 
to its mass, the terminal speed will be relatively high; for a light­
weight, spread-out object like a feather, the opposite is true. 
Another issue is that the higher the terminal velocity is, the 
longer it takes for a falling object to reach terminal velocity: 
from this, the feather will reach its terminal velocity quickly, 
19. 
21 . 
23. 
25. 
27. 
29. 
31 . 
33. 
35. 
37 . 
39. 
41. 
43. 
45. 
47. 
Chapter 5 A-7 
and fall at an almost constant speed very soon after being 
dropped; a rock, if not dropped from a great height, will have 
almost the same acceleration as if it were in free-faU for the dura­
tion of its fall, and thus be continually speeding up as it falls. 
(a) M/T; kg/s 
(b) M/L; kg/m 
(c) ML/T2 
(d) 56.9 m/s 
(e) 86.9 m/s 
(b) 
(a) 15 .0 N 
(b) 12.0 N 
500 N 
(a) -5.89 m/s2 
(b) 76.4 m 
(a) 49.1 N 
(b) 123 N 
4.57° 
(a) 0.667 
(b) 2.16 m/s2; 1 .36 5 
(a) 
2.36 m/s2; 37.2 N 
(n) 0.599 
(b) 9.25 m 
(c) 4.73 m/s 
(a) 2.75 m/s2 
(b) 10.1 s 
(a) 0.965 m/s2 directed up the incline 
(b) 0.184 N 
(a) 25.0° 
(b) 0.118 N 
(a) 
(b) 
The static-frictional force opposes the motion of the object, 
and the maximum value of the static-frictional force is pro­
portional to the normal force FN • The normal force is equal 
to the weight min us the vertical camp on en t F v of the force 
F. Keeping the magnitude F constant while increasing 0 
from 0 results ill a decreasein F" and thus a corresponding 
decrease in the maximum static-frictional force Im.x' The 
object will begin to move if the horizontal component FI-I 
of the force F exceeds Imax' An increase ill 0 results in a 
decrease in FI-I' As 0 increases from 0, the decrease in FN is 
larger than the decrease in FH , so the object is more 
and more likely to slip. However, as 0 approaches 90°, 
FH approaches 0, and no movement will be initiated. If F is 
large enough and if 0 increases from 0, then at some value 
of 0, the block will start to move. 
240 .----,----,----,-----------,-----,-, 
235 .\\ .. ......... � ............... :. . " ...... + .. ....... . + ......... . : ........ . � \\ .. . + ....... ...... :.......... c · · · , ................ :.. Ii 230 ' \ L � 225 
� :: \"'\" ...... .. --!-........ . ... . f ......... ..... : ............ 
J',f! 
210 + . . ...... + .... '\ . .. ,.. .......... ,..... .. ..j// .......... : .............. . f"--�! 205 +-----�---+----�----�--��--� o 10 20 30 40 50 60 
o (degrees) 
A-8 
49. 
51 . 
53. 
55. 
57. 
59. 
61 . 
63. 
65. 
67. 
Answers 
(b) 1400 
1 200 
'1000 
� 800 ..... 
600 
400 
200 a 10 
(a) 0.238 
(b) 1 .40 In/s2 (n) 17.7 N 
(b) 1 .47 In /s2; 5.88 N 
(e) 1 .96 In/s2; 7.87 m/s2 
(n) 0.163 In /s2 
(b) 0.0381 In 
( e) -0.254 In/S2 
-8.41 glapp/plipp2; 0 .191 
(n) - 1 .57 N; 83.8 N 
(b) 6.49 N; 37.5 N 
(n) -2.60 In/s2; 19.2 m 
(b) -2.11 m/s2; 23.7 m (n) 0.297 
(b) 2.82 In/s 
(e) 
(n) 1 .41 Ill /S 
(b) 8.50 N 
69. (n) 8.33 In/s2; upward 
(b) 542 N; upward 
(e) 1 . 18 kN; upward 
15 20 25 30 35 40 45 
o (degrees) 
71 . T2 = [m2(LJ + L2)le;t TJ = [m2(L J + L2) + 1I"1 JL J 1(2;)2 
73. 53.3°; 410 N 
75. (n) 0.395 N 
(b) 0.644 
77. 3.44 x 1O-3g; 6.07 x 1O-4g 
79. 
81 . 
83. 
85. 
(n) 
V6 ( 1 )2 nc = � l + (IL;Vo} 
(b) a, = -ILknc 
(e) n = nc \h + IL� 
12.8 Ill / S (n) 7.25 m/s 
(b) 0.536 
87. 21 .7° 
89. (n) 7.832 kN 
(b) -766 kN 
91. vm;n = 20. 1 kln/h; vOlax = 56.1 kln/h 
93. 2.79 x 10-4 kg/In 
95. 88.2 kln/h 
50 
97. 
99. 
101 . 
103. 
105. 
107. 
109. 
111 . 
113. 
115. 
117. 
3.31 s; 100 
y(3.5 s) = 60.4 Ill; Ymax = 60.6 m @ t = 3.3 s; t lUghl = 7 s; The ball 
spends a little longer coming down than it does going up. 
0.511 
(n) 0.289 
(b) 600 N 
1 .49 kN n = g(sin 01 - tan 00 cos 01 ) 
(n) 49.4 m/s2 
(b) 4.49 s (n) 193 N 
(b) 51 .8 N 
(e) The sled does not move. 
(d) ILk is undetermined. 
(e) 536 N 
0.433 
23.6 rev / min 
(n) Toward the earth's axis. 
(b) A stone dropped from a hand at a location on the earth. 
The effective weight of the stone is equal to 11 last, surl' 
where aSI, surl is the acceleration of the falling stone 
(neglecting air resistance) relative to the local surface of 
the earth. The gravitational force on the stone is equal to 
rnast, ;ner' where aSI, ;ner is the acceleration of the stone rela­tive to an inertial reference frame. These accelerations are 
related by as!, surf + asurf, iner = as!, iner' where asurf, iner is the 
acceleration of the local surface of the earth relative to the 
inertial frame (the acceleration of the surface due to the ro­
tation of the earth). Multiplying through this equation by J1l. 
and rearranging gives 1nasI, surf = 111asl. iner - tnasurf, iller' 
which relates the apparent weight to the acceleration due 
to gravity and the acceleration due to the earth's rotation. 
A vector addition diagram can be used to show that the 
magnitude of mast, surl is slightly less than that of maSI, ;ner' 
(e) 983 cm/s2 
Chapter 6 
1 . (n) False 
(b) True 
(e) True 
3. 
5. 
7. 
9. 
11 . 
13. 
15. 
False 
No. TI::.e work done on any o!?Ject by any force F is defined as 
dW == F ' df. The direction of Fnel is toward the center of the cir­
cle in which the object is traveling, and df is taJ�&ent to the cir­
cle. No work is done by the net force because Fnel and df are 
perpendicular, so the dot product is zero. 
Because W IX x2, doubling the distance the spring is stretched 
will require four times as much work. 
(d) (n) False 
(b) False 
(e) True 
(n) False 
(b) False (n) 0.245 In 
(b) 120 J 
17. "" 1% 
19 . 20.8 kJ 
21. (n) 147 J 
(b) 266 J 
23. 10.6 kJ 
25. (n) 6.00 J 
(b) 12.0 ] 
(e) 3.46 m/s 
27. W = -�kx12 - l axf 
29. (a) m(y) = 40 kg - (1 kg/m)y 
(b) 5.89 kJ 
31 . (a) 4.17 N 
35. 
37. 
39. 
(b) T, Fg, and Fn; Because all of these forces act perpendicu­
larly to the direction of motion of the object, none of them 
do any work. 
1800 
(a) -24 
(b) - 10 
(e) 0 
(n) 1 .00 J 
(b) 0.213 N 
43. No. Let A = i, B = 3i + 4J and C = 3i - 4] and form A . B 
and A · C. 
45. (b) The results of (a) and (b) tell us that a is perpendicular to v 
and parallel (or antiparallel) to r. 
47. (a) 98.1 W 
(b) 392 J 
49. (a) v = (� m/s2)1 
(b) P = 3.13I W/s 
(e) 9.38 W 
51 . 445 W 
53. v = Vvl + 2gH 
55. 4.71 kJ 
57. (a) 392 J 
59. 
61 . 
63. 
(b) 2.45 m; 4.91 m/s 
(e) 
(d) 
(n) 
(b) 
(n) 
(b) 
(e) 
(a) 
(b) 
(e) 
(d) 
24.1 J; 368 J 
392 J; 19.8 m/s 
0.100 m 
0.141 m 
U(O) = (m2C2 - 111 1 C1)g sin 0 
U is a minimum at 0 = - 7T/2 and a maximum at 0 = 7T/2 
U = 0 independently of 0 
( Fx = -2 X 
F" is positive for x '* 0 and therefore F is directed away 
from the origin. 
U(x) decreases with increasing x. 
F" is negative for x '* 0 and therefore F is directed toward 
the origin. U(x) increases with increasing x. 
a 65. U(x) = - + Uo x 
67. (a) F" = 4x(x + 2)(x - 2) 
(b) -2 m, 0, 2 m 
(e) Unstable equilibrium at x = -2 m; stable equilibrium at 
x = 0; unstable equilibrium at x = 2 111 
Chapter 6 
69. (a) 0 and 2 m; neutral equilibrium for x > 3 m 
(b) 4. a .------,_--,----,'"'"'---, 
3.5 
3.0 
2.5 
....-., 2.0 
:::; 1 .5 
1 .0 
0.5 
0.0 +----".;L--i--;-.---I 
-1 a 1 2 3 
x (m) 
A-9 
(e) Stable equilibrium at x = 0; unstable equilibrium at x = 2 m 
(d) 2.00 m/s 
71. (a) U(y) = -/JIgy - 2Mg(L - Vy2 + d2) 
! 1112 
(b) y = d \f 4M2 _ 1112 
(e) Stable equilibrium 
73. (a) 706 MJ 
(b) 11 .8 MW 
75. 0.500 m 
77. (a) 34.4 N 
(b) 
T 
\_ 4 nun 
/ 3 f0. \_ -1.6..:.-- - y 
79. 
81 . 
1 .68 N 
(e) 3.38 mJ 
,?-- -
(a) F(x) = rn(2x 
(b) W = 1 111(2X12 
In the following, if I is in seconds and 111 is in kilograms, then 
v is in m/s, a is in m /s2, P is in W, and W is in J. 
(n) v = (612 - 81); a = (121 - 8) 
(b) P = 8rlll(9t2 - 181 + 8) 
(e) W = 21111� (311 - 4)2 
83. 5.74 km 
85. (a) x 
(m) 
-4 
-3 
-2 
- 1 
a 
2 
3 
4 
(b) 
W 
(J) 
6 
4 
2 
0.5 
0 
0.5 
1.5 
2.5 
3 
� 
::J 
a 
-1 
-2 
-3 
-4 
-5 
-3 -2 -1 0 1 2 3 4 
x (m) 
A- l 0 Answers 
87. (b) W = (107T m)Fo if the rotation is clockwise; -(107T m)Fo if 
the rotation is counterclockwise. Because W '" 0 for a com­
plete circuit, F is not conservative. 
89. (a) "6-1 2" Potential 
0.120 ,..--,------,-----,---..,----,----.,---,------,----, 
(b) 
(c) 
0.100 
0.080 
0.060 
0.040 
0.020 
0.000 1-=1;k;�:±=±�rJ[1=J 
-0.020 
0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 
r (nm) 
The minimum value is about -0.0107 eV, occurring at a 
separation of approximately 0.380 nm. Because the func­
tion is concave upward at this separation, this separation 
is one of stable equilibrium, although very shallow. 
-6.69 X 10-12 N; 7.49 X 10-11 N 
Ch a p t er 7 
1. (a) 
3. (a) False 
(b) False 
5. As she starts pedaling, chemical energy inside her body is con­
verted into kinetic energy as the bike picks up speed. As she rides it 
up the hill, chemical energy is converted into gravitational poten­
tial energy. While freewheeling down the hill, potential energy is 
converted to kinetic energy, and while braking to a stop, kinetic 
energy is converted into thermal energy (a more random form of 
kinetic energy) by the frictional forces acting on the bike. 
7. (d) 
9. No. From the work-kinetic energy theorem, no total work is 
being done on the rock, as its kinetic energy is constant. How­
ever, the rod must exert a tangential force on the rock to keep 
the speed constant. The effect of this force is to cancel the com­
ponent of the force of gravity that is tangential to the trajectoryof the rock. 
11 . 33.6 s 
13. 3.04 X 1019 J/y; = 6% 
15. 1 . 10 x 106 Lis 
17. (c) 
19. 3.89 m 
21 . 5.05 m 
23. 25.6° 
25. 
[111g(Sin 8 + f.L cos 8)]2 
U = 5 
27. 
29. 
31 . 
33. 
6mg 
(c) 
6mg 
(a) 31.0 m 
(b) -31.7 J 
(c) 33.7 mls 
2k 
35. 
37. 
39. 
41 . 
43. 
45. 
47. 
(a) 151 m 
(b) 45.3 mls 
(a) � mgL 
(b) 6mg 
(a) 20.2° 
(b) 6.39 mls 
V = L)2t ( 1 - COS 8) + � (V¥ - 3 COS 8 - 1r 
(a) 94.2 kJ 
(b) The energy required to do this work comes from chemical 
energy stored in the body. 
(c) 471 kJ 
(a) 104 J 
(b) 70.2 J 
(c) 33.8 J 
(d) 2.91 m/s 
(a) 7.67 m/s 
(b) 58.9 J 
(c) 0.333 
49. (a) 
(b) 
(c) 
Wf = (13.7 N)y 
Emeeh = -(13.7 N)y 
1 .98 m/s 
51 . 0.875 m; 2.49 mls 
53. (a) 9.00 x 1013 J 
(b) $2.5 X 106 
(c) 28,400 y 
55. 1 .88 X 10-28 kg 
57. 3.56 X 1014 reactions 
59. 0.782 MeV 
61 . (a) 3.16 kg 
(b) 8.04 X 109 kg 
63. (b) 
65. 57.6 MJ 
67. (a) 0.208 
(b) 3.45 MJ 
69. (a) From the FBD we can see that the forces acting on the box 
are the normal force exerted by the inclined plane, kinetic 
friction force, and the gravitational force (the weight of 
the box) exerted by the earth. 
(b) 0.451 m 
(c) l .33 J 
(d) 2.52 mls 
71 . 11 .3 kW; -6.77 kW 
73. (a) 1 .60 kJ 
(b) 619 J 
(c) 23.4 m/s 
75. (a) 147 J 
(b) The energy is transferred to the girder from its surround­
ings, which are warmer than the girder. As the tempera­
ture of the girder rises, the atoms in the girder vibrate 
with a greater average kinetic energy, leading to a larger 
average separation, which causes the girder's expansion. 
77. 
79. 
81 . 
83. 
87. 
89. 
91. 
93. 
(n) 0.8 
0.6 
0.4 
::::0 0.2 ::::; 
0 
-0 .2 
-0.4 
0 
Yeq 
(b) F = -ky + Ing 
(c) 
21ng 
Yrnnx = T 
(d) 
mg 
Yeq = T 
(e) 
11l2g2 
Wf = --2k 
(n) 1 7.3 m 
(b) 4.91 kN 
(c) 4.91 m/s2 
(d) 13.4 kN, upward 
(e) 5.46 kN; 63.9° 
(j) 1 .44 kN 
(a) 491 N; 981 N 
(b) 9.82 kW; 29.4 kW 
(c) 8.85° 
(d) 6.36 kmlL 
(n) 17.4 MJ 
(b) 1 .39 x 1010 J 
(c) 9.73 x 109 J 
(d) 1 .59 MW 
(n) 
2111gY 
v = M + In 
(b) v = 
0 = 2VHL(1 - cos 0) 
(n) 
(b) 
(c) 
(n) 
(b) 
11l2g2 
K",ax = II/gh + U 
mg 
x = - + IllilX k 
Ing 
x = T
+ 
246 
245 
244 
g 243 
::::; 242 
241 
240 
239 
238 
5.39 kJ 
0 
11l2g2 2111gh 
-- + -­
F k 
Potential Energy 
50 100 
s (m) 
150 200 
Chapter 8 A- l l 
Chapter 8 
1 . A doughnut. 
3. (b) 
5. No. Consider a I-kg block with a speed of 1 mls and a 2-kg 
block with a speed of 0.707 m/s. The blocks have equal 
kinetic energies but momenta of magnitude 1 kg·m/s and 
1 .414 kg-m/s, respectively. 
7. Precoil = Prine = -Pbullet or Prine + Pbullet = 0 
9. Conservation of momentum requires only that the net external 
force acting on the system be zero. It does not require the pres­
ence of a medium such as air. 
11 . Think of someone pushing a box across a floor. Her push on the 
box is equal but opposite to the push of the box on her, but the 
action and reaction forces act on different objects. You can only 
add forces when they act on the same object. 
13 . The problem i s that the comic situations violate the conserva­
tion of momentum! To move forward requires pushing some­
thing backward, which Superman doesn't appear to be doing 
when flying around. In a similar manner, if Superman picks up 
a train and throws it at Lex Luthor, he (Superman) ought to be 
tossed backward at a pretty high speed to satisfy the conserva­
tion of momentum. 
15 . The friction of the tire against the road causes the car to slow 
down. This is rather subtle, as the tire is in contact with the 
ground without slipping at all times, so as you push on the 
brakes harder, the force of static friction of the road against 
the tires must increase. Also, of course, the brakes heat up, and 
not the tires. 
17. Assume that the ball travels at 80 milh = 35 m/s. The ball 
stops in a distance of about 1 cm, so the distance traveled is 
about 2 cm at an average speed of about 18 m/s. The collision 
0.02 m time is --1- = 1 ms. 18 m s 
19. (n) False 
(b) True 
(c) True 
21. (a) The loss of kinetic energy is the same in both cases. 
(b) The percentage loss is greatest for the case in which the 
two objects have oppositely directed velocities of magni­
tude � v. 
23. (b) is correct because all of l 's kinetic energy is transferred to 
2 when /1'12 = 111 1 , 
25. The water is changing direction when it rounds the corner in 
the nozzle. Therefore, the nozzle must exert a force on the 
stream of water to change its direction, and, from Newton's 3rd 
law, the water exerts an equal but opposite force on the nozzle. 
27. No. FeXl,net = dp I dt defines the relationship between the net 
force acting on a system and the rate at which its momentum 
changes. The net external force acting on the pendulum bob is 
the sum of the force of gravity and the tension in the string and 
these forces do not add to zero. 
29. Think of the stream of air molecules hitting the sail. Imagine 
that they bounce off the sail elastically-their net change in mo­
mentulll is then roughly twice the change in momentum that 
they experienced going through the fan. Another way of look­
ing at it: ini tially, the air is at rest, but after passing through the 
fan and bouncing off the sail, it is moving backward; therefore, 
the boat must exert a net force on the air pushiJlg it backward, 
and there must be a force on the boat pushing it forward. 
A- 1 2 Answers 
31 . (a) 2.33 s 
(b) 6.74 m/s 
33. (0.233 m, 0) 
35. (2 .00 m , 1 .40 m) 
37. (1 .50 m, 1 .36 m) 
41. zem = � R 
43. vcm = (3 m/s)f - (1.5 m/s)! 
45. acm = (2.4 m/s2)1 
47. (a) Fn = (rl'lp + 11Ib)g 
(b) Fn = (l11p + 2l11b)g 
(c) Fn = m� 
49. (a) Fn = (11'lp + I11b)g 
(b) F = 11'1 g + In g(l + n -p b 
51 . Vl O = (4 m/s)f 
53. v' = 2vf - v! 
55. -Jii- f 
57. (a) 43.5 J 
(b) vcm = (1 .50 m/s)f 
(c) v\rcl = (3.50 m/s)f and V2•rel = (-3.50 m/s)f 
(d) 36.8 J 
(e) Kcm = 6.75 J = K - Krel 
59. (a) 10.8 N·s 
(b) 1 .34 kN 
61. 1 .81 MN·s; 10.602 MN 
63. 230 N 
65. (a) 1 = (1 .08 N·s)f (directed into wall) 
67. 
69. 
71 . 
73. 
(b) 360 N, into wall 
(c) 
(d) 
(a) 
(b) 
(a) 
(b) 
(a) 
(b) 
(a) 
(b) 
(c) 
0.480 N·s, away from wall 
3.84 N, away from wall 
20.0 m/s 
20% of the initial kinetic energy is transformed into ther­
mal energy, sound, and the deformation of metal. 
-2.00 m/s 
The collision was inelastic. 
von = (23.1 m/s)f 
-254 m/s 
5.00 m/s 
0.250 m 
Vlf = 0; V2f = 7.00 m/s 
75. (a) 0.200vo 
(b) 0.400vo 
77. 450 m/s 
81. h = - -v
2 ( 111,)2 8g 1112 
83. 0.0529 
85. 1 .50 X 106 m/s 
87. (a) v, = (312 m/s)f + (66.6 m/s)! 
(b) 5.61 km 
(c) 35.8 kJ 
89. 0.913 
91. (a) 20% of its mechanical energy is lost. 
(b) 0.894 
93. 
95. 
97. 
99. 
101. 
(a) 1 . 70 m/s 
(b) 0.833 
(a) 60° 
(b) 2.50 m/s; 4.33 m/s 
(a) 1 .00 m/s; 1 .73 m/s 
(b) The collision was elastic. 
v, = 8.66 ml s; v2 = 5.00 ml s 
In an elastic collision 
. = = p,
2 [ 11l� + 6m,11I2 + In� ] = p ' f [m�. + 6111111.12 +. m� ] K, Kr 2 2 ? ? 2 111, 1112 + 11111112 2 rl'l,I1l2 + mimi 
If p{ = +p" the particles do not collide. 
103. (a) vcm = 0 
(b) 
(c) 
(d) 
(e) 
105. (a) 
(b) 
(c) 
107. (a) 
(b) 
(c) 
u3 = (-5 m/s)f; Us = (3 m/s)f 
u; = (5 m/s)f; u� = 0.75 m/s 
v; = (5 m/s)f; v� = (-3 m/s)f 
60.0 J; 60.0 J 
360 kN 
120 s 
1 .72 km/s 
ao 
7 = 1 + -o g ( m 1 ( 111 )) vr = gIsp ln� - - 1 - ----.!. 11Ir 70 mo 
3 
en 2 + ................ : .... . '-E C 
" 
O +---i---;--------;------+---I 
o 2 
(d) 28. 1 
109. 0.192 m/s; 31.3 mJ; 12.0 mJ 
111 . 0.462 m/s 
4 6 8 10 
113. (a) p = - (1 .10 X 105 kg·km/h)f + (1 .05 x 105 kg·km/h)! 
(b) 43.4 km/h; 46.3° west of north 
1 15. (a) 6.26 m/s 
(b) 20.0 m 
117. 3.72 m 
119. (a) The velocity of the basketball will be equal in magnitude 
but opposite in direction to the velocity of the baseball. 
(b) Vlf = 0 
(c) v2r = 2v 
121 . (a) 29.6 km/s 
(b) 8.10; The energy comes from an immeasurably small 
slowing of Saturn. 
123. 3.00 x 105 m/s 
125. (a) 0.600 m/s2 
(b) 960 N 
127. No. The driver was traveling at 23.3 km/h. 
129.8.85 kg 
131. � r 
133. (a) 0.716'\1 Eo 
(b) 55 
135. (a) 
(b) 
(c) 
v2 
Ycm = 2L 1
2 
v2 nCm = L 
VI ( v ) F = L gt + 1 Mg 
V l f = 2 V (m1 + I1lb ) (/112 + I11b) 
139. -0.960 m/s2 
141 . v = ( 1 .70 mI/2/s) VL 
Chapter 9 
1. Because r is greater for the point on the rim, it moves the greater 
distance. Both turn through the same angle. Because r is greater 
for the point on the rim, it has the greater speed . .Both have the 
same angular velocity. Both have zero tangential acceleration. 
Both have zero angular acceleration. Because r is greater for the 
point on the rim, it has the greater centripetal acceleration. 
39. 
41 . 
43. 
45. 
47. 
49. 
51 . 
55. 
57. 
(b) 124 rad/ S2 
(c) 620 rad/s 
(a) g sin 8 
Chapter 9 A- 1 3 
(b) Because the line-of-action of the tension passes through 
the pendulum'S pivot point, its lever arm is zero and it 
causes no torque. 
(c) g sin8 
(a) 
(b) 
(c) 
2i-<kMg 
dT = --- r2dr f R2 
Tf = � MRi-<kg 
3Rw M = --
4i-<kg 
56.0 kgom2 
(a) 28.0 kgom2 
(b) 28.0 kg·m2 
2.60 kg·m2 
(b) [em = 12/11 (a2 + b2) 
5.41 X 10- 47 kg·m2 
1 = foMR2 (H2 R2) J = 3M - + ­
x 5 20 
3. � �. (a) 84.6 mJ 
5. (d) (b) 347 rev/min 
7. No. A net torque is required to change the rotational state of an 63. (a) 19.6 kN 
object. A net torque may decrease the angular speed of an ob-
ject. AU we can say for sure is that a net torque will change the 
angular speed of an object. 
9 . (b) 
11. (b) 
13. For a given applied force, this increases the torque about the 
hinges of the door, which increases the door 's angular accelera­
tion, leading to the door being opened more quickly. It is clear 
that putting the knob far from the hinges means that the door 
can be opened with less effort (force). However, i t also means 
that the hand on the knob must move through the greatest dis­
tance to open the door, so it may not be the quickest way to 
open the door. Also, if the knob were at the center of the door, 
you would have to walk aroLmd the door after opening it, 
assuming the door is opening toward you. 
15. (b) 
17. (b) 
19. (a) 
21 . True. If the sphere is slipping, then there is kinetic friction that 
dissipates the mechanical energy of the sphere. 
23. 1 0.3% 
25. 
27. 
29. 
31 . 
6.42 
(a) 
(b) 
(c) 
(d) 
(a) 
(b) 
(a) 
(b) 
33. (d) 
15.6 rad/s 
46.8 rad 
7.45 rev 
73.0 m/s2 
40.0 rad/s 
0.960 m/s2; 192 m/s2 
0.589 rad/s2 
4.71 [ad 
35. 1 .04 rad ls; 9.92 rev/min 
37. (a) 1 .87 Nom 
(b) 5.89 kN-m 
(c) 0.267 rad/s 
(d) 1 .57 kN 
65. (a) 3.62 rad/s 
(b) 3.62 rad/s 
67. Unless M, the mass of the ladder, is zero, v, > vf. It is better to 
let go and fall to the ground. 
69. 3.11 m/s2; T1 = 12.5 N; T2 = 13 .4 N 
71. 8.23 m/s 
73. (a) 
g 
a = ---
2M 1 + -5rn 
(b) 
211lMg 
T = 
5 111 + 2M 
75. (a) 72.0 kg 
(b) 1 .37 rad/s2; TI = 294 N; T2 = 746 N 
77. (a) g sin 8 a = ---1111 
1 + -2m2 
(b) T = 
� 1111 g sin 8 
1111 
1 + -
21'/"1 2 
(c) E = 1112gh 
(d) Ebo!lom = 1112gh 
(e) v = � 1111 
\ 1 + -2m2 
(j) For 8 = 0: a = T = 0 
For 8 = 90°· a = --
g
-- T = " 111 a and v = 
. ' 1111 ' 2 l ' 1 + -
2m2 
For 1'1 '1 1 = 0: a = g sin 8, T = 0, and v = V2ih 
A- 1 4 Answers 
79. 
81 . 
83. 
85. 
87. 
89. 
9l . 
93. 
97. 
99. 
101 . 
103. 
105. 
107. 
111 . 
113. 
115 . 
117. 
119. 
121 . 
123. 
125. 
0.0864 m/s2; 3.14 m/s 
0.192 m/s2; 0.962 N 
1 .13 kJ 
45.9 m 
19 .5° 
(a) a = �g sin 0 
(b) Is = � Illg sin 0 
(c) Omax = tan - , (3,us) 
v' = �v 
223 J 
(a) 
2F 
a = ; counterclockwise R (M + 3111) 
(b) 
(c) 
(a) 
(b) 
(a) 
(b) 
(c) 
F 
a = ---
C M + 3m 
2F 
aCB = - M + 3m 
0.400 rad/ S2; 0.200 rad/ S2 
4.00 N 
12 v6 2 va s, = - -- , tl = - --, and v 49 ,ukg 7 ,ukg 
5/7 
26.6 m; 3.88 s ; 5 .71 m/s 
2rwo 
v = --7 
(a) 360 kN 
(b) 120 s 
(c) 1 .72 km/s 
(a) v = 1 .57 va 
(b) 
4 va !It = - -7 ,ukg 
v2 
(c) �x = 0.735-° 
,ukg 
1 = 2mR2 
0.134 m 
(a) 7.36 m/s2 
(b) 14.7 m/ S2 
(c) 2.43 m/s 
(a) 780 kJ 
(b) 90.3 N·m; 150 N 
(c) 
(a) 
(b) 
(a) 
(b) 
(c) 
(a) 
(b) 
1380 rev 
15 .0 m 
15.4 rad/s 
51 
S3 
S5 
w - f4i \}3; 
F = � Mg 
,------:---
(a) v = 2MgD sin 0 
I M + -
(b) f = 
Mg sin 0 
s R 1 + ­
r 
r2 
5 
1 = "7vo 
127. (a) 14.7 m/s2 
(b) 66.7 cm 
129. 41 .7 J 
131 . The solid line on the graph shown below shows the position y 
of the bucket when it is in free fall and the dashed line shows y 
under the conditions modeled in this problem. 
20 r---�---�--�--� 
15 
]: 10 ;:,., 
5 
0 
0 0.5 1 .0 1 .5 2.0 
t (s) 
133 . (a) 25.7 N 
(b) 3.21 kg 
(c) 1 . 10 m/s2 
Chapter 1 0 
1 . (a) True 
3. 
5. 
7. 
9. 
11. 
13. 
15 . 
17. 
19. 
21 . 
23. 
25. 
27. 
(b) True 
(c) True 
90° 
(a) Doubling p doubles L. 
(b) Doubling r doubles L. 
False 
(e) 
It is easier to crawl radially outward. In fact, a radially inward 
force is required just to prevent you from sliding outward. 
The hardboiled egg is solid inside, so everything rotates with a 
uniform velocity. By contrast, it is difficult to get the viscous 
fluid inside a raw egg to start rotating; however, once it is rotat­
ing, stopping the shell will not stop the motion of the interior 
fl uid, and, for this reason, the egg may start rotating again after 
momentarily stopping. 
(b) 
(b) 
(a) The lifting of the nose of the plane rotates the angular mo­
mentum vector upward. It veers to the right in response to 
the torque associated with the lifting of the nose. 
(b) The angular momentum vector is rotated to the right 
when the plane turns to the right. In turning to the right, 
the torque points down. The nose will move downward . 
(b) 
The center of mass of the rod-and-putty system moves in a 
straight line, and the system rotates about its center of mass. 
4.17 rev/s 
(a) 2.40 X 10-8 kg·m2/s 
(b) 5.22 X 1052; 2.29 X 1026 
(c) The quantization of angular momentum is not noticed in 
macroscopic physics because no experiment can differen­
tiate between e = 2 X 1026 and e = 2 X 1026 + 1 . 
29. (a) 0.331 
(b) Because experimentally C < 2/5 = 0.4, the mass density 
must be greater near the center of the earth. 
31. 10.1 rad/s 
33. T = FRk 
35. (n) 24k 
(b) - 24J 
(c) -51( 
39. B = 4J + 3k 
45. (a) 54.0 kgom2/s 
(b) w increases as the particle approaches the point and de­
creases as it recedes. 
47. (a) 1 .33 x 10-5 kgom2/s 
(b) 1 .33 x 10-5 kgom2/s 
(c) 1 .33 x 10-5 kgom2/s 
(d) 8.83 x 10-5 kgom2/s; -6.17 X 10-5 kgom2/s 
49. (n) 4.00 Nom 
(b) (0.192 rad/s2) t 
51 . (a) Tnet = Rg(m2 sin e - m,) 
(b) L = VR(;2 + 111, + 1112) 
g(m2 sin e - 111 , ) (c) a = =----=-----'-I - + 111, + 1112 R2 
55. (a) 5.00 rev / s 
(b) 622 J 
(c) The energy comes from your internal energy. 
57. 
59. 
61 . 
63. 
65. 
67. 
69. 
71 . 
9.67 mm/s 
(n) 
(b) 
(c) 
54.7° 
(n) 
(b) 
Lo = rolrlvo 
Ko = � II/V 2 0 
v2 3 
T = F = 111�; W = --mv2 
e ,.o 2 0 
3.46 X 10-47 kgom2 
1 .99 meV; 5.98 meV; 12.0 meV 
82.5 m/s 
mv mMvd v = --- ' w = em M + 111 ' tzMU(M + 11'1 ) + Mmd2 
(0.5 M + 0.8111 ) (� Md2 + 0.64md2)g v = 0.32dm2 
(n) 
K Vem = M 
(b) 
4K 
M 
(c) 
2K --
M 
(d) X = � e 
73. 0.349 
75. 12 rad/ s; 10 .8 J 
77. (n) 18.1 Jos 
(b) 0.414 rad/s 
(c) 15.2 s 
(d) 0.0791 Jos 
79. (n) r = - (47.7 kgom2/s)k 
(b) T = (15.9 Nom)k 
81 . (a) 243 J oS 
(b) 306 J 
83. (a) No, L decreases. 
Chapter R 
(b) Its kinetic energy is constant. 
(c) Vo (The kinetic energy remains constant.) 
85. Yes. 
87 v = e
w , I(U - (2) . , 2L v 
91 . (n) 0.228 rad/ s 
(b) 0.192 rad/s 
93. 4.47 x 1022 Nom 
95. 12.5 rad/ s 
97. (a) 26.5 rad/s 
(b) r = (0.303 kgom2/s)e (1 .4' S-')1 
Chapter R 
1 . The friend in the car. 
A- 1 5 
3. Yes. If two events occur at the same time and place in one 
reference frame they occur at the same time and place in aU ref­
erence frames. (Any pair of events that occur at the same time 
and at the same place in one reference frame are called a space­
time coincidence.) 
5. 
7. 
9. 
11 . 
13. 
15. 
17. 
19. 
21 . 
23. 
25. 
27. 
29. 
31 . 
33. 
(a) 
1 + (8.61 x 10- 11 ) 
6.00 ns 
(n) 6.63 m 
(b) 12 .6 m 
(a)599 m 
(b) 13.4 km 
(a) 129 y 
(b) 87.6 Y 
(a) 0.600 m 
(b) 2.50 ns 
0.800e 
(a) 4.50 x 10-10% 
(b) 0.142 ms 
25.0 min; 25.0 min 
60.0 min 
0.400c; event B can precede event A provided II > 0.400e 
(a) 11 .3 y 
(b) 40.0 Y 
(a) 1 .005 
(b) 1 . 155 
(c) 1 .667 
(d) 7.089 
(a) 0.155Eo 
(b) 1 .29Eo 
(c) 6.09Eo 
35. 2.97 GeV 
39. (b) 0.866c 
(c) 0.999c 
A- 1 6 Answers 
41 . (a) 0.794% 
(b) 68.7% 
43. (a) 0.943 
45. (a) 617 eV 
(b) 79.6 eV 
(e) 7.96 eV 
47. (a) 0.745 
(b) 5.00 ft 
(e) No. In Keisha's rest frame, the back end of the ladder will 
clear the door before the front end hits the wall of the shed, 
while in Ernie's rest frame, the front end will hit the wall of 
the shed while the back end has yet to reach the door. 
Chapter 1 1 
1 . (a) False 
(b) True 
3. (d) 
5 . (a) 
7. The gravitational field is proportional to the mass within the 
sphere of radius r and inversely proportional to the square of r, 
i .e., proportional to ,.3/ r2 = 1". 
9. (d) 
11 . 1.08 X 1011 Ms 
13 . (a) 2.78 h 
(b) 19.3 X 1042 kgom2/s; 7.85 X 1042 kgom2/s; 0.703% 
(e) 4.80 X 10-4 rad/s 
15 . 84.0 Y 
17. (a) 1 .59 X 1011 m 
(b) 2.71 X 1010 m; 2.91 X 1011 m 
19. (a) 90° 
(b) 0.731 AU 
21. (a) 1 .90 X 1027 kg 
(b) 0.282 m/s2; 0.0356 m/s2 
23. (a) 8.18 X 104 s 
(b) 1 .22 X 109 m 
25. 1 .99 X 1030 kg 
27. lOw, where w is your weight on earth. 
29. 2.27 X 104 m/s 
31 . 1 .43 
33. (a) 7.37 m 
(b) 0.0319 mm 
35. 0.605 
37. 
39. 
41 . 
(a) 2.27 kg 
(b) It is the inertial mass of 1112, 
109 m 
GMmo 
W = --R 
43. 6.94 km/s 
45. (a) 
(b) 
--> GMmo , 
F outside = --r-Z - r 
GMrl"lo GMmo 
U(r) = ---; U(R) = ---r R 
GMmo (d) U(r) = U(R) = --
R
-
47. 
49. 
51 . 
53. 
55. 
57. 
59. 
61 . 
63. 
65. 
(e) 
2.38 km/s R 
19.4 km/s 
(a) 62.7 MJ 
(b) 1 7.4 kWoh 
(e) $139 
(a) 7.31 h 
(b) 1 .04 GJ 
(e) 8.72 X 1012 J-s 
11.1 GJ 
g = (4 N/kg)1 
(a) 
� Gm < Gm < 
g = U I + U ' 
(b) v'2
Gm 
g = 2 U 
(a) g = (- 1 .67 X 10-11 N/kg)1 
(b) g = (-8.34 X 1O-12 N/kg)1 
(e) 
(a) 
(b) 
(a) 
(b) 
(e) 
2.48 m 
M = 1 CU 
� 2GM [ ( Xo ) g = -- In -- -U Xo - L 
0 
0 
3.20 X 10-9 N / kg 
GrnM, (b) F = 3.61a2 
(e) 0 
Co: L)] I 
71 . (a) 
mg F = - r g R 
(b) FN = C� - mUlZ}-
(e) The change in mass between you and the center of the 
earth as you move away from the center is more impor­
tant than the rotational effect. 
73. g(x) = GC7T�oR3)[� - 8(x � 1R)Z ] 
75. Ul = �47T;oG 
77. 0.104 mm/s 
79. (a) --> GMm [ � l ' F = ----;r- 1 - {
d2 + �2 y/2 i 
-, GMm , 
(b) F(R) = -0.821 J?2i 
81 . 249 Y 
83. (n) 
(b) W = IIlgR�(� _ __ 1_) . RE RE + h 
85. 8.96 x 107 m 
87. 1 .70 Mm 
(GM 91 . v = 1 .64 y ----;--n-
GM 
93. For r < R I , g(r) =0; For r > R" g(r) = -, ; - r-
GM( r3 - Rt) 
For R, < r < R2, g(r) = r2(R� _ R�) 
2GA 
95. g = -}-. 
0.12 -,--,--.,---,--.,----,--.,---,-----, 
0.10 
0.08 
:s 0.06 bl) 
0.04 
002 
o +--+-+--+-�-+-��� o 2 3 4 5 6 7 8 
I I 
Rl R2 
GMmo (Xo + L/2) 97. (b) U = --- In L Xo - L/2 
99. 33.5 pN 
1 0 1 . (n) The gravitational force i s greater on the lower robot, so if 
it were not for the cable its acceleration would be greater 
than that of the upper robot, and they would separate. In 
opposing this separation the cable is stressed. 
(b) 220 km 
Chapter 1 2 
----------------------
1 . (n) False 
(b) True 
(c) True 
(d) False 
3. No. The definition of the center of gravity does not require that 
there be any material at its location. 
5. Th.is technique works because the center of mass must be di­
rectly under the balance point. Hence the intersection of the 
two lines must be at the center of mass. 
7. (b) 
9. (c) 
1 1 . The tensile strengths o f stone and concrete are a t least a n order 
of magnitude lower than their compressive strengths, so you 
want to build compressive structures to match their properties. 
13. (b) 200 N/m 
15. 318 N 
Chapter 1 2 A- 1 7 
1 7. (b) Taking long strides requires a larger coefficient of static 
friction because 0 is then large. 
(c) If ILs is smal l , that is, there is ice on the surface, 0 must be 
smal l to avoid slipping. 
19 . 
. _ (�n2b - 7TnR2 + 7TR3 � ) 21 . (xcg' Ycg) - ab _ 7TR2 ' 2 b 
23. 692 N; 900; 2.54 kN; No block is required to prevent the mast 
from moving. 
25. 0.728 m 
1 V3 27. F2 = "2W; FI = -2-W 
29. (n) 5.00 III 
(b) 4.87 m 
� MgV�h(-2-R---J--l ) A A 31 . FI = h - R i + Mgj 
33. (n) F = (30.0 N)[ + (30.0 N)J 
(b) F = (35.0 N)l + (45.0 N)J 
35. (n) N-F = Mg - F - -n h 
(b) FC•h = F 
(c) _ )2 R - h Fe." - F h 
37. (n) 6.87 N 
(b) 1 .65 Nom 
(c) -8.26 N; 15 .1 N 
39. 636 N; 21 .50 
41. (n) 70.7 N 
(b) 1 .77 m 
(c) 3.54 m 
(d) 497 N 
43. 7net = (69.3 N)b - (40.0 N)n 
45. 0 = � ( V3b - n) 
35.7 m - 30.4x 47. Y = 3.57 m - (294m-l )x 
12 .0 
10.0 
8.0 
I 6.0 
;0-, 
4.0 
2.0 
0 
0.0 0.2 0.4 0.6 
x (111) 
0.8 1 .0 1 .2 
A- 1 8 Answers 
49. 
51 . 
53. 
55. 
57. 
61. 
63. 
65. 
69. 
71. 
73. 
75. 
77. 
79. 
81 . 
83. 
85. 
h = f.LsL tan e sin e 
21-1 f.Ls = -L
-
t
-
an-e
-
s
-
il
-
l
-
e 
59.00 
(a) 41.6 N 
(b) 0.136% 
5.010 
(a) 1 .82 x 106 N/m2 
(b) 6.62 mJ 
0.686 
It will not support the elevator. 
FL = 117 N; FR = 333 N 
WI = 1 .50 N; w2 = 7.00 N; W3 = 3.50 N 
0.148 
f.Ls < 0.500 
f.Ls = 1 (cot e - 1 ) 
(a) 147 N 
(b) 3.62 m 
The block will tip before it slides. 
f.Ls < 0.500 
(a) The stick remains balanced as long as the center of mass is 
between the two fingers. For a balanced stick the normal 
force exerted by tbe finger nearest the center of mass is 
greater than that exerted by the other finger. Consequently, 
a larger static-frictional force can be exerted by the finger 
closer to tbe center of mass, which means the slipping 
occurs at tbe otber finger. 
(b) The finger farthest from the center of mass will slide in­
ward until the normal force it exerts on the stick is suffi­
ciently large to produce a kinetic-frictional force exceed­
ing the maximum static-frictional force exerted by the 
other finger. At that point the finger that was not sliding 
begins to slide, the finger that was sliding stops sliding, 
and the process is reversed. When one finger is slipping 
the other is not. 
87. (a) 23.0 m/s 
(b) 29.1 m/s 
89. (c) Cs = 1 . 142 m; Clo = 1 .464 m; Cwo = 2.594 m 
91. 
93. 
(d) Increasing N in the spreadsheet solution suggests that the 
sum of the individual offsets continues to grow as N in­
creases without bound. The series is, in fact, divergent 
and the stack of bricks has no maximum offset or length. 
566 N 
mg R - r Fn = 2111g; F = -- ; Fw = mg---:==== cos e YR(2r - R) 
Chapter 1 3 
1 . (e) 
3. (d) 
5. Nothing. The fish is in neutral buoyancy, so the upward accel­
eration of the fish is balanced by the downward acceleration of 
the displaced water. 
7. (b) 
9. It blows over the ball, reducing the pressure above the ball to 
below atmospheric pressure. 
11 . 
13 . 
15. 
17. 
19. 
21. 
23. 
25. 
27. 
29. 
31 . 
33. 
35. 
False 
The buoyant force acting on the ice cubes equals the weight of 
the water they displace (i .e., B = wr = PrVrg) . When the ice 
melts, the volume of water displaced by the ice cubes will oc­
cupy the space previously occupied by the submerged part of 
the ice cubes. Therefore the water level remains constant. 
Because the pressure increases with depth, the object will be 
compressed and its density will increase. Thus it will sink to 
the bottom. 
The drawing shows the beaker and a strip within the water. As 
is readily established by a simple demonstration, the surface of 
the water is not level while the beaker is accelerated, showing 
that there is a pressure gradient. That pressure gradient results 
in a net force on the small element shown in the figure. 
From Bernoulli's principle, the opening above which the air 
flows faster will be at a lower pressure than the other one, 
which will cause a circulation of air in tbe tWlllel from opening 
1 toward opening 2. It bas been shown tbat enough air will cir­
culate inside the tunnel even with the slightest breeze outside. 
0.673kg 
103 kg 
29.8 inHg 
230 N 
198 atm 
(a) 14.8 kN 
(b) 0.339 kg 
0.453 m 
pga3 
F = -8 
37. 4.36 N 
39. (a) 1 1 . 1 x 103 kg/m3 
(b) lead 
41 . 800 kg/m3; 1 .11 
43. 250 kg/m3 
45. 3.89 kg 
47. 2.46 x 107 kg 
49. 491 kN 
51 . (a) 9.28 cm/s 
(b) 0.331 cm 
(c) 8.31 cm, in reasonable agreement witb everyday experience. 
53. (a) 12.0 m/ s 
(b) 133 kPa 
55. 
(c) The volume flow rates are equal. 
(a) 4.58 L/min 
(b) 763 cm2 
57. 144 kPa 
59. (a) 21.2 kg/ s 
(b) 636 kg·m/s 
(c) 899 kg'm/s; 899 N 
61 . (a) x = 2 Yh(H - 1-1) 
(b) h = 1 H ::': � YH2 - x2 
63. (b) Ptop = Pot'" - pgd 
65. 1 .43 mm 
67. 93.4 mi/h; Since most major league pitchers can throw a fast­
ball in the low-to-mid-90s, this drag crisis may very well play a 
role in the game. 
69. 0.0137; 0.0115 
71 . The net force is zero. Neglecting the thickness of the table, the 
atmospheric pressure is the same above and below the surface 
of the table. 
73. 
75. 
77. 
79. 
81 . 
83. 
1061 kg/m3 
65.7% 
If you are floating, the density (or specific gravity) of the liquid 
in which you are floating is immaterial as you are in transla­
tional equilibrium under the influence of your weight and the 
buoyant force on your body. Thus the buoyant force on your 
body is your weight in both (a) and (b). 
rn 
V = --0.96pw 
11 .8 cm 
1 m is a reasonable diameter for the pipeline. 
85. hA = 12.6 m; hs = 9 .78 m 
87. (a) 64.6% 
(b) 10.7 kN 
(c) 17.9 m/s2 
Chapter 1 3 
89. 3.31 X 10-3 mmHg or 3.31 /LmHg 
91 . 1 .37 
93. (a) 70.0 m3 
(b) 7.47 m/s2 
95. (c) 0.126 km-1 
97. (a) 33.9 kN 
(b) 39.8 kN; 36.1 kN 
99. (c) 11 = Vii - -- Vig t 
( 
A, Y 2A1 
(d) 1 h 46 min 
A- 1 9 
A-20 Answers 
Chapter 1 4 
1 . 0; 47T2j2A 
3. (a) False 
(b) True 
(e) True 
5. (a) 
7. False 
9. Assume that the first cart is given an initial velocity v by the 
blow. After the initial blow, there are no external forces acting on 
the carts, so their center of mass moves at a constant velocity v12. 
The two carts will oscillate about their center of mass in simple 
harmonic motion where the amplitude of their velocity is v12. 
Therefore, when one cart has velocity vl2 with respect to the 
center of mass, the other will have velocity -v/2. So, the velocity 
with respect to the laboratory frame of reference will be +v 
and 0, respectively. Half a period later, the situation is reversed; 
so, one will appear to move as the other stops, and vice-versa. 
1 1 . True 
13. Examples of driven oscillators include the pendulum of a clock, 
31. (a) 7.85 m/s; 24.7 m/s2 
(b) -6.28 m/s; - 14.8 m/s2 
33. (a) 0.313 Hz 
(b) 3.14 s 
(e) x = (40 cm)cos[(2 s-1 ) 1 + 8] 
35. 22.5 J 
37. (a) 0.368 J 
(b) 3.84 cm 
39. 1 .38 kN/m 
41. (a) 6.89 Hz 
(b) 0.145 s 
(e) 0.100 m 
(d) 4.33 m/s 
(e) 187 m/s2 
(j) 36.3 ms; 0 
43. (a) 682 N/m 
(b) 0.417 s 
(e) 1 .5 1 m/s 
(d) 22.7 m/s2 
a bowed violin string, and the membrane of any loudspeaker. 45. (a) 3.08 kN/m 
15. Because f' varies inversely with the square root of 1n, taking 
into account the effective mass of the spring predicts that the 
frequency will be reduced. 
1 7. (d) 
19. (b) 
21. 87T 
23. (a) 3.00 Hz 
(b) 0.333 s 
(e) 7.00 cm 
(d) 0.0833 s; Because v < 0, the particle is moving in the nega­
tive direction at I = 0.0833 s. 
25. (a) x = (25 cm)cos[(4.19 s-1 ) 1] 
(b) v = - (105 cm/s)sin[(4.19 s-1 ) 1 ] 
(e) a = -(439 cm/s2)cos[(4 .19 s-1 ) 1 ] 
27. (a) x = (27.7 cm)cos[(4.19 s-1 ) 1 - 0.445] 
(b) v = - (116 cm/s)sin[(4.19 s-1 ) 1 - 0.445] 
(e) a = -(486 cm/s2)cos[(4.1 9 s-1 ) 1 - 0.445] 
29. (a) 10 ",",--,--�--,------,------,--,----�...,..., 
(b) If 
(s) 
1 
2 
3 
4 
I; 
6 
2 
-2 
-6 
-10 +-�����L-�-+-�� 0 1 2 3 4 5 6 7 8 
I (s) 
LlX 
(s) (cm) 
0 1 2.93 1 
1 1 7.07 1 
2 1 7.07 1 
3 1 2.93 / 
47. 
49. 
51 . 
53. 
55. 
57. 
59. 
(b) 4.16 Hz 
(e) 0.240 5 
(a) 0.438 m/s 
(b) 0.379 m/s; 120 m/s2 
(e) 95.5 ms 
0.262 s 
10 .1 kJ 
(a) 0.997 Hz 
(b) 0.502 s 
(e) 0.294 N 
(a) 46.66 cm 
(b) 0.261 s 
(e) 0.767 m/s 
(a) 0.270 J 
(b) -0.736 J 
(e) 1 .01 J 
(d) 0.270 J 
(a) 1 .90 cm 
(b) 0.0542 J 
(e) ::':0.224 J 
(d) 0.334 J 
61 . 12.2 5 
63. 11 .7 s 
65. T = 27T I L \j g(l - sin 0) 
67. 1 .10 5 
69. 0.504 kg·m2 
71. (b) 3.17 s 
73. 21 . 1 cm from the center of the meter stick 
77. (a) 1 .63572 m 
(b) 14.5 mm, upward 
79. 13S 
81. 
85. 
87. 
89 
91 . 
93. 
95. 
97. 
99. 
3.14% 
(a) 0.314 
(b) -3.13 X 10-2 percent 
(a) 1 .57% 
(c) 0.430£0 
(a) 1 .01 Hz 
(b) 2.01 Hz 
(c) 0.352 Hz 
(a) 4.98 cm 
(b) 14 .1 rad /s 
(c) 35.4 cm 
(d) 1 .00 rad/s 
(a) 0 
(b) 4.00 m/ s 
(a) 14.1 cm; 0.444 s 
(b) 23.1 cm; 0.363 s 
(c) (14.1 cm)sin [(14 .1 S-l) t]; (23.1 cm)sin[(17.3 S-l ) t ] 
(a) v = - (1 .2 m/s)sin [ (3 rad/s)t + �] 
(b) -0.849 m/s 
(c) 1 .20 m/ s 
(d) 1.31 s 
(a) The normal force is identical to the tension in a string of 
length r that keeps the particle moving in a circular path 
and a component of mg provides, for small displacements 
00 or 52' the linear restoring force required for oscillatory 
motion. 
(b) The particles meet at the bottom. Because 51 and 52 are both much smaller than r, the particles behave like the 
bobs of simple pendulums of equal length and, therefore, 
have the same periods. 
101 . 1 .62 s 
103. 3.86 X 1O-7 N'm/rad 
105. g' is closer to g than is goo. Thus the error is greater if the clock is 
elevated. 
107. (a) Ak f-L = -,------
5 (11'1 1 + m)g 
(b) A is unchanged. £ is unchanged since £ = ! kN. w is re­
duced by increasing the total mass of the system and T is 
increased. 
109. (b) 2.04 cm/s2 
113. (a) x = 0 
(b) v = x � 5 . u \j � 
(c) Xf = Xo 
11'1 P 
m + 11'1 b P 
115. (a) 10 TT-��-�-�-�--
8 
:; 6 
4 
0.5 1 .0 1 .5 2.0 2.5 3.0 
x/a 
(b) Xo = a or ao = 1 
Chapter 1 5 
(c) U(xo + c:) = Uo[l + f3 + (1 + f3)-1] 
(d) 8
2 
U(Xo + 8) = constant + Uo? a-
119. 6 .44 X 101 3 rad/s 
121. 7.78.Jff 
123. (a) 0.0478 
(b) 0.00228 
127. (a) 
A-2 1 
0.6 ,-------------------,---------,------�---/----" .
.
..
.
.
.. . � �:: + ................... . -j .............................. j .. .. ................. +. ..................... , .. · ...... ··· .. · .. 7· ....
..
..
.. 
·
.
· 
.
..
..
..
.. 
+/ ..
.. 
· ..
..
..
..
..
.. 
··· 
.. 
-;/' 
.... .. 
::J 0.2 + ........................ , .............................. ii .. . ............ . . ! ...... � ............. . :;7'C;7 ...................... .. . +- ........... ............ . 1 
0.1 .... � 
0.0 
0.0 0.5 1.0 1.5 
(b) r = ro; k = 2f320 
(c) w = 2f3\IQ / -;;; 
Chapter 1 5 
r (nm) 
2.0 2.5 3.0 
1 . The speed o f a transverse wave on a rope i s given by v = 
� where F is the tension in the rope and f-L is its linear den­
sity. The waves on the rope move faster as they move up be­
cause the tension increases due to the weight of the rope below. 
3. True 
5. The speed of the wave v on the bullwhip varies with the ten­
sion F in the whip and its linear density f-L according to v = 
�. As the whip tapers, the wave speed in the tapered end 
increases due to the decrease in the mass density, so the wave 
travels faster. 
7. No; Because the source and receiver are at rest relative to each 
other, there is no relative motion of the source and receiver and 
there will be no Doppler shift in frequency. 
9 . The light from the companion star will be shifted about its 
mean frequency periodically due to the relative approach to 
and recession from the earth of the companion star as it re­
volves about the black hole. 
11. (a) True 
(b) False 
(c) False 
13 . There was only one explosion. Sound travels faster in water 
than air. Abel heard the sound wave in the water first, then, 
surfacing, heard the sound wave traveling through the air, 
which took longer to reach him. 
15. 
vy 
4 6 8 
X (cm) 
A-22 Answers 
17. Path C. Because the wave speed is highest in the water, and 
more of path C is underwater than A or B, the sound wave will 
spend the least time on path C. 
19. (a) 78.5 m 
(b) 69.7 m 
(c) 70.5 m . . . about 1% larger than our result in part (b) and 
11 %smaller than our first approximation in (a). 
21. 270 m/s; 20.6% 
23. 1 .32 km/s 
25. 1 9.6 g 
27. (a) 265 m/s 
(b) 15 .0 g 
29. (b) 40.0 N 
33. The lightning struck 680 m from the ball park, 58.4° W (or E) of 
north. 
39. (a) y(x,t) = A sin k(x - vt) 
(b) y(x,t) = A sin 2 7TG - ft) 
(c) C 1 ) y(x,t) = A sin 2 7T i - y t 
(d) 
27i 
y(x,t) = A sinA (x - vt) 
(e) y(x,t) = A sin 27Tf (; - t) 
41. 9.87 W 
43. (a) The wave is traveling in the -x direction.; 5.00 m/s 
(b) 10.0 cm; 50.0 Hz; 0.0200 s 
(c) 0.314 m/s 
45. (a) 6.82 J 
(b) 44.0 W 
47. (a) 79.0 mW 
(b) Increasingfby a factor of 10 would increase Pa" by a factor 
of 100. Increasing A by a factor of 10 would increase Pa" by 
a factor of 100. Increasing F by a factor of 104 would in­
crease v by a factor of 100 and Pa" by a factor of 100. 
(e) Depending on the adjustability of the power source, in­
creasing for A would be the easiest. 
49. (a) 0.750 Pa 
(b) 4.00 m 
(e) 85.0 Hz 
(d) 340 m/s 
51 . (a) 
(b) 
53. (a) 
(b) 
55. (a) 
(b) 
(e) 
57. (a) 
(b) 
3.68 X 10-5 m 
8.27 X 10-2 Pa 
The displacement s is zero. 
3.68 fLm 
138 Pa 
21 .7 W 1m2 
0.217 W 
50.3 W 
2.00 m 
(e) 4.45 x 10-3 W 1 m2 
59. (a) 20.0 dB 
(b) 100 dB 
61 . 90.0 dB 
65. (a) 100 m 
(b) 0.126 W 
67. (a) 100 dB 
(b) 50.3 W 
(e) 2.00 m 
(d) 96.5 dB 
69. (a) 81 .1 dB 
(b) 80.0 dB; Eliminating the two least intense sources does not 
reduce the intensity level significantly. 
71 . 87.8 dB 
73. 57.0 dB 
75. (a) 260 m/s 
(b) 1 .30 m 
(e) 262 Hz 
77. (a) 1 .70 m 
(b) 247 Hz 
79. 153 Hz 
81. 1021 Hz or a fraction increase of 2.06%; Because this fractional 
change in frequency is less than the 3% criterion for recognition 
of a change in frequency, it would be impossible to use your 
sense of pitch to estimate your running speed. 
83. 349 mi/h 
85. 7.78 kHz 
87. 15.0 km west of P 
89. (a) f' = (1 - u,/v)(l - usIV)-l fo 
91 . 1 .33 m/s 
93. (a) 824 Hz 
(b) 849 Hz 
95. 184 m 
97. -2.07 x 10-5 nm; 99 2.25 x 108 m/s 
99. 2.25 x 101\8 m/s . . . where the upper arrow means the 8 is an 
exponent. 
101 . 20.8 cm 
103. 3.42 m/s 
105. 529 Hz; 474 Hz 
107. 7.99 m 
109. (a) 55.1 N I m2 
(b) 3.46 W 1m2 
(e) 0.109 W 
111 . 77.0 kN 
113. 
115. 
117. 
119. 
204 m 
24.0 cm 
(b) Vo = Jf; 
(e) As seen by an observer at rest, the pulse remains at the 
same position because its speed along the chain is the 
same as the speed of the chain. With respect to a fixed 
point on the chain, the pulse travels through 360°. 
(b) 2.21 s 
Chapter 1 6 
1 . 
t = 1 �I--------T--,..-----'If-------r-----!RI-------+------""'----' 
1 = 2 1 � 
1 = 3 rl --,--.--�R--�--r--+--.--.--�-. 
3. (c) 
5. (b) 
7. (a) 
9. since v C/. T, increasing the temperature increases resonant 
frequencies. 
11 . No; the wavelength of a wave is related to its frequency and 
speed of propagation (,\ = vlj). The frequency of the plucked 
string will be the same as the wave it produces in air, but the 
speeds of the waves depend on the media in which they are 
propagating. Since the velocities of propagation differ, the 
wavelengths will not be the same. 
13. When the edges of the glass vibrate, sound waves are produced 
in the air in the glass. The resonance frequency of the air 
columns depends on the length of the air column, which de­
pends on how much water is in the glass. 
15. (b) 
17. The pitch is determined mostly by the resonant cavity of the 
mouth, and the frequency of sounds he makes is directly pro­
pOI·tional to their speed. Since vHe > va;r (see Equation 15-5), the 
resonance frequency is higher if helium is the gas in the cavity. 
19. Pianos are tuned by ringing the tuning fork and the piano note 
simultaneously and tuning the piano string until the beats are 
far apart (i.e., the time between beats is very long). If we assume 
that 2 s is the maximum detectable period for the beats, then 
one should be able to tune the piano string to at least 0.5 Hz. 
21 . 34.0 Hz; Because v C/. T, the frequency will be somewhat higher 
in the summer. 
23. 7.07 em 
25. (a) 90.0° 
(b) V2A 
27. (a) 0 
(b) 2 10 
(c) 410 
29. (a) � ,\ 
(b) � ,\ 
31 . (a) 60.0 cm 
(b) 2'71" -5 
(c) 24.0 mls 
33. 4726 Hz; 9452 Hz 
35. (b) Y 
(c) 0.500 mls 
37. 1 .81; 51 .5" 
39. (a) 0.279 m 
(b) 1 .22 kHz 
(c) 111 
3 
8m 
(rad) 
0.432 
4 0.592 
5 0.772 
6 0.992 
7 1 .354 
8 undefined 
(d) 0.0698 rad 
41 . 1 .98 rad or 113° 
43. (a) 70.5 Hz 
Chapter 1 6 A-23 
x (s) 
-1 = 0.0 
-I = 0.53 
1 = 1 .05 
(b) The person on the street hears no beat frequency as the 
sirens of both ambulances are Doppler shifted up by the 
same amount (approximately 35 Hz). 
45. (a) 2.00 m; 25.0 Hz 
(b) Y3(x,l) = (4 mm)sin kx cos wl, where k = '71"m-1 and w = 
50'71"S-1 
47. (a) 521 mls 
(b) 2.80 m; 186 Hz 
(c) 372 Hz; 558 Hz 
49. 141 Hz 
51 . (a) 31.4 cm; 47.7 Hz 
(b) 15.0 mls 
(c) 62.8 cm 
A-24 Answers 
53. (a) 
4 . ... 4 
2 2 
E 0 E 0 "" "" 
"., "., -2 -2 
-4 
0 0.5 1.0 1.5 2.0 2.5 2.0 2.5 
x lm) 
4 
2 
E 0 "" 
'" 
-2 
-4 . 
0 0.5 1.0 1.5 2.0 2.5 
x (m) 
(b) 12 .6 ms 
(c) Since the string is moving either upward or downward 
when y(x) = 0 for all x, the energy of the wave is entirely 
kinetic energy. 
55. (a) 70.8 Hz 
(b) 4.89 Hz 
(c) 35 
57. 452 Hz; I t would be better to have the pipe expand so that vlL, 
where L is the length of the pipe, is independent of temperature. 
59. (a) 80 cm 
(b) 480 N 
(c) You should place your finger 9.23 cm from the scroll 
bridge. 
61. (a) 75.0 Hz 
(b) The harmonics are the 5th and 6th. 
(c) 2.00 m 
63. (a) 0.574 glm 
(b) 1 .29 g/m; 2.91 g/m; 6.55 glm 
65. (a) The two sounds produce a beat because the th.ird harmonic 
of the A string equals the second harmonjc of the E string, 
and the original frequency of the E string is sligh.tly 
greater than 660 Hz. If fE = (660 + Llf)Hz, a beat of 2Llf 
will be heard. 
(b) 661 .5 Hz 
(c) 79.6 N 
69. 76.8 N; 19.2 N; 8.53 N 
71. (a) Nlfa 
(b) Llx/N 
(c) 27TNI LlX 
(d) N is uncertain because the waveform dies out gradually 
rather than stopping abruptly at some time; hence, where 
the pulse starts and stops is not well defined. 
73. (a) 3.40 kHz; 10 .2 kHz; 1 7.0 kHz 
(b) Frequencies near 3400 Hz will be most readily perceived. 
75. ! A 
77. 6.62 m 
79. (a) 1 .90 cm; 3.59 mls 
(b) 0; 0 
(c) 1 .18 cm; 2.22 m/s 
(d) 0 ; 0 
81 . (a) At resonance, standing waves are set up in the tube. At a 
displacement antinode, the powder is moved about; at a 
node the powder is stationary, and so it collects at the 
nodes. 
(b) 2fD 
(c) If we let the length L of the tube be 1 .2 m and assume that 
Va;, = 344 mls (the speed of sound in air at 20°C), then the 
10th harmonic corresponds to 0 = 25.3 cm and a driving 
frequency of 680 Hz. 
(d) I ff = 2 kHz and vHe = 1008 mls (the speed of sound in 
helium at 20°), then 0 for the 1 0th harmonic in helium 
would be 25.3 em, and 0 for the 10th harmonic in air 
would be 8.60 cm. Hence, neglecting end effects at the 
driven end, a tube whose length is the least common 
multiple of 8.60 cm and 25.3 cm (218 cm) would work 
well for the measurement of the speed of sound in either 
air or helium. 
83. (a) The pipe is closed at one end. 
(b) 262 Hz 
(c) 32.4 cm 
85. (a) Y' (X,t) = (0.01 m)Sin [(� m-')x - (407TS- ' ) t ] ; 
)/2(x,t) = (0.01 m)Sin [ (� m-I)x + (407T S-I) t ] ; 
(b) 2.00 m 
(c) vy{l m,t) = - (2.51 m/s)sin(40 7T s-' ) t 
(d) ay(l m,t) = - (316 m/s2)cos(407T s-' ) t 
87. Y,es(x,t) = 0.1 sin(kx - wt) 
89. (b) 203 Hz 
91 . (a) What you hear is the fundamental mode of the tube and 
its overtones. A more physical explanation is that the echo 
of the finger snap moves back and forth along the tube 
with a characteristic time of 2L/c, leading to a series of 
clicks from each echo. Since the clicks happen with a fre­
quency of e/2L, the ear interprets this as a musical note of 
that frequency. 
(b) 
93. (a) 
(b) 
95. (a) 
(b) 
(c) 
97. (a) 
1 .5 
38.6 cm 
Since no conditions were placed on its derivation, this ex­
pression is valid for all harmonics. 
1 .54% 
vy(x,t)= - w,AI sin wIt sin k,x - w2A2 sin w2t sin k2x 
dK = � fL[ w� Ai sin2 WI t sin2 k,x + 2w,w2A,A2 sin wIt 
sin k,x sin w2t sin k2x 
+ wi Ai sin2 W2t sin2 k2x Jdx 
K = ! l11wiAi sin2 WIt + ! l11w� A� sin2 wi 
-1.5L-----�----�----�----�----�----�----� 
(b) f(27T) = 1 which is equivalent to the Liebnitz formula. 
99. (b) 0.014 .,------r--------,,------,---------, 
0.012 
0.010 t ······························,········ r.--c ............. ,L--............................... , .... ·················.···· ···· 1 / 
0 008 · · · · ·· · · ··1 
<i 0.006 / ........ ; . .......................... , ................................ ;....................... . . 1 
0.00. //-
···
.
·
.
·
.
·
.
· 
.............. , ............................... � ................................ ; ........... ····················1 
0.002 17 
0.000 +------i----�c----+-----I 
o 50 100 
1/ 
150 200 
(c) The frequency heard at any time is l /Llt", so because Llt" 
increases over time, the frequency of the culvert whistler 
decreases.; 7.65 kHz 
Chap ter 1 7 
1 . (n) False 
(b) False 
(c) True 
(d) False 
3. Mert's room was colder. 
5 . From the ideal-gas law we have P = nRTN. In the process de­
picted, both the temperature and the volume increase but the 
temperature increases faster than does the volume. Hence the 
pressure increases. 
7. True 
9. Kav increases by a factor of 2; Kav is reduced by a factor of � . 
11 . False 
13. Since 107 » 273, it does not matter. 
15. (b) 
17. (d) 
19 . The ratio of the rms speeds is inversely proportional to the 
square root of the ratio of the molecular masses. The kinetic 
energies of the molecules are the same. 
21. Because the temperature remains constant, the average speed 
of the molecules remains constant. When the volume decreases, 
the molecules travel less distance between collisions, so the 
pressure increases because the frequency of collisions increases. 
23. The average molecular speed of He gas at 300 K is about 
1 .4 km/s, so a significant fraction of He molecules have speeds 
in excess of earth's escape velocity (11 .2 km/s), and thus "leak" 
away into space. Over time, the He content of the atmosphere 
decreases to almost nothing. 
25. (n) 3.61 X 103 K 
(b) 225 K 
(c) I f Ven" > ! ve or T 2:: 25T"m' H2 molecules escape. Therefore, the more energetiC Hz molecules escape from the upper at­
mosphere. 
(d) 164 K; 10.3 K; If we assume that the temperature on the 
moon with an atmosphere would have been approximately 
1000 K, then all O2 and H2 would have escaped during the 
time since the formation of the moon to the present. 
27. (a) 1 .24 km/s 
(b) 310 m/s 
29. 
31 . 
33. 
35. 
37. 
39. 
41. 
Chapter 1 8 
(c) 264 m/s 
(d) O2, CO2, and H2 should be found on Jupiter. 
1063°C 
(n) 8.40 cm 
(b) 107°C 
-319°F 
(n) 54.9 torr 
(b) 3704 K 
-40°C = -40°F 
-183°C; -297°F 
(n) B = 3.94 X 103 K; Ro = 3.97 X 10-3 fl 
(b) 1 .31 kO 
(c) -389 fl/K; -433 fl/K 
A-2S 
(d) The thermistor is more sensitive (i.e., has greater sensitiv­
ity, at lower temperatures). 
43. 1 .79 mol; 1 .08 X 1024 molecules 
45. -83.2 glips 
47. (n) 3.66 X 103 mol 
(b) 60.0 mol 
49. 10.0 atm 
51 . 1 .19 kg/m3 
53. 2.56 N 
55. (n) 276 m/s 
(b) 872 m/s 
57. 499 km/ s; 2.07 X 10-16 J 
61 . K/ LlU = 7.95 X 104 
65. (n) 0.142 s 
(b) 0.143 s 
67. (n) 122 K 
(b) 244 K 
(c) 1 .43 atm 
69. 111 mol; 55.5 mol 
71. 711'1H 
73. 400.49 K 
75. (n) 4.10 X 10-26 m 
(b) 4.28 nm; The mean free path is larger by approximately a 
factor of 1000. 
77. (a) 48.9% 
(b) 70.6% 
Chapter 1 8 
1 . LlTB = 4LlT" 
3. (c) 
5. Yes, if the heat adsorbed by the system is equal to the work 
done by the system. 
7. Won + Qm = LlE;nt; For an ideal gas, LlE;nt is a hmction of T only. 
Since W = 0 and Q = 0 in a free expansion, LlE;nt = 0 and T is 
constant. For a real gas, LlE;nt depends on the density of the gas 
because the molecules exert weak attractive forces on each other. 
In a free expansion, these forces reduce the average kinetic 
energy of the molecules and, consequently, the temperature. 
9 . The temperature of the gas increases. The average kinetic en­
ergy increases with increasing volume due to the repulsive in­
teraction between the ions. 
A-26 
11 . 
13 . 
15. 
(a) 
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(d) 
Answers 
False 
False 
True 
True 
True 
True 
True 
17. If V decreases, the temperature decreases. 
19. The heat capacity of a substance is proportional to the number of 
degrees of freedom per molecule associated with the molecule. 
Since there are 6 degrees of freedom per molecule in a solid, and 
only 3 per molecule (translational) for a monatomic liquid, you 
would expect the solid to have the higher heat capacity. 
21 . 1 .63 min, an elapsed time that seems to be consistent with 
experience. 
23. ep = (1 .01 %)ew.,., 
25. (a) 10.5 MJ 
(b) 121 W 
27. 7.48 kcal 
29. 48.8 mg 
31 . 365°C 
33. 20.8°C 
35. 453 kg 
37. (a) O°C 
(b) 125 g 
39. (a) 4.94°C 
(b) No ice is left. 
41. (a) 2.99°C 
(b) 199.8 g 
(c) The answer would be the same. 
43. 618°C 
45. 2.21 kJ 
47. 176°C 
49. 53.7 J 
51 . 
53. 
55. 
57. 
59. 
61 . 
63. 
(a) 6.13 W 
(b) 19 .0 min 
(a) 405 J 
(b) 861 J 
(a) 507 J 
(b) 963 J 
�PoVo 
(a) 555 J 
(b) 555 J 
(a) 55.7 g/mol 
(b) Fe 
(a) 0; 6.24 kJ; 6.24 kJ 
(b) 8.73 kJ; 6.24 kJ; 2.49 kJ 
(c) 2.49 kJ 
65. 59.6 L 
67. �Cp = -!iNk 
69. Cv,w.'er = 5Nk 
71 . (a) 465 K 
(b) 387 K 
73. (a) 300 K; 7.80 L; 1 . 14 kJ; 1 . 14 kJ 
(b) 208 K; 5.41 L; 574 J; 0 
75. (a) 263 K 
(b) 10.8 L 
(c) -1 .48 kJ 
(d) 1 .48 kJ 
79. -142 J 
81 . QO-;A = 8.98 kJ; QA-;B = 13.2 kJ; QB-;C = -8.98 kJ; 
Qc-;o = -6.56 kJ; Wcycle = 6.62 kJ 
83. (a) p 
, \ , P2 - \ -, 
" 
P4 " - - -' 4 ' 
P3 - - -, - ,- - -, , , 
Vj V4 V2 
85. 180 kJ 
87. (a) 65.2 K; 81.2 K 
(b) 1 .62 kJ 
(c) 2.22 kJ 
89. (a) 65.2 K; 81 .2 K 
(b) 2.65 kJ 
(c) 3.25 kJ 
91 . (a) 9.20 X 10-2 J/kg'K 
(b) 0.0584 J/kg 
93. 47.6 kPa; 51.5 K; 71 .2 K; 148 kPa 
95. (a) 2.49 kJ 
(b) 3.20 kJ 
97. 171 K 
99. (a) W = 0; Q = 3.74 kJ 
T,; 
T ' 
c 
, 3 
V3 V 
(b) �U = 3.74 kJ; Q = 6.24 kJ; W = 2.50 kJ 
101 . 
110 - - - - -
-10 " � ______________________ L--
103. 4RT 
105. 396 K 
107. (a) 1 Po 
20353.5 771.5 
(b) diatomic 
3028.5 3048.5 
t (s) 
(c) In the isothermal process, T is constant, and the transla­
tional kinetic energy is unchanged. In the adiabatic process, 
T3 = 1 .32To, and the translational kinetic energy increases 
by a factor of 1 .32. 
109. (a) 93.5 kPa 
(b) 6266 K; 1 .30 MPa 
(c) 56.7 kPa 
111. (b) flU = 4621 J, a result in good agreement with the result of 
Problem 106. 
Chapter 1 9 
1 . Friction reduces the efficiency of the engine. 
3. Increasing the temperature of the steam increases the Carnot effi­
ciency, and generally increases the efficiency of any heat engine. 
5. (c) 
7. (d) 
9. Note that A--7B is an adiabatic expansion. B--7C is a constant 
volume process in which the entropy decreases; therefore heat 
is released. C--7D is an adiabatic compression. D--7A is a con­
stant volume process that returns the gas to its original state. 
The cycle is that of the Otto engine (see Figure 1 9-3). 
1 1 . 5 
� __ ---, 3 
1 
V 
13. p 
C 
"-----�A 
�---------V 
15. 56.5% 
17. (a) 1 .66 x 1017 W 
(b) 5.66 x 1014 J IK·s 
(c) 3.09 x 1013 J/K·s 
19. 29.8 kJ/K 
21. (a) 500 J 
(b) 400 J 
23. (a) 40.0% 
(b) 80.0 W 
25. (a) 
2 
tJ 
1 4 
10 20 30 40 50 
V (L) 
27. 
Chapter 20 
W1 .... 2 = 0; Ql .... 2 = 3.74 kJ 
W2 .... 3 = 4.99 kJ; Q2 .... 2 = 12.5 kJ 
W3 .... 4 = 0; Q3 .... 4 = -7.48 kJ 
W4 .... 1 = 2.49 kJ; Q4 .... 1 = -6.24 kJ 
(b) 15.4% 
2 \ 
\ 
1.5 \ ]' - -�-� � , 1 , 
P.., 4, 2 
0.5 - - - - , - - - .... .... _400 K 
A-27 
, 3 .... - - -, - - 300 K , 
0 0 10 20 30 40 50 60 
V (L) 
13 .1% 
29. (a) 600 K; 1800 K; 600 K 
(b) 15.4% 
31 . (a) 5 .16%; The fact that this efficiency is considerably less 
than the actual efficiency of a human body does not con­
tradict the Second Law of Thermodynamics. The applica­
tion of the second law to chemical reactionssuch as the 
ones that supply the body with energy have not been 
discussed in the text. 
(b) 
35. (a) 
(b) 
(c) 
(d) 
Most warm-blooded animals survive under roughly the 
same conditions as humans. To make a heat engine work 
with appreciable efficiency, internal body temperatures 
would have to be maintained at an unreasonably high level. 
33.3% 
33.3 J 
66.7 J 
2.00 
37. Let the first engine be run as a refrigerator. Then it will remove 
140 J from the cold reservoir, deliver 200 J to the hot reservoir, 
and require 60 J of energy to operate. Now take the second 
engine and run it between the same reservoirs, and let it eject 
140 J into the cold reservoir, thus replacing the heat removed 
by the refrigerator. If 82, the efficiency of this engine, is greater 
than 30%, then Qh2' the heat removed from the hot reservoir by 
this engine, is 140 J 1 (1 - 82) > 200 J, and the work done by this 
engine is W = 82Qh2 > 200 J. The end result of all this is that the 
second engine can run the refrigerator, replacing the heat taken 
from the cold reservoir, and do additional mechanical work. 
The two systems working together then convert heat into me­
chanical energy without rejecting any heat to a cold reservoir, 
in violation of the second law. 
39. (a) 33.3% 
(b) If COP > 2, then 50 J of work will remove more than 100 J 
of heat from the cold reservoir and put more than 150 J of 
heat into the hot reservoir. So running engine (a) to oper­
ate the refrigerator with a COP > 2 will result in the trans­
fer of heat from the cold to the hot reservoir without doing 
any net mechanical work in violation of the second law. 
41. (a) 100°C 
(b) Ql .... 2 = 3.12 kJ; Q2 .... 3 = 0; Q3 .... 1 = -2.91 kJ 
(c) 6.73% 
(d) 35.5% 
43. (a) 5.26 
(b) 3.l9 kW 
(c) 4.81 kW 
A-28 Answers 
45. (a) 173 kJ 
(b) 121 kJ 
47. t.Su = 2.40 J IK 
49. (a) 11 .5 J IK 
(b) Since the process is not quasi-static, it is nonreversible and 
the entropy of the universe must increase. 
51 . 1 .22 kJ IK 
53. 
55. 
57. 
59. 
61. 
63. 
65. 
67. 
(a) 0 
(b) 267 K 
(a) 244 kJ/K 
(b) -244 kJ/K 
(c) t.Su > 0 
(a) - 117 J/K 
(b) 137 J/K 
(c) 20.3 J/K 
1 .97 kJ/K 
(a) 0.417 J/K 
(b) 125 J 
(a) 20.0 J 
(b) 66.7 J; 46.7 J 
(a) 51 .0% 
(b) 102 kJ 
(c) 98.0 kJ 
113 W/K 
69. (a) Process (1 ) is more wasteful of mechanical energy. Process 
(2) is more wasteful of total energy. 
(b) 1 .67 J IK; 0.833 J IK 
71 . 313 K 
73. 10.0 W 
75. (a) 253 kPa 
(b) 462 K 
(c) 6.96 kJ; 25.9% 
77. (a) 253 kPa 
(b) 416 K 
(c) 6.58 kJ; 34.8% 
79. 180 J 
83. =10478 
-
Chapter 2 0 
1 . The glass bulb warms and expands first, before the mercury 
warms and expands. 
3. (c) 
5. (a) With increasing altitude P decreases; from curve OF, T of 
the liquid-gas interface diminishes, so the boiling temper­
ature decreases. Likewise, from curve OH, the melting 
temperature increases with increasing altitude. 
(b) Boiling at a lower temperature means that the cooking 
time will have to be increased. 
7. The thermal conductivity of metal and marble is much greater 
than that of wood; consequently, heat transfer from the hand is 
more rapid. 
9. (c) 
11 . In the absence of matter to support conduction and convection, 
radiation is the only mechanism. 
13. (a) 
15. The temperature of an object is inversely proportional to the 
maximum wavelength at which the object radiates (Wein's dis­
placement law) . Since blue light has a shorter wavelength than 
red light, an object for which the wavelength of the peak of 
thermal emission is blue is hotter than one which is red. 
17. 18.1 mW l (m·K) 
19. 2.90 nm 
21 . (a) t.AIA y = � 
(b) y = 2at.T 
23. 217°C 
25. 15.4 x 10-6 K-l 
27. 5.24 m 
29. 0.255 mm 
31 . (a) The clock runs slow. 
(b) 8.21 s 
33. 3.68 x 10-12 N 1m2 
35. (a) 90°C 
(b) 82°C 
(c) 170 kPa 
37. (b) (pr + �2 }3Vr - 1) = 8Tr 
39. 2.07 kBtu/h 
41 . (a) leu = 962 W; fAl = 569 W 
(b) 1 .53 kW 
(c) 0.0523 K/W 
43. (a) Conservation of energy requires that the thermal current 
through each shell be the same. 
2 1TkL 
f = I ( I ) (T2 - TI ) n 1'1 1'2 
45. 
47. 9.35 X 10-3 m2 
49. 1598°C 
51 . 2.10 km 
53. 5767 K 
55. 
57. 
59. 
1 .18 cm 
f3exp - 13th � 
(b) < � f3t11 
1 .26 X 1010 kW; <0.002% 
61 . 132 W ignoring the cylindrical insulation; 142 W taking the in­
sulation into account. 
63. L2 = LI; W2 = (1 - 2at.T)wI; E2 = EI (l - 2at.T) 
65. (a) 0.698 cm/h 
(b) 11 .9 d 
67. (b) 40.5 min 
600 ,-------------------------. 
�o -- - - -- - - - - - - - ---- - - - - - -
� 400 � 
h 300 
200 
100 
O +----,-----,----,----.--� o 500 1000 1500 
t (h) 
2000 2500 
	FISICA PARA LA CIENCIA Y LA TECNOLOGIA - PAUL TIPLER. 5 EDICION
	INDICE ANALITICO
	1 - Sistemas de medida
	1.1 Unidades
	1.2 Conversion de unidades
	1.3 Dimensiones de las magnitudes fisicas
	1.4 Notacion cientifica
	1.5 Cifras significativas y ordenes de magnitud
	Resumen del capitulo
	Problemas capitulo 1
	2 - El movimiento en una dimensión
	2.1 Despazamiento, velocidad y modulo de la velocidad
	2.2 Aceleracion
	2.3 Movimiento con aceleracion constante
	2.4 Integracion
	Resumen del capitulo
	Problemas del capitulo 2
	3 - Movimiento en dos y tres dimensiones
	3.1 El vector desplazamiento
	3.2 Propiedades generales de los vectores
	3.3 Posicion, velocidad y aceleracion
	3.4 Primer caso particular: movimiento de proyectiles
	3.5 Segundo caso particular: movimiento circular
	Resumen del capitulo
	Problemas del Capitulo 3
	4 - Leyes de Newton
	4.1 Primera ley de Newton: Ley de la inercia
	4.2 Fuerza, masa y segunda ley de Newton
	4.3 Fuerza debida a la gravedad: el peso
	4.4 Las fuerzas de la naturaleza
	4.5 Resolucion de problemas: diagramas de fuerzas de sistemas aislados
	4.6 La tercera ley de Newton
	4.7 Problemas con dos o mas objetos
	Resumen del capitulo
	Problemas del capitulo 4
	5 - Aplicaciones de las leyes de Newton
	5.1 Rozamiento
	5-2 Movimiento a lo largo de una trayectoria curva
	*5.3 Fuerzas de arrastre
	*5.4 Integracion numerica: el metodo de Euler
	Resumen del Capitulo
	Problemas del Capitulo 5
	6 - Trabajo y energía
	6.1 Trabajo y energia cinetica
	6.2 Producto escalar (producto punto)
	6.3 Trabajo y energia en tres dimensiones
	6.4 Energia potencial
	Resumen del capitulo
	Problemas del capitulo 6
	7 - Conservación de la energía
	7.1 Conservacion de la energia mecanica
	7.2 Conservacion de la energia
	7.3 Masa y energia
	7.4 Cuantizacion de la energia
	Resumen del capitulo
	Problemas del capitulo 7
	8 - Sistemas de partículas y conservación del momento lineal
	8.1 Centro de masas
	*8.2 Determinacion del centro de masas por integracion
	8.3 Movimientos del centro de masas
	8.4 Conservacion del momento lineal
	8.5 Energia cinetica de un sistema
	8.6 Colisiones
	*8.7 Sistema de referencia del centro de masas
	*8.8 Sistemas de masa variable: la propulsion de los cohetes
	Resumen del capitulo
	Problemas del capitulo 8
	9 - Rotación
	9.1 Cinematica de la rotacion: velocidad angular y aceleracion angular
	9.2 Energia cinetica de rotacion
	9.3 Calculo del momento de inercia
	9.4 La segunda ley de Newton en la rotacion
	9.5 Aplicaciones de la segunda ley de Newton a la rotacion
	9.6 Objetos rodantes
	Resumen del capitulo
	Problemas del capitulo 9
	10 - Conservación del momento angular
	10.1 Naturaleza vectorial de la rotacion
	10.2 Momento angular 
	10.3 Conservacion del momento angular
	OPCIONAL: 10.4 Cuantizacion del momento angular
	Resumen del capitulo
	Problemas del Capitulo 10
	11 - Gravedad
	11.1 Leyes de Kepler
	11.2 Ley de la gravitacion de Newton
	11.3 Energia potencial gravitatoria
	11.4 El campo gravitatorio g
	* 11.5 Calculo de la ecuacion correspondiente al campo gravitatorio de una corteza esferica por integracion
	Resumen del capitulo
	Problemas del Capitulo 11
	12 - Equilibrio estático y elasticidad
	12.1 Condiciones de equilibrio
	12.2 Centro de gravedad
	12.3 Ejemplos de equilibrio estatico
	12.4 Par de fuerzas
	12.5 Equilibrio estatico en un sistema acelerado
	12.6 Estabilidad del equilibrio de rotacion
	12.7 Problemas indeterminados