,Student Solution Manual For Fundamentals Of Physics 8th Edition
By Jearl Walker, Halliday, Resnik, J. Richard Christman
TABLE OF CONTENTS
Chapterl .., . .. . .... . o. .. . I Chapter 23 .. .. .. . .140
Chapter?. .i . . . . . . . . . . . . . .4 Chapter 24 o.o.o..146
Chapter 3 . . . . . . . . . . . 0 . . . . . 10 Chapter 25 .......154
Chaptet 4 . .. . . . . . . . . . . . . . .I4 Chapter 26 . . . . . . . 159
Chaptgr5 . . . . . . . . . . . . . . . . .21 Chapter 27 .......162
Chapter6. . . . . . . . . . . . . . . . .28 Chapter 28 ...... .170
ChapterT o.......... o.... .37 Chapter 29 . . . . . . .175
ChapterS . . . . . . . . . . . . . . . . .42 Chapter 30 . . . . . . . 193
Chapter9. ............... 50 Chapter 31 . . . . . . , l9l
Chapter 10 . . . . . . . . . . . . . . . . 58 Chapter 32 ....,..199
Chapter 11 .......... r... ..63 Chapter 33 ...... .205
Chapter 12.. .. . .. . .. .. .. ..71 Chapter 34 ...... .213
Chapter 13 ... o... . .. . ... ..77 Chapter 35 ...... .221
Chapter 14 .. . . . . r . . . . . . . . . 84 Chapter 36 ...... .229
Chaptgr15 . . . . . . . . . . . . . . . o 89 Chapter 37 ...... .235
Chapter16.. .. ... . .. . . . .. .95 Chapter 38 ...... .239
Chaptgr17 . . . . . . . . . . . . . . , . 101 Chapter 39 ...... .243
Chapter18 . . . . . . . . . . . . . . r . 109 Chapter 40 ...... .247
Chaptgr19 . . . , . . . . . . . . . . . . 115 Chapter 4l ...... .251
Chaptgr20. . . . . . . . . . . . . . . . 122 Chapter 42 ...... .254
Chapter2l ............... a I28 Chapter 43 ...... .260
Chaptet 22. . . . . . . . . . . . . . . . 134 Chapter 44 ...... .264
,Chapter L
1yd - (0.9144mX106 pmlm) - 9.144 x 105 pm.
3
use the given conversion factors.
(a) The distance d, in rods is
(4.0 furlongsX2Ol . 168 m/furlong)
d - 4.0 furlongs - 5.0292mf rod
(b) The distance in chains is
(4'0 furlongsX20l ' 168 m/furlong)
d - 4.Lfurrongs - : 4|chains .
20.L7 mlchain
I-
(a) The circumference of a sphere of radius R is given by 2r R. Substitute R - (6.37 x
106mX10-tk^lm)
should obtain 4.00 x 104km.
(b) The surface area of a sphere is given by 4trR2, so the surface area of Earth is 4n(6.37 x
103 k*)'
(c) The volume of a sphere is given by (4nlrR3, so the volume of Earth is G"13X6.37 x
103 k*)3 _ 1.08 x 1012 km3.
t7
None of the clocks advance by exactly 24h in a 24-h period but this is not the most important
criterion for judging their quality for measuring time intervals. What is important is that the
clock advance by the same amount in each 24-h period. The clock reading can then easily be
adjusted to give the correct interval. If the clock reading jumps around from one 24-h period to
another, it cannot be coffected since it would impossible to tell what the coffection should be.
The followittg table gives the coffections (in seconds) that must be applied to the reading on
each clock for each 24-h period. The entries were determined by subtracting the clock reading
at the end of the interval from the clock reading at the beginning.
Chapter I
, CLOCK Sun. Mon. Tues. Wed. Thurs. Fri.
-Mon. -Tues. -Wed. -Thurs. -Fri. -Sat
A -16 -16 -15 -17 -15 -15
B -3 +5 -10 +5 +6 -7
C -58 -58 -s8 -s8 -58 -58
D +67 +67 +67 +67 +67 +67
E +70 +55 +2 +20 +10 +10
Clocks C and D are the most consistent. For each clock the same coffection must be applied
for each period. The coffection for clock C is less than the coffection for clock D, so we judge
clock C to be the best and clock D to be the next best. The coffection that must be applied to
clock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to *10 s, for clock
E it is in the range from -70 s to -2s. After C and D, A has the smallest range of correction,
B has the next smallest range, and E has the greatest range. From best the worst, the ranking of
the clocks is C, D, A, B, E.
21
(a) Convert grams to kilograms and cubic centimeters to cubic meters: 1 g _ I x 10-' kg and
1cm3 - (1 x To-2 m)3
rs-(1 e)(#)
\t (,%) :rx1o3kg
(b) Divide the mass (in kilograms) of the water by the time (in seconds) taken to drain it. The
mass is the product of the volume of water and its density: M - (5700m3X1 x 103 kg/nt')
5.70 x 106kg. The time is t - (10.0hX3600s/h) - 3.60 x 104 s, so the mass flow rate n is
M- 5.70 x 106kg-
R - t 3.60 x 104 s
1 58 kg/s "
3s
(a) The amount of fuel she believes she needs is (750mi)l(0mif gal): 18.8gal. This is actually
the number of IJ.K. gallons she needs although she believes it is the number of IJ.S. gallons.
(b) The ratio of the U.K. gallon to the U.S. gallon is (4.545963 IL)l(3.785 3060L) : 1.201.
The number of U.S. gallons she actually needs is
(18.8 IJ.K. galx 1.201 U.S gallu.K. gal) : 22.5 IJ.S. gal .
39
The volume of a cord of wood is V - (8 ftx 4 tt)(4 ft) : I28ft3. IJse l ft
Appendix D) to obtain V - I28ft3X0 .3048 ft)3 - 3.62m3. Thus 1.0 m3 of wood coffesponds
to (l 13.62) cord - 0 .28 cord.
^l
2 Chapter I
By Jearl Walker, Halliday, Resnik, J. Richard Christman
TABLE OF CONTENTS
Chapterl .., . .. . .... . o. .. . I Chapter 23 .. .. .. . .140
Chapter?. .i . . . . . . . . . . . . . .4 Chapter 24 o.o.o..146
Chapter 3 . . . . . . . . . . . 0 . . . . . 10 Chapter 25 .......154
Chaptet 4 . .. . . . . . . . . . . . . . .I4 Chapter 26 . . . . . . . 159
Chaptgr5 . . . . . . . . . . . . . . . . .21 Chapter 27 .......162
Chapter6. . . . . . . . . . . . . . . . .28 Chapter 28 ...... .170
ChapterT o.......... o.... .37 Chapter 29 . . . . . . .175
ChapterS . . . . . . . . . . . . . . . . .42 Chapter 30 . . . . . . . 193
Chapter9. ............... 50 Chapter 31 . . . . . . , l9l
Chapter 10 . . . . . . . . . . . . . . . . 58 Chapter 32 ....,..199
Chapter 11 .......... r... ..63 Chapter 33 ...... .205
Chapter 12.. .. . .. . .. .. .. ..71 Chapter 34 ...... .213
Chapter 13 ... o... . .. . ... ..77 Chapter 35 ...... .221
Chapter 14 .. . . . . r . . . . . . . . . 84 Chapter 36 ...... .229
Chaptgr15 . . . . . . . . . . . . . . . o 89 Chapter 37 ...... .235
Chapter16.. .. ... . .. . . . .. .95 Chapter 38 ...... .239
Chaptgr17 . . . . . . . . . . . . . . , . 101 Chapter 39 ...... .243
Chapter18 . . . . . . . . . . . . . . r . 109 Chapter 40 ...... .247
Chaptgr19 . . . , . . . . . . . . . . . . 115 Chapter 4l ...... .251
Chaptgr20. . . . . . . . . . . . . . . . 122 Chapter 42 ...... .254
Chapter2l ............... a I28 Chapter 43 ...... .260
Chaptet 22. . . . . . . . . . . . . . . . 134 Chapter 44 ...... .264
,Chapter L
1yd - (0.9144mX106 pmlm) - 9.144 x 105 pm.
3
use the given conversion factors.
(a) The distance d, in rods is
(4.0 furlongsX2Ol . 168 m/furlong)
d - 4.0 furlongs - 5.0292mf rod
(b) The distance in chains is
(4'0 furlongsX20l ' 168 m/furlong)
d - 4.Lfurrongs - : 4|chains .
20.L7 mlchain
I-
(a) The circumference of a sphere of radius R is given by 2r R. Substitute R - (6.37 x
106mX10-tk^lm)
should obtain 4.00 x 104km.
(b) The surface area of a sphere is given by 4trR2, so the surface area of Earth is 4n(6.37 x
103 k*)'
(c) The volume of a sphere is given by (4nlrR3, so the volume of Earth is G"13X6.37 x
103 k*)3 _ 1.08 x 1012 km3.
t7
None of the clocks advance by exactly 24h in a 24-h period but this is not the most important
criterion for judging their quality for measuring time intervals. What is important is that the
clock advance by the same amount in each 24-h period. The clock reading can then easily be
adjusted to give the correct interval. If the clock reading jumps around from one 24-h period to
another, it cannot be coffected since it would impossible to tell what the coffection should be.
The followittg table gives the coffections (in seconds) that must be applied to the reading on
each clock for each 24-h period. The entries were determined by subtracting the clock reading
at the end of the interval from the clock reading at the beginning.
Chapter I
, CLOCK Sun. Mon. Tues. Wed. Thurs. Fri.
-Mon. -Tues. -Wed. -Thurs. -Fri. -Sat
A -16 -16 -15 -17 -15 -15
B -3 +5 -10 +5 +6 -7
C -58 -58 -s8 -s8 -58 -58
D +67 +67 +67 +67 +67 +67
E +70 +55 +2 +20 +10 +10
Clocks C and D are the most consistent. For each clock the same coffection must be applied
for each period. The coffection for clock C is less than the coffection for clock D, so we judge
clock C to be the best and clock D to be the next best. The coffection that must be applied to
clock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to *10 s, for clock
E it is in the range from -70 s to -2s. After C and D, A has the smallest range of correction,
B has the next smallest range, and E has the greatest range. From best the worst, the ranking of
the clocks is C, D, A, B, E.
21
(a) Convert grams to kilograms and cubic centimeters to cubic meters: 1 g _ I x 10-' kg and
1cm3 - (1 x To-2 m)3
rs-(1 e)(#)
\t (,%) :rx1o3kg
(b) Divide the mass (in kilograms) of the water by the time (in seconds) taken to drain it. The
mass is the product of the volume of water and its density: M - (5700m3X1 x 103 kg/nt')
5.70 x 106kg. The time is t - (10.0hX3600s/h) - 3.60 x 104 s, so the mass flow rate n is
M- 5.70 x 106kg-
R - t 3.60 x 104 s
1 58 kg/s "
3s
(a) The amount of fuel she believes she needs is (750mi)l(0mif gal): 18.8gal. This is actually
the number of IJ.K. gallons she needs although she believes it is the number of IJ.S. gallons.
(b) The ratio of the U.K. gallon to the U.S. gallon is (4.545963 IL)l(3.785 3060L) : 1.201.
The number of U.S. gallons she actually needs is
(18.8 IJ.K. galx 1.201 U.S gallu.K. gal) : 22.5 IJ.S. gal .
39
The volume of a cord of wood is V - (8 ftx 4 tt)(4 ft) : I28ft3. IJse l ft
Appendix D) to obtain V - I28ft3X0 .3048 ft)3 - 3.62m3. Thus 1.0 m3 of wood coffesponds
to (l 13.62) cord - 0 .28 cord.
^l
2 Chapter I
