Solutions Manual for
Fundamentals of Physics
(Extended) 12th Edition By David
Halliday, Robert Resnick, Jearl
Walker (All Chapters 1-44, 100%
Original Verified, A+ Grade)
This is The Only Original and
Complete Solutions Manual for
Extended 12th Edition, All Other
Files in the Market are
Fake/Old/Wrong Edition.
Extra Supplement Files Download
Link is Added at The End of PDF
File.
,Chapter 1
1. THINK In this problem we’re given the radius of Earth and asked to compute its
circumference, surface area, and volume.
EXPRESS Assuming Earth to be a sphere of radius
RE = ( 6.37 × 106 m )(10 −3 km m ) = 6.37 × 103 km,
we find that the corresponding circumference, surface area, and volume are
4π 3
C = 2π RE , A = 4π RE2 , V= RE .
3
These geometric formulas are given in Appendix E.
ANALYZE Using the formulas, we find (a) the circumference to be
C = 2π RE = 2π (6.37 × 103 km) = 4.00 ×10 4 km,
(b) the surface area to be
A = 4π RE2 = 4π ( 6.37 × 103 km ) = 5.10 × 108 km 2 ,
2
and (c) the volume to be
4π 3 4π
( 6.37 × 103 km ) = 1.08 × 1012 km3 .
3
V= RE =
3 3
LEARN From the formulas, we see that C RE , A RE2 , and V RE3 . The ratios of
volume to surface area and surface area to circumference are V /A = RE /3 and
A / C = 2RE .
2. The conversion factors are: 1 gry = 1/10 line, 1 line = 1/12 inch, and 1 point = 1/72
inch. The factors imply that
1 gry = (1/10)(1/12)(72 points) = 0.60 point.
Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50 gry 2 = 0.18 point 2 .
3. The metric prefixes (micro, pico, nano, …) are given in Table 1.1.2.
1
,2 CHAPTER 1
(a) Since 1 km = 1 × 103 m and 1 m = 1 × 106 μm,
1 km = 103 m = (103 m )(106 μ m m ) = 109 μ m.
The given measurement is 1.0 km (two significant figures), which implies our result
should be written as 1.0 × 109 μm.
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10–2 m,
1 cm = 10 −2 m = (10 −2 m )(106 μ m m ) = 10 4 μ m.
We conclude that the fraction of one centimeter equal to 1.0 μm is 1.0 × 10–4.
(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
1.0 yd = ( 0.91 m ) (106 μ m m ) = 9.1× 105 μ m.
4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we
obtain
1 inch 6 picas
0.80 cm = ( 0.80 cm ) ≈ 1.9 picas.
2.54 cm 1 inch
(b) With 12 points = 1 pica, we have
1 inch 6 picas 12 points
0.80 cm = ( 0.80 cm ) ≈ 23 points.
2.54 cm 1 inch 1 pica
5. THINK This problem deals with conversion of furlongs to rods and chains, all of
which are units for distance.
EXPRESS Given that 1 furlong = 201.168 m, 1 rod = 5.0292 m, and
1 chain = 20.117 m, the relevant conversion factors are
1 rod
1.0 furlong = 201.168 m = (201.168 m ) = 40 rods
5.0292 m
and
1 chain
1.0 furlong = 201.168 m = (201.168 m ) = 10 chains.
20.117 m
Note the cancellation of m (meters), the unwanted unit.
, CHAPTER 1 3
ANALYZE Using the above conversion factors, we find the distance d (a) in rods to be
(
d = 4.0 furlongs = 4.0 furlongs ) 1 40furlong
rods
= 160 rods
and (b) in chains to be
(
d = 4.0 furlongs = 4.0 furlongs ) 110furlong
chains
= 40 chains.
LEARN Since 4 furlongs is about 800 m, this distance is approximately equal to 160
rods (1 rod ≈ 5 m) and 40 chains (1 chain ≈ 20 m). So our results make sense.
6. We make use of Table 1.1.
(a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of
cahiz? We note from the already completed part of the table that 1 cahiz equals a
1
dozen fanega. Thus, 1 fanega = cahiz, or 8.33 × 10–2 cahiz. Similarly, “1 cahiz =
12
1
48 cuartilla” (in the already completed part) implies that 1 cuartilla = cahiz, or
48
2.08 × 10–2 cahiz. Continuing in this way, the remaining entries in the first column
are 6.94 × 10−3 and 3.47 × 10 −3 .
(b) In the second (“fanega”) column, we find 0.250, 8.33 × 10–2, and 4.17 × 10–2 for
the last three entries.
(c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two
entries.
1
(d) Finally, in the fourth (“almude”) column, we get = 0.500 for the last entry.
2
(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our
amount of 7.00 almudes must be equal to 14.0 medios.
(f) Using the value (1 almude = 6.94 × 10–3 cahiz) found in part (a), we conclude that
7.00 almudes is equivalent to 4.86 × 10–2 cahiz.
(g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055 501 m3
7.00 7.00
or 55 501 cm3. Thus, 7.00 almudes = fanega = (55 501 cm3) = 3.24 × 104 cm3.
12 12
Fundamentals of Physics
(Extended) 12th Edition By David
Halliday, Robert Resnick, Jearl
Walker (All Chapters 1-44, 100%
Original Verified, A+ Grade)
This is The Only Original and
Complete Solutions Manual for
Extended 12th Edition, All Other
Files in the Market are
Fake/Old/Wrong Edition.
Extra Supplement Files Download
Link is Added at The End of PDF
File.
,Chapter 1
1. THINK In this problem we’re given the radius of Earth and asked to compute its
circumference, surface area, and volume.
EXPRESS Assuming Earth to be a sphere of radius
RE = ( 6.37 × 106 m )(10 −3 km m ) = 6.37 × 103 km,
we find that the corresponding circumference, surface area, and volume are
4π 3
C = 2π RE , A = 4π RE2 , V= RE .
3
These geometric formulas are given in Appendix E.
ANALYZE Using the formulas, we find (a) the circumference to be
C = 2π RE = 2π (6.37 × 103 km) = 4.00 ×10 4 km,
(b) the surface area to be
A = 4π RE2 = 4π ( 6.37 × 103 km ) = 5.10 × 108 km 2 ,
2
and (c) the volume to be
4π 3 4π
( 6.37 × 103 km ) = 1.08 × 1012 km3 .
3
V= RE =
3 3
LEARN From the formulas, we see that C RE , A RE2 , and V RE3 . The ratios of
volume to surface area and surface area to circumference are V /A = RE /3 and
A / C = 2RE .
2. The conversion factors are: 1 gry = 1/10 line, 1 line = 1/12 inch, and 1 point = 1/72
inch. The factors imply that
1 gry = (1/10)(1/12)(72 points) = 0.60 point.
Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50 gry 2 = 0.18 point 2 .
3. The metric prefixes (micro, pico, nano, …) are given in Table 1.1.2.
1
,2 CHAPTER 1
(a) Since 1 km = 1 × 103 m and 1 m = 1 × 106 μm,
1 km = 103 m = (103 m )(106 μ m m ) = 109 μ m.
The given measurement is 1.0 km (two significant figures), which implies our result
should be written as 1.0 × 109 μm.
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10–2 m,
1 cm = 10 −2 m = (10 −2 m )(106 μ m m ) = 10 4 μ m.
We conclude that the fraction of one centimeter equal to 1.0 μm is 1.0 × 10–4.
(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
1.0 yd = ( 0.91 m ) (106 μ m m ) = 9.1× 105 μ m.
4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we
obtain
1 inch 6 picas
0.80 cm = ( 0.80 cm ) ≈ 1.9 picas.
2.54 cm 1 inch
(b) With 12 points = 1 pica, we have
1 inch 6 picas 12 points
0.80 cm = ( 0.80 cm ) ≈ 23 points.
2.54 cm 1 inch 1 pica
5. THINK This problem deals with conversion of furlongs to rods and chains, all of
which are units for distance.
EXPRESS Given that 1 furlong = 201.168 m, 1 rod = 5.0292 m, and
1 chain = 20.117 m, the relevant conversion factors are
1 rod
1.0 furlong = 201.168 m = (201.168 m ) = 40 rods
5.0292 m
and
1 chain
1.0 furlong = 201.168 m = (201.168 m ) = 10 chains.
20.117 m
Note the cancellation of m (meters), the unwanted unit.
, CHAPTER 1 3
ANALYZE Using the above conversion factors, we find the distance d (a) in rods to be
(
d = 4.0 furlongs = 4.0 furlongs ) 1 40furlong
rods
= 160 rods
and (b) in chains to be
(
d = 4.0 furlongs = 4.0 furlongs ) 110furlong
chains
= 40 chains.
LEARN Since 4 furlongs is about 800 m, this distance is approximately equal to 160
rods (1 rod ≈ 5 m) and 40 chains (1 chain ≈ 20 m). So our results make sense.
6. We make use of Table 1.1.
(a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of
cahiz? We note from the already completed part of the table that 1 cahiz equals a
1
dozen fanega. Thus, 1 fanega = cahiz, or 8.33 × 10–2 cahiz. Similarly, “1 cahiz =
12
1
48 cuartilla” (in the already completed part) implies that 1 cuartilla = cahiz, or
48
2.08 × 10–2 cahiz. Continuing in this way, the remaining entries in the first column
are 6.94 × 10−3 and 3.47 × 10 −3 .
(b) In the second (“fanega”) column, we find 0.250, 8.33 × 10–2, and 4.17 × 10–2 for
the last three entries.
(c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two
entries.
1
(d) Finally, in the fourth (“almude”) column, we get = 0.500 for the last entry.
2
(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our
amount of 7.00 almudes must be equal to 14.0 medios.
(f) Using the value (1 almude = 6.94 × 10–3 cahiz) found in part (a), we conclude that
7.00 almudes is equivalent to 4.86 × 10–2 cahiz.
(g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055 501 m3
7.00 7.00
or 55 501 cm3. Thus, 7.00 almudes = fanega = (55 501 cm3) = 3.24 × 104 cm3.
12 12
