1
Soluton_ManualFundamentals_of_Physics_Extended_10th_E
dition
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,
the corresponding circumference, surface area and volume are:
4 3
C = 2 RE , A = 4 RE2 , V= RE .
3
The geometric formulas are given in Appendix E.
ANALYZE (a) Using the formulas given above, we find the
circumference to be
C = 2 RE = 2 (6.37 103 km) = 4.00 104 km.
(b) Similarly, the surface area of Earth is
( )
2
A = 4 RE2 = 4 6.37 103 km = 5.10 108 km 2 ,
(c) and its volume is
,2
4 3 4
( )
3
V= RE = 6.37 103 km = 1.08 1012 km3 .
3 3
LEARN From the formulas given, 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 gry2 = 0.18 point 2 .
3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front
cover of the textbook (see also Table 1–2).
(a) Since 1 km = 1 103 m and 1 m = 1 106 m,
( )( )
1km = 103 m = 103 m 106 m m = 109 m.
The given measurement is 1.0 km (two significant figures), which implies our result should be
written as 1.0 109 m.
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10 −2 m,
( )( )
1cm = 10−2 m = 10−2 m 106 m m = 104 m.
We conclude that the fraction of one centimeter equal to 1.0 m is 1.0 10−4.
, 3
(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
( )
1.0 yd = ( 0.91m ) 106 m m = 9.1 105 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.
ANALYZE Using the above conversion factors, we find
40 rods
(a) the distance d in rods to be d = 4.0 furlongs = ( 4.0 furlongs ) = 160 rods,
1 furlong
10 chains
(b) and in chains to be d = 4.0 furlongs = ( 4.0 furlongs ) = 40 chains.
1 furlong
, 4
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 1-6.
(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 dozen fanega. Thus, 1
1
fanega = 12 cahiz, or 8.33 10−2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already
1
completed part) implies that 1 cuartilla = 48 cahiz, or 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 2 = 0.500 for the last entry.
(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 0.055501 m 3 or
7.00 7.00
55501 cm3. Thus, 7.00 almudes = 12 fanega = 12 (55501 cm3) = 3.24 104 cm3.
Soluton_ManualFundamentals_of_Physics_Extended_10th_E
dition
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,
the corresponding circumference, surface area and volume are:
4 3
C = 2 RE , A = 4 RE2 , V= RE .
3
The geometric formulas are given in Appendix E.
ANALYZE (a) Using the formulas given above, we find the
circumference to be
C = 2 RE = 2 (6.37 103 km) = 4.00 104 km.
(b) Similarly, the surface area of Earth is
( )
2
A = 4 RE2 = 4 6.37 103 km = 5.10 108 km 2 ,
(c) and its volume is
,2
4 3 4
( )
3
V= RE = 6.37 103 km = 1.08 1012 km3 .
3 3
LEARN From the formulas given, 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 gry2 = 0.18 point 2 .
3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front
cover of the textbook (see also Table 1–2).
(a) Since 1 km = 1 103 m and 1 m = 1 106 m,
( )( )
1km = 103 m = 103 m 106 m m = 109 m.
The given measurement is 1.0 km (two significant figures), which implies our result should be
written as 1.0 109 m.
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10 −2 m,
( )( )
1cm = 10−2 m = 10−2 m 106 m m = 104 m.
We conclude that the fraction of one centimeter equal to 1.0 m is 1.0 10−4.
, 3
(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
( )
1.0 yd = ( 0.91m ) 106 m m = 9.1 105 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.
ANALYZE Using the above conversion factors, we find
40 rods
(a) the distance d in rods to be d = 4.0 furlongs = ( 4.0 furlongs ) = 160 rods,
1 furlong
10 chains
(b) and in chains to be d = 4.0 furlongs = ( 4.0 furlongs ) = 40 chains.
1 furlong
, 4
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 1-6.
(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 dozen fanega. Thus, 1
1
fanega = 12 cahiz, or 8.33 10−2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already
1
completed part) implies that 1 cuartilla = 48 cahiz, or 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 2 = 0.500 for the last entry.
(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 0.055501 m 3 or
7.00 7.00
55501 cm3. Thus, 7.00 almudes = 12 fanega = 12 (55501 cm3) = 3.24 104 cm3.
