,Chapter 1
1. Various geometric formulas are given in Appendix E.
(a) Expressing the radius of the Earth as
R = ( 6.37 × 106 m )(10−3 km m ) = 6.37 × 103 km,
its circumference is s = 2π R = 2π (6.37 × 103 km) = 4.00 × 104 km.
(b) The surface area of Earth is A = 4π R 2 = 4π ( 6.37 × 103 km ) = 5.10 × 108 km 2 .
2
4 π 3 4π
( )
3
(c) The volume of Earth is V = R = 6.37 × 103 km = 1.08 × 1012 km3 .
3 3
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 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 μm,
( )( )
1km = 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,
( )( )
1cm = 10−2 m = 10−2 m 106 μ m m = 104 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
,2 CHAPTER 1
( )
1.0 yd = ( 0.91m ) 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. Given that 1 furlong = 201.168 m , 1 rod = 5.0292 m and 1 chain = 20.117 m , we find
the relevant conversion factors to be
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. Using the given conversion
factors, we find
(a) the distance d in rods to be
40 rods
d = 4.0 furlongs = ( 4.0 furlongs ) = 160 rods,
1 furlong
(b) and that distance in chains to be
10 chains
d = 4.0 furlongs = ( 4.0 furlongs ) = 40 chains.
1 furlong
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.
1
Thus, 1 fanega = 12 cahiz, or 8.33 × 10−2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the
1
already 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 .
, 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
7.00 7.00
m3 or 55501 cm3. Thus, 7.00 almudes = 12 fanega = 12 (55501 cm3) = 3.24 × 104 cm3.
7. We use the conversion factors found in Appendix D.
1 acre ⋅ ft = (43,560 ft 2 ) ⋅ ft = 43,560 ft 3
Since 2 in. = (1/6) ft, the volume of water that fell during the storm is
V = (26 km 2 )(1/6 ft) = (26 km 2 )(3281ft/km) 2 (1/6 ft) = 4.66 × 107 ft 3 .
Thus,
4.66 × 107 ft 3
V = = 11
. × 103 acre ⋅ ft.
4.3560 × 10 ft acre ⋅ ft
4 3
8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is
equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z.
(a) In units of W, we have
⎛ 258 W ⎞
50.0 S = ( 50.0 S) ⎜ ⎟ = 60.8 W
⎝ 212 S ⎠
(b) In units of Z, we have
⎛ 156 Z ⎞
50.0 S = ( 50.0 S ) ⎜ ⎟ = 43.3 Z
⎝ 180 S ⎠
9. The volume of ice is given by the product of the semicircular surface area and the
thickness. The area of the semicircle is A = πr2/2, where r is the radius. Therefore, the
volume is
1. Various geometric formulas are given in Appendix E.
(a) Expressing the radius of the Earth as
R = ( 6.37 × 106 m )(10−3 km m ) = 6.37 × 103 km,
its circumference is s = 2π R = 2π (6.37 × 103 km) = 4.00 × 104 km.
(b) The surface area of Earth is A = 4π R 2 = 4π ( 6.37 × 103 km ) = 5.10 × 108 km 2 .
2
4 π 3 4π
( )
3
(c) The volume of Earth is V = R = 6.37 × 103 km = 1.08 × 1012 km3 .
3 3
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 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 μm,
( )( )
1km = 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,
( )( )
1cm = 10−2 m = 10−2 m 106 μ m m = 104 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
,2 CHAPTER 1
( )
1.0 yd = ( 0.91m ) 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. Given that 1 furlong = 201.168 m , 1 rod = 5.0292 m and 1 chain = 20.117 m , we find
the relevant conversion factors to be
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. Using the given conversion
factors, we find
(a) the distance d in rods to be
40 rods
d = 4.0 furlongs = ( 4.0 furlongs ) = 160 rods,
1 furlong
(b) and that distance in chains to be
10 chains
d = 4.0 furlongs = ( 4.0 furlongs ) = 40 chains.
1 furlong
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.
1
Thus, 1 fanega = 12 cahiz, or 8.33 × 10−2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the
1
already 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 .
, 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
7.00 7.00
m3 or 55501 cm3. Thus, 7.00 almudes = 12 fanega = 12 (55501 cm3) = 3.24 × 104 cm3.
7. We use the conversion factors found in Appendix D.
1 acre ⋅ ft = (43,560 ft 2 ) ⋅ ft = 43,560 ft 3
Since 2 in. = (1/6) ft, the volume of water that fell during the storm is
V = (26 km 2 )(1/6 ft) = (26 km 2 )(3281ft/km) 2 (1/6 ft) = 4.66 × 107 ft 3 .
Thus,
4.66 × 107 ft 3
V = = 11
. × 103 acre ⋅ ft.
4.3560 × 10 ft acre ⋅ ft
4 3
8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is
equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z.
(a) In units of W, we have
⎛ 258 W ⎞
50.0 S = ( 50.0 S) ⎜ ⎟ = 60.8 W
⎝ 212 S ⎠
(b) In units of Z, we have
⎛ 156 Z ⎞
50.0 S = ( 50.0 S ) ⎜ ⎟ = 43.3 Z
⎝ 180 S ⎠
9. The volume of ice is given by the product of the semicircular surface area and the
thickness. The area of the semicircle is A = πr2/2, where r is the radius. Therefore, the
volume is
