Leciv
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1. Using the given conversion factors, we find(a) the distance d in rods to be d = 4.0 furlongs =
( 4.0 furlongs ) ( 201.168 m furlong )
5.0292 m rod
= 160 rods,
(b) and that distance in chains to be d= ( 4.0 furlongs ) ( 201.168 m furlong )
20.117 m chain
= 40 chains.
2. The conversion factors 1 g ry = 1/10 line , 1 line=1/12 inch and 1 point = 1/72 inch 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,
(
)
1.0 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 0.80 cm = ( 0.80 cm )
1 inch
2.54 cm
6 picas
1 inch
≈ 1.9 picas.
(b) With 12 points = 1 pica, we have
0.80 cm = ( 0.80 cm )
1 inch
2.54 cm
6 picas
1 inch
12 points
≈ 23 points.
1 pica
5. Various geometric formulas are given in Appendix E.
(a) Substituting
c
hc
h
R = 6.37 × 10 6 m 10 −3 km m = 6.37 × 10 3 km
into circumference = 2πR, we obtain 4.00 × 104 km.
(b) The surface area of Earth is
(
A = 4π R 2 = 4π 6.37 × 103 km
)
2
= 5.10 × 108 km 2 .
(c) The volume of Earth is
V=
4π 3 4π
R=
6.37 × 103 km
3
3
(
)
3
= 1.08 × 1012 km3 .
6. We make use of Table 1-6.