Halliday resolvido 8ed vol3

1. 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.

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2. (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
0.80 cm = ( 0.80 cm ) 1 inch 2.54 cm 6 picas 1 inch 12 points ≈ 23 points. 1 pica

3. 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.

4. The conversion factors 1 gry = 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 .

5. 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 .
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(c) The volume of Earth is V =

4 π 3 4π R = 6.37 × 103 km 3 3

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= 1.08 × 1012 km3 .

6. From Figure 1.6, we see that 212 S is equivalent to 258 W and 212 – 32 =