Exercicio resolvidos do halliday 8º ed
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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 =