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Unformatted text preview: 1 Chapter 1 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 × 10 3 m and 1 m = 1 × 10 6 μ m, ( 29 ( 29 3 3 6 9 1km 10 m 10 m 10 m m 10 m. = = = μ μ The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 10 9 μ m. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10 2 m, ( 29 ( 29 2 2 6 4 1cm = 10 m = 10 m 10 m m 10 m. = μ μ We conclude that the fraction of one centimeter equal to 1.0 μ m is 1.0 × 10 4 . (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m, ( 29 ( 29 6 5 1.0 yd = 0.91m 10 m m 9.1 10 m. = × μ μ 2. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain ( 29 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 ( 29 1 inch 6 picas 12 points 0.80 cm = 0.80 cm 23 points. 2.54 cm 1 inch 1 pica ≈ 3. Using the given conversion factors, we find (a) the distance d in rods to be ( 29 ( 29 4.0 furlongs 201.168 m furlong 4.0 furlongs = 160 rods, 5.0292 m rod d = = CHAPTER 1 2 (b) and that distance in chains to be ( 29 ( 29 4.0 furlongs 201.168 m furlong 40 chains. 20.117 m chain d = = 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 gry 2 = (0.60 point) 2 = 0.36 point 2 , which means that 2 2 0.50 gry = 0.18 point . 5. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earth as ( 29 ( 29 6 3 3 6.37 10 m 10 km m 6.37 10 km, R = × = × its circumference is 3 4 2 2 (6.37 10 km) 4.00 10 km. s R π π = = × = × (b) The surface area of Earth is ( 29 2 2 3 8 2 4 4 6.37 10 km 5.10 10 km . A R = π = π × = × (c) The volume of Earth is ( 29 3 3 3 12 3 4 4 6.37 10 km 1.08 10 km . 3 3 V R π π = = × = × 6. From Figure 1.6, 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 ( 29 258 W 50.0 S 50.0 S 60.8 W 212 S = = (b) In units of Z, we have ( 29 156 Z 50.0 S 50.0 S 43.3 Z 180 S = = 7. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = π r 2 /2, where r is the radius. Therefore, the volume is 3 2 2 V r z π = where z is the ice thickness. Since there are 10 3 m in 1 km and 10 2 cm in 1 m, we have ( 29 3 2 5 10 m 10 cm 2000km 2000 10 cm....
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This note was uploaded on 05/24/2010 for the course PHYS 502 taught by Professor Yang during the Spring '10 term at National Taiwan University.
 Spring '10
 YANG
 Physics

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