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Chapter01

# Chapter01 - 1 The metric prefixes(micro pico nano are given...

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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 × 10 3 m and 1 m = 1 × 10 6 μ m, ( ) ( ) 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, ( ) ( ) 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, ( ) ( ) 6 5 1.0 yd = 0.91m 10 m m 9.1 10 m. = × μ μ

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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 § ·§ · ¨ ¸¨ ¸ © ¹© ¹ (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 § ·§ ·§ · ¨ ¸¨ ¸¨ ¸ © ¹© ¹© ¹
(b) and that distance in chains to be ( ) ( ) 4.0 furlongs 201.168 m furlong 40 chains. 20.117 m chain d = = 3. Using the given conversion factors, we find (a) the distance d in rods to be ( ) ( ) 4.0 furlongs 201.168 m furlong 4.0 furlongs = 160 rods, 5.0292 m rod d = =

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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 ( ) ( ) 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 ( ) 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 ( ) 3 3 3 12 3 4 4 6.37 10 km 1.08 10 km . 3 3 V R π π = = × = ×

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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 ( ) 258 W 50.0 S 50.0 S 60.8 W 212 S § · = = ¨ ¸ © ¹ (b) In units of Z, we have ( ) 156 Z 50.0 S 50.0 S 43.3 Z 180 S § · = = ¨ ¸ © ¹
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 ( ) 3 2 5 10 m 10 cm 2000km 2000 10 cm. 1km 1m r § · § · = = × ¨ ¸ ¨ ¸ © ¹ © ¹ In these units, the thickness becomes ( ) 2 2 10 cm 3000m 3000m 3000 10 cm 1m z § · = = = × ¨ ¸ © ¹ which yields ( ) ( ) 2 5 2 22 3 2000 10 cm 3000 10 cm 1.9 10 cm . 2 V π = × × = × 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

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8. 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. Thus, 1 fanega = 1 12 cahiz, or 8.33 × 10 2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = 1 48 cahiz, or 2.08 × 10 2 cahiz.
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