Chapter-01 - 1 Chapter 1 1. The metric prefixes (micro,...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Page1 / 20

Chapter-01 - 1 Chapter 1 1. The metric prefixes (micro,...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online