Exam (elaborations) TEST BANK FOR Fundamentals of Physics 7th Edition David Halliday, Robert Resnick, Jearl Walker (Solution Manual)

1. 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= = (b) and that distance in chains to be (4.0 furlongs)(201.168 m furlong) 40 chains. 20.117 m chain d= = 2. 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 gry2= 0.18 point2 . 3. 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 μm. 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 = (10−2m)(106 μ m m) = 104 μm. 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 μm. 4. (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