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

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