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1. Using the given conversion factors, we find (a) the distance d in rods to be d = 4.0 furlongs = ( 4.0 furlongs )( 201.168 m furlong ) 5.0292 m rod = 160 rods, (b) and that distance in chains to be d = ( 4.0 furlongs )( 201.168 m furlong ) 20.117 m chain = 40 chains. .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 gry 2 = 0.18 point 2 . .3 | 1. Using the given conversion factors we find a the distance d in rods to be . . 4.0 furlongs 201.168 m furlong A d 4.0 furlongs . ------- 160 rods 5.0292 m rod b and that distance in chains to be 4.0 furlongs 201.168 m furlong d ---------- --------------------- 40 chains. 20.117 m chain 2. The conversion factors 1 gry 1 10 line 1 line 1 12 inchand 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 X 103 m and 1 m 1 X 106 pm 1km 103 m 103 m 106 pm m 109 pm. The given measurement is 1.0 km two significant figures which implies our result should be written as 1.0 X 109 pm. b We calculate the number of microns in 1 centimeter. Since 1 cm 10-2 m 1cm 10-2m 10-2m 106pm m 104 pm. We conclude that the fraction of one centimeter equal to 1.0 pm is 1.0 X 10-4. c Since 1 yd 3 ft 0.3048 m ft 0.9144 m 1.0yd 0.91m 106 pm m 9.1 X 105 .