Elementary Principles 17 - CHAPTER THREE 3.1 (a) m = 16 6 2...

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3- 1 CHAPTER THREE 3.1 (a) m = ×× ≈× 1 662 1 0 0 0 2 1 0521 0 0 35 m k g m kg 3 3 bg b g b g di (b) ± m =≈ × × 81 0 23 2 41 0 31 01 0 11 0 6 6 3 2 oz 1 qt cm 1 g s oz 1056.68 qt cm g/s 3 3 (c) Weight of a boxer 220 lb m W max × 12 220 220 stones lb 1 stone 14 lb m m ( d ) dictionary V DL == ××× × × × × π 2 2 7 4 314 45 4 345 81 0 51 0 7 441 0 0 . . ft 800 miles 5880 ft 7.4805 gal 1 barrel 1 mile 1 ft 42 gal barrels 2 3 (e) ( i ) V ≈×× 6 331 0 11 0 45 ft 1 ft 0.5 ft 28,317 cm 1 ft cm 3 3 3 ( ii ) V ≈≈ 150 28 317 150 3 10 60 0 4 5 lb 1 ft cm 62.4 lb 1 ft cm m 33 m 3 3 , (f) SG 105 . 3.2 (a) (i) 995 1 0 028317 0 45359 1 6212 kg lb m k g f t lb / ft m 3 m 3 . . . = (ii) 995 62 43 1000 kg / m lb / ft kg / m lb / ft 3 m 3 3 m
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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