4_Physics ProblemsTechnical Physics

# 4_Physics ProblemsTechnical Physics - 4 Physics and...

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4 Physics and Measurement P1.9 Mass of gold abraded: m =−== F H G I K J 3 80 3 35 0 45 0 45 1 45 10 4 ... . . g g g kg 10 g kg 3 bg . Each atom has mass m 0 27 25 197 197 166 10 1 327 10 == × F H G I K J u u kg u kg . .. Now, ∆∆ mN m = 0 , and the number of atoms missing is N m m × × 0 4 25 21 138 10 . . . kg kg atoms. The rate of loss is N t N t = × F H G I K J F H G I K J F H G I K J F H G I K J 50 1 11 1 872 10 21 11 . atoms yr yr 365.25 d d 24 h h 60 min min 60 s atoms s P1.10 (a) mL × = × −− ρ 33 6 3 16 19 7 86 5 00 10 9 83 10 9 83 10 . gc m c m g k g ej e j . (b) N m m × × 0 19 27 7 983 10 55 9 1 66 10 106 10 . . kg k g 1 u atoms P1.11 (a) The cross-sectional area is A =+ 2 0 150 0 010 0 340 0 010 640 10 3 m m m m 2 a fa f a fa f . The volume of the beam is VA L × = × 6 4 01 0 1 5 0 9 6 0 . m m 23 af . Thus, its mass is
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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