15_Physics ProblemsTechnical Physics

# 15_Physics ProblemsTechnical Physics - Chapter 1*P1.57...

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Chapter 1 15 *P1.57 Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has mass m 0 27 25 197 3 27 10 = × F H G I K J u 1.66 10 kg 1 u kg af .. So, the number of atoms in the cube is N = × 19 300 590 10 25 28 kg 3.27 10 kg The imagined cubical volume of each atom is d 3 28 29 1 169 10 = × m m 3 3 . So d 257 10 10 . m . P1.58 AN A V V A V r r total drop total drop drop total 4 == F H G I K J = F H G G I K J J ej π 3 3 2 4 A V r total total 3 2 m m m = F H G I K J = × × F H G I K J = 3 3 30 0 10 200 10 450 6 5 . . . P1.59 One month is 1 30 24 3 600 2 592 10 6 mo day h day s h s × bg b g b g Applying units to the equation, Vt t =+ 150 000800
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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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