3_Physics ProblemsTechnical Physics

3_Physics ProblemsTechnical Physics - Chapter 1 P1.6 3 4 43...

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Chapter 1 3 P1.6 For either sphere the volume is Vr = 4 3 3 π and the mass is mV r == ρρ 4 3 3 . We divide this equation for the larger sphere by the same equation for the smaller: m m r r r r s ss A A A = ρπ 43 5 3 3 3 3 . Then rr s A = 5 4 50 1 71 7 69 3 ... cm cm af . P1.7 Use 1 u . g 166 10 24 . (a) For He, m 0 24 4 00 6 64 10 = × F H G I K J .. u 1.66 10 g 1 u g -24 . (b) For Fe, m 0 23 55 9 9 29 10 = × F H G I K J u 1.66 10 g 1 u g -24 . (c) For Pb, m 0 24 22 207 344 10 = × F H G I K J u g 1 u g . *P1.8 (a) The mass of any sample is the number of atoms in the sample times the mass m 0 of one atom: mN m = 0 . The first assertion is that the mass of one aluminum atom is m 0 27 26 2 7 0 2 7 0 1 6 61 0 1 4 4 81 0 ==× × = × −− . . u u k g u k g . Then the mass of 6 02 10 23 . × atoms is m × × × = = 0 23 26 6 02 10 4 48 10 0 027 0 27 0 . . kg kg g . Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be written m = 0 . 0 027 0 6 02 10 23 0 kg m , so m 0 23 26 0027 602 10 448 10 = × . . . kg kg, in agreement with the first assertion. (b) The general equation m = 0 applied to one mole of any substance gives
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