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567_Physics ProblemsTechnical Physics

# 567_Physics ProblemsTechnical Physics - Chapter 19 With...

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Chapter 19 569 P19.68 With piston alone: T = constant, so PV P V = 0 0 or P Ah P Ah i b g b g = 0 0 With A = constant, P P h h i = F H G I K J 0 0 But, P P m g A p = + 0 where m p is the mass of the piston. Thus, P m g A P h h p i 0 0 0 + = F H G I K J FIG. P19.68 which reduces to h h i m g P A p = + = + = × 0 20 0 1.013 10 1 50 0 49 81 0 5 2 . . . cm 1 cm kg 9.80 m s Pa 0.400 m 2 e j a f π With the man of mass M on the piston, a very similar calculation (replacing m p by m M p + ) gives: ′ = + = + = + × h h m M g P A p 0 1.013 10 1 50 0 49 10 0 5 2 e j e j a f . . cm 1 cm 95.0 kg 9.80 m s Pa 0.400 m 2 π Thus, when the man steps on the piston, it moves downward by h h h i = ′ = = = 49 81 49 10 0 706 7 06 . . . . cm cm cm mm . (b) P = const, so V T V T i = or Ah T Ah T i i = giving T T h h i i = F H G I K J = F H G I K J = 293 297 K 49.81 49.10 K (or 24°C) P19.69 (a) dL L dT = α : α α α dT dL L L L T L L e T T L L f i f i T i i i i z z = F H G I
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