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 = 00 or PAh P Ah i bg b g = With A = constant, PP h h i = F H G I K J 0 0 But, mg A p =+ 0 where m p is the mass of the piston. Thus, P A P h h p i 0 += F H G I K J FIG. P19.68 which reduces to h h i PA 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 ej af π With the man of mass M on the piston, a very similar calculation (replacing m p by mM p + ) gives: ′= + = + = + × h h g p 0 1.013 10 1 50 0 49 10 0 5 2 e j . . cm 1 cm 95.0 kg 9.80 m s Pa 0.400 m 2 Thus, when the man steps on the piston, it moves downward by hh 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 TT 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 : αα dT dL L L L TL L e T T L L f i fi
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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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