{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

567_Physics ProblemsTechnical Physics

567_Physics ProblemsTechnical Physics - Chapter 19 With...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern