553_Physics ProblemsTechnical Physics

553_Physics ProblemsTechnical Physics - Chapter 19 P19.20...

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Chapter 19 555 P19.20 (a),(b) The material would expand by ∆∆ LL T i = α , L L T F A YL L YT i i = == × ° ° , but instead feels stress Nm C C N m . This will not break concrete. 2 2 700 10 120 10 300 252 10 96 1 6 .. . . ej a f a f P19.21 (a) VV TV T VT tt t i =− = −×° ° −− ββ β Al Al Al 3 C c mC 3 9 00 10 0 720 10 2 000 60 0 44 1 bg e j af . V = 99 4 . cm 3 overflows. (b) The whole new volume of turpentine is 2 000 9 00 10 2 000 60 0 2 108 41 cm C cm C cm 33 3 +×° ° = so the fraction lost is 99 4 2108 471 10 2 . . cm cm 3 3 and this fraction of the cylinder’s depth will be empty upon cooling: 200 0943 2 . ×= cm cm a f . *P19.22 The volume of the sphere is Vr Pb 3 cm cm = 4 3 4 3 23 3 5 3 3 ππ a f The amount of mercury overflowing is overflow Hg Pb glass Hg Hg Pb Pb glass glass =+ = + VVV V V where VV V glass Hg
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