Midterm - Winter 2007 (Solutions Only)

Midterm - Winter 2007 (Solutions Only) - ME320 Exam 1...

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ME320 Exam 1 Solutions Winter 2007 Solution to Problem 1 (10 points) (a) P B = 1 h 1 = (13.55)(10 3 kg/m 3 )(9.8 m/s 2 )(0.12 m) = 1.59 × 10 4 Pa. P A = P B 2 h 2 = 1.59 × 10 4 – (0.96) )(10 3 kg/m 3 )(9.8 m/s 2 )(0.06 m) = 1.54 × 10 4 Pa ( 4 points ) (Minor calculation error: -1 point, otherwise: -4)) (b) Because of the geometrical symmetry, the net force acting on the cap has a vertical component , Fv, only. Fv = P A r 2 + weight of champagne with 20 cm height – weight of champagne replaced by the cap ( 4 points ) = P A r 2 + 2 ( r 2 H -4/3 r 3 ×1/2) = (1.54 × 10 4 Pa) (0.04m) 2 + (0.96)(10 3 kg/m 3 )(9.8 m/s 2 )[ (0.04m) 2 (0.2m) - 2/3 (0.04m) 3 ]= 85.5 N ( 2 points ) (Minor calculation error: -2)
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Solution to Problem 2 (15 points) (a) The pressure at the bottom of the tank is: ) ( t gh p B where h(t) is a function of time ( 2 points ) From conservation of mass: k A dt dh Q tan (where dh/dt is the velocity of the of the free surface of water) (
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This note was uploaded on 02/09/2009 for the course MECHENG 320 taught by Professor Im during the Fall '08 term at University of Michigan.

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Midterm - Winter 2007 (Solutions Only) - ME320 Exam 1...

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