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Unformatted text preview: zhou (cz3574) HW12 mackie (10611) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Note were back to FRI deadline 001 (part 1 of 2) 10.0 points In a car lift used in a service station, com pressed air exerts a force on a small piston of circular crosssection having a radius of 3 . 87 cm. This pressure is transmitted by a liquid to a second piston of radius 18 . 8 cm. What force must the compressed air exert in order to lift a car weighing 13600 N? Correct answer: 576 . 295 N. Explanation: Let : r 1 = 3 . 87 cm , r 2 = 18 . 8 cm , and F 2 = 13600 N . Because the pressure exerted by the compressed air is transmitted undiminished throughout the fluid, F 1 A 1 = F 2 A 2 F 1 = parenleftbigg A 1 A 2 parenrightbigg F 2 = parenleftbigg r 2 1 r 2 2 parenrightbigg F 2 = (3 . 87 cm) 2 (18 . 8 cm) 2 (13600 N) = 576 . 295 N . 002 (part 2 of 2) 10.0 points What air pressure will produce this force? Correct answer: 1 . 22482 10 5 Pa. Explanation: P = F 1 A 1 = F 1 r 2 1 = 576 . 295 N (3 . 87 cm) 2 parenleftbigg 100 cm 1 m parenrightbigg 2 = 1 . 22482 10 5 Pa . 003 (part 1 of 2) 10.0 points A person rides up a lift to a mountain top, but the persons ears fail to pop; that is the pressure of the inner ear does not equalize with the outside atmosphere. The radius of each eardrum is 0 . 358 cm. The pressure of the atmosphere drops from 1 . 01 10 5 Pa at the bottom of the lift to 98700 Pa at the top. What is the net pressure on the inner ear at the top of the mountain? Correct answer: 2300 Pa. Explanation: Let : P b = 1 . 01 10 5 Pa and P t = 98700 Pa . P net = P b P t = 1 . 01 10 5 Pa 98700 Pa = 2300 Pa . 004 (part 2 of 2) 10.0 points What is the magnitude of the net force on each eardrum? Correct answer: 0 . 092607 N. Explanation: Let : r = 0 . 358 cm = 0 . 00358 m . F net = P net A = P net ( r 2 ) = (2300 Pa) (0 . 00358 m) 2 = . 092607 N . 005 10.0 points The air pressure above the liquid in figure is . 472 atm . The depth of the air bubble in the zhou (cz3574) HW12 mackie (10611) 2 liquid is 38 . 1 cm and the liquids density is 847 kg / m 3 . 38 . 1 cm air air Determine the air pressure in the bubble suspended in the liquid. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 50976 . 1 Pa. Explanation: Let : = 847 kg / m 3 , P = 0 . 472 atm , g = 9 . 8 m / s 2 , and h = 38 . 1 cm = 0 . 381 m . P = P + g h = (0 . 472 atm) 1 . 013 10 5 Pa atm + (847 kg / m 3 ) (9 . 8 m / s 2 ) (0 . 381 m) = 50976 . 1 Pa . 006 10.0 points A test tube standing vertically in a test tube rack contains 2 . 4 cm of oil, whose density is . 81 g / cm 3 and 6 cm of water....
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This note was uploaded on 03/25/2011 for the course PHYS 122 taught by Professor Hiyishi during the Spring '10 term at Temple College.
 Spring '10
 Hiyishi

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