hw12 - zhou(cz3574 HW12 mackie(10611 This print-out should...

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zhou (cz3574) – HW12 – mackie – (10611) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Note we’re 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 cross-section 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 π r 2 1 π 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 π r 2 1 = 576 . 295 N π (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 person’s 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 ( π r 2 ) = (2300 Pa) π (0 . 00358 m) 2 = 0 . 092607 N . 005 10.0 points The air pressure above the liquid in figure is 0 . 472 atm . The depth of the air bubble in the
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zhou (cz3574) – HW12 – mackie – (10611) 2 liquid is 38 . 1 cm and the liquid’s 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 : ρ = 847 kg / m 3 , P 0 = 0 . 472 atm , g = 9 . 8 m / s 2 , and h = 38 . 1 cm = 0 . 381 m . P = P 0 + ρ 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 0 . 81 g / cm 3 and 6 cm of water. What is the gauge pressure on the bottom of the tube? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 778 . 512 Pa. Explanation: Let : ρ 1 = 0 . 81 g / cm 3 = 810 kg / m 3 , h = 2 . 4 cm = 0 . 024 m , ρ w = 1 g / cm 3 = 1000 kg / m 3 , h w = 6 cm = 0 . 06 m , and g = 9 . 8 m / s 2 . P = P o + P w = ρ 1 h g + ρ w h w g = (810 kg / m 3 ) (0 . 024 m) (9 . 8 m / s 2 ) + (1000 kg / m 3 ) (0 . 06 m) (9 . 8 m / s 2 ) = 778 . 512 Pa .
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