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HW12 - boland(kbb544 HW12 mackie(10611 This print-out...

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boland (kbb544) – 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. 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 4 . 44 cm. This pressure is transmitted by a liquid to a second piston of radius 26 . 9 cm. What force must the compressed air exert in order to lift a car weighing 11200 N? Correct answer: 305 . 126 N. Explanation: Let : r 1 = 4 . 44 cm , r 2 = 26 . 9 cm , and F 2 = 11200 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 = (4 . 44 cm) 2 (26 . 9 cm) 2 (11200 N) = 305 . 126 N . 002 (part 2 of 2) 10.0 points What air pressure will produce this force? Correct answer: 49267 . 8 Pa. Explanation: P = F 1 A 1 = F 1 π r 2 1 = 305 . 126 N π (4 . 44 cm) 2 parenleftbigg 100 cm 1 m parenrightbigg 2 = 49267 . 8 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 . 349 cm. The pressure of the atmosphere drops from 1 . 002 × 10 5 Pa at the bottom of the lift to 99700 Pa at the top. What is the net pressure on the inner ear at the top of the mountain? Correct answer: 500 Pa. Explanation: Let : P b = 1 . 002 × 10 5 Pa and P t = 99700 Pa . P net = P b - P t = 1 . 002 × 10 5 Pa - 99700 Pa = 500 Pa . 004 (part 2 of 2) 10.0 points What is the magnitude of the net force on each eardrum? Correct answer: 0 . 0191325 N. Explanation: Let : r = 0 . 349 cm = 0 . 00349 m . F net = P net A = P net ( π r 2 ) = (500 Pa) π (0 . 00349 m) 2 = 0 . 0191325 N . 005 10.0 points The air pressure above the liquid in figure is 1 . 25 atm . The depth of the air bubble in the liquid is 31 . 3 cm and the liquid’s density is 837 kg / m 3 .
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boland (kbb544) – HW12 – mackie – (10611) 2 31 . 3 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: 1 . 29192 × 10 5 Pa. Explanation: Let : ρ = 837 kg / m 3 , P 0 = 1 . 25 atm , g = 9 . 8 m / s 2 , and h = 31 . 3 cm = 0 . 313 m . P = P 0 + ρ g h = (1 . 25 atm) · 1 . 013 × 10 5 Pa atm + (837 kg / m 3 ) (9 . 8 m / s 2 ) (0 . 313 m) = 1 . 29192 × 10 5 Pa . 006 10.0 points A test tube standing vertically in a test tube rack contains 4 . 3 cm of oil, whose density is 0 . 81 g / cm 3 and 7 . 3 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: 1056 . 73 Pa. Explanation: Let : ρ 1 = 0 . 81 g / cm 3 = 810 kg / m 3 , h = 4 . 3 cm = 0 . 043 m , ρ w = 1 g / cm 3 = 1000 kg / m 3 , h w = 7 . 3 cm = 0 . 073 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 . 043 m) (9 . 8 m / s 2 ) + (1000 kg / m 3 ) (0 . 073 m) (9 . 8 m / s 2 ) = 1056 . 73 Pa .
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