hw41 - hinojosa(jlh3938 – homework 41 – Turner...

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Unformatted text preview: hinojosa (jlh3938) – homework 41 – Turner – (58185) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The small piston of a hydraulic lift has a cross-sectional area of 5 . 4 cm 2 and the large piston has an area of 25 cm 2 , as in the figure below. F 5 . 4 cm 2 area 25 cm 2 What force F must be applied to the small piston to maintain the load of 24 kN at a constant elevation? Correct answer: 5184 N. Explanation: Let : A 1 = 5 . 4 cm 2 , A 2 = 25 cm 2 , and W = 24 kN . According to Pascal’s law, the pressure ex- erted on A 1 must be equal to the one exerted on A 2 . The pressure P 1 = F A 1 must be equal to the pressure P 2 = W A 2 due to the load. F A 1 = W A 2 , so F = A 1 A 2 W = (5 . 4 cm 2 ) (25 cm 2 ) (24000 N) = 5184 N . 002 10.0 points The air pressure above the liquid in figure is 1 . 64 atm . The depth of the air bubble in the liquid is 39 . 8 cm and the liquid’s density is 805 kg / m 3 . 39 . 8 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 . 69272 × 10 5 Pa. Explanation: Let : ρ = 805 kg / m 3 , P = 1 . 64 atm , g = 9 . 8 m / s 2 , and h = 39 . 8 cm = 0 . 398 m . P = P + ρ g h = (1 . 64 atm) · 1 . 013 × 10 5 Pa atm + (805 kg / m 3 ) (9 . 8 m / s 2 ) (0 . 398 m) = 1 . 69272 × 10 5 Pa . 003 10.0 points A 37 . 5 kg woman balances on one heel of a pair of high-heeled shoes. If the heel is circular with radius 0 . 591 cm, what pressure does she exert on the floor? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 3 . 34913 × 10 6 N / m 2 . Explanation: Let : m = 37 . 5 kg , g = 9 . 8 m / s 2 , and r = 0 . 591 cm . The pressure is equal to the force she exerts on the floor (which is equal to her weight) divided by the area of the heel: hinojosa (jlh3938) – homework 41 – Turner – (58185) 2 P = mg π r 2 = (37 . 5 kg) (9 . 8 m / s 2 ) π (0 . 591 cm) 2 · parenleftbigg 100 cm 1 m parenrightbigg 2 = 3 . 34913 × 10 6 N / m 2 ....
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This note was uploaded on 03/26/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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hw41 - hinojosa(jlh3938 – homework 41 – Turner...

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