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HW09-solutions

# HW09-solutions - markowitz(am45362 HW09 distler(56295 This...

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markowitz (am45362) – HW09 – distler – (56295) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. For the siphon problem, note that the pres- sure in the hose must be everywhere non- negative 001 10.0points A constriction in a pipe reduces its diameter from 4 . 7 cm to 2 . 3 cm . Where the pipe is wider, the fluid velocity is 8 m / s . Find the fluid velocity where the pipe is narrow. Correct answer: 33 . 4064 m / s. Explanation: Let : r 1 = 2 . 35 cm , r 2 = 1 . 15 cm , and v 1 = 8 m / s . A 1 v 1 = A 2 v 2 = constant π r 2 1 v 1 = π r 2 2 v 2 v 2 = v 1 parenleftbigg r 1 r 2 parenrightbigg 2 = (8 . 0 m / s) parenleftbigg 2 . 35 cm 1 . 15 cm parenrightbigg 2 = 33 . 4064 m / s . 002 10.0points A small metal sphere weighs 0 . 287 N in air and has a volume of 16 . 1 cm 3 . What is the acceleration of the sphere as it falls through water? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 4 . 41239 m / s 2 . Explanation: Let : W = 0 . 287 N , V = 16 . 1 cm 3 , and g = 9 . 8 m / s 2 . The apparent weight is the net downward force. When completely submerged, the sphere displaces a volume of water equal to its own volume. The buoyant force of water is B = ρ w V g , so B = W − W w ρ w V g = W − W w W w = W − ρ w V g W g a = W − ρ w V g a = g ρ w V g 2 W = 9 . 8 m / s 2 (1000 kg / m 3 ) 0 . 287 N × (16 . 1 cm 3 )(9 . 8 m / s 2 ) 2 = 4 . 41239 m / s 2 . 003(part1of2)10.0points In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The sys- tem has a small piston with a cross-sectional area of 0 . 066 m 2 and a large piston with a cross-sectional area of 0 . 171 m 2 . An engine weighing 2200 N rests on the large piston. What force must be applied to the small piston in order to lift the engine? Correct answer: 849 . 123 N. Explanation: Let : A 1 = 0 . 066 m 2 , A 2 = 0 . 171 m 2 , and F 2 = 2200 N . The pressure is the same, so F 1 A 1 = F 2 A 2 F 1 = A 1 F 2 A 2 = (0 . 066 m 2 ) (2200 N) 0 . 171 m 2 = 849 . 123 N .

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markowitz (am45362) – HW09 – distler – (56295) 2 004(part2of2)10.0points If the engine rises 0 . 151 m, how far does the smaller piston move? Correct answer: 0 . 391227 m. Explanation: Let : d 2 = 0 . 151 m . A hydraulic lift is effectively 100% efficient, so W in = W out F 1 d 1 = F 2 d 2 d 1 = F 2 d 2 F 1 = (2200 N) (0 . 151 m) 849 . 123 N = 0 . 391227 m . 005 10.0points Consider a steel ax and an aluminum piston. (Note that steel is denser than aluminum.) When weighed in water, the ax and the piston have the same apparent weight. But when the same ax and the same piston are weighed in air, 1. the wider object weighs more than the other. 2. they again have equal weights, but both are heavier than they are in water. 3. the ax is heavier than the piston. 4. they again have equal weights, but both are lighter than they are in water. 5. the longer object weighs more than the other. 6. the piston is heavier than the ax. correct Explanation: Archimedes’ principle tells us that the buoy- ancy force acting on a submerged object is equal to the weight of water that was diplaced when the object was submerged. Since the objects are completely submerged in this ex- periment, we have B = ρ w V g .
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