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Unformatted text preview: fryer (vdf96) – Fluid Mechanics – graves – (6) 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In a deep dive, a whale is appreciably com pressed by the pressure of the surrounding water. What happens to the whale’s density? 1. Its density increases. correct 2. It cannot be determined. 3. Its density remains the same as before. 4. Its density decreases. Explanation: Since the mass is unchanged and the volume decreases, the density of the whale increases. 002 10.0 points An engineer weighs a sample of mercury ( ρ = 13 . 6 × 10 3 kg / m 3 ) and finds that the weight of the sample is 7 . 0 N. What is the sample’s volume? The acceler ation of gravity is 9 . 81 m / s 2 . Correct answer: 5 . 24675 × 10 5 m 3 . Explanation: Let : W = 7 . 0 N , ρ = 13 . 6 × 10 3 kg / m 3 , and g = 9 . 81 m / s 2 . m = W g so V = m ρ = W g ρ = 7 N (9 . 81 m / s 2 )(13 . 6 × 10 3 kg / m 3 ) = 5 . 24675 × 10 5 m 3 . 003 10.0 points Air within the funnel of a large tornado may have a pressure of only 0 . 56 atm. What is the approximate outward force F on a 7 . 6 m × 22 m wall if a tornado suddenly envelopes the house? Atmospheric pressure is 1 . 013 × 10 5 Pa. Correct answer: 7 . 45244 × 10 6 N. Explanation: Let : P = 1 atm = 1 . 013 × 10 5 Pa , P = 0 . 56 atm , a = 7 . 6 m , and b = 22 m . The inside pressure is 1 atm since the tor nado envelopes the house suddenly. The force due to the air inside the house (pointing out ward) is F outward = P A = P a b = (1 . 013 × 10 5 Pa) (7 . 6 m) (22 m) = 1 . 69374 × 10 7 N . The inward force due to the air outside the house is F inward = P A = P a b = (0 . 56 atm) (7 . 6 m) (22 m) × 1 . 013 × 10 5 Pa atm = 9 . 48492 × 10 6 N , so the net force on the house is F = F outward − F inward = 1 . 69374 × 10 7 N − 9 . 48492 × 10 6 N = 7 . 45244 × 10 6 N . 004 (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 fryer (vdf96) – Fluid Mechanics – graves – (6) 2 4 . 26 cm. This pressure is transmitted by a liquid to a second piston of radius 16 . 6 cm. What force must the compressed air exert in order to lift a car weighing 13200 N? Correct answer: 869 . 315 N. Explanation: Let : r 1 = 4 . 26 cm , r 2 = 16 . 6 cm , and F 2 = 13200 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 . 26 cm) 2 (16 . 6 cm) 2 (13200 N) = 869 . 315 N ....
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 Fall '10
 Graves
 Physics, mechanics

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