This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 whales 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 ( = 13 . 6 10 3 kg / m 3 ) and finds that the weight of the sample is 7 . 0 N. What is the samples 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 , = 13 . 6 10 3 kg / m 3 , and g = 9 . 81 m / s 2 . m = W g so V = m = W g = 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 r 2 1 r 2 2 parenrightbigg F 2 = (4 . 26 cm) 2 (16 . 6 cm) 2 (13200 N) = 869 . 315 N ....
View
Full
Document
 Fall '10
 Graves
 Physics, mechanics

Click to edit the document details