cw19 - × 10 5 N , so the net Force on the house is F = F...

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patel (kp9583) – CW19 – Mackie – (20211) 1 This print-out should have 2 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Air within the Funnel oF a large tornado may have a pressure oF only 0 . 074 atm. What is the approximate outward Force F on a 5 . 9 m × 11 m wall iF a tornado suddenly envelopes the house? Atmospheric pressure is 1 . 013 × 10 5 Pa. Correct answer: 6 . 08787 × 10 6 N. Explanation: Let : P 0 = 1 atm = 1 . 013 × 10 5 Pa , P = 0 . 074 atm , a = 5 . 9 m , and b = 11 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 0 A = P 0 a b = (1 . 013 × 10 5 Pa) (5 . 9 m) (11 m) = 6 . 57437 × 10 6 N . The inward Force due to the air outside the house is F inward = P A = P a b = (0 . 074 atm) (5 . 9 m) (11 m) × 1 . 013 × 10 5 Pa atm = 4 . 86503
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Unformatted text preview: × 10 5 N , so the net Force on the house is F = F outward-F inward = 6 . 57437 × 10 6 N-4 . 86503 × 10 5 N = 6 . 08787 × 10 6 N . 002 10.0 points Determine the absolute pressure at the bot-tom oF a lake that is 44 . 2 m deep. The accel-eration oF gravity is 9 . 8 m / s 2 and atmospheric pressure is 1 . 01 × 10 5 Pa . Correct answer: 5 . 3416 × 10 5 Pa. Explanation: Let : g = 9 . 8 m / s 2 , P atm = 1 . 01 × 10 5 Pa , h = 44 . 2 m , and ρ w = 1000 kg / m 3 . The pressure at the bottom oF the lake is equal to the pressure at the surFace plus the pressure given by the mass oF water above, so P = P atm + ρ w g h = 1 . 01 × 10 5 Pa + (1000 kg / m 3 ) (9 . 8 m / s 2 ) (44 . 2 m) = 5 . 3416 × 10 5 Pa ....
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This note was uploaded on 12/10/2009 for the course PHYSICS 2021 taught by Professor Mihalisin during the Fall '09 term at Temple.

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