Exam 3-solutions - Version 014 Exam 3 chelikowsky (59005)...

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Version 014 – Exam 3 – chelikowsky – (59005) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points An attack helicopter is equipped with a 20- mm cannon that fres 79 . 2 g shells in the Forward direction with a muzzle speed oF 1100 m / s. The Fully loaded helicopter has a mass oF 5170 kg. A burst oF 109 shells is fred in a 5 . 52 s interval. What is the resulting average Force on the helicopter? 1. 1720.3 2. 1657.99 3. 2528.9 4. 3775.45 5. 2622.24 6. 2144.55 7. 944.368 8. 3508.4 9. 1774.68 10. 3261.05 Correct answer: 1720 . 3 N. Explanation: The impulse imparted to the shells equals the change in momentum: F av Δ t = Δ mv The mass change is Δ m = nm = (109 shells) (79 . 2 g) = 8 . 6328 kg , so the average Force is F = v Δ m t = (1100 m / s) (8 . 6328 kg) 5 . 52 s = 1720 . 3 N . Since the velocity oF the shells is much greater than the velocity oF the helicopter, there is no need to use relative velocity. 002 (part 2 oF 2) 10.0 points By what amount is its Forward speed reduced? 1. 0.813897 2. 6.30422 3. 5.95452 4. 3.16342 5. 7.75076 6. 2.03055 7. 6.08547 8. 5.1961 9. 10.7575 10. 1.83677 Correct answer: 1 . 83677 m / s. Explanation: ±rom conservation oF momentum Δ p = 0, so Δ v = Δ mv M = (8 . 6328 kg) (1100 m / s) 5170 kg = 1 . 83677 m / s , 003 (part 1 oF 3) 10.0 points A car accelerates uniFormly From rest and covers a distance oF 57 m in 6 . 5 s. IF the diameter oF a tire is 38 cm, fnd the angular acceleration oF the wheel. 1. 10.519 2. 6.39166 3. 14.0499 4. 14.2012 5. 16.2432 6. 10.3618 7. 19.3488 8. 6.46383 9. 11.4368 10. 21.1177 Correct answer: 14 . 2012 rad / s 2 . Explanation: Let : Δ x = 57 m , Δ t = 6 . 5 s , and R = 19 cm = 0 . 19 m .
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Version 014 – Exam 3 – chelikowsky – (59005) 2 From kinematics, Δ x = 1 2 a Δ t 2 a = 2 Δ x Δ t 2 , so a = α = a R = 2 Δ x R t ) 2 = 2 (57 m) (0 . 19 m) (6 . 5 s) 2 = 14 . 2012 rad / s 2 . 004 (part 2 of 3) 10.0 points Find the ±nal angular velocity of one of the car’s wheels. 1. 63.6605 2. 62.607 3. 86.5874 4. 55.1572 5. 130.8 6. 109.501 7. 81.5742 8. 137.691 9. 92.3077 10. 82.7211 Correct answer: 92 . 3077 rad / s. Explanation: Since the acceleration is uniform, the ±nal angular velocity is ω f = ω 0 + αt = αt = (14 . 2012 rad / s 2 ) (6 . 5 s) = 92 . 3077 rad / s . 005 (part 3 of 3) 10.0 points Find total number of rotations of the wheel. 1. 43.3443 2. 59.1147 3. 63.662 4. 54.0387 5. 54.684 6. 47.7465 7. 35.079 8. 42.9026 9. 70.414 10. 48.6307 Correct answer: 47 . 7465. Explanation: x = nC = n (2 π r ) n = x 2 π r = 57 m 2 π (0 . 19 m) = 47 . 7465 .
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This note was uploaded on 07/07/2010 for the course PHY 302 taught by Professor Staff during the Summer '08 term at University of Texas at Austin.

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Exam 3-solutions - Version 014 Exam 3 chelikowsky (59005)...

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