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Exam 3-solutions

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 – find all choices before answering. 001 (part 1 of 2) 10.0 points An attack helicopter is equipped with a 20- mm cannon that fires 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 fired 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 = n m = (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: From conservation of momentum Δ p = 0, so Δ v = Δ m v 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, find 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) 3 What is the kinetic energy of a satellite of mass m that orbits the Earth of mass M in a circular orbit of radius R ? 1. K = 1 4 G M m R 2. K = 0 3. K = 1 2 G M m R 2 4. K = G M m R 2 5. K = 1 2 G M m R correct Explanation: The gravitational force on the satellite pro- vides the centripetal force needed to keep it in circular orbit: G M m R 2 = F G = F c = m v 2 R m v 2 = G M m R .

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

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