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Online Homework Answers 13 (Dec. 7, 2006)

Online Homework Answers 13 (Dec. 7, 2006) - Hood Charles...

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Unformatted text preview: Hood, Charles – Homework 13 – Due: Dec 7 2006, 11:00 pm – Inst: Jan Opyrchal 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A 1 kg steel ball strikes a wall with a speed of 9 . 78 m / s at an angle of 59 ◦ with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 9 . 7 8 m / s 1 kg 9 . 7 8 m / s 1 kg 59 ◦ 59 ◦ If the ball is in contact with the wall for . 315 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 31 . 9814 N. Explanation: Let : M = 1 kg , v = 9 . 78 m / s , and θ = 59 ◦ . The y component of the momentum is un- changed. The x component of the momentum is changed by Δ P x =- 2 M v cos θ . Therefore, using impulse formula, F = Δ P Δ t =- 2 M v cos θ Δ t =- 2(1 kg) (9 . 78 m / s) cos 59 ◦ . 315 s k ~ F k = 31 . 9814 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign. keywords: 002 (part 1 of 1) 10 points A 0.023 kg golf ball moving at 16.2 m/s crashes through the window of a house in 5 . 2 × 10- 4 s. After the crash, the ball contin- ues in the same direction with a speed of 12.0 m/s....
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Online Homework Answers 13 (Dec. 7, 2006) - Hood Charles...

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