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Homework 9 - Gozick Brandon Homework 9 Due 7:00 pm Inst D...

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Gozick, Brandon – Homework 9 – Due: Oct 10 2006, 7:00 pm – Inst: D Weathers 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 2 . 4 g bullet moving at 268 m / s penetrates a tree to a depth of 1 . 63 cm. Use energy considerations to find the aver- age frictional force that stops the bullet. Correct answer: 5287 . 66 N. Explanation: We can use conservation of energy to relate the initial kinetic energy of the bullet to the work done by the frictional force. 1 2 m v 2 = f · s Solving for the frictional force, f , f = m v 2 2 s = (0 . 0024 kg) (268 m / s) 2 2 (0 . 0163 m) = 5287 . 66 N . keywords: 002 (part 1 of 1) 10 points A car of weight 2050 N operating at power 195 kW develops a maximum speed of 26 . 5 m / s on a level, horizontal road. Assume: The resistive force remains con- stant under the conditions mentioned below. What is the maximum speed of the car up an incline of 7 . 92 relative to the horizontal? Correct answer: 25 . 5204 m / s. Explanation: Basic Concepts: P = d W dt = ~ F · ~v . Maximum Speed and Terminal Velocity Resistive forces. Let : W = 2050 N , p = 195 kW , v 1 = 26 . 5 m / s , and θ = 7 . 92 . Solution: On level ground, the only forces acting on the car in the horizontal direction are the applied force F 1 and the resistive force R . (Gravity and the normal force act perpen- dicular to this and add vectorially to zero.) Since the car has reached it maximum speed,
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