Homework 9 - Gozick, Brandon Homework 9 Due: Oct 10 2006,...

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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 – fnd 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 fnd 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, it is no longer accelerating, so the net Force is zero. That is F 1 - R = 0 so R =
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This note was uploaded on 09/12/2008 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.

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Homework 9 - Gozick, Brandon Homework 9 Due: Oct 10 2006,...

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