{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 9

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

This preview shows pages 1–2. Sign up to view the full content.

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,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}