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Unformatted text preview: originally compressed from its equilibrium length? Correct answer: 0.27618 m. 4) A child bounces a 47 g superball on the side walk. The velocity change of the superball is from 22 m/s downward to 12 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Answer in units of N. ANSWER : (0.047)*(22+12)=1.598 =1.598*800=1278.4 5) A 1 kg steel ball strikes a wall with a speed of 14.1 m/s at an angle of 38◦ with the normal to the wall. It bounces off with the same speed and angle as shown in the figure. If the ball is in contact with the wall for 0.143 s, what is the magnitude of the average force exerted on the ball by the wall? ANSWER : (1)(14.1)sin(52)=11.11 kg m/s 11.11 x 2 = 22.22 m/s = 22.22/0.143=155.398 6) m2) immediately after the collison and before the pendulum starts to swing upwards. Answer in units of m/s ANSWER : max height= 0.056 Max GPE= mgh= (795)(9.81)(.056)=436.7412 J ½(795)v 2 =436.7412 V=1.048...
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This note was uploaded on 02/23/2012 for the course PHYS 2425 taught by Professor Sherman during the Summer '08 term at Collin College.
 Summer '08
 Sherman
 Physics, Force, Friction, Work

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