Physics 115 Test 2A with key

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Unformatted text preview: Because you’re not moving, your velocity is constant (it’s zero all the time), so your acceleratio n is zero, and so the net force acting on you is zero. Vector addition: weight (down) + spring force from scale (up) + and force f rom table (up) must be zero. Using “+” to indicate “up” and “–” to indicate “down”, we have W – Fspring – Ftable = 0 so that (add + Fspring on each side of this equality) W – Ftable = Fspring, s o Fspring is less than W, so it becomes less distorted than when there is no table force. You could say that “The ta ble carries part of your weight now so the scale has to carry less now”. PHYS115 – Spring 2013 7 Test 2 - 19 April 2013 Question 16 – The rebound en ergy is always less than the collision energy because … a. ... momentum is not co nserved in inelastic collisions b. ... some energy is lost as thermal energy c. ... the ball’s coefficient of restitution is greater than 1 d. ... collisions cannot cha ange the ball’s kinetic energy Answer (b) See book Fig. 3.2.1 and its discussion. Question 17 – The truck’s weight is 4,000 lbf, and the bridge sags 1 inch under its weight. W hat is the bridge’s spring constant? a. b. c. d. 700,000 N/m 157,000 N/m 3,400,000 N/m 140,000 N/m Answer (a) spring constant force 4,000 lbf 4,000 1 lb g 0.454 kg 1 inc...
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This document was uploaded on 02/23/2014 for the course PHYSICS 115 at UNL.

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