PE; = mgy;%3DKE, = mv,?%3DPE = 0%3D If the mass of the swimmer is 50kg.a. Potential Energy at the Top is =Jb Velocity of the swimmer midway down is =m/sc. Kinetic Energy at the bottom is =d. Velocity impact is =m/se Assuming the swimmer falls directly downward and the motion is the work of Gravity, thenpower is =W15.00 mKE, = 0PE;= mgyi%3D

Name: PE; = mgy;%3DKE, = mv,?%3DPE = 0%3D If the mass of the swimmer is 50k
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Description: PE; = mgy;%3DKE, = mv,?%3DPE = 0%3D If the mass of the swimmer is 50kg.a. Potential Energy at the Top is =Jb Velocity of the swimmer midway down is =m/sc. Kinetic Energy at the bottom is =d. Velocity impact is =m/se Assuming the swimmer falls direct

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If the mass of the swimmer is 50kg. a. Potential Energy at the Top is = b. Velocity of the swimmer midway down is = + m/s c. Kinetic Energy at the bottom is %3D d. Velocity impact is = m/s e. Assuming the swimmer falls directly downward and the motion is the work of Gravity, then power is = 15.00 m KE, = 0 PE;= mgyi %3D %3D

If the mass of the swimmer is 50kg. a. Potential Energy at the Top is = b. Velocity of the swimmer midway down is = + m/s c. Kinetic Energy at the bottom is %3D d. Velocity impact is = m/s e. Assuming the swimmer falls directly downward and the motion is the work of Gravity, then power is = 15.00 m KE, = 0 PE;= mgyi %3D %3D

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter5: Energy

Section: Chapter Questions

Problem 63AP: A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves...

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