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(I) A constant friction force of 25 $\mathrm{N}$ acts on a 65 -kg skier for 15. S. What is the skier's change in velocity?
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Physics 101 Mechanics
Chapter 9
Linear Momentum
Motion Along a Straight Line
Kinetic Energy
Potential Energy
Energy Conservation
Moment, Impulse, and Collisions
Cornell University
University of Washington
University of Winnipeg
McMaster University
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the average force is given by the rate of change momentum. Delta P A riddle to tea on DSO Delta P is just mass times the raid times of change in velocity on That's divided by Delta two. Okay, so in this problem were given force and mass in time. We want adult of you so full the friction saw it will opposed will be in the opposite direction. To the direction of motion will be a negative negative. 25 Newton's That's equal to Mass, which is 65 kilograms times Delta V, which we don't know over Delta T, which is 15 seconds. Therefore, Delta V is just 25 times 15 over 65. This will be in meters per second and we get, um, are negative 25 times now and we get minus 5.8 meters per second. Therefore velocity was 5.8 meters per second. Was a velocity lost
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