anonymous
  • anonymous
After scoring a touchdown, an 84.0 kg wide receiver celebrates by leaping 1.20m off the ground. - what is the energy he gained by leaping forward?
Physics
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anonymous
  • anonymous
After scoring a touchdown, an 84.0 kg wide receiver celebrates by leaping 1.20m off the ground. - what is the energy he gained by leaping forward?
Physics
jamiebookeater
  • jamiebookeater
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
A horizontal force of 5.0N acts on a 3.0kh mass over a distance of 6.0m along a horizontal frictionless surface. What is the change in the KE of the mass?
anonymous
  • anonymous
W= F x d
anonymous
  • anonymous
it would be equal to w = mgd?

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anonymous
  • anonymous
|dw:1327642068642:dw||dw:1327642099129:dw|
anonymous
  • anonymous
A 14, 700 N car is traveling at 25.0 m/s. The brakes are applied to stop as per attached figure . The average breaking force is 7100N - how much energy does the car lose in order to stop?
anonymous
  • anonymous
- how much work is done on the car in order to stop it? - how far will it take the car to stop?
anonymous
  • anonymous
All of these make use of the work-energy theorem. \[W = \Delta U\]where \(\Delta U = \Delta KE + \Delta PE\)

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