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Wewaii [24]
3 years ago
7

a 0.05-kg starts from rest at a height of 0.95m. assuming no friction, what is the kinetic energy of the car when it reaches the

bottom of the hill? (Assume g = 9.81 m/s)
Physics
1 answer:
Alex73 [517]3 years ago
3 0
Fhubugukhkbjhvlfipy0iy ft
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A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

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D) human overpopulation
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If a sled has at the top of 10 m hill had 1000 J of potential energy what would happen to the PE if the sled were to moved to a
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What is the relationship between the atoms of elements,its isotopes and its ions
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