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Feliz [49]
3 years ago
10

A 0.5 kg ball rolls down a frictionless inclined plane. The height of the inclined plane is 0.5m. What is the kinetic energy of

the ball when it reaches the bottom of the inclined plane?
Chemistry
1 answer:
joja [24]3 years ago
8 0

Answer:

2.45Joules

Explanation:

since kinetic energy is K.E.=1/2mv^{2}

lost in potential energy is gain of kinetic energy

mgh=1/2mv^2

9.8*2*0.5=v^2

v^{2}=9.8

v=\sqrt{9.8} =3.13m/s

now k.e. = 0.5*m*v^{2} =0.5*0.5*9.8 =2.45J

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