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A 0.05-kg car starts from rest at a height of 0.95 m. Assuming no friction, what is the kinetic energy of the car when it reaches the bottom of the hill? (Assume g = 9.81 m/s2.)

Sagot :

Giovi
Considering conservation of mechanical energy we have:
K+U (start)= K+U (end)
Where:
K=kinetic energy=[tex] \frac{1}{2}m v^{2} [/tex]
U=potential energy=[tex]mgh[/tex]
So:
0.05*0.95*9.81+0 (start) = 0+K (end)
K (end)=0.5 Joules

Answer: What is the answer?

Explanation:

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