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A spring with spring constant 25N/m is stretched 10cm. Using the equation Ee=1/2 k e2, the stored elastic potential energy is ...

0.125 J
12.5 J
25 J
1250 J

Respuesta :

MathPhys

Answer:

0.125 J

Explanation:

E = ½ kx²

E = ½ (25 N/m) (0.10 m)²

E = 0.125 J