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Zigmanuir [339]
3 years ago
10

2)

Physics
1 answer:
earnstyle [38]3 years ago
8 0

Answer:

0.0312J

Explanation:

Let x be the distance the staple moves:

x=0.150m-0.115m=0.035m

And spring constant is k=51.0N/m

PE=0.5kx^2\\=0.5\times 51.0\times 0.035^2\\\\=0.312

Hence, the potential energy is 0.0312J

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