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What distance with a 6 N weight expand a spring with a spring constant of 15 N/m?


Sagot :

Answer:

-0.4 m

Explanation:

F = 6 N

k = 15 N/m

Plug those values into Hooke's Law:

F = -kx

6 N = -(15 N/m)x

---> x = -0.4 m

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