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emmainna [20.7K]
3 years ago
6

A 3500 N is force is applied to a spring that has a spring of constant of k= 14000 N/m. How far from equilibrium will the spring

be displaced?
Physics
1 answer:
Taya2010 [7]3 years ago
4 0

Answer:

the spring be displaced by 25.0 cm

Explanation:

The computation is shown below:

As we know that

F= -K × x

So,

x = \frac{-F}{K}

Now  

x = \frac{-3500}{14000} \\\\

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm

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