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Hooke's Law Question - Will mark brainliest

A spring of natural length 10cm and spring constant 4.00Ncm −1 has a load of 22.0N placed on it. What is its new length?


Sagot :

Explanation:

Hooke's Law: Fe = kx

x = Fe / k = (22.0N) / (4.00N/cm) = 5.50cm.

10.00cm + 5.50cm = 15.50cm

The new length of the spring is 15.50cm.

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