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15. A spring has a constant of 152 when a force of 50.92 N is applied to it. How far did the spring stretch?
O=0.321 m
O z=0.418 m
O z=0.289 m
Oz=0.335 m
O None of these is correct.

Respuesta :

quasarJose

Answer:

Oz=0.335 m

Explanation:

Given parameters;

Spring constant  = 152 N/m

Force applied  = 50.92N

Unknown:

How far did the spring stretch  =  ?

Solution:

For springs;

      F = ke

F is the force

K is the spring constant

e is the extension

  Now insert the parameters and find e;

   50.92  = 152 x e

      e = 0.335m

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