pilarrios6715 pilarrios6715
  • 17-07-2020
  • Physics
contestada

A spring with a spring constant of 110N/m is stretched by 12 cm. How much force is applied to the spring to stretch it

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nobleswift
nobleswift nobleswift
  • 24-07-2020

Answer:

force = 1320N

Explanation:

spring constant = force/extension

110 = F/12

F= 110 x 12 = 1320N

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pstnonsonjoku
pstnonsonjoku pstnonsonjoku
  • 04-11-2021

The force on the spring is 13.2 N.

The force acting on an elastic material is directly proportional to the extension of the material. This is given by Hooke's law. Mathematically, we can write; Fαe.

Introducing the constant of proportionality k;

F = ke

F = force on the spring

k = force constant

e = extension

Substituting the values we have;

F = 110N/m (0.12m)

F = 13.2 N

Learn more: https://brainly.com/question/2673886

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