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11111nata11111 [884]
3 years ago
8

What is the restoring force of a spring stretched 0.35 meters with a spring contact of 55 newtons?

Physics
2 answers:
Vesna [10]3 years ago
6 0
F=-kx = -19,25 answer D
Bogdan [553]3 years ago
6 0

Answer:

F=19.25 N

Explanation:

The restoring force for a spring is given by:

\vec{F}=-k \vec{x}

Where k is the spring constant and \vec{x} is the spring deformation.

The minus sign indicates that force is opposed to the deformation direction.

According to the problem:

k=55 \;N/m and  x=0.35 \;m

Thus, in magnitude, the restoring force is:

F=55\frac{N}{m} *0.35m=19.25\; N

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we calculate the refractive index

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