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STatiana [176]
2 years ago
15

What is the stretch when you pull with a force of 2.3 N on a spring with a spring constant of 19 N/m?

Physics
1 answer:
Neporo4naja [7]2 years ago
8 0

Answer:

x=0.12 m

Explanation:

Use the equation F=kx and isolate x because that's what you're solving for (x=F/k). Plug in your given values for F and k and solve for x [x=(2.3 N)/(19 N/m)] and you get x=0.12 m.

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Hi there!

We can use the rotational equivalent of Newton's Second Law:

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We can plug in the given values to solve.

\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}

4 0
2 years ago
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5 0
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