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beks73 [17]
3 years ago
9

The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm

from its unstressed position. Find the force constant of the spring. Answer in units of N/m. Find the work done in stretching the spring. Answer in units of J.
Physics
1 answer:
Alika [10]3 years ago
8 0

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force,  F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by F=kx

So 63.5=k\times 0.0531

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J

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2 years ago
A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?
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Answer:

<h2>6.36 cm</h2>

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

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u = object distance

v = image distance

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1/v = 1/16 - 1/24.8

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2 years ago
The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig
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Explanation:

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A ball is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equa
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Answer:

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The explanation is attached as a picture with this answer

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