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netineya [11]
3 years ago
9

When a 4.10-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.90 cm. If the

4.10-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?
Physics
1 answer:
kumpel [21]3 years ago
4 0

Answer:

Explanation:

Using Hooke's law,

F = -kx

Where,

F = force of the mass

k = spring constant/stiffness

x = length of the spring

Given:

F1 = m1*a

= 4.1 * 9.81

= 40.221 N

F2 = m2*a

= 1.5 * 9.81

= 14.715 N

x1 = 2.9 cm

F1/x1 = F2/x2

Therefore,

x2 = (14.715 * 2.9)/40.221

= 1.06 cm.

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A wooden piece is made in different shapes take length (l) = radius (r) = 2m Calculate its volume as a:
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A skydiver of mass 87 kg falls straight down. If air resistance exerts 355 N of force on him as he falls, what is the net force
alukav5142 [94]

Answer:

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From the question,

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