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german
3 years ago
12

A rock with a mass of 0.3 kg falls from the top of a cliff. If it takes the rock 2.5 s to reach the ground, what was the impulse

on the rock during the fall?
Physics
1 answer:
kaheart [24]3 years ago
4 0
Impulse=force x time
force=mass x acceleration due to gravity
force=
300 \times 10 = 3000
impulse =3000 x 2.5= ( sorry i don't have a calculator right now so you must calculate this yourself)
I converted from kg to g because it is the standard.
Hope this helps you.
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Explanation:

It is known that relation between pressure and density is as follows.

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Putting the given values into the above formula as follows.

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To stretch a certain nonlinear spring by an amount x requires a force F given by F = 40 x − 6 x 2 , where F is in Newtons and x
strojnjashka [21]

Answer:

64 J

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The potential energy change of the spring ∆U = -W where W = work done by force, F.

Now W = ∫F.dx

So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)

∆U = - ∫-Fdx

=  ∫F.dx

Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U =  ∫₀²F.dx

=  ∫₀²(40x - 6x²).dx

=  ∫₀²(40xdx - 6x²dx)

=  ∫₀²(40x²/2 - 6x³/3)

=  ∫₀²(20x² - 2x³)

= [20x² - 2x³]₀²

= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)

= [(20(4) - 2(8)) - (0 - 0))

= [80 - 16 - 0]

= 64 J

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