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Mumz [18]
3 years ago
11

Question 15

Physics
1 answer:
Butoxors [25]3 years ago
5 0
Jjdjdjfjififififkjfndndj
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What is the force required accelerate a 5.0 kg object at 3.0m/s^ 2?
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The equation to find force is f=ma. So, if you plug in the information that you have you'll get F=5x3 and that'll equal F=15N
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The total amount of kinetic energy and potential energy within a system is called Question 4 options: A. thermal energy B.electr
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I'm pretty sure the answer is C. :)
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Find the instantaneous velocity at 1 s . can anyone help with c-h!!!
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Rise over run at 1 second
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Physics question, help please!
vova2212 [387]

The change in potential energy when the block falls to ground is -480J.

The maximum change in kinetic energy of the ball is 480 J.

The initial kinetic energy of the ball is 0 J.

The final  kinetic energy of the ball is 0.148J.

The initial potential energy of the ball is 0.187 J.

The final  potential energy of the ball is 0 J.

The work done by the air resistance is 0.039 J.

<h3>Change in potential energy when the block falls to ground</h3>

ΔP.E = -mgh

ΔP.E = -Wh

ΔP.E = - 40 x 12

ΔP.E = -480 J

<h3>Maximum change in kinetic energy of the ball</h3>

ΔK.E = - ΔP.E

ΔK.E = - (-480 J)

ΔK.E = 480 J

<h3>Initial kinetic energy of the ball</h3>

K.Ei = 0.5mv²

where;

  • v is zero since it is initially at rest

K.Ei = 0.5m(0) = 0

<h3>Final kinetic energy</h3>

K.Ef =  0.5mv²

K.Ef = 0.5(0.0091)(5.7)²

K.Ef = 0.148 J

<h3>Initial potential energy of the ball</h3>

P.Ei = mghi

P.Ei = 0.0091 x 9.8 x 2.1

P.Ei = 0.187 J

<h3>Final potential energy</h3>

P.Ef = mghf

P.Ef = 0.0091 x 9.8 x 0

P.Ef = 0

<h3>Work done by the air resistance</h3>

W = ΔE

W = P.E - K.E

W = 0.187 J - 0.148 J

W = 0.039 J

Learn more about potential energy here: brainly.com/question/1242059

#SPJ1

<h3 />
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1 year ago
What is the law of variation of the period T of a simple pendulum
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The period of a simple pendulum is given by:
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