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1 year ago

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A freely falling object eventually stops on reaching the ground what happens to its kinetic energy.

1 answer:

Evgesh-ka [11]1 year ago

5 0

When a freely falling object reaches the ground, its kinetic energy gets converted to heat and sound energy.

As an object falls, its kinetic energy continues to increase as its potential energy continues to decrease as potential energy gets converted to kinetic energy.

The kinetic energy of an object depends on the mass of the object and the velocity at which it is traveling. It is maximum when the object is about to hit the ground. When the object reaches the ground, the kinetic energy gets converted to heat and sound energy. That is why a sound is heard whenever an object strikes the ground after falling from a height.

Learn more about kinetic energy here:

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Pls help me pls<br>pls use the points as my thank you

Ivanshal [37]

Answer:

A

Explanation:

In section A, she is moving in constant speed because there's a flat line which indicates speed remained same as time passed. In section c, there's a flat line but look at y axis for this one, it is 0 which means there's no speed during this part. B and D have varying speeds as there is a change in y axis variables as time passes.

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3 years ago

Read 2 more answers

A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming

Monica [59]

Answer:

The deceleration is $a = - 76.27 m/s^2$

Explanation:

From the question we are told that

The height above firefighter safety net is $H = 14 \ m$

The length by which the net is stretched is $s = 1.8 \ m$

From the law of energy conservation

$KE_T + PE_T = KE_B + PE_B$

Where $KE_T$ is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

and $PE_T$ is the potential energy of the before jumping which is mathematically represented at

$PE_T = mg H$

and $KE_B$ is the kinetic energy of the person just before landing on the safety net which is mathematically represented at

$KE_B = \frac{1}{2} m v^2$

and $PE_B$ is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

So the above equation becomes

$mgH = \frac{1}{2} m v^2$

=> $v = \sqrt{2 gH }$

substituting values

$v = 16.57 m/s$

Applying the equation o motion

$v_f = v + 2 a s$

Now the final velocity is zero because the person comes to rest

So

$0 = 16.57 + 2 * a * 1.8$

$a = - \frac{16.57^2 }{2 * 1.8}$

$a = - 76.27 m/s^2$

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3 years ago

A 0.3-m-radius automobile tire rotates how many revolutions after starting from rest and accelerating at a constant 2.13 rad/s2

Aloiza [94]

Answer:

The automobile tire rotates 91 revolutions

Explanation:

Given;

angular acceleration of the automobile, α = 2.13 rad/s²

time interval, t = 23.2-s

To calculate the number of revolutions, we apply the first kinematic equation;

$\theta = \omega_i \ + \frac{1}{2} \alpha t^2$

the initial angular velocity is zero,

$\theta =0\ + \frac{1}{2} (2.13) (23.2)^2\\\\\theta = 573.2256 \ Rad$

Find how many revolutions that are in 573.2256 Rad

$N = \frac{\theta}{2 \pi} = \frac{573.2256}{2\pi} \\\\N = 91 \ revolutions$

Therefore, the automobile tire rotates 91 revolutions

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On a speed-versus-time graph, a straight line that slopes down toward
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Answer:

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Explanation:

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