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MariettaO [177]
2 years ago
9

A spring is compressed so that it has 7.2 J of elastic potential energy. A 0.3 kg ball is placed on top of the spring. When the

spring is released, how high will the ball go ?
Physics
1 answer:
VLD [36.1K]2 years ago
5 0

The height of the ball is 2.45 m.

To calculate the height of the ball, we use the formula below.

<h3>Formula</h3>
  • E = mgh................ Equation 1

Where:

  • E = Elastic energy of the spring
  • m = mass of the ball
  • h = height of the ball
  • g = acceleration due to gravity.

make h the subject of the equation

  • h = E/mg...............Equation 2

From the question,

Given:

  • E = 7.2 J
  • m = 0.3 kg
  • g = 9.8 m/s²

Substitute these  values into equation 2

  • h = 7.2/(0.3×9.8)
  • h = 7.2/2.94
  • h = 2.45 m

Hence, The height of the ball is 2.45 m.

Learn more about height here: brainly.com/question/1739912


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Answer:

wrong statement :  Momentum is not conserved for a system of objects in a head-on collision.

Explanation:

In a head on collision of two objects , two equal and opposite forces are created at the point of collision . These two forces create two impulses in opposite direction which results in equal and opposite changes in momentum in each of them . Hence net change in momentum is zero. In this way momentum is conserved in head on collision of two objects.

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6.93 km/h

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
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Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

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v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

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That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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What is Power?

Electric power is the rate at which an electric circuit transfers electrical energy per unit time.

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To learn more about Power from the given link:

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