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jonny [76]
3 years ago
8

A child drops a ball. The ball hits the ground and bounces. The graph to the left shows the velocity-time graph for the ball fro

m when the ball is dropped until when the ball reaches the top of its first bounce. Air resistance has been ignored. When the ball hits the ground, energy is transferred from the ball to the Earth. Explain how the data in the graph above shows this energy transfer.

Physics
1 answer:
WINSTONCH [101]3 years ago
4 0

Answer:

Consider the velocity-time graph attached below.

The velocity-time graph represents the acceleration of a body under a force.

We can see that is the graph that if a child release the ball above the ground at A, it hits the ground at B. Bounces back with a reaches the top again at C, and hits the ground again at D.

The slope of velocity time graph represents acceleration. From A to B, velocity in increasing constantly with respect to time, which means constant acceleration from A to B. AS velocity increase, momentum of the ball also increases, which results in the increase of Kinetic energy.

At B, the ball hits the ground, the velocity decreases, momentum decrease s, because kinetic energy is transferred from the ball to the ground, due to which the ball would not attain the same height after the bounce.

Then the velocity remains negative at C, which means that now the ball is moving in opposite direction till C. It reaches its new at height at C, which is not the same as that of A because of lost in Kinetic Energy, and fall again.

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Resultant displacement is 29.2 km at 83.1^{\circ} north of west

Explanation:

To solve the problem, we have to use the rules of vector addition, resolving first each vector along the x- and y- direction.

Taking east as positive x direction and north as positive y- direction, we have:

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- Second displacement is 41.0 km northwest, so its components are

B_x = (41.0)cos(135^{\circ})=-29.0 km\\B_y =(41.0)sin(135^{\circ})=29.0 km

So, the components of the resultant displacement are

R_x=A_x+B_x=25.5+(-29.0)=-3.5 km\\R_y=A_y+B_y=0+29.0=29.0 km

And so, the magnitude is calculated using Pythagorean's theorem:

R=\sqrt{R_x^2+R_y^2}=\sqrt{(-3.5)^2+(29.0)^2}=29.2 km

And the direction is given by

\theta=tan^{-1}(\frac{R_y}{|R_x|})=tan^{-1}(\frac{29.0}{3.5})=83.1^{\circ}

Where the angle is measured from the west direction, since Rx is negative.

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Answer:

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Explanation:

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Explanation:

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It's been a while since I've studied this, but my answers would be:

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