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mezya [45]
2 years ago
15

A ball with a mass of 0.25 kg hits a gym ceiling with a force of 78.0 n. what happens next?

Physics
1 answer:
QveST [7]2 years ago
6 0
<span>a. The ball accelerates downward with a force of 80.5 N. This is a rather badly worded question since the answer depends upon whether or not the impact with the gym ceiling was elastic or non-elastic. With an elastic collision, the ball will accelerate downward with it's original force plus the acceleration due to gravity. With a non-elastic collision (the energy in the ball being used to damage the ceiling of the gym), then the initial energy the ball has would be expended while causing damage to the gym ceiling and then the ball would accelerate downward solely due to the force of gravity. In either case, we need to take into consideration the force of gravity. So multiply the mass of the ball by the gravitational acceleration, giving F = 0.25 kg * 9.8 m/s^2 = 2.45 kg*m/s^2 = 2.45 N Since the initial force is 78.0 newtons, let's add them 78.0 N + 2.45 N = 80.45 N and after rounding to 3 figures, gives 80.5 N So we have a possible answer of 2.45N or 80.5N depending upon if the collision is elastic or not. And unfortunately, both possible answers are available. Since no mention of the ceiling being damaged is made in the question, and to be honest a 100% non-elastic collision is highly unlikely, I will assume the collision is elastic, so the answer is "a".</span>
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Explanation:

Using Coulomb's Law we know that the electric field E at point \vec{r} is:

\vec{E(\vec{r})} = k_e \frac{q}{d^2} \frac{\vec{r}-\vec{r'}}{d}

where  k_e is the Coulomb's Constant, q is the source charge, d is the distance between point and position of the source point charge, and \vec{r}' is the position of the source point charge.

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For our problem, \vec{r'} = (0,0), as the charge is located at the origin.

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\hat{r} =\frac{\vec{r}}{d}

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Now, we can take the values for each point.

<h3>a.</h3>

\vec{r}= (0,-1.35 \ m)

and, the magnitude of the vector is

|\vec{r}| = \sqrt{r_x^2 + r_y^2}

|\vec{r}| = \sqrt{(0 \ m)^2 + (-1.35 \ m )^2}

|\vec{r}| =1.35 \ m

So, the unit vector is:

\hat{r} =\frac{(0,-1.35 \ m)}{1.35 \ m}

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<h3>b.</h3>

\vec{r}= (12 \ cm,12 \ cm)

and, the magnitude of the vector is

|\vec{r}| = \sqrt{r_x^2 + r_y^2}

|\vec{r}| = \sqrt{(12 \ cm)^2 + (12 \ cm )^2}

|\vec{r}| = \sqrt{2} \ 12 \ cm

So, the unit vector is:

\hat{r} =\frac{(12 \ cm,12 \ cm)}{\sqrt{2} \ 12 \ cm}

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<h3>c.</h3>

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|\vec{r}| = \sqrt{r_x^2 + r_y^2}

|\vec{r}| = \sqrt{(-1.10 \ m)^2 + (2.60 \ m)^2}

|\vec{r}| = 2.8415 \ m

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\hat{r} =\frac{ (-1.10 \ m, 2.60 \ m)}{2.8415 \ m}

\hat{r} =(-0.3871 ,0.91501)

\hat{r} = \ -0.3871 \ \hat{i} + \ 0.91501\ \hat{j}

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