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skelet666 [1.2K]
3 years ago
7

II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically st

arts from rest and is in contact with the tennis racquet for 30.00 ms. Assuming constant acceleration, what was the average force exerted on the tennis ball during this record serve, expressed in terms of the ball’s weight W?
Physics
1 answer:
AveGali [126]3 years ago
8 0

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}

Since we start counting at 0s and the ball departs from rest, this is just a=\frac{v}{t}

So we can write:

F=ma=\frac{mv}{t}=\frac{gmv}{gt}

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

F=\frac{Wv}{gt}=\frac{v}{gt}W

Substituting our values:

F=\frac{(73.14m/s)}{(9.81m/s^2)(30\times10^{-3}s)}W=248.5W N

Where the average force exerted has been written it terms of the tennis ball's weight W.

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Explanation:

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A 20×10⁹charge is moved between two points A andB that are 30mm apart and have an electric potential difference of 600v between
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Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

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By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

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in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

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