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Alexus [3.1K]
3 years ago
5

An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in t

he air, she runs to and from a table 5.50 m away at an average speed of 3.00 m/s, returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?
Physics
1 answer:
sp2606 [1]3 years ago
3 0

Answer:

a) 17.97 m/s

b) 16.5 m

Explanation:

The total distance that she runs is twice the distance between her and the table, which is 5.5 * 2 = 11m

So the time she would need to run that distance at a rate of 3m/s is

t = 11/3 = 3.67s

This is also the time it takes for the ball to be thrown up and down, which is twice the time it needs to reach the top, its maximum height at v = 0 m/s

So the time for the ball to reach maximum height is 3.67 / 2 = 1.83 s

Let g = 9.8m/s2. For the ball to get to 0 within 1.83s at the deceleration of 9.8m/s. Its initial speed must be

v = gt = 9.8*1.83 = 17.97 m/s

As she reaches the table within half of the total time (3.67 /2 = 1.83s), the position of the ball would be

h = vt + gt^2/2 = 17.97*1.83 - 9.8*1.83^2/2 = 16.5 m

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Answer:

The height of the water column = 1.62405\overline{30} × 10⁻¹ m

Explanation:

The air cavity in the Coke bottle = 0.220 m deep

The fundamental (frequency) it plays when water is added to shorten the column and it is blown across the top, f = 528 Hz

The given speed of sound in air, v = 343 m/s

We note that the air cavity in the coke bottle is equivalent to a tube closed at one end

The fundamental frequency for a tube closed at one end, 'f', is given as follows;

f = v/(4·L) = v/λ

Where;

L = The height of the water column

λ = The wavelength of the wave

∴ 4·L = v/f = (343 m/s)/(528 Hz) = 0.6496\overline{21} m

∴ L = 0.6496\overline{21} m/4 = 0.162405\overline{30} m

The height of the water column = 1.62405\overline{30} × 10⁻¹ m.

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2 years ago
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gtnhenbr [62]

Answer: 5.06\times10^{9}

At sea level, there is one standard atmospheric pressure which is equal to 101.325 kilopascals.

The pressure of faintest sound that a human ear can hear is 20 micro-pascals.

taking the ratio of two:

\frac {101.325 kilopascals. }{20 micropascals} = \frac {101.325 \times 10^{3}Pa}{20 \times 10^{-6}Pa}

=5.06\times10^{9}

Hence, the atmospheric pressure at sea level is 5.06\times10^{9} times greater than the faintest sound that a human ear can hear.



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<u>Harmful</u><u />

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Answer:

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n_e=1.935\times 10^{20} is the no. of electrons

Explanation:

Given:

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<u>No. of electrons in the given amount of charge:</u>

As we have charge on one electron 1.602\times 10^{-19}\ C

so,

n_e=\frac{Q}{e}

n_e=\frac{31}{1.602\times 10^{-19}}

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<u>Now the no. of moles in this many molecules:</u>

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where:

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