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AleksAgata [21]
4 years ago
7

A rifle is fired in a valley with parallel vertical walls. The echo from one wall is heard 1.55 s after the rifle was fired. The

echo from the other wall is heard 2.5 s after the first echo. How wide is the valley? The velocity of sound is 343 m/s. Answer in units of m.
Physics
1 answer:
elena55 [62]4 years ago
6 0

Answer:

w=694.575\ m

Explanation:

Given:

velocity of the sound, v=343\ m.s^{-1}

time lag in echo form one wall of the valley, t= 1.55\ s

time lag in echo form the other wall of the valley, t'=2.5\ s

<u>distance travelled by the sound in the first case:</u>

d=v.t

d=343\times 1.55

d=531.65\ m

Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.

s=\frac{d}{2}

s=\frac{531.65}{2}

s=265.825\ m

<u>distance travelled by the sound in the second case:</u>

d'=v.t'

d'=343\times 2.5

d'=857.5\ m

Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.

s'=\frac{d'}{2}

s'=\frac{857.5}{2}

s'=428.75\ m

<u>Now the width of valley:</u>

w=s+s'

w=265.825+428.75

w=694.575\ m

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answer:

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Explanation:

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first step : calculate the area

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                    = [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]

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8 0
3 years ago
The largest and the smallest balls used in the experiment are with diameter 9.52 mm, and 2.38 mm respectively. For a glycerin wi
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6\pi \eta r V_t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{t}

t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{6\pi \eta r V_t}

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t = 0.033 ms

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t = 0.531 ms

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