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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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A 50.0 mg sample of an unknown radioactive substance was placed in storage and its mass measured periodically. After 19.7 days,
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Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula: </u>

<u></u>

A=A_{o}.2^{\frac{-t}{h}} (1)

Where:

A=3.13mg is the final amount of the material

A_{o}=50mg is the initial amount of the material

t=19.7days is the time elapsed

h is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find h from (1):

3.13mg=(50mg)2^{\frac{-19.7days}{h}} (2)

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