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Andreyy89
3 years ago
14

A train sounds its whistle as it approaches a tunnel in a cliff. The whistle produces a tone of 650 Hz and the train travels wit

h a speed of 21.2 m/s. Find the frequency heard by an observer standing near the tunnel entrance.
Physics
1 answer:
Alex777 [14]3 years ago
5 0

Answer:f_{0} = 692.82Hz

Explanation: This question of ours is based on Doppler effect.The Doppler effect states that<em> if a source is producing sound at a specific frequency to an observer and there is a relative motion between the observer and the source, the frequency of sound perceived by the observer  will be different from that of the original from the source. </em>

This represented mathematically below as

f_{0} =\frac{v +v_{0} }{v -v_{s} }  * f

where

f_{0} = observed frequency = ?

f = original frequency of sound source = 650Hz

v = speed of sound in air = 343m/s

v_{0} = velocity of observer relative to the source = 0m/s ( the observer is standing, thus meaning he is not moving)

v_{s} = velocity of sound source relative to the observer = 21.2m/s

by putting all of these in the formulae, we have that

f_{0} = \frac{343 + 0}{343 -21.2} *650\\\\\\f_{0} = \frac{343}{321.8} * 650\\\\\\f_{0} = 692.98Hz

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E. greater than the angle of incidence.

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Answer:

A) V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}

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Explanation:

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W_{d} = W_{a} - W_{w}

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W_{a}: is the weight of the block in the air = 20.1 N

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Now, the mass of the water displaced is:

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The volume of the block can be found using the mass of water displaced and the density of the water:

V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}

B) The density of the block can be found as follows:

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I hope it helps you!            

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