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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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Answer:

They are 7.4m apart.

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An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f
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Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

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a = F/m = 4 m/s

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Given that

mass m_1 = 30 kg

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m_{2} a _ \ {net} }= m_2g - T   ----- equation (1)

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