Describe how air particles differ in the way they move when air begins to cool.
1 answer:
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Answer:
v_s = 34.269 m / s
Explanation:
This is a Doppler effect exercise, in this case the observer is fixed and the sound source is moving.
f ’= f
where the negative sign is used for when the source approaches the observer and the positive sign for when the source moves away from the observer
In this case when f ’= 5500 Hz approaches and when f’ = 4500 Hz moves away, let's write the two expressions together
5500 = f ()
4500 = f ( )
let's solve these two equations
1.222 (v-v_s) = v + v_s
v_s (1+ 1.22) = v (1.222 -1)
v_s = v
the speed of sound in air is v = 343 m / s
v_s = 343 0.09990
v_s = 34.269 m / s