(1) The speed in trial U is less than that in Q.
(2) The speed in trial T is greater than that in other trials.
(3) The distance of Amy from the path in R is equal to distance in S.
(4) The speed in trial U is less than that in other trials.
(5) When the car is far away, the average of the frequencies moving in and out is equal to the frequency at rest.
(6) The speed in trial R is equal to that in S.
(7) The speed of the car in trial T is 58.6 m/s.
<h3>
Speed of wave</h3>
The speed of a wave is directly proportional to its frequency and wavelength.
v = fλ
where;
- v is speed of the wave
- f is frequency of the wave
- λ is wave length
The speed in trial U is less than that in Q, due to greater frequency observed in trial Q.
The speed in trial T is greater than that in other trials, due to greater frequency observed in trial T.
The distance of Amy from the path in R is equal to distance in S.
The speed in trial U is less than that in other trials.
<h3>Average of the frequencies when the car is far away</h3>
f(avg) = (1,055 + 1,020 + 986 + 985 + 940 + 862 + 830 + 828 + 808 + 730)/10
= 904.3 ≈ 900 Hz
When the car is far away, the average of the frequencies moving in and out is equal to the frequency at rest.
The speed in trial R is equal to that in S.
<h3>Speed of the car in trial T</h3>
f = f₀(v / v - vs)
where;
- f is frequency of the car at zero position
- f₀ is the original frequency
- vs is the speed of the car
- v is speed of sound
900 = 1055(340 / 340 - vs)
900/1055 = 340/340 - vs
0.853 = 340/340 - vs
340 - vs = 340/0.853
340 - vs = 398.6
vs = 398.6 - 340
vs = 58.6 m/s
Learn more about speed of wave here: brainly.com/question/2142871
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