A railroad train is traveling at 32 m/s in still air. The frequency of the train whistle is 350 Hz. What is the wavelength of th
e sound and the frequency of the sound heard by the listener in front of the locomotive?
2 answers:
<h3>Answer:</h3>
40. 383 Hz, 0.9 m
6. 20.78 m/s
<h3>Explanation:</h3>40. You know the pitch will be increased when the train is moving toward you. The only answer choice that is greater than 350 Hz is the one shown above.
Using the formula shown in the attachment, we compute the observed frequency to be ...
... f' = (343 m/s)/(343 m/s -32 m/s)×(350 Hz) ≈ 1.102893×350 Hz ≈ 386 Hz
The wavelength is given by ...
... λ = v/f = (343 m/s)/(386 Hz) ≈ 0.8886 m ≈ 0.9 m
_____
<em>Comment on the answer choice</em>
If you erroneously use the frequency modification formula f'=(v+vo)/v·f, then you get 382.7 Hz ≈ 383 Hz. We suspect an error in the answer key. Have your teacher show you how to work this problem.
__________
6. The velocity can be computed from the acceleration and distance covered by ...
... v = √(2ad)
... v = √(2·9.8 m/s²·22 m) = (√431.2) m/s ≈ 20.77 m/s
The closest answer among those offered is 20.78 m/s. Again, you should have your teacher show you how to work this problem to arrive at the offered answer.
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Answer:
8. acceleration = = 1 unit .
9. acceleration = = -1 unit.
10. acceleration = = 0 units.
Explanation:
8. i) acceleration = velocity / time
ii) In this figure velocity = time
iii) therefore acceleration = = 1 unit .
9. i) acceleration = velocity / time
ii) In this figure 4 = m + 5, therefore m = -1
therefore velocity = (-0.5 time) + 5
iii) therefore acceleration = = -1 units.
10.) velocity is constant at 2
therefore acceleration = = 0 units