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FrozenT [24]
4 years ago
5

You are standing at a subway platform when a subway passes, sounding its whistle. If you observe the whistle to be at 12750Hz wh

en it is approaching you and at 10750Hz after it passes,
Physics
1 answer:
adell [148]4 years ago
5 0

As per doppler's Effect of sound we can say when subway is approaching the platform we will have

f_1 = f_o* \frac{v}{v- v_s}

12750 = f_o* \frac{340}{340 - v_s}

Similarly we can find the frequency when subway is passing away with same speed from us

f_2 = f_o* \frac{v}{v + v_s}

10750 * f_o* \frac{340}{340 + v_s}

now we can find the ratio of two

\frac{12750}{10750} = \frac{340 + v_s}{340 - v_s}

1.186*(340 - v_s) = 340 + v_s

403.26 - 1.186 v_s = 340 + v_s

63.26 = 2.186 v_s

v_s = \frac{63.26}{2.186}

v_s = 28.94 m/s

So the speed of subway is 28.94 m/s

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Answer:

Hey

I am sure from what i know that your answer should be the following:

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3 years ago
Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o
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Answer:

V = 20 miles /sec

Explanation:

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1.5*V = 96/t₁      (1)

V*t₁  + 16*V = 384     (2)

We solve  from equation    (1)      t₁  = 64/V

And by substitution   in equation (2)

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6 0
3 years ago
A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,
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Answer:

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m is the mass of the train

v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

m'=2m

therefore, the new momentum is

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so, the momentum has also doubled.

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