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Effectus [21]
4 years ago
10

If a fast-moving car making a loud noise approaches and moves past the person, what will happen as the distance between the two

increases?
A. The pitch of the sound being heard by the person will appear to be higher than the pitch of the source.
B. The frequency of the sound waves reaching the person's ear will be greater than the frequency of the waves leaving the car.
C. The pitch of the sound being heard by the person will appear to be lower than the pitch of the source.
D. The pitch and frequency of the sound waves reaching the person's ear will remain unchanged.
Physics
2 answers:
densk [106]4 years ago
5 0
C.  The Pitch of the sound being heard by the person will appear to be lower than the pitch of the source.  I hope this is right

vladimir2022 [97]4 years ago
3 0
The answer is B.
The frequency will be greater


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soldier1979 [14.2K]

Answer:

B

Explanation:

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Since all options have the mass and speed in the same units, there is no need for conversion.

A. 20 x 500 = 10000,  B. 200 x 60 = 12000

The same goes for the rest!

6 0
3 years ago
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A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed
gregori [183]

Answer:

the rate of the change of the length of the shadow is - 0.8625 m/s.

The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.

Explanation:

Given the data in the question;

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initially x = 1m2

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so we determine dy/dt when x is 4m from the building

form the image description of the problem, we see two-like triangles with the same base and height ratios

so

 2 / (12-x) = y / 12

24 = y(12 - x )

y = 24 / (12-x)

dy/dt = 24/(12-x)² × dx/dt

Now at x = 4,

we substitute

dy/dt will be;

⇒ 24/(12 - 4)² × -2.3

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dy/dt = - 0.8625 m/s

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The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.

5 0
3 years ago
Can someone tell me if I’m correct or not
Anika [276]
You are correct because nothing is being done to the cake
6 0
3 years ago
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\lambda=\dfrac{dq}{dx}

dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx

Integrating equation (1) from x = x₀ to x = infinity

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5 0
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tester [92]

Answer:

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7 0
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