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saw5 [17]
3 years ago
6

You hear the engine roaring on a race car at the starting line. Predict the changes in the sound as the race starts and the car

approaches the camera. A) The sound of the engine will get louder and the pitch lower. Eliminate B) The sound of the engine will get louder and the pitch higher. C) The sound is determined by the car and it will not change in any way. D) The sound of the engine will get louder but the pitch will not change.
Physics
2 answers:
gogolik [260]3 years ago
4 0

B. The sound of the engine will get louder and the pitch higher.

Mariana [72]3 years ago
3 0

Answer: The correct option is B.

Explanation:

Doppler effect is the phenomenon in which there is an apparent change in the frequency when there is relative motion between source and listener.

When the source and the listener are approaching each other then the frequency increases. The loudness depends on the amplitude. The energy of the wave is directly proportional to the square of the amplitude.

In the given problem, the engine is roaring on a race car at the starting line. There is change in the sound as the race starts and the car approaches the camera.

Pitch depends on the frequency.

Therefore, the sound of the wave will get louder and the pitch will get higher.

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An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
Katen [24]
<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

Where:

y  is the instrument's final position  

y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

t is the time the parabolic movement lasts

g=2.5\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of planet X.

As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

Finally:

t=12s

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