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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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The answer is A.

Explanation:

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A force of 45 newtons is applied on an object, moving it 12 meters away in the same direction as the force. What is the magnitud
klio [65]

If the applied force is in the same direction as the object's displacement, the work done on the object is:

W = Fd

W = work, F = force, d = displacement

Given values:

F = 45N

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Plug in and solve for W:

W = 45(12)

W = 540J

5 0
3 years ago
Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45 m diamete
Readme [11.4K]

Answer:

In m/s^2:

a=11.3778 m/s^2

In units of g:

a=1.161 g

Explanation:

Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.

a=\frac{v^2}{r}

where:

v is the speed

r is the radius

Now,

a=\frac{16^2}{45/2}\\ a=11.3778 m/s^2

In g units:

a=\frac{11.3778\ g}{9.8}\\ a=1.161\ g

7 0
3 years ago
The depth of the Pacific Ocean in the Mariana Trench is 36,198 ft. What is the gauge pressure at this depth
FinnZ [79.3K]

Answer:

the pressure at the depth is 1.08 × 10^{8} Pa

Explanation:

The pressure at the depth is given by,

P = h \rho g

Where, P = pressure at the depth

h = depth of the Pacific Ocean in the Mariana Trench = 36,198 ft = 11033.15 meter

\rho = density of water = 1000 \frac{kg}{m^{3} }

g = acceleration due to gravity ≈ 9.8 \frac{m}{s^{2} }

P = 11033.15 × 9.8 × 1000

P = 1.08 × 10^{8} Pa

Thus, the pressure at the depth is 1.08 × 10^{8} Pa

4 0
3 years ago
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