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Dominik [7]
3 years ago
9

What is the magnitude of the detected sound frequency shift from 170 Hz during the projectile flight described in the passage?

Physics
1 answer:
Serhud [2]3 years ago
5 0

Answer:

The magnitude of the frequency shift (that is the change in frequency) first falls to zero, then increases. This is so because as the projectile moves upward, the position of the sound source is constantly changing and as a result there would be a difference between the frequency of the sound emitted and the sound received by the microphone (that is the frequency shift). A point will be reached in the projectile's motion when the projectile will be momentarily at rest. (At the highest point reached in its flight h).

Explanation:

At this point, the frequency emitted by the source becomes equal to that received by the microphone. This can be seen from the equation 2 for Doppler effect contained in the attachment below.

This moment is short lives and the projectile soon starts a return journey downward under the influence of gravity. The velocity of the projectile is negative ( v < 0 ) and from the equation can be confirmed that the frequency received by the microphone increases as the projectile fall under gravity. This is so because as the projectile falls downward it velocity becomes more and more negative. And from the equivalent in the picture below, the frequency received by the microphone increases

Thank you very much and I hope this was helpful.

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A mass of .1539 kg moves down a 5 meter ramp in 2 seconds. What
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In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
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Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

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