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Elodia [21]
3 years ago
6

A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe

n this fellow, clutching his radio, jumps off. Describe the Doppler Effect heard by (a) a person left behind on the platform and (b) a person down below floating on a rubber raft. In each case, specify (1) whether the frequency is greater or smaller than the frequency produced by the radio, (2) whether the observed frequency is constant, and (3) how the observed frequency changes during the fall, if it does change. Give your reasoning.
Physics
1 answer:
joja [24]3 years ago
7 0

Answer:

The Doppler Effect is given by the following relation;

f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

v_o = The velocity of the observer

v_s = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(ii) When the source is receding from the observer, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

(1) Given that (v + v_s) > v, therefore, v < (v + v_s), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(1) Given that (v - v_s) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

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wlad13 [49]

Answer:

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2 years ago
An open organ pipe is 1.6m long. If the speed of sound is 343m/s, what are the pipes: a) fundamental , b) 1st overtone , &amp; c
Yakvenalex [24]

Answer:

a) 107.1875 Hz

b) 214.375 Hz

c) 321.5625 Hz

Explanation:

L = length of the open organ pipe = 1.6 m

v = speed of sound = 343 m/s

f = fundamental frequency

fundamental frequency is given as

f = \frac{v}{2L}

inserting the values

f = \frac{343}{2(1.6)}

f = \frac{343}{2(1.6)}

f = 107.1875 Hz

b)

first overtone is given as

f' = 2f

f' = 2 (107.1875)

f' = 214.375 Hz

c)

first overtone is given as

f'' = 3f

f'' = 3 (107.1875)

f'' = 321.5625 Hz

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3 years ago
Water is a compound because it A. cannot be broken down into simpler substances. B. always has two hydrogen atoms for each oxyge
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I'm almost positive it's b
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3 years ago
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A tiger started running when it saw a deer running at uniform velocity 2m/s at 15m far from it ..if the tiger ran at acceleratio
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You need to set their position functions equal to one another and so for the time t when that is true. That is when the tiger and the deer are in the same place meaning the tiger catches the dear
Xdear= 2t+15 deer position function.
(I integrated the velocity function )
To get the Tigers position function you must integrate the acceleration twice. This becomes
Xtiger=t^2
Now t^2=2t+15
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t^2-2t-15=0
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7 0
3 years ago
A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it
anzhelika [568]

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = \frac{sum of coefficient of static friction}{number of trials}

                                              = \frac{0.779}{6}

                                              = 0.12983

The average coefficient of static friction is 0.130

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