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hodyreva [135]
3 years ago
13

A whistle of frequency 592 Hz moves in a circle of radius 64.7 cm at an angular speed of 13.8 rad/s. What are (a) the lowest and

(b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle?
Physics
1 answer:
Ugo [173]3 years ago
3 0

Answer:

(a) lowest frequency=577 Hz

(b) highest frequency=608 Hz

Explanation:

Given data

f(whistle frequency)=592 Hz

ω(angular speed)=13.8 rad/s

r(radius)=64.7 cm=0.647 m

To find

(a) Lowest frequency

(b) highest frequency

Solution

From Doppler effect

f=f×{(v±vd)/(v±vs)}

Where

v is speed of sound

Vd is speed detector relative to the medium(vd=0)

Vs is the speed of the source

Since

v=rω

For (a) lowest frequency

f^{i}=f(\frac{v}{v+rw} )\\f^{i}=(592Hz)(\frac{343m/s}{343m/s+(0.647m)(13.8rad/s)} )\\f^{i}=577Hz

For (b) highest frequency

f^{i}=f(\frac{v}{v-rw} )\\f^{i}=(592Hz)(\frac{343m/s}{343m/s-(0.647m)(13.8rad/s)} )\\f^{i}=608Hz

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Answer:

No

Explanation:

Please let me know if my answer is correct

7 0
3 years ago
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A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave,
Oliga [24]

The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.

<u>Given the following data:</u>

  • Period = 645 μs

Note: μs represents microseconds.

<u>Conversion:</u>

1 μs = 1 × 10^-6 seconds

645 μs = 645 × 10^-6 seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

Frequency = \frac{1}{Period}

Substituting the value into the formula, we have;

Frequency = \frac{1}{645 * 10^-6}

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

<u>Conversion:</u>

1 hertz = 0.001 kilohertz

1550.39 hertz = X kilohertz

Cross-multiplying, we have;

X = 0.001 × 1550.39

X = 155039 kHz

To 3 significant figures;

<em>Frequency = 155 kHz</em>

Therefore, the corresponding frequency of this sinewave, in kHz is 155.

Find more information: brainly.com/question/23460034

6 0
3 years ago
Because not all stars are the same distance from the earth, it can be difficult to determine
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Whats the question here?
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What will the temperature of the cosmic microwave background be when the average distances between galaxies are 1.7 times as lar
Aliun [14]

Answer:

T'=1.58K

Explanation:

As the Universe expands, the photons of the cosmic microwave background increase its wavelength, making its temperature inversely proportional to the scale factor of the Universe. That is, as the average distances between galaxies increase, the temperature of the cosmic microwave background decreases by the same factor. Therefore, the temperature when the distances between galaxies are 1.7 times as large as they are today will be:

T'=\frac{T_{now}}{1.73}\\T'=\frac{2.725K}{1.73}\\T'=1.58K

4 0
3 years ago
A man pulls on a (massless) rope tied to a tree with a force of 500 N. Later, two men pull on opposite ends of the same rope wit
qwelly [4]

Answer:

  • <em>In both cases the tension in the rope is </em><u>equal to 500N</u>

Explanation:

It may be that in the case of the <em>tree</em>, the result is more intuitive, because you can think that there is only one force. But this is misleading.

To find the <em>tension in the rope</em>, you should draw a free body diagram. By doing so, you would find that the rope is static because there are two opposite forces. Assuming, for simplicity, that the rope is horizontal,  a force of 500N is pulling to one direction (let's say to the right) and a force of 500N is pulling to the opposite direction (to the left). Else, the rope would not be static.

That analysys is the same for the<em> rope tied to the tree</em> ( the tree is pulling with 500N, such as the man, but in opposite direction) and when the rope is pulled  by <em>two men</em> on opposite ends, each with<em> forces of 500N.</em>

Hence, the tension is the same and equal to 500N.

7 0
3 years ago
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