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leonid [27]
2 years ago
6

A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from i

t at 21.8 m/s, what is the received sound frequency?
f= ? Hz
Physics
1 answer:
murzikaleks [220]2 years ago
4 0

Answer:

f" = 40779.61 Hz

Explanation:

From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;

from the Doppler effect equation, we can calculate the initial observed frequency as:

f' = f(1 - (v_o/v))

We are given;

f = 46.2 kHz = 46200 Hz

v_o = 21.8 m/s

v is speed of sound = 343 m/s

Thus;

f' = 46200(1 - (21/343))

f' = 43371.4285 Hz

In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;

Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;

f" = f'/(1 + (v_o/v))

f" = 43371.4285/(1 + (21.8/343))

f" = 40779.61 Hz

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Explanation:

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3 years ago
A- 1000 m/s2<br> Xi-0m<br> Xf-0.75m<br> Vf-?
sleet_krkn [62]

Answer:

The final velocity of the object is,  v_{f} = 27 m/s    

Explanation:

Given,

The acceleration of the object, a = 1000 m/s²

The initial displacement of the object, x_{i} = 0 m

The final displacement of the object,  x_{f} = 0.75 m

The initial velocity of the object will be, v_{i} = o m/s

The final velocity of the object, v_{f} = ?

The average velocity of the object,

                                    v = ( x_{f} - x_{i} )/ t

                                      = 0.75 / t

The acceleration is given by the relation

                                     a = v / t

                                   1000 m/s² = 0.75 / t²

                                            t² = 7.5 x 10⁻⁴

                                            t = 0.027 s

Using the I equation of motion,

                                  v_{f} = u + at

Substituting the values

                                   v_{f} = 0 + 1000 x 0.027

                                                           = 27 m/s

Hence, the final velocity of the object is,  v_{f} = 27 m/s          

8 0
3 years ago
Pls answer I will make brainlist
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3 0
3 years ago
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3 0
3 years ago
You measure a watch's hour and minute hands to be 7.4 mm and 12.1 mm long, respectively. Part A In one day, by how much does the
pychu [463]

Answer:

109.385m

Explanation:

In 1 day, the hour hand travels 2 circles, or 4π rad in angular. The distance it travels is its angle times the radius

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12.1 * 2880π = 109478 mm

So the distance traveled by the tip of the minute hand that exceed the distance traveled by the tip of the hour hand is

109478 - 93 = 109385 mm or 109.385 m

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3 years ago
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