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IgorLugansk [536]
2 years ago
14

A galloping pony speeds past you at 5 m/s. The frequency of the sound produced by the hooves on the dirt is 221 Hz. Assume the s

peed of sound to be 342 m/s. What is the observed frequency of the hooves after the pony has passed your position?
1. 218 Hz
2. 226 Hz
3. 216 Hz
4. 224 Hz
Physics
2 answers:
joja [24]2 years ago
7 0
Given:
speed of passing pony 5 m/s
frequency of the sound produced: 221 Hz
speed of sound 342 m/s

Let us use the Doppler Shift Formula:
Where the <span>source is moving away from the observer at rest
</span>
f' = (v / v+vs) f
Where, vs<span> = Velocity of the Source,</span>
           v = Velocity of sound or light in medium,
           f = Real frequency,
           f' = Apparent frequency.

f'= [342 m/s / (342 m/s+5m/s)] * 221 Hz
f' = 0.9856 * 221Hz
f' = 217.8176 Hz or 218 Hz

The observed frequency <span>of the hooves after the pony has passed your position is 218 Hz.</span>
Nat2105 [25]2 years ago
5 0
The answer is A) 218 Hz
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Question 5 At 12:00 pm, a spaceship is at position ⎡⎣324⎤⎦ km ⎣ ⎢ ⎡ ​ 3 2 4 ​ ⎦ ⎥ ⎤ ​ km away from the origin with respect to so
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the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

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By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

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I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

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