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ICE Princess25 [194]
2 years ago
6

(4%) Problem 10: The standard frequency of the musical pitch called "middle C" is 261.6 Hz. show answer No Attempt How fast, in

meters per second, would a source emitting a tone of middle C have to move toward you so that you would hear the pitch C# (C sharp) above middle C, with a frequency of 277.2 Hz? Assume that takes place in air at a high altitude, where the speed of sound is 298 m/s.
Physics
1 answer:
bearhunter [10]2 years ago
6 0

Answer:

16.77 m/s

Explanation:

Given that

Frequency of middle pitch, Fo = 261.6 Hz

Frequency of C sharp, f = 277.2 Hz

Velocity of sound in air, v = 298 m/s

Speed of sound from the source, Vs = ? m/s

Using the formula

f = Fo•(V + Vr)/(V + Vs)

← Doppler

Vr would be +ve if the receiver is moving toward source;

Vs would be -ve if source is moving toward the receiver

277.2 Hz = 261.6Hz * (298 + 0) / (298 - Vs)

277.2 = 77956.8 / (298 - Vs)

298 - Vs = 77956.8 / 277.2

298 - Vs = 281.23

Vs = 298 - 281.23

Vs = 16.77 m/s

Thus, the speed needed is 16.77 m/s

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Answer: 6 m

Explanation:

30 = 5 * d

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1 year ago
A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di
Vikentia [17]

Answer:

option B

Explanation:

given,

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diameter of piston, d₁ = 0.02 m

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diameter of second piston,  d₂ = 0.15 m

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now,

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    \dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}

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5 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

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K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

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Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

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iren2701 [21]
I would have to say that it is Y
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