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scZoUnD [109]
3 years ago
9

A speaker at the front of a room and an identical speaker at the rear of the room are being driven at 456 Hz by the same sound s

ource. A student walks at a uniform rate of 1.12 m/s away from one speaker and toward the other. How many beats does the student hear per second? (Take the speed of sound to be 343 m/s.) Hint: The Doppler effect causes both frequencies to be shifted. The difference between those two frequencies is what causes the beats.
Physics
1 answer:
vampirchik [111]3 years ago
6 0

Answer:

2.97795 Hz

Explanation:

v = Speed of sound in air = 343 m/s

v_r = Relative speed between the speakers and the student = 1.12 m/s

f' = Actual frequency of sound = 456 Hz

Frequency of sound heard as the student moves away from one speaker

f_1=f'\dfrac{v-v_r}{v}\\\Rightarrow f_1=456\dfrac{343-1.12}{343}\\\Rightarrow f_1=454.51102\ Hz

Frequency of sound heard as the student moves closer to the other speaker

f_2=f'\dfrac{v+v_r}{v}\\\Rightarrow f_2=456\dfrac{343+1.12}{343}\\\Rightarrow f_2=457.48897\ Hz

The difference in the frequencies is

f_2-f_1=457.48897-454.51102=2.97795\ Hz

The student hears 2.97795 Hz

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Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

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\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

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\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

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6 0
3 years ago
If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far
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Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

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d=\frac{N2}{mg}*l * tan∅

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As f₁= чN₁

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