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Andrej [43]
3 years ago
7

A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 235 Hz . A person on the p

latform right next to the speaker detects the sound waves reflected off the wall and those emitted by the speaker. How fast should the platform move for the person to detect a beat frequency of 6.00 Hz ?
Take the speed of sound to be 344 m/s .
Physics
1 answer:
Anvisha [2.4K]3 years ago
5 0

Answer:

vr = 4.336 m/s

Explanation:

We are given that;

Beat frequency is 6Hz

Speed of sound = 344 m/s

Now,

Reflective Doppler frequency; f = 235 + 6 = 241 Hz

We can calculate the observed frequency if both the source sound and the observer are moving towards each other. In this case, the formula is:

f = fo[(c + vr)/(c + vs)]

Where;

ƒ = observed frequency

c = speed of sound

vs = velocity of source (negative if it’s moving toward the observer)

ƒ0 = emitted frequency of source

Since it’s moving toward the observer, thus we can rewrite equation as;

f = fo[(c + vr)/(c - vs)]

We have that;

fo = 235 , f = 241 , c = 344 , vr=vs

Thus,

241 = 235[(344+vr)/(344-vr)]

241(344 - vr)= 235(344 + vr)

82904 - 241vr = 80840 + 235vr

82904 - 80840 = 241vr + 235vr

2064 = 476 vr

vr = 2064/476

vr = 4.336 m/s

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While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an ident
arsen [322]

Answer:

         v_s =7.74\ m/s

Explanation:

given,

Speed of sound = 343 m/s

frequency of horn = 260 Hz

the friend is approaching, the frequency is increased by the Doppler Effect. The frequency is 266 Hz

using formula

         f' = \dfrac{v}{v-v_s}f_0

         266= \dfrac{343}{343 - v_s}(260)

         1.023= \dfrac{343}{343 - v_s}

         343 - v_s = 335.26

         v_s =7.74\ m/s

the speed of friends approaching is equal to v_s =7.74\ m/s

7 0
3 years ago
a fixed amount of ideal gas is held in a rigid container that expands negligibly when heated. at 20 the gas pressure is p. if we
pantera1 [17]

Answer:

When the temperature of the gas is increased from 20 to 40, the pressure will be 2p

Explanation:

Given;

initial temperature of the gas, T₁ = 20 K

final temperature of the gas, T₂ = 40 k

initial pressure of the gas, P₁ = P

final pressure of the gas, P₂ = ?

Apply pressure law of gases;

\frac{P_1}{T_1} = \frac{P_2}{T_2} \\\\P_2 = \frac{P_1T_2}{T_1} \\\\P_2 = \frac{40P}{20} \\\\P_2 = 2P

Therefore, when the temperature of the gas is increased from 20 to 40, the pressure will be 2p

5 0
3 years ago
These questions are from science subject
timama [110]

Answer:

why we follow ur order..........

8 0
3 years ago
Which of the following is not a function of PACs?
Katena32 [7]

Answer:

i think D

hope this helps

let me know if i'm wrong i will change the answer

Explanation:

5 0
3 years ago
Read 2 more answers
Question # 40
jenyasd209 [6]

Answer:

Potential Energy = x = m g h

Kinetic energy = 1/2 m v^2

Assuming the mass fall from rest

1/2 m v^2 = m g h

v^2 = 2 g h

So the speed attained is independent of the mass

Also, x / v   does not have the units of mass

So the solution is none of the above.

7 0
3 years ago
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