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rjkz [21]
3 years ago
5

A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 205 Hz. A person on the pl

atform 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, vp , for the person to detect a beat frequency of 5.00 Hz?
Take the speed of sound to be 344 m/s.
Physics
1 answer:
Arlecino [84]3 years ago
6 0

Answer:

Explanation:

The question relates to Doppler effect and beat.

The observer is moving towards the reflected sound so apparent frequency will be increased

f = f₀ x (V + v₁) / (V - v₂)

f is apparent frequency , f₀ is real frequency , V is velocity of sound , v₁ is velocity of observer and v₂ is velocity of source . Here

v₁ = v₂ = vp as both observer and source have same velocity

f = f₀ x (V + v₁) / (V - v₂)

205 +5 = 205 x (344 +vp)/ ( 344 - vp)

1.0234 = (344 +vp)/ ( 344 - vp)

= 352 - 1.0234vp = 340+vp

12 / 2.0234vp

vp = 6 m /s approx.

You might be interested in
Which of the following will be the best condition for bread fermentation?
dezoksy [38]

Answer:

a. Wet, soft dough at 85 degrees Fahrenheit

Explanation:

Fermentation is an anaerobic process that transforms starches into simpler substances. The rising of dough is due to fermentation.

According to Harold McGee, 85°F (29°C) is the best temperature for fermenting bread dough. Temperatures below 85°F (29°C) take  much longer to ferment, and temperatures higher than that result into unpleasant flavors in the dough.

Wet, soft dough is usually more preferable because it produces a softer bread.

3 0
3 years ago
A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2
stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

                     s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2

For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

We also have

                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

7 0
3 years ago
Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp
grandymaker [24]

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

5 0
3 years ago
Need help! Photo says all! Will mark brainliest :)
AVprozaik [17]

Answer:

C-D

Explanation:

As you can see from the graph, the distance from A to B was from 0 m to 6 m in a duration of 3 seconds.

Divide 6 meters by 3 seconds to find the speed:

6 ÷ 3 = 2 m/s

B-C is not moving due to a straight line as said in the graph, so speed is

0 m/s.

There is also C-D since the car traveled from a distance of 9 meters

(6 -(-3) = 9) in 3 seconds too. (NOTE: The graph line going down does not mean it is slowing down, but rather going to a certain distance like going backwards)

Divide 9 meters by 3 seconds to get the speed:

9 ÷ 3 = 3 m/s

Between A-B, B-C, and C-D, C-D has the fastest speed recorded with 3 m/s.

A-D does not count here as the line has no connection between point A and point D.

Cheers!

6 0
2 years ago
The hydraulic oil in a car lift has a density of 8.30 102 kg/m³. The weight of the input piston is negligible. The radii of the
denpristay [2]

Answer:

(a) F_i=68.58\ N

(b) F_i=69.903\ N

Explanation:

Given:

  • density of hydraulic oil, \rho=830\ kg.m^{-3}
  • radius of input piston, r_i=6.3\times 10^{-3}\ m
  • radius of output plunger, r_o=0.125\ m
  • force to be supported, F_o=27000\ N

(a)

<u><em>Condition:</em></u><em>The bottom surfaces of piston and plunger at the same level.</em>

According to Pascal's law the pressure of a fluid is exerted equally in all directions against the walls of its container.

Mathematically:

\frac{F_i}{A_i} =\frac{F_o}{A_o}

putting respective values

\frac{F_i}{\pi\times r_i^2} =\frac{27000}{\pi\times r_o^2}

\frac{F_i}{\pi\times (6.3\times 10^{-3})^2} =\frac{27000}{\pi\times 0.125^2}

F_i=68.58\ N

(b)

<u><em>Condition:</em></u><em>The bottom surface of the output plunger is 1.30 m above that of the input piston.</em>

Given:

h=1.3\ m

Now,

P_i=P_o+\rho.g.h

\frac{F_i}{\pi\times (6.3\times 10^{-3})^2} =\frac{27000}{\pi\times 0.125^2} +830\times 9.8\times 1.3

F_i=69.903\ N

7 0
3 years ago
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