A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s, how high is the cliff, if it took 12 s for the
2 answers:
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Answer:
<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>
<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>
<em>c. The beat frequency produced by the direct and reflected sound is </em>
<em> 11.62 Hz</em>
Explanation:
Part A
The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;
..............................1
where is the frequency of the wave as it strikes the hill;
f is the frequency of the produced by the horn of the car = 231 Hz;
is the speed of sound = 340 m/s;
v is the speed of the car = 36.4 mph
Converting the speed of the car from mph to m/s we have ;
hint (1 mile = 1609 m, 1 hr = 3600 secs)
v =
v = 16.27 m/s
Substituting into equation 1 we have
=
= 242.61 Hz.
Therefore, the frequency which the wave strikes the hill is 242.61 Hz.
Part B
At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;
where is the frequency of the reflected sound;
is the frequency which the wave strikes the hill = 242.61 Hz;
is the speed of sound = 340 m/s;
v is the speed of the car = 16.27 m/s.
Substituting our values into equation 1 we have;
= 254.23 Hz.
Therefore, the frequency of the reflected sound is 254.23 Hz.
Part C
The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;
Δf = -
The parameters as specified in Part A and B;
Δf = 254.23 Hz - 242.61 Hz
Δf = 11.62 Hz
Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
where
is the charge density
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
where
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate: