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a_sh-v [17]
1 year ago
12

The unknown quantities to be determined are (a) the capacitive reactance, (b) the maximum and the rms voltages from the source,

and (c) the rms current in the capacitor.
Evaluate the capacitive or inductive reactance, XC or XL.

The unknown quantity to be determined in part (a) is the circuit's capacitive reactance
XC defined as
XC ≡ 1/2fC
,
where, in this problem, C = 4.48 F and the AC source frequency f must be determined.

What is the frequency f of the AC source?
Physics
1 answer:
Digiron [165]1 year ago
8 0

The frequency f of the AC source is determined as 0.446 Xc.

<h3>Frequency of the AC source</h3>

The frequency of the AC source is calculated as follows;

Xc ≡ 1/2fC

where;

  • Xc is the capacitive reactance
  • f is frequency
  • C is capacitance

fC = 2Xc

f = 2Xc / C

f = (2Xc)/4.48

f = 0.446 Xc

Thus, the frequency f of the AC source is determined as 0.446 Xc.

Learn more about frequency here: brainly.com/question/10728818

#SPJ1

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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>

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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;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

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f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

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v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

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Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 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;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} 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;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 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 = f_{2} -  f_{1}  

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

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