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Pepsi [2]
2 years ago
9

Even in the most advanced circuits, we cannot oscillate electrons back and forth at that rate through wires. But we can oscillat

e charges back and forth quickly enough to broadcast TV using radio wave signals. At what frequency do the electronics at the TV station need to have the charges oscillate back and forth on a TV broadcast antenna to transmit a typical TV signal (say a radio wave transmission signal with a wavelength of 1 meter)
Physics
1 answer:
den301095 [7]2 years ago
5 0

Answer:

the oscillations of the electrons must be in the 10⁸ Hz = 100 MHz range

Explanation:

The speed of a wave of radio, television, light, heat, all are manifestations of electromagnetic waves that are oscillations of electric and magnetic fields that support each other, the speed of all these waves is the same and the vacuum is equal to c = 3 108 m / s

All waves have a relationship between the speed of the wave, its frequency and wavelength

          c = λ f

          f = c /λ

for this case lam = 1 m

          f = 3 10⁸/1

          f = 3 10⁸ Hz

the oscillations of the electrons must be in the MHz range

It should be clarified that the speed of light in air is a little lower

          n = c / v

          v = c / n

the refractive index of vacuum is n = 1 and the refractive index of air is n = 1.000002

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A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
In what type of circuit are all the resistances of all the circuit elements added linearly?
ycow [4]
<em>It is a series circuit.</em> 
5 0
3 years ago
A tiger runs at 58 km/h [S]. What is the displacement of the tiger in 38 s?
Artyom0805 [142]
58 K/h = 58000/3600= 16.1 m/s
In 38 s displacement is 38x16.1= 612.2 m
3 0
3 years ago
You are driving on a freeway and notice a small car (very small) cut off a big truck. Explain why it is a bad idea for the small
Leno4ka [110]

Answer: small cars can stop and go fast big trucks can not

Explanation:

7 0
3 years ago
An open pipe of length 0.58 m vibrates in the third harmonic with a frequency of 939Hz. What is the speed of sound through the a
vaieri [72.5K]

363 m/s is the speed of sound through the air in the pipe.

Answer: Option B

<u>Explanation:</u>

The formula used to calculate the wavelength given as below,

      Wavelength (\lambda)=\frac{\text { wave velocity }(v)}{\text { frequency }(f)}

      v=\lambda \times f   --------> eq. 1

In power system, harmonics define by positive integers of the fundamental frequency. So the third order harmonic is a multiple of the third fundamental frequency. Each harmonic creates an additional node and an opposite node, as well as an additional half wave within the string.

If the number of waves in the circuit is known, the comparison between standing wavelength and circuit length can be calculated algebraically. The general expression for this given as,

         L=\frac{n \lambda}{2}

For first harmonic, n =1

         L=\frac{\lambda}{2}

For second harmonic, n =2

         L=\frac{2 \lambda}{2}=\lambda

For third harmonic, n =3

         L=\frac{3 \lambda}{2}

         \lambda=\frac{2 L}{3}   -------> eq. 2

Here given f = 939 Hz, L = 0.58 m...And, substitute eq 2 in eq 1 and values, we get

   v=\frac{2 \times 0.58 \times 989}{3}=\frac{1089.24}{3}=363.08 \mathrm{m} / \mathrm{s}

3 0
3 years ago
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