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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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According to the principle of flotation:

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The plastic will float due to the fact the average density of the total volume of the plastic and the air inside it is less than the same volume of water it is floating in

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The electric field strength in the space between two closely spaced parallel disks is 1.0 10^5 N/C. This field is the result of
alex41 [277]

To solve this problem it is necessary to apply the concepts related to the capacitance in the disks, the difference of the potential and the load in the disc.

The capacitance can be expressed in terms of the Area, the permeability constant and the diameter:

C = \frac{\epsilon_0 A}{d}

Where,

\epsilon_0 = Permeability constant

A = Cross-sectional Area

d = Diameter

Potential difference between the two disks,

V = Ed

Where,

E = Electric field

d = diameter

Q = Charge on the disk equal to \rightarrow Q=ne=(3.9*10^9)(1.6*10^{-19})= 6.24*10^{-10}C

Through the value found and the expression given for capacitance and potential, we can define the electric charge as

Q = CV

Q = \frac{\epsilon A}{d}(Ed)

Q = \epsilon_0 AE

Q = \epsilon_0 \pi(\frac{d}{2})^2E

Q = \frac{\epsilon \pi d^2E}{4}

Re-arranging the equation to find the diameter of the disks, the equation will be:

d = \sqrt{\frac{4D}{\epsilon_0 \pi E}}

Replacing,

d = \sqrt{\frac{4(6.24*10^{-10})}{(8.85*10^{-12})\pi(1*10^{5})}}

d = 0.0299m

Therefore the diameter of the disks is 0.03m

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