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Ganezh [65]
3 years ago
15

Which of the following statement is false

Physics
2 answers:
guajiro [1.7K]3 years ago
8 0
There is nothing to choose from
Nesterboy [21]3 years ago
7 0
There is not options.
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In order to effectively radiate a radio signal, an antenna must be at least one-half the wavelength of transmission in length, i
Virty [35]
<span>The equation that relates Wavelength & frequency of electromagnetic waves is : Velocity (c) = Wavelength (λ) * frequency (f) ------ (1) Electromagnetic wave velocity (c) = Speed of Light = 3 * 10^8 m/s From (1), Wavelength , λ = c / f λ= (3*10^8)/(980*10^3) λ = 306.12 m Minimum height of the antenna for effective transmission = λ * 0.5 = 306.12*0.5 Answer : 153 m (rounded off)</span>
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2 years ago
What is the rate of vibration measured in
jek_recluse [69]

Answer:

I'm not 100% sure tbh but the only thing I think makes sense to represent vibration would be frequency which is measure in Hertz (Hz)

Explanation:

7 0
2 years ago
Read 2 more answers
What is CGS and SI unit of weight?​
Marta_Voda [28]

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The system of measurements taking the centimeters ,gram and second as its fundamental units , now replaced by the SI unit System

4 0
3 years ago
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
Plz i need help please help​
Gemiola [76]

Answer:

if you look it up i think u can find it it could be a mealltiod  Co 27

Explanation:

3 0
2 years ago
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