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victus00 [196]
3 years ago
9

What are three types of electromagnetic waves that are used to transmit information? What types of devices are used to receive e

ach type of wave?
Type of electromagnetic wave:


Receiver devices:
Physics
1 answer:
MrMuchimi3 years ago
3 0

Answer:

Three types of electromagnetic waves, used to transmit various information

Explanation:

A form of energy waves having both electric & magnetic fields are Electromagnetic waves. Three types -

Radio Waves - These have longest wavelengths & transmit data through radio, satellites, radar .

Micro Waves - These have shorter wavelengths & are used in cooking appliances & predicting weather.

X rays - These have more short wavelength and can penerate soft tissues like skin & muscle, hence are used for medical examining

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if jane is trying to escape a dragon and runs 34m north then must run 38m east, what total distance does jane run
Andrej [43]
You would just add them, 34+38=72m
4 0
3 years ago
What is the magnitude of a point charge in coulombs whose electric field 58 cm away has the magnitude 2.8 N/C?
Yuliya22 [10]

Answer:

q = 1.05 × 10^-10 C

Explanation:

E = 2.8 N/C

r = 58cm = 0.58m

k = 9×10^9

q = ?

E = f/q

E = kq/r²

q = E*r²/k

q = (2.8×0.58²)÷9×10^9

q = 1.05 × 10^-10 C

8 0
3 years ago
The electric field strength is 5.50×10^4 N/C inside a parallel-plate capacitor with a 2.50 mm spacing. A proton is released from
valentina_108 [34]

Answer:

v=1.6\times10^{5}m/s

Explanation:

The equation that relates electric field strength and force that a charge experiments by it is F=qE.

Newton's 2nd Law states F=ma, which for our case will mean:

a=\frac{F}{m}=\frac{qE}{m}

We know from accelerated motion that v^2=v_0^2+2ad

In our case the proton is released from rest, so v_0=0m/s and we get v=\sqrt{2ad}

Substituting, we get our final velocity equation:

v=\sqrt{\frac{2qEd}{m}}

For the <em>proton </em>we know that q=1.6\times10^{-19}C and m=1.67\times10^{-27}kg. Writing in S.I. d=0.0025m, we obtain:

v=\sqrt{\frac{2(1.6\times10^{-19}C)(5.5\times10^{4}N/C)(0.0025m)}{(1.67\times10^{-27}kg)}}=162318.5m/s=1.6\times10^{5}m/s

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3 years ago
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Answer:

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6 0
3 years ago
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At a certain elevation, the ________ , the air becomes saturated and water-vapor molecules ________.
andriy [413]

At certain altitude, the temperature of air decrease, The air becomes saturated and water vapour molecules starts condensing.


As the altitude of air increase, the atmospheric pressure decrease due to which the temperature of the air decrease. The water molecules in the atmosphere start condensing, which saturate the air (that is air can no hold water molecules), due to which the water vapour molecules starts condensing and falls on the earth in the form of rain.


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4 years ago
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