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mezya [45]
3 years ago
13

An amplitude modulation transmitter radiates 10 KW power with the modulation percentage of 75 %. Find the magnitude of the carri

er power?
Physics
1 answer:
Lostsunrise [7]3 years ago
7 0

Answer:

carrier power is 7.8 kW

Explanation:

given data

power = 10 kW

modulation percentage = 75 %

to find out

carrier power

solution

we will use here power transmitted equation that is

power = carrier power * ( 1+  \frac{modulation}{2})   .................1

put here value in equation 1  we get carrier power

10 = carrier power *(1+ \frac{0.75}{2})

carrier power = 7.8

so carrier power is 7.8 kW

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When someone yells in it
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URNOW Lidl llIIS 15 TIUta WiCluie U water!<br> ify-What two kinds of matter are pure substances?
Arada [10]

Elements and compounds

Explanation:

  • All pure substances are distinct substances.
  • All their parts are the same throughout i.e they are homogeneous.
  • They have a definite composition.
  • They cannot be easily separated or broken down into simpler substances by physical means.
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Elements are distinct substances that cannot be split up into simpler ones.

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Learn more:

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6 0
3 years ago
A projectile is launched with an initial velocity 60m/s at an angle 60° to the vertical. What the magnitude of it's displacement
emmasim [6.3K]

Answer:

the magnitude of the displacement after 5s is 137.31 m.

Explanation:

Given;

initial velocity of the projectile, u = 60 m/s

angle of projection, θ = 60°

time of motion, t = 5s

the vertical component of the velocity, u_y= u\ sin \theta = 60sin(60^0)

The magnitude of the displacement after 5s is calculated as;

h = u_yt -\frac{1}{2} gt^2\\\\h = 60sin (60^0)\times 5 - \frac{1}{2} (9.8)(5)^2\\\\h = 259.81-122.5\\\\h = 137.31 \ m

Therefore, the magnitude of the displacement after 5s is 137.31 m.

3 0
3 years ago
Help ASAP PLEASE THANKS
seraphim [82]

Answer:

a

Explanation:

5 0
3 years ago
Read 2 more answers
An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11
Arada [10]

Answer:

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

8 0
3 years ago
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