A 71 kW radio station broadcasts its signal uniformly in all directions. - What is the average intensity of its signal at a dist
1 answer:
Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
=
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.
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Answer:
The image distance is 20.0 cm.
Explanation:
Given that,
Power = 1.55 dp
Distance between book to eye = 26.0+3.00=29.0 cm
We need to calculate the focal length
Using formula of focal length
Put the value into the formula
We need to calculate the image distance
Using lens formula
Put the value into the formula
Hence, The image distance is 20.0 cm.
Let us consider body moves a distance S due to the force F.
Hence the work by the body W = FS
If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -
where
is the component of the force along the direction of displacement.
As per the question the power P is given as -
Hence alternative definition of power P = F.V
The gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.
<h3>What is the gravitational force of the earth on the person?</h3>The gravitational force exerted by the earth on a person standing on the earth's surface is given below as follows:
where
G = 6.67 * 10⁻¹¹
m¹ = 62 kg
m² = 5.97 * 10²⁷ kg
r = 6.4 * 10⁶ m
Therefore, the gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.
Learn more about gravitational force at: brainly.com/question/940770
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