Explanation:
It is given that,
Power of EM waves, P = 1800 W
We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.
Intensity,
The formula that is used to find the rms value of the electric field is as follows :
c is speed of light and 
is permittivity of free space
So,
Hence, this is the required solution.
Let x = the angle of elevation for shooting the arrow.
Assume
g = 9.8 m/s²
No wind resistance
The vertical launch velocity is 25.1 sin(x) m/s
The horizontal velocity is 25.1 cos(x) m/s
The time of flight is
24/[25.1 cos(x)] s = 0.9562 sec(x) s
Therefore
0.5*[0.9562 sec(x)]*(9.8) = 25.1 sin(x)
4.6854 = 25.1* sin(x)cos(x)
sin(2x) = 0.3733
2x = sin⁻¹ 0.3733 = 21.92 deg
x = 10.96 deg
Answer: 11 degrees (nearest integer)
Surrounding every charge (or group of charges) is a thing, called an electric field. (it is a vector thing). E. Definition: The electric field at a point. E in empty space is a vector quantity which can be measured by the following procedure: place a small test charge q at that point, measure the force on q due to all other charges.
8 electron are needed for bonding
Answer:
it is 12.15 km
Explanation:
1215 km ÷1000
=12.15
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