Given the equation for the Speed of a Satellite
v = SqRt{Gravitational Constant}{Mass of Earth} divided by the radius given in your problem
we have:
(square root whole term on right side)
v = G Me
———
r
so. (6.67x10^-11)(5.97x10^24)
___________________
(8.0x10^6)
v = 7055 m/s (which is reasonable)
so utilize the Kinetic Energy Formula
KE = 1/2mv^2
KE = 1/2(200)(7055)^2
KE = 4.977x10^9 J
<span>electromagnetic.........</span>
The answer is 1,600 J.
A work (W) can be expressed as a product of a force (F) and a
distance (d):
W = F · d<span>
We have:
W = ?
F = 20 N = 20 kg*m/s</span>²
d = 80 m
_____
W = 20 kg*m/s² * 80 m
W = 20 * 80 kg*m/s² * m
W = 1600 kg*m²/s²
W = 1600 J
Answer:
10 feet.
Explanation:
Conductors are considered outside the building when they installed. The requirement of NEC is above 10 feet for the clearance after the final grade for service conductors and cables where they are connected with the building, above the sidewalks and the to other areas which are accessible to the pedestrians.
