If do 50 J of Work in 20 seconds, what is my Power ?
1 answer:
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Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² =
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
Now, plug in for 'σ' and
for
and solve for the electric field. This gives,
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
1) 3 miles/Hour
The speed is defined as the distance covered divided by the time taken:
where
d = 1.5 mi is the distance
t = 0.5 h is the time taken
Substituting,
2) 1.34 m/s south
Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:
is the displacement in meters
is the time taken in seconds
Substituting,
And the direction of the velocity is the same as the displacement, so it is south.