Complete question:
A uniform electric field is created by two parallel plates separated by a
distance of 0.04 m. What is the magnitude of the electric field established
between the plates if the potential of the first plate is +40V and the second
one is -40V?
Answer:
The magnitude of the electric field established between the plates is 2,000 V/m
Explanation:
Given;
distance between two parallel plates, d = 0.04 m
potential between first and second plate, = +40V and -40V respectively
The magnitude of the electric field established between the plates is calculated as;
E = ΔV / d
where;
ΔV is change in potential between two parallel plates;
d is the distance between the plates
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m
Answer:
Acceleration is the change in velocity divided by time
Explanation:
This is the correct answer because distance divided by time is the position. Speed multiplied by time is the distance. And acceleration is not just velocity, but the change in velocity over time.
Question: What is the frequency of a wave that has a wave speed of 120 m/s and a wavelength of 0.40 m?
Answer: The equation that relates frequency of a wave to a waves speed and wavelength is Speed of Wave= Frequency X Wavelength. Since you are given speed and wavelength, you plug those two known numbers into the equation, 120= Frequency X 0.40. You then divide 120 by .4 to get your frequency of 300.
Explanation: this might help for
