Complete question:
The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge -Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 x 10⁻³ m², and the spacing between the plates is 6 x 10⁻³ m. There is no dielectric between the plates. What is the charge on the capacitor?
Answer:
The charge on the capacitor is 1.77nC
Explanation:
Given;
magnitude of electric field between the plates, E = 100 kV/m
Area of each plate, A = 2 x 10⁻³ m²
Distance between the plates, d = 6 x 10⁻³ m
Charge on the capacitor is calculated as;
Q = CV
V = Ed
Therefore, the charge on the capacitor is 1.77nC
The force exerted on the bottle is 532 N
Explanation:
When the air is removed from inside the bottle, vacuum is created, therefore the external pressure (the atmospheric pressure) is no longer balanced, and it creates a net force downward on the can.
The pressure is related to the force by the equation:
where
p is the pressure
F is the force
A is the area of the can
Here we have:
is the atmospheric pressure
r = 0.0410 m is the radius of the can, so the area is
And solving for F, we can find the force on the can:
Learn more about pressure:
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The net force with be 10 N east. As both boys are pushing a wagon to the same direction, you just add the numbers together. If the directions were opposite like if it was the boy pushing 6 N east and the other boy pushing 4 N west...then it would come out as 2 N east (you subtract and go to the direction with the biggest number) if you have any confusions, feel free to ask
Nitrogen
oxygen
argon
carbon dioxide
I assume L=120 yards as the length of the football field.
1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is
And the time taken is t=22.4 s, so the average speed of the player is
2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is
and so, the average velocity is
