The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance 
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q =
= 
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .
The direction is the same as the direction the ball is moving in. Because of the rolling of the ball, the direction of movement of the surface of the ball is opposite the overall direction of the ball. Since friction will oppose the direction of movement of the surface of the ball, it is in the same direction as the net direction of movement of the ball.
Answer:
-3 m
Explanation:
Displacement is the final position minus the initial position.
Δx = x − x₀
Δx = -3 m − 0 m
Δx = -3 m
Answer:
Explanation:
= Tension force in string A = 403 N
= Tension force in string B = 800 N
= diameter of string A = 0.513 mm
= diameter of string B = 1.29 mm
= wave speed of string A
= wave speed of string B
Ratio of the wave speeds is given as
