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
.
Answer:
0.8214 m/s^2
Explanation:
Fnet= Fpushed - Ffriction
Fpushed = 12.7N Ffriction = 8.33N
Fnet = 12.7N - 8.33N = 4.37N
Fnet= mass(acceleration)
Fnet = 4.37N mass = 5.32 kg
4.37N = 5.32 kg(acceleration)
acceleration= 0.8214 m/s^2
Answer:
58.33 Hz
175 Hz
291.67 Hz
Explanation:
L = Length of tube = 1.5 m
v = Speed of sound in air = 350 m/s
The first resonant frequency is given by

The first resonant frequency is 58.33 Hz
The second resonant frequency is given by

The first resonant frequency is 175 Hz
The third resonant frequency is given by

The first resonant frequency is 291.67 Hz