For radio broadcasting, in electricity meters, in any generator.
Answer:
887.1Hz
Explanation:
Given parameters:
Speed of sound wave = 330m/s
Wavelength = 0.372m
Unknown:
Frequency = ?
Solution:
To solve this problem, we use the expression below:
Speed = Frequency x wavelength
330 = Frequency x 0.372
Frequency = 887.1Hz
Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive
Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:
Replacing the dat we obtain F=82 nN.
The force is repulsive because the points charged have the same sign.
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 .