There's nothing mysterious about it at all. "Frequency" simply means
"often-ness" ... how often or how frequently something happens.
-- The frequency of traditional meals is 3 per day.
-- The frequency of an equinox is 2 per year.
-- The frequency of my sleeping really late is 1 per week.
-- The frequency of my intense desire to sleep late is 30 per month.
etc.
-- The standard unit of frequency in the SI system is "per second".
The special name for that unit is "Hertz". (Hz)
Answer:
T = 0.017s
Explanation:
period is the time it takes a particle to make one oscillation
An electric current is periodic in nature
The current reaches 3.8A ten times.
So there must have been 10 cycles (10 periods) in 0.17s. let 'T' be the period:

t is the total time interval
n is the number of oscillations

10T = 0.17
T = 0.17/10 = 0.017s
This problem uses the relationships among current
I, current density
J, and drift speed
vd. We are given the total of electrons that pass through the wire in
t = 3s and the area
A, so we use the following equation to to find
vd, from
J and the known electron density
n,
so:

<span>The current
I is any motion of charge from one region to another, so this is given by:
</span>

The magnitude of the current density is:

Being:

<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd