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
Given:
u = 0, initial speed (sprinter starts from rest)
v = 11.5 m/s, final speed
s = 15 m, distance traveled to attain final speed.
Let
a = average acceleration,
t = time taken to attain final speed.
Then
v² = u² + 2as
or
(11.5 m/s)² = 2*(a m/s²)*(15 m)
a = 11.5²/(2*15) = 4.408 m/s²
Also
v = u +a t
or
(11.5 m/s) = (4.408 m/s²)*(t s)
t = 11.5/4.408 = 2.609 s
Answer:
The average acceleration is 4.41 m/s² (nearest hundredth).
The time required is 2.61 s (nearest hundredth).
It is 20 voltage per hour
Answer:
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