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
Answer:
2, High mass stars.
Add-on:
i hope this helped at all. im sorry if its incorrect.
Answer:
Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.
P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0
Explanation:
P1 = P2 = atmospheric pressure so, P1 - P2 = 0
h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²
From the continuity equation for fluids
A1v1 = A2v2
v2 = A1v1/A2
Substituting into the equation above
(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho
Making A2² the subject of the formula,
A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}
The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.
Thank you for reading.
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