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:
Uh No thanks but make me brainiest!
Explanation:
Velocity is a vector. Therefore, it depends on the direction. Pilots need to know the direction of wind, not just the speed. If the pilot is going South, and there's 5 mph wind going South, they'll be happy, but if the wind is going 5 mph North, they'll be going against the wind.
Answer:
6318 N
Explanation:
From the question given above, the following data were obtained:
Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²
Mass (m) of container = 650 kg
Weight (W) of container on the earth =.?
Next, we shall determine the acceleration due to gravity of the earth. This can be obtained as follow:
Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²
Acceleration due to gravity of the earth (gₑ) =.?
gₘ = 1/6 × gₑ
1.62 = 1/6 × gₑ
1.62 = gₑ /6
Cross multiply
gₑ = 1.62 × 6
gₑ = 9.72 m/s²
Finally, we shall determine the weight of the container on the earth as follow:
Mass (m) of container = 650 kg
Acceleration due to gravity of the earth (gₑ) = 9.72 m/s²
Weight (W) of container on the earth =.?
W = m × gₑ
W = 650 × 9.72
W = 6318 N
Therefore, the weight of the container on earth is 6318 N