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:
ФE = 9.403W
Explanation:
In order to calculate the magnitude of the electric flux trough the sheet, you use the following formula:
(1)
A: area of the rectangular sheet = (0.400m)(0.600m) = 0.24m^2
E: magnitude of the electric field = 95.0N/C
θ: angle between the direction of the electric field and the normal to the surface of the sheet
You replace the values of the parameters in the equation (1):
The magnitude of the electric flux is trough the sheet is 9.403W
Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
Answer:
E0
Explanation:
Yes. The "0" indicates that it is spherical
Answer:
Explanation:
The remaining amount of a radioactive isotope is found as the product of the original amount by (1/2) raised to the number of half-lives elapsed. The formla is:
Where M is the remaining amoun, M₀ i s the initial amount, and n is the number of half-lives.
Here, 1/64 of the original carbon-14 remained, meaning that M/M₀ = 1/64.
Then, you can substitute in the equation and solve:
Then, 6 half-lives elapsed since the mastodon died and the remains were dated.
Then, you must multiply 6 by the <em>half-life </em>time:
- 6 × 5730 years = 34,380 years ← answer
