Answer:
n = 756.25 giga electrons
Explanation:
It is given that,
If the charge on the negative plate of the capacitor, 
Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

e is the charge on electron

or
n = 756.25 giga electrons
So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.
Answer:


Explanation:
The Newton's law in this case is:

Here,
is the air temperture, C and k are constants.
We have
in
So:

And we have
in
, So:

Now, we have:

Applying (1) for
:

Applying (1) for
:

It decreases because it gave its momentum to the other car.
Mimicry. This is when oraganisms imitate another oraganism whether it be physical or behavoioral.
Answer:
B = 0.546 T, F = 2.59 10⁻¹² N
Explanation:
The magnetic force is
F = q v x B
We can calculate the magnitude of the force and find the direction by the right hand rule
F = q v B sin θ
Let's use Newton's second law
F = m a
Acceleration is centripetal
a = v² / r
We substitute
q v B sin θ = m v² / r
The angle between the field and the radius of the circle is 90º so sin 90 = 1
q B = m v / r
B = m v / q r
Let's calculate ’
B = 1.67 10⁻²⁷ 2.97 10⁷ / (1.60 10⁻¹⁹ 0.568)
B = 0.546 T
The foce is
F = q v B
F = 1.60 10⁻¹⁹ 2.97 10⁷ 0.546
F = 2.59 10⁻¹² N