Ok, I think this is right but I am not sure:
Q = ϵ
0AE
A= π π
r^2
=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons
Answer:

Explanation:
For answer this we will use the law of the conservation of the angular momentum.

so:

where
is the moment of inertia of the merry-go-round,
is the initial angular velocity of the merry-go-round,
is the moment of inertia of the merry-go-round and the child together and
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:



Finally we replace all the data:

Solving for
:

Answer:
the two factors are the mass of the objects and the coefficient of friction between them
Explanation:
internet :)
Answer:
As Per Provided Information
Moving body has 2m/s² acceleration
Time taken by body is 4 second
We are asked to find the 'change in velocity' ( ∆V) by the body.
<u>Formula Used here</u>

<u>Substituting </u><u>the </u><u>given </u><u>value</u>
<u>
</u>
<u>Therefore</u><u>,</u>
- <u>Change </u><u>in </u><u>velocity </u><u>is </u><u>8</u><u> </u><u>m/</u><u>s</u>
Answer:
Explanation:
For the first case , the expression for electrostatic force can be given by the following .
F = K x 8Q x 2Q / r² where k is a constant .
F = K 16 Q² / r²
When they touch , some charge is neutralized . Net charge remaining
= 8Q - 2 Q = 6 Q
Charge on each sphere = 6Q/2 = 3 Q .
Force between them
F₁ = k 3Q x 3 Q / r² = k 9 Q² / r²
F₁ / F = 9 / 16
F₁ = 9 F / 16 .