Answer:
Fnet = F√2
Fnet = kq²/r² √2
Explanation:
A exerts a force F on B, and C exerts an equal force F on B perpendicular to that. The net force can be found with Pythagorean theorem:
Fnet = √(F² + F²)
Fnet = F√2
The force between two charges particles is:
F = k q₁ q₂ / r²
where
k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.
If we say the charge of each particle is q, then:
F = kq²/r²
Substituting:
Fnet = kq²/r² √2
To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.
Here,
M = Mass
R = Radius of the hoop
The precession frequency is given as
Here,
M = Mass
g= Acceleration due to gravity
d = Distance of center of mass from pivot
I = Moment of inertia
= Angular velocity
Replacing the value for moment of inertia
The value for our angular velocity is not in SI, then
Replacing our values we have that
The precession frequency is
Therefore the precession period is 5.4s