Particles q1 = +8.0 UC, 92 = +3.5 uc, and
2 answers:
Answer:
F_total = 29.4 N, directed to the right of particle 2
Explanation:
We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.
Let's use Coulomb's law to calculate each force
F =
particles 1 and 2
q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m
F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²
F₁₂ = 2.59 10¹ N
Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.
particles 2 and 3
q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m
we calculate
F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²
F₂₃ = 3.5 N
as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2
Now we add the forces as vectors
F_total = ∑ F = F₁₂ + F₂₃
F_total = 25.2 +3.5
F_total = 29.4 N
directed to the right of particle 2
You might be interested in
To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.
The angular velocity can be described as
Where,
Final Angular Velocity
Initial Angular velocity
Angular acceleration
t = time
The relation between the tangential acceleration is given as,
where,
r = radius.
PART A ) Using our values and replacing at the previous equation we have that
Replacing the previous equation with our values we have,
The tangential velocity then would be,
Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation
Replacing with our values and re-arrange to find
That is equal in revolution to
The linear displacement of the system is,