Answer:
<h2>Magnitude of the second charge isExplanation:
According to columbs law;
F =
F is the attractive or repulsive force between the charges = 12N
q1 and q2 are the charges
let q1 = - 8.0 x 10^-6 C
q2=?
r is the distance between the charges = 0.050m
k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²
On substituting the given values
12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²
Cross multiplying
Explanation:
Given that,
Charge 1,
Charge 2,
The distance between charges, r = 1.99 m
To find,
The electrostatic force and its nature
Solution,
(a) The electric force between two charges is given by :
(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.
Therefore, this is the required solution.
Answer:
A. 1.4 m/s to the left
Explanation:
To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:
where:
M = momentum [kg*m/s]
M = m*v
where:
m = mass [kg]
v = velocity [m/s]
where:
m1 = mass of the basketball = 0.5 [kg]
v1 = velocity of the basketball before the collision = 5 [m/s]
m2 = mass of the tennis ball = 0.05 [kg]
v2 = velocity of the tennis ball before the collision = - 30 [m/s]
v3 = velocity of the basketball after the collision [m/s]
v4 = velocity of the tennis ball after the collision = 34 [m/s]
Now replacing and solving:
(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)
1 - (0.05*34) = 0.5*v3
- 0.7 = 0.5*v
v = - 1.4 [m/s]
The negative sign means that the movement is towards left