Answer:
The electrical force between the given charges remains the same.
Explanation:
The expression for the electrical force is as follows as;
Here, k is the constant, are the charges, F is the electrical force and R is the distance between the charges.
It is given in the problem that the magnitudes of the charges and the magnitudes of the separation between the charges are doubled.
Then, the expression of the electrical force becomes as;
Therefore, the electrical force between the given charges remains the same.
Cleaning your area and wearing safety gear
Answer: F = 3.28 × 10^-7 N
Explanation:
Alpha particle charge (q)= 2e
e = 1.6 × 10^-19
q = 2e = 2 × (1.6 *10^-19) = 3.2 × 10^-19
1/4πEo = 9 × 10^9
Distance(r) = 5.3 × 10^-11m
The force of attraction between the two particles is given by:
F = (1/4πEo) (q1q2 / r^2)
F = [(9 × 10^9) (3.2 × 10^-19) (3.2 × 10^-19)] / (5.3 × 10^-11)^2
F = (92.16 × 10^(9 - 19 - 19)) / 28.09 × 10^-22
F = (92.16 × 10^-29) / 28.09 × 10^-22
F = 3.28 × 10^(-29 + 22)
F = 3.28 × 10^-7 N
Answer:
The correct order of events over time
Answer:
Explanation:
first mass, m1
second mass, m2
initial velocity of m1 is v1.
initial velocity of m2 is zero.
Let the final velocity of m1 is v' and the final velocity of m2 is v''.
Collision is elastic so the coefficient of restitution is 1.
By using the conservation of momentum
m1 x v1 + m2 x 0 = m1 x v + m2 x v''
m1 v1 = m1 v' + m2 v'' .... (1)
by the formula of coefficient of restitution
v1 = v'' - v' .... (2)
So, v' = v'' - v1 put in equation (1)
m1 v1 = m1 x (v'' - v1) + m2 v''
m1 v1 = m1 v'' - m1 v1 + m2 v''
2 m1 v1 = (m1 + m2) v''
so,