how does the electric force between two charged particles change if the distance between them is increased by a factor of 3?
a. it is reduced by a factor of 3
<h2>
Speed with which it return to its initial level is 100 m/s</h2>
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 100 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = ?
Displacement, s = 0 m
Substituting
v² = u² + 2as
v² = 100² + 2 x -9.81 x 0
v² = 100²
v = ±100 m/s
+100 m/s is initial velocity and -100 m/s is final velocity.
Speed with which it return to its initial level is 100 m/s
Answer:
has units of distance
has units of distance over time
has units of distance over 
has units of distance over 
Explanation:
Since the expression for the distance is:

then:
has units of distance
has units of distance over time
has units of distance over 
has units of distance over 
because we are supposed to be able to add all of the terms and get a distance. So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.