For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:

The second equation is going to the total force acting on the charge q3:

k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:

Let's solve this for r_1^2:

Now we have a quadratic equation with following parameter:

We know that two solutions are:

We need a positive solution. When we plug in all the numbers we get:
1) In the reference frame of one electron: 0.38c
To find the relative velocity of one electron with respect to the other, we must use the following formula:

where
u is the velocity of one electron
v is the velocity of the second electron
c is the speed of light
In this problem:
u = 0.2c
v = -0.2c (since the second electron is moving towards the first one, so in the opposite direction)
Substituting, we find:

2) In the reference frame of the laboratory: -0.2c and +0.2c
In this case, there is no calculation to be done. In fact, we are already given the speed of the two electrons; we are also told that they travel in opposite direction, so their velocities are
+0.2c
-0.2c
Answer:
His final velocity is 15.8 m/s.
Step-by-step explanation:
Given:
Initial velocity of the driver is,
m/s
Acceleration of the driver is,
m/s²
Time taken to reach final velocity is,
s.
The final velocity is given using the Newton's equations of motion as:
, where,
is the final velocity.
Now, plug in the given values and solve for
.

Therefore, his final velocity is 15.8 m/s.
Check the current weather map for 2 different times, and see where the center of the storm is. That tells you what direction it's moving. With its speed and direction, you have its velocity.
Answer:
√(6ax)
Explanation:
Hi!
The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:
x=(1/2)at^2
Then the we need to find the time t' for which the displacement is 3x
3x=(1/2)a(t')^2
Solving for t':
t'=√(6x/a)
Now, the velocity of the motorcycle as a function of time is:
v(t)=a*t
Evaluating at t=t'
v(t')=a*√(6x/a)=√(6*x*a)
Which is the final velocity
Have a nice day!