Answer:
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Explanation:
please check your previose question
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.
So,
F = 
Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.
Let's suppose, q1 = q
So,
if q1 = q
then
q2 = Q-q
Product of Charges = q1 x q2
Now, it is:
Product of Charges = q x (Q-q)
So,
Product of Charges = qQ - 
And the expression qQ -
is clearly a quadratic expression. And clearly its roots are 0 and Q.
So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.
So, the midpoint is:
q =
q = Q/2 and it is the highest value of each charge in order to get the greatest force.
Explanation:
First, we will calculate the electric potential energy of two charges at a distance R as follows.
R = 2r
= 
= 0.2 m
where, R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.
U = 
= 
= 0.081 J
As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.
Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is,
.
Hence, K.E = 
= 
= 0.0405 J
Now, we will calculate the speed of balls as follows.
V = 
= 
= 0.142 m/s
Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.
Answer:
Change in Velocity because

Explanation:
Remeber area is length times Width. In this case, the area under a accleraton vs time graph is Accleration Times Time. Which is velocity
Hipparcos estimated its distance at roughly 96 parsecs from Earth, or 310 ± 20 light years away.