Answer:
8F_i = 3F_f
Explanation:
When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.
Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.
Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.
The electrostatic force, Fi, in the initial configuration can be calculated as follows.
![F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f](https://tex.z-dn.net/?f=F_i%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7Bq_Aq_B%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7BQ%5E2%7D%7Br%5E2%7D%5B%2Ftex%7D%3C%2Fp%3E%3Cp%3EThe%20electrostatic%20force%2C%20Ff%2C%20in%20the%20final%20configuration%20is%20%3C%2Fp%3E%3Cp%3E%5Btex%5DF_f%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7Bq_Aq_B%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B3Q%5E2%2F8%7D%7Br%5E2%7D%5B%2Ftex%7D%3C%2Fp%3E%3Cp%3ETherefore%2C%20the%20relation%20between%20Fi%20and%20Ff%20is%20as%20follows%3C%2Fp%3E%3Cp%3E%5Btex%5DF_i%20%3D%20F_f%5Cfrac%7B3%7D%7B8%7D%5C%5C8F_i%20%3D%203F_f)
1. circle graph
2. Bar graph
3. line graph
hope this helps
Answer:
angular speed of both the children will be same
Explanation:
Rate of revolution of the merry go round is given as
f = 4.04 rev/min
so here we have

here we know that angular frequency is given as



now this is the angular speed of the disc and this speed will remain same for all points lying on the disc
Angular speed do not depends on the distance from the center but it will be same for all positions of the disc
To calculate the initial velocity of the bike, we use the following equation
.
or

Here, u is initial velocity, v is final velocity, t is the time and d is the distance covered by bike.
Given,
,
and
.
Substituting these values in above equation, we get
.
Thus, the initial velocity of the bike is 1.2 m/s.
For a constant-velocity object, the average and instantaneous are the same. So the answer is no. It's like taking a running average of a string of numbers that are all the same number. The average is always the sum of the numbers divided by how many have accumulated, which will always equate to the repeated number.