If the distance between two charges is halved, the electrical force between them increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:
![F= k \frac{q_1 q_2}{r^2}](https://tex.z-dn.net/?f=F%3D%20k%20%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D%20)
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
![r'= \frac{r}{2}](https://tex.z-dn.net/?f=r%27%3D%20%5Cfrac%7Br%7D%7B2%7D%20)
the magnitude of the force changes as follows:
![F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k \frac{q_1 q_2}{r^2}=4 F](https://tex.z-dn.net/?f=F%27%3Dk%20%5Cfrac%7Bq_1%20q_2%7D%7Br%27%5E2%7D%3Dk%20%5Cfrac%7Bq_1%20q_2%7D%7B%28%20%5Cfrac%7Br%7D%7B2%7D%29%5E2%20%7D%3Dk%20%5Cfrac%7Bq_1%20q_2%7D%7B%20%5Cfrac%7Br%5E2%7D%7B4%7D%20%7D%20%3D4k%20%20%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D%3D4%20F%20%20%20)
so, the force increases by a factor 4.
The work is path independent since we have a conservative force.
Thus
Answer (1)
Answer: Gravity
Explanation: Gravity is pulling down on the ball, making it stay on the floor
Answer:
2.32 s
Explanation:
Using the equation of motion,
s = ut+g't²/2............................ Equation 1
Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.
Note: Since Onur drops the basket ball from a height, u = 0 m/s
Then,
s = g't²/2
make t the subject of the equation,
t = √(2s/g')...................... Equation 2
Given: s = 10 m, g' = 3.7 m/s²
Substitute this value into equation 2
t = √(2×10/3.7)
t = √(20/3.7)
t = √(5.405)
t = 2.32 s.