Answer:
A. 5.4 * 10^(-4) m
B. 500V
Explanation:
A. Electric potential, V is given as:
V = kq/r
This means that radius, r is
r = kq/V
r = (9 * 10^9 * 30 * 10^(-12))/500
r = (270 * 10^(-3))/500
r = 5.4 * 10^(-4) m
B. Now the radius is doubled and the charge is doubled,
V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)
V = 500V
The two situations are similar because in both you are trying to minimize the damage and make the best out of a bad situation
6.0 m longer because the player ran 3 and came back 3 at the very end, which looks like he went nowhere but in reality he ran 6.
Given:
Distance = 50 yard = 45.72 meter
Speed = 40 km/hr = 11.11 m/s
To find:
Time required by ball to reach the receiver = ?
Formula used:
speed = 
Solution:
The speed of the ball is given by,
speed = 
Thus,
Time = 
Distance = 50 yard = 45.72 meter
Speed = 40 km/hr = 11.11 m/s
Time = 4.12 second
Hence, ball reaches the receiver in 4.12 second.
Answer:

Explanation:
Given that,
The radius of sphere, r = 5 cm = 0.05 m
Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C
We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

So, the surface charge density on the sphere is
.