Answer:
clockwise and counter clockwise
it depends wich way you start paddling
The coldest part of the atmosphere is the mesosphere
The potential at a distance r from a charge Q is given by

where ke is the Coulomb's constant.
The charge in our problem is

; for the point at

, the potential is

For the point at infinity, we immediately see that the potential is zero, because

and so

.
Therefore, the potential difference between the two points is
Answer:
4.53 second, 195 ft/s
Explanation:
u = 50 ft/s
h = 555 ft
g = 32 ft/s^2
Let the time taken by the ball to reach the ground is t and the velocity of the ball as it hits the ground is v.
Use third equation of motion


v = 195 ft /s
Thus, the ball strikes the ground with velocity 195 ft/s.
Use first equation of motion
v = u + at
195 = 50 + 32 t
32 t = 145
t = 4.53 second
Thus, the ball reach the ground in 4.53 second.