Answer:
U2 = KAε0V2 / (2d)
Explanation:
The dielectric constant K just replaces the "3″ from Part B.
Answer:
False
Explanation:
Most comets are located outside the solar system, in part of the cloud that originated from dust and gas that has remained virtually untouchable for billions of years. The orbit of these comets can reach the order of a light year. Thus, they are called long-period comets.
Answer:
2.0 m/s/s
Explanation:
The acceleration of an object is the rate of change of velocity of the object.
Mathematically, it is given by:

where
u is the initial velocity
v is the final velocity
t is the time taken for the velocity to change from u to v
Acceleration is a vector, so it has both a magnitude and a direction.
For the runner in this problem, we have:
u = 0 is the initial velocity (he starts from rest)
v = 8.0 m/s is the final velocity
t = 4.0 s is the time taken
Substituting, we find

Answer:

Explanation:
The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

We have
. Thus, solving for t:

Neither set of choices is correct.
If the distance is tripled, then the forces decrease to
1/9 Fg. and. 1/9 Fe.
Note. When the objects are charged, the gravitational force Fg can almost always be ignored, since Fe is like 10^40 greater when the quantities of mass and charge are similar.