<span>the speed of a direction</span>
Answer:
Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;
The closest (minimum) distance of Neptune from the Earth is 29 AU
The farthest (maximum) distance of Neptune fro the Earth is 31 AU
Explanation:
The following parameters are given;
The distance from the Earth to the Sun = 1 AU
The distance of Neptune from the Earth = 30 AU
We have;
When the Sun is between the Earth and Neptune, the distance is found by the relation;
Distance from the Earth to Neptune = 30 + 1 = 31 AU
When the Earth is between the Sun and Neptune, the distance is found by the relation;
Distance from the Earth to Neptune = 30 - 1 = 29 AU
Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU
The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.
Assuming the driver starts slamming the brakes immediately, the car moves by uniformly decelerated motion, so we can use the following relationship

(1)
where
a is the deleceration
S is the distance covered after a time t

is the velocity at time t

is the initial speed of the car
The accident is 80 m ahead of the car, so the minimum deceleration required to avoid the accident is the value of a such that S=80 m and

(the car should stop exactly at S=80 m to avoid the accident). Using these data, we can solve the equation (1) to find a:

And the negative sign means it is a deceleration.
Answer:
<em>-2 units of charge</em>
Explanation:
charge on A = Qa = -6 units
charge on B = Qb = 2 units
if the spheres are brought in contact with each other, the resultant charge will be evenly distributed on the spheres when they are finally separated.
charge on each sphere will be = 
charge on each sphere =
=
= <em>-2 units of charge</em>