Two charged dust particles exert a force of 9.0 ✕ 10-2 n on each other. what will be the force if they are moved so they are only one-ninth as far apart?

The new force would be F' = 7.29x10 N

Explanation: We know that the force between the charges is:

9.0x10^-2 N

Now, the force between two charged particles can be obtained by the Coulomb's law:

F = k*q1*q2/r^2

where k is a constant, q1 and q2 are the charges of the particles and r is the distance between the particles.

then we have that:

k*q1*q2/r^2 = 9.0x10^-2 N

if we move tha particles only one ninth as far apart, then we have a new force:

F´= k*q1*q2/(r/9)^2 = 181k*q1*q2/r^2 = 81*F = 81*9.0x10^-2 N = 729x10^-2

F' = 7.29x10 N

answer; the correct answer is technician a; ///100%good luck

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.