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: option c.

explanation:

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards