I believe that the most likely answer is conductivity.
The electrostatic forces between two charges are
F = (9 x 10⁹) x (charge-1) x (charge-2) / (distance between them)² .
The charges in this question are each (1 x 10⁻⁶ Coulomb).
Force = (9 x 10⁹) (1 x 10⁻⁶) (1 x 10⁻⁶) / (4.0)
= (9 x 10⁹) (1 x 10⁻¹²) / (4.0)
= 0.00225 Newton .
-- The force on each mass is the same.
Each mass is acted on by a force of 0.00225 newton,
directed either toward the other one or away from it.
-- If the charges both have the same sign, then the forces
are repulsive ... driving the masses apart.
-- If the charges have different signs, then the forces
are attractive ... drawing the masses together.
-- For as far as you've gone in copying this question ... asking
only for the forces between the charged objects ... you don't
need to know their masses, or about friction with the table, or
any of that stuff. None of that has any effect on the forces
between the charged masses.
You'll need that information when you get down to the next part
of the question, where it's probably going to ask you for the
acceleration of each mass due to the electrostatic force on it.
You know the force on each mass now. All you have to remember
is F = (mass) x (acceleration) , and grind it out.
Don't you dare post the next part on Brainly ! You have everything
you need to do it yourself now, and you'll learn a lot more that way.
<span>Let's make a few assumptions.
1. The paratrooper's lowest speed will be just prior to impact.
2. Since the jump was from a relatively low altitude, the paratrooper used a static line and the parachute should have opened almost immediately upon jumping.
So let's convert 100 mi/h to ft/s
100 mi/h * 5280 ft/mi / 3600 s/h = 146.67 ft/sec
Given the 1st assumption above, the MAXIMUM distance the paratrooper would have fallen would be 8 seconds at 146.67 ft/s, so
8 s * 146.67 ft/s = 1173.36 ft
The calculated distance is close to the jump distance, which agrees with both assumptions 1 and 2. So this account does seem reasonable.
Additionally, looking for the speed of a parachutist doing a freefall in the belly-to-earth position with arms and legs outspread, they will generally reach a terminal velocity of 120 mi/h which is slightly faster than the 100 mi/h in the article. This too is in agreement with the defective parachute flapping and causing some extra air resistance.</span>
