A parallel-plate capacitor is charged by a 14.0 V battery, then the battery is removed. You may want to review (Pages 692 - 695)
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As per the question the mass of two falling sky drivers is 132 kg.
First we have to calculate their acceleration.
Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.
The earth pulls the object with a force equal to the weight of the body.
Hence the force gravity F=W= mg [ here m is mass of the body]
Here m =132 kg.
Hence force of gravity F= mg
=132 kg ×9.8 m/s^2
=1293.6 kg m/s^2
=1293.6 N [ here N[newton] is the unit of force.]
As per the question the air resistance is one fourth of weight of the bodies.
Hence air resistance F' =1/4 mg
Here F acts in vertically downward direction while F' acts in vertically upward direction.
Hence the net force acting on the particle is F-F'.
From Newton's second law of motion we know that net force is the product of mass and acceleration i.e
[Here a is the acceleration]
In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.
we know that F= ma
=m×0
=0 N
Hence they will not get any force when they will descend with a uniform speed.