Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach the surface of a 15-fmfm-diameter 238U238U nucleus

Explanation:

GIVEN

diameter = 15 fm  =m

we use here energy conservation

there will be some initial kinetic  energy but after collision kinetic energy will zero

on solving these equations we get kinetic energy initial

J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

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.