I would say option D, it depends on the size of the star
Electric field = potential difference
-----------------------------
distance between plates
Distance between plates = 45
----------
500
= 0.09 meters.
Answer:
Hold on Ill answer it..When do u need it by
Explanation:
Answer:
The work done by the drag force is given by 29.96 J
Explanation:
Given :
Thrust force
N
Displacement
m
Mass of rocket
Kg
From work energy theorem,


Where
thrust work
gravitational work

After cutoff kinetic energy is converted into potential energy,

Put value of KE

Work done by drag force is given by,

J
Therefore, the work done by the drag force is given by 29.96 J