Answer:
the force would increase 4 times more
Explanation
more force results more mass or acceleration
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
Answer:
29.412m/s
Explanation:
where F= force, m= mass, and a=acceleration
we also know that,
a = Δv / t where Δv = change in velocity and t = time
thus F = m ( Δv / t)

Δv
29.412m/s=Δv
The magnitude of the electric field at the proton's location is 10,437.5 N/C.
<h3>What the magnitude of the
electric field?</h3>
The size of the electric field is basically characterized as the power per charge on the test charge. On the off chance that the electric field strength is meant by the image E. Very much like gravity, electric fields work the same way. In any case, while gravity generally draws in, an electric field, then again, can either rebuff or draw in. By and large, the Electric Field submits to the super-position guideline. the all out Electric Field from various charges is equivalent to the amount of the electric fields from each charge separately. An electric field is the actual field that encompasses electrically charged particles and applies force on any remaining charged particles in the field, either drawing in or repulsing them.
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