Answer:
The force of the impact would be smaller
Explanation: Examples:
If the force is big then the time would be small (2500N of Force = 10 seconds)
If the force is small then the time would be big (250N of Force = 50 seconds)
Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum
Density
is a value for mass, such as kg, divided by a value for volume, such as m3.
Density is a physical property of a substance that represents the mass of that
substance per unit volume. It is a property that can be used to describe a
substance.<span> </span><span>It has standard units of
kg/m^3 or g/mL.
So, the best answer is option C.</span>
The coefficient of static friction between the puck and the surface.
In fact, that coefficient describes exactly how "hard" it is to cause the puck to start moving, if it starts from an idle condition.
Answer:
Part a)

Part b)

Explanation:
Part a)
Electric field due to large sheet is given as


now the electric field is given as


Now acceleration of an electron due to this electric field is given as



Now work done on the electron due to this electric field



So work done is given as



Part b)
Now we know that work done by all forces = change in kinetic energy of the electron
so we will have


