The velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
<h3>What is Electric field?</h3>
Electric field is the physical field that surrounds a charge.
<h3>How to find final velocity of the electron when it moves some distance in a certain electric field?</h3>
- From Newton's second law, the acceleration the electron will be
a=F/m=qE/m
- where q= charge of electron
E= electric field
m= mass of electron
=(−1.60×10^−19C)(3×10³N/C)/(9.11×10^-31kg)
=10¹⁵×0.526m/s²
- The kinematics equation v²=v0²+2a(Δx)
- where v=final velocity of the electron
v0=initial velocity of the electron =5×10⁶m/s
a=acceleration of the electron =10¹⁵×0.526m/s²
Δx=distance moved by the electron in east direction =1cm=10^-2m
- Now v^2=(5×10⁶)²+2×10¹⁵×0.526×10^-2
=25×10¹²+10.52×10¹²
=35.52×10¹²
- Now velocity of electron=5.95×10⁶m/s.
Thus , we can conclude that the velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
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F - False.
Its greatest kinetic energy is at the point of release.
It has the least kinetic energy, zero, at its highest point in its path.
Answer:
The net force = 0
Explanation:
The given information includes;
The mass of the crate = 250 kg
The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)
In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.
The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate
The weight of the crate, 
↓ = 250 kg × 9.81 m/s² = 2,452.5 N
The force the helicopter should provide to just lift the crate, 
↑ = The weight of the crate = 2,452.5 N
The net force, 
= ↑ - ↓ = 2,452.5 N - 2,452.5 N = 0
The net force = 0.
F= ma; a= F/m
a = 26.4 N/60 kg= 0.44 m/s^2
