Complete question:
At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.
[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]
Answer:
The time it will take the particle to pass through point (P) again is 1.639 ns.
Explanation:
F = qvB
Also;

solving this two equations together;

where;
m is the mass of electron = 9.11 x 10⁻³¹ kg
q is the charge of electron = 1.602 x 10⁻¹⁹ C
B is the strength of the magnetic field = 3.47 x 10⁻³ T
substitute these values and solve for t

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.
Answer:
Change in potential energy = 7350 Joules
Explanation:
It is given that,
Side of cube, a = 0.5 m
Density of cube, 
The cube is lifted vertically by a crane to a height of 3 m
We know that, density 
So, m = d × V (V = volume of cube = a³)

m = 250 kg
We have to find the change in potential energy of the cube. At ground level, the potential energy is equal to 0.
Potential energy at height h is given by :

PE = 250 kg × 9.8 m/s² ×3 m
PE = 7350 Joules
So, change in potential energy of the cube is 7350 Joules.
Mass of object/source
Gravity
Fg=mg
Answer:
Rotational inertia decreases proportional to the decrease in the radius of rotation.
Explanation: