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 with explanation:
We are given that
Mass of ball,
75 g=
1 kg=1000 g
Height,

Horizontal velocity,
Mass of plate
a.Initial velocity of plate,
Velocity before impact=
Where 
Velocity after impact,
According to law of conservation of momentum

Substitute the values



Velocity of plate=1.69 m/s
b.Initial energy=
Final energy=
Final energy=
Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J
Answer:
Density = Mass/volume. D= 60/30.Divide it and you'll get ur answer as 2