N, or Neutons are a common way to measure force
Apply Newton's second law to the person's motion:
F = ma
F = net force, m = mass, a = acceleration
Given values:
m = 50.8kg, a = 3.50m/s²
Plug in and solve for F:
F = 50.8(3.50)
F = 178N
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:
105 m/s
Explanation:
Given that the speed of train A,
= 45 m/s from west to east.
Speed of train B,
= 60 m/s from east to west.
Train B is moving in the opposite direction with respect to the speed of train A. Assuming that the speed from east to west direction is positive.
So, the speed of train A from east to west= - 45 m/s
The speed of train B w.r.t train A
m/s
Hence, the speed of train B w.r.t train A is 105 m/s from east to west.