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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.
If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.
The graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.
- A student could use the graph to determine the net work done on the object by Calculating the area bound by the line of best fit and the horizontal axis from 0m to 5m
For more information on work done, visit
Answer:
mass of box 1 = 2.20 kg
mass of box 2 = 5.93 kg
Explanation:
Let the mass of box 1 and box 2 is respectively
and
so we will have
Force applied on box 1 then acceleration
Now we know that contact force between them in above case is given as
now we have