One end of a steel rod of radius R = 10 mm and length L = 80cm is held in a vise. A force of magnitude F = 60kN is then applied
1 answer:
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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.
Answer:
Inductive reactance is 125.7 Ω
Explanation:
It is given that,
Inductance,
Voltage source, V = 15 volt
Frequency, f = 400 Hz
The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :
So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.