450 J / 3 s = 150 J/s = 150 watts.
The speed of the particle when it enters the magnetic-field region is 120,000 m/s.
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Speed of the particle</h3>
Magnetic force on the particle is given as;
F = qvB
where;
- q is magnitude of the charge
- v is speed of the particle
- B is magnetic field strength
From Newton's second law, force on an object is given as;
F = ma
where;
- m is mass of the particle
- a is acceleration
ma = qvB
v = ma/qB
v = (3 x 10⁻² x 12)/(3 x 10⁻⁶ x 1)
v = 120,000 m/s
Thus, the speed of the particle when it enters the magnetic-field region is 120,000 m/s.
Learn more about speed here: brainly.com/question/6504879
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Answer:
The potential energy of the electron in the field of the positive proton point charge is U(r) = -qeV(r) = - keqe2/r. The total energy is the sum of the electron's kinetic energy and its potential energy.
Explanation:
In each case the origin is located at the intersection of the axes. The electric potential from a single charge is defined to be zero an infinite distance from the charge, and the electric potential associated with two charges is also defined to be zero when the charges are infinitely far apart.
Answer:
It would be 2.2 if you have to round the five to the one but if your not rounding the number, it'd be 2.1.
Explanation: