A proton moves through a magnetic field at 21.5% of the speed of light. At a location where the field has a magnitude of 0.00627 T and the proton's velocity makes an angle of 147∘ with the field, what is the magnitude FB of the magnetic force acting on the proton? Use c=2.998×108 m/s for the speed of light and e=1.602×10−19 C as the elementary charge.

The charge of an ion with 55 protons and 52 electrons is +3. the reason is because protons are positive, and there are three more than the amount of negative electrons.

Well the magnitude of its acceleration related to that car starts with the oil change like when you have to change the oil in your car you have to make sure that your precious car is starting really propley and very well

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W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j