True is the correct answer
Answer:
D
Explanation:
From the formula of coulombs law F = Kq1q2/square of r, we can say the electric force is indirectly related to square of r
Acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units
Answer:In the case of two proton beams the protons repel one another because they have the same sign of electrical charge. There is also an attractive magnetic force between the protons, but in the proton frame of reference this force must be zero! Clearly then the attractive magnetic force that reduces the net force between protons in the two beams as seen in our frame of reference is relativistic. In particular the apparent magnetic forces or fields are relativistic modifications of the electrical forces or fields. As such modifications, they cannot be stronger than the electrical forces and fields that produce them. This follows from the fact that switching frames of reference can reduce forces, but it can’t turn what is attractive in one frame into a repulsive force in another frame.
In the case of wires the net charges in two wires are zero everywhere along the wires. That makes the net electrical forces between the wires very nearly zero. Yet the relativistic magnetic forces and fields will be of the same sort as in the case of two beams of charges of a single sign. This is true even in the frame of reference of what we think as the moving charges, that is, the electrons. In the frame of reference moving at the drift velocity of these current-carrying electrons, it is the protons or positively charged ions that are moving in the other direction. Consequently in any frame of reference for current-carrying wires in parallel, the net electrical force will be essentially zero, and there will be a net attractive magnetic force
Explanation:
