.... I don’t know but, he will be able to make smarter choices, he will be able to think before he does something, honestly don’t know
3. In a uniform electric field, the equation for the magnitude of the magnetic field is E=(V/d). V= voltage d= distance. If the magnetic field magnitude is
constant , as stated in your problem, then the voltage must stay the same otherwise the value of "E" would change". And the problem already told us the "E" is uniform and so, not changing. Does that make sense?
4a. If the magnetic field lines are equally spaced apart, in other words share the same
density. Then we know that the magnitude of the magnetic field is unchanging. This is because the density of of the magnetic field lines(how many are in a certain area) is related to the magnitude being expressed by the electric field. Greater magnitude is expressed by the presence of more lines (higher line density)
4b. The electric potential is measured in Volts(V) and is uniform along the same equipotential line. What is an equipotential line(gray)? It is a line drawn perpendicular(forms a right angle with) to the magnetic field lines(black) to show the changes in electric potential. One space where electric potential will always be the same because it will always be equal to 0 Volts is exactly in between a positive and negative charges of equal charge value I have pointed to this line with a purple arrow in my picture.
I really hope this makes sense to you and that my pictures help! :)
-- As far as we know, the forces on the wheelbarrow are balanced.
-- That tells us that the net force on the wheelbarrow is zero, just
as if there were no forces acting on it at all.
-- That tells us that the wheelbarrow's acceleration is zero ... its
speed and direction of motion are not changing.
-- That tells us that the wheelbarrow is moving in a straight line
at a constant speed. It's very possible that relative to us, the speed
may be zero, but we can't tell that from the given information.
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Answer:
d. 5 ohms
Explanation:
For resistors in parallel, the equivalent resistance is found with:
1/Req = ∑(1/R)
1/R = 1/15 + 1/15 + 1/15
1/R = 3/15
R = 15/3
R = 5