(a)Two charges are placed on the x-axis: one is placed at x = 3 m and the other is at x = -3 m. The magnitude of both charges is 2.3 µC but the blue one (at x = -3 m) is positive while the red one (at x = +3 m) is negative.
What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?
(b) Now the positive and negative charge switch places. The magnitude of the charges is still 2.3 µC where the blue one (now at x = +3 m) is positive and the red one (now at x = -3 m) is negative.
What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?
(c) Now both charges (still at x = -3 m and x = +3 m) are positive. The magnitude of both charges is still 2.3 µC.
What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?
(d) Finally, both charges (still at x = -3 m and x = +3 m) are negative. The magnitude of both charges is still 2.3 µC.
Answer:
C?
Step-by-step explanation:
Answer:
C) (y - 1)(y - 5)
Step-by-step explanation:
Hi there!
First you separate your expression into groups.
Now we take the common multiple out of each.
Since we have two (y - 5)s, we can pull it out of the equation as its own common multiple.
After we do that, we are left with y - 1. (we got this from taking the common multiples out)
Now we are just left with (y - 5)(y - 1)
Hope this helped!
Answer:
4 units to the right & 1 unit up.
20x^2+50 = -40x^2+110x [ Taking x as the unknown positive integer ]
