Answer:
6!
Step-by-step explanation:
chord x chord = chord x chord
(5) (n+8) = (7) (n+4)
i just started plugging in numbers
(5) (6+8) = (7) (6+4)
5 x 14 = 7 x 10
70=70
The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.
The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = <span>-7.041 N/C
</span>
Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*<span>7.041 N/C </span>= 14.082 N/C
Answer:
Ooh man, this was back in geometry I think, I think the answer is 4 if my calculations are correct.
i think the bigger side is twice the size of the smaller side, so 2 times 8 is 16,
16 = 5x - 4
x = 4
The correct answer is C. i think
