<span>9v^2=25
v^2 = 25/9
v^2 = (5/3)^2
v = + 5/3 and v = -5/3
hope it helps</span>
The answer is 2 and negative 9
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
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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:
Step-by-step explanation:
1.) y= 1/5x - 2
2.) y = 1x + 2
3.) y = 2x + 4
4.) y = 1x + 4
5.) y = 1/3x - 4
6.) y = 1/2x - 3