X-5+5x-12=25
x+5x=25+5+12
6x=42
x=42/6
x=7
Q1: 14
Q3: 37
Interquartile: 37-14 = 23
THE ANSWER IS 23
(Brainliest please?)
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:
(13.8,1.42) (18,3.7)(16.7,3.21) and so on REMEMBER ITS (X,Y)
Step-by-step explanation:
Answer:
25.92
Step-by-step explanation: