Solve within parenthesis first so .49 +.045 =.535 so then divide by five which is .107
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:
660.08
Step-by-step explanation:
1. first you will need to line up the decimals and make sure the places are correct
647.2
+ 12.88
2. Then you just add them
The answer is 34%
30+33%=39.9
So round up two 34%
I hope that is right but I think the answer is 34%
Answer:
Umm i think yu but it... I would hae to physcally be with you to help you.
Step-by-step explanation:
Sorry. Its hard to explain...
