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:
144
Step-by-step explanation:
108 plus 36 then you will get 144
Answer:
752.95
Step-by-step explanation:
Data provided in the question
The standard deviation of population = 210
The Margin of error = 15
The confidence level is 75%, so the z value is 1.96
Now the required sample size is
= 752.95
Hence, the number of college students spends on the internet each month is 752.95
Simply we considered the above values so that the n could come
The answer is the second option given
Answer:
it 1/3
Step-by-step explanation:
2/4 is just 1/2, 3/6 is also 1/2, 7/9 is over 1/2 and 1/3 is the only one less then 1/2 so it's the lowest fraction. This is not the proper way to solve it but it works and it's faster
