Answer:
x plus 4 divided by 5 minus 9
x + 4 / 5 - 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
For Independent Events, P(A) × P(B) = P(A∩B)
so we have, P(A∩B) = 0.4×0.1 = 0.04
P(A') = 1 - 0.4 = 0.6
This information can be represented on a Venn diagram as shown below
P(A'∪B) means the union of everything that is not A with everything that is B
P(A'∪B) = 0.06 + 0.54 + 0.04 = 0.64
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Answer:
£75.53
Step-by-step explanation:
3tshirt=£17.82
2pauntbrush=£13.94
4bags of pasta=£3.52
toral=£17.82+£13.94+£3.52+£41.25=75.53