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
Kent can paint the room 1/5
luannda can paint the room 1/3
(3+5) (5*3) = 8/15 or 0.53 th of the room together but just for kent would be 1/5
9514 1404 393
Answer:
31.243 units
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you of the relationships between sides and angles in a right triangle. Using the attached figure, it is convenient to find the length of BE as an intermediate step in the solution.
Sin = Opposite/Hypotenuse
sin(30°) = BE/100
BE = 100·sin(30°)
Then ...
Tan = Opposite/Adjacent
tan(58°) = BE/x
x = BE/tan(58°) = 100·sin(30°)/tan(58°)
x ≈ 31.243 . . . . units
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<em>Comment on the figure</em>
The intermediate problem in creating the figure was to locate point D. That was accomplished by locating point C on a line at an angle of 58° CCW from the horizontal, using point B as a center. Then D is the intersection of BC with the x-axis. BE is drawn perpendicular to the x-axis.
Answer:
no they are not
Step-by-step explanation:
priyas paint would be darker then mais paint because priya added an extra cup of red paint.
Answer:
-4
Step-by-step explanation:
If a = -4 then you can eliminate hte -4x and 4x by adding them, you get 0x, so you eliminate the x variable
