Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0
Explanation:
- a) First of all find the distance between the two charges;
- x = 0, y = 0.30 and x = 0.40 m, y = 0
hence, the force F = 2Kq1q2cosθ /r²...............equation 1
but cosθ = y/r = 0.3/0.5
cosθ = 0.6
plugging back to equation 1;
F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2
F = 540 x 10^-3
Magnitude of Force = 0.54N
b) Direction is at angle 90
Explanation:
Given that,
Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.
For series combination,
For parallel combination,
When 6 ohm and 3 ohm are in series,
When 6 ohm and 3 ohm are in paralle,
So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.
Answer:
Explanation:the atom consists of a tiny nucleus at its center which is surrounded by a moving electrons. The nucleus contains a positively charged proton equal in size with the negatively charged electrons . The nucleus also may contain neutrons which have the same mass with the protons but no charge is neutral.
The distance between object P1 and its image formed is determined as 36 m.
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Distance of the image</h3>
The distance of the image formed by object P1 is calculated as follows;
In a plane mirror; object distance = image distance
image distance of P1 = 18 m
distance between object and image = 18m + 18 m = 36 m
Thus, the distance between object P1 and its image formed is determined as 36 m.
Learn more about plane mirrors here: brainly.com/question/1126858
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