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The answer is 18000 J
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Answer:
i'm not sure if you are asking as a personal question or a book question so i'm taking it personal.
Explanation:
I was doing a simple task that was handed to me to test my responsibility and I agreed (knowing i am responsible :3). my first thought was "man , this is easy!" but then i started seeing the other kids slaking off and quiting their tasks. I thought that was against the rules, but then i saw my bff doing it too and i thought "this should be ok then!" so i did the same. other kids where still doing it. the teacher came, saw the ones still working and smiled... but when the teacher looked at the ones slaking off omg... his face was like * im gonna kill yall* we took one big gulp and whined. the teacher awarded the ones who completed the task... the others , we had to do our original task but doubled... for 3 weeks!!! it was awful!!!
I WOULD NEVER DO THAT AGAIN!!!
Answer:
a. cosθ b. E.A
Explanation:
a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for 90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function
b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A