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Salsk061 [2.6K]
3 years ago
7

An experiment is being done. Name the independent variable, and the dependent variable for the following question.

Physics
1 answer:
Nata [24]3 years ago
3 0

Answer:

Independent variable: Surrounding color

Dependent Variable: Eating Habits

Explanation:

Dependent variables are dependent on Independent variables which means that in this question Eating habits are affected by the Surrounding color.

Hope it Helps!!

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Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

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- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

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                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

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                                 cos(Q) = sqrt(3)*a / 2*r

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                                 F_1,2 = 2*F_x*cos(Q)

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                                F_net = F_1,2 + kq / y^2

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                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

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