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3 years ago

The image of the lemon is at point I. What is the size of the image compared to the size of the lemon?

Physics

2 answers:

sp2606 [1]3 years ago

6 0

Answer:

The same size.

Explanation:

I just took the test

nasty-shy [4]3 years ago

6 0

Answer:

Same Size

Explanation:

The person above me got it correct

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Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

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Maksim231197 [3]

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I think it's amplitude

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andreev551 [17]

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Explanation:

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