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2 years ago

I got the correct answer, but I don't know what I did. Why is the answer E?

Physics

1 answer:

MariettaO [177]2 years ago

6 0

Explanation:

Displacement is the straight line distance from the starting position to the final position.

Person X walks halfway around circle. So her displacement is 100 m.

Person Y walks 3/4 of the way around. So his displacement is 50√2 m ≈ 70.7 m.

Person Z walks completely around the circle, so their displacement is 0 m.

Therefore:

Z < Y < X

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

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

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

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

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

4 0

3 years ago

What is pulling force? Give any two examples.,

Dima020 [189]

Answer:

Push or Pull Forces - example

When you push against a wall the force that you exert is an example of a push force. When you pull a trolley car the force that you exert is an example of pull force.

8 0

3 years ago

A bus is moving with an acceleration of 4 m/s^2. If it was initially at rest,

shusha [124]

Probably b) 6 m/s
Not sure

6 0

3 years ago

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a_______ is a region where the magnetic fields of a large number of atoms are lined up parallel to a magnets field.

patriot [66]

Answer:

Magnetic Domain