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SVEN [57.7K]
3 years ago
5

Capacitances of 10uF and 20uF are connected in parallel,

Physics
1 answer:
jarptica [38.1K]3 years ago
5 0

Answer:

The equivalent capacitance will be 15\mu F  

Explanation:

We have given two capacitance C_=10\mu F\ and\ C_2=20\mu F

They are connected in parallel

So equivalent capacitance C=C_1+C_2=10+20=30\mu F

This equivalent capacitance is now connected in series with 30\mu F

In series combination of capacitors the equivalent capacitance is given by \frac{1}{C}=\frac{1}{30}+\frac{1}{30}

C=\frac{30}{2}=15\mu F

So the equivalent capacitance will be 15\mu F  

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B is strength of magnetic field in T.
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I is current in A.
r is distance from conductor in m.

<span>So I = 2πrB /μ_0</span>
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2 years ago
While watching a planet orbiting a star, you notice that the doppler shift in the light of the star is becoming increasingly blu
Lyrx [107]

Answer:

The star must be moving "closer" to the earth because the wave fronts that are being emitted are shifted toward the blue end of the spectrum (they are closer together)

6 0
2 years ago
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
Is this right or they wrong definitions which ones are the right ones someone !!!!!
Alex73 [517]

Answer:

They are right.

Explanation:

4 0
2 years ago
Read 2 more answers
I need help with this
Fed [463]
HEY THERE !!

It is called SUBLIMATION.

so the correct answer is (D).
3 0
2 years ago
Read 2 more answers
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