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sveta [45]
4 years ago
13

Two capacitors A and B are connected in

Physics
1 answer:
fredd [130]4 years ago
5 0

Answer:

Capacitor A = 0.24\mu F\\Capacitor B = 0.16\mu F

Explanation:

I'll upload my work shortly as an attachment, but here is my process in words:

  1. In our first situation we have two capacitors in parallel, which means the charge distribution on both of them is the same. With that, we can find a ratio between the values of Capacitors A and B.
  2. In  our second situation , we add a capacitor parallel to A (I called it C). Because A and C are in parallel, we know that they must have the same potential difference; which should come to be 10V since 90V of the total 100V is on B. Also, the equivalent charge distribution across A and C must be equal to that of the charge distribution at B, because A&C are in series with B. So I added the charges on A&C and set that equal to the charge on B.
  3. Next, I used the ratio from the first situation to substitute Capacitor A out of the equation. This allows us to solve for B's capacitance. (Note: You could have also substituted B for A and solved for A first if you wanted to.)
  4. Finally, I used B's capacitance to plug back into the ratio from the first situation to find A's capacitance. And they wanted the answer in micro-Farads, so I went ahead and converted each answer to micro.

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How many seconds are in 28 hours?
kumpel [21]

Answer:

1680 seconds

Explanation:

28 hrs * \frac{60 s}{1 hr} =1680s

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2 years ago
Two children are pulling and pushing a 30.0 kg sled. The child pulling the sled is exerting a force of 12.0 N at a 45o angle. Th
djverab [1.8K]

Answer:

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3 years ago
A uniform electric field exists everywhere in the x,y plane. The electric field has a magnitude of 3500 N/coil, and is directed
alexandr402 [8]

Answer:

5525 N/C

Explanation:

Magnitude of electric field ( E ) = 3500 N/c

Direction of electric field : positive X axis

point charge ( q ) = -9.0 * 10^-9

<u>Calculate the Magnitude of the net electric field  at (a) x = -0.20 m</u>

Magnitude =  5525 N/C

Electric field due to q = ( 9 * 10^9 * 9 * 10^-9 ) / ( -0.2 )^2  

                                    = 81 / 0.04 = 2025 N/c

<em>Therefore the magnitude of the net electric field </em>

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= 5525 N/C

6 0
4 years ago
A 100 A current circulates around a 2.00-mm-diameter superconducting ring.
igomit [66]

Answer:

0.000314 Am²

6.049*10^-7 T

Explanation:

A

From the definitions of magnetic dipole moment, we can establish that

= , where

= the magnetic dipole moment in itself

= Current, 100 A

= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²

On solving, we have

= ,

= 100 * 3.14 * 0.000001

= 0.000314 Am²

B

= (0)/4 * 2 / ³, where

(0) = constant of permeability = 1.256*10^-6

z = 4.7 cm = 0.047 m

B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]

B = 1*10^-7 * 0.000628/1.038*10^-4

B = 1*10^-7 * 6.049

B = 6.049*10^-7 T

8 0
3 years ago
A simple generator has a square armature 5.0 cm on a side. The armature has 95 turns of 0.59-mm-diameter copper wire and rotates
guajiro [1.7K]

Answer:

25.39Hz

Explanation:

#First, we need to determine the actual emf required. The generator's internal resistance will cause a voltage drop inside the generator/

Internal resistance is defined using the formula:

R=\rho L/A\\\rho=1.68\times10^-^8 \Omega m\\L=0.050m\times4\times 60=12.0m\\A=\pi r^2=\pi d^2/4=\pi(0.00059m)^2/4=2.734\times10^-^7m^2\\\\R=(1.68\times10^-^8 \Omega \ m\times 12.0m)/2.734\times10^-^7m^2\\=0.7374\ \Omega

#The bulb is rated 12.0V,25.0W

Current, I=25.0W/12.0V=2.083A

Therefore, the voltage drop in the generator is calculated as:

2.083A\times0.7374\Omega=1.5360V  

Actual EMF required is thus   1.536V+12.0V=13.536V

#peak voltage is 13.536V\sqrt 2=19.143V

#For a generator, by Faraday's Law

E_m_a_x=NBA\ \omega\\19.143=60\times 0.800T\times (0.05m)^2\ \omega\\\\\omega=159.525rad/s

f=\omega/2\pi\\=159.525/2\pi=25.39Hz

#The rate of the generator is 25.39Hz

7 0
4 years ago
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