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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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BartSMP [9]
I believe the answer is A.

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5 0
4 years ago
Does anyone knows the answer??​
Tamiku [17]

Answer:

yaa

Explanation:

8 0
3 years ago
A .422 kg coffee mug rests on a table top 0.63 m above the floor. What is the potential energy of the mug with respect to the fl
kondaur [170]

Answer:

2.61 J

Explanation:

Since potential energy U = mgy where m = mass of object, g = acceleration due to gravity = 9.8 m/s² and y = height of object above the ground.

Now for the coffee mug, m= 0.422 kg and it is 0.63 m on a table, so it is 0.63 m above the ground. Thus, y = 0.63 m.

We compute U

U = mgy

= 0.422 kg × 9.8 m/s² × 0.63 m

= 2.605 J

≅ 2.61 J

So, the potential energy of the mug with respect to the floor is 2.61 J

4 0
3 years ago
A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
4 years ago
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melisa1 [442]

Answer:

Weight = 8.162 Newton.

Explanation:

Given the following data;

Mass = 2.2 kg

Acceleration due to gravity = 3.71 N/kg

To find the weight of the textbook;

Weight = mass * acceleration due to gravity

Weight = 2.2 * 3.71

Weight = 8.162 N

Therefore, the weight of the science textbook in mars is 8.162 Newton.

3 0
3 years ago
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