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Anuta_ua [19.1K]
1 year ago
13

Three identical capacitors are connected in parallel to a battery. if a total charge of q flows from the battery, how much charg

e does each capacitor carry? group of answer choices
Physics
1 answer:
muminat1 year ago
7 0

Each capacitor carry the same charge 'q'.

Discussion:
The voltage from the battery is distributed equally across all of the capacitors when they are linked in series. The three identical capacitors' combined voltage is computed as follows:

V_{T} = V₁ +V₂ +V₃

This voltage may also be calculated using capacitance and charge;

V = Q/ C

V_{T} = V₁ +V₂ +V₃

Provided that the total charge is 'q', hence the total voltage can be expressed as:

V_{T} = (Q/C₁) + (Q/C₂) + (Q/C₃) = Q(1/C₁ +1/C₂ +1/C₃)

Therefore from the above explanation, it is concluded that each and every capacitor carry same charge 'q'.

Learn more about the capacitor here:

brainly.com/question/17176550

#SPJ4

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The three vectors in Fig. 3-33 have magnitudes a = 3.00 m, b = 4.00 m, and c = 10.0 m and angle θ = 30.0°. What are
aliya0001 [1]

Answer:

a) a_{x} =3              b) a_{y} =0

c) b_{x} =3.46        d) b_{y} =2

e) c_{x} =0              f) c_{y} =10

g) p = -5.77                            h) q=5

Explanation:

Diagram for given question is attached below in fig 1

<h3>Part (a) (b)</h3>

for vector \vec{a}

θ = 0°

          a_{x} = 3 cos (0)\\a_{x} = 3\\a_{y} = 3 sin (0)\\a_{y} = 0

<h3>Part (c) (d)</h3>

for vector \vec{b}

θ = 30°

      b_{x} = 4 cos (30)\\b_{x} = 3.46\\b_{y} = 4 sin (30)\\b_{y} = 2

<h3>Part (e) (f)</h3>

for vector \vec{c}

θ = 90°

    c_{x} = 10 cos (90)\\c_{x} = 0\\c_{y} = 10 sin (90)\\c_{y} = 10

<h3>Part (g) (h)</h3>

                       \vec{c} = p\vec{a} + q\vec{b}

c =c_{x} \hat{i} + c_{y}\hat{j}\\as a_{y} =0\\c_{x} \hat{i} + c_{y}\hat{j} = pa_{x} \hat{i} +q(b_{x} \hat{i}  +b_{y} \hat{j} )\\c_{x} \hat{i}  = pa_{x} \hat{i} +qb_{x} \hat{i}\\c_{y}\hat{j}  =qb_{y}\hat{j}

                  q=\frac{c_{y}}{b_{y}} \\q=\frac{10}{2}\\q=5

c_{x} \hat{i}  = pa_{x} \hat{i} +qb_{x} \hat{i}\\0 = p(3) + (5)(3.46)\\p = -5.77

5 0
3 years ago
The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. the radi
natali 33 [55]
Total number of Photons = 92. 
Radius of the Nucleus = 7.4 x 10^-15 m 
Charge of the nucleus = 1.6*10^-19. 
Total charge q = 92 x 1.6*10^-19 = 147.2 x 10^-19 
k = 9 x 10^9 N m^2 / C^2 
Electric field charge E = kq / r^2 
=> E = (9 x 10^9 x 147.2 x 10^-19) / (7.4 x 10^-15) ^2  
=> (1324.8 x 10^-10) / 54.76 x 10^-30 => 24.19 x 10^20 N/C
5 0
3 years ago
Read 2 more answers
A 10-cm-thick aluminum plate (α = 97.1 × 10−6 m2/s) is being heated in liquid with temperature of 550°C. The aluminum plate has
Zarrin [17]

Answer:

356°C.

Explanation:

(1). The first step to the solution to this particular Question/problem is to determine the Biot number, and after that to check the equivalent value of the Biot number with plate constants.

That is, Biot number = (length × ∞)÷ thermal conductivity. Which gives us the answer as ∞. Therefore, the equivalent value of the ∞ on the plates constant = 1.2732 for A and 1.5708 for λ.

(2). The next thing to do is to determine the fourier number.

fourier number = [α = 97.1 × 10−6 m2/s × 15 s] ÷ (.05m)^2 = 0.5826.

(3). The next thing is to determine the temperature at the center plane after 15 s of heating.

The temperature at the center plane after 15 s of heating = 500°C [ 25°C - 500°C ] [1.2732] × e^(-1.5708)^2 ( 0.5826).

The temperature at the center plane after 15 s of heating = 356°C.

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3 years ago
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