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Paha777 [63]
3 years ago
13

Part AIf the potential of plate 1 is V, then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the negativ

e terminal of the battery is at zero potential.Part BIf the charge of the first capacitor (the one with capacitance C) is Q, then what are the charges of the second and third capacitors?Part CSuppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the charges of the three individual capacitors calculated in the previous part, find the total charge Qtot for this equivalent capacitor.Part DUsing the value of Qtot, find the equivalent capacitance Ceq for this combination of capacitors.Express your answer in terms of C.

Physics
1 answer:
Nana76 [90]3 years ago
6 0

Answer:

Part a: The potential at point 3 and point 6 are V and 0 respectively.

Part b: The charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c: The net charge is  6CV.

Part d: The equivalent capacitance is 6C

Explanation:

As the question is not given here ,the complete question is found online and is attached herewith.

Part a:

V1 = V, V3 = V1 = V

and, V6 = V1-V = 0

The potential at point 3 and point 6 are V and 0 respectively

Part b

Q1 =CV = Q,

Q2 = 2C *V = 2Q

Q3 = 3 C*V = 3Q

So the charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c

Total charge of the system,

Q_net =Q1+Q2+Q3= (1+2+3) CV = 6 CV

So the net charge is  6CV.

Part d:

Equivalent Capacitance = Net charge / Voltage

Eq. C = 6CV/V = 6C

So the equivalent capacitance is 6C

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Dimas [21]

Answer:

n_f=2

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

R = Rydberg constant, R=1.097\times 10^7\ m^{-1}

\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

Solving above equation we get the value of final n is,

n_f=2.04

or

n_f=2

So, it will relax in the n = 2. Hence, this is the required solution.        

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3 years ago
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the answer is a: galaxies. galaxy clusters are groups of galaxies that like to hang out together out there in space. the milky way's own galaxy cluster is called the local group.

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-- Bob covered a distance of (32m + 45m) = 77 meters.

-- His displacement is the straight-line distance and direction
from his starting point to his ending point.

The straight-line distance is

D = √(32² + 45²)
D = √(1,024 + 2,025)
D = √3,049 = 55.22 meters

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3 years ago
A playground merry-go-round has radius 2.10m and moment of inertia 2500kg*m^2 about a vertical axle through its center, and it t
vladimir2022 [97]

Answer

Radius of the wheel r = 2.1 m

Moment of inertia I = 2500 Kg m²

Tangential force applied F = 18 N

Time interval t = 16 s

Initial angular speed ω1 = 0

Final angular speed ω2 = ?

Let α be the angular acceleration.

Torque applied τ = Iα

                         F r = Iα

Angular acceleration α = F r/I

                                    = \dfrac{18\times 2.1}{2500}

                                    = 0.015 rad/s²

(a)From rotational kinematic relation

            Final angular speed ω₂ = ω₁ + αt

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                                                 = 0.24 rad/s

(b) Work done W = 0.5 Iω₂² - (1/2)Iω₁²

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(c) Average power supplied by the child P = W/t = \dfrac{72}{16}

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8 0
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IrinaVladis [17]

Answer:

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Explanation:

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If motion is in only two direction (x and y directions) then it is know as two directional motion.

If motion is in only three direction (x , y and z directions) then it is know as three directional motion.

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A leaf is falling from the tree - This is two directional motion.

A train is running on track - This is  one directional motion

A lady is crawling  on the surface of the ball - This is three directional motion.

Therefore the answer is  ---

A leaf is falling from the tree.

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