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ZanzabumX [31]
3 years ago
12

Suppose two parallel-plate capacitors have the same charge Q, but the area of capacitor 1 is A and the area of capacitor 2 is 2

A.If the spacing between the plates, d, is the same in both capacitors, and the voltage across capacitor 1 is V, what is the voltage across capacitor 2?Express your answer in terms of V but do not type in the symbol "V"
Physics
1 answer:
almond37 [142]3 years ago
8 0

Answer:

V' = V/2

Explanation:

The voltage across a parallel plate capacitor is given as follows:

V = Q/C

where,

V = Voltage across capacitor

Q = Charge on Capacitor

C = Capacitance of Capacitor = A∈₀/d

Therefore,

V = Qd/A∈₀

where,

A = Area of plate

d = distance between plates

∈₀ = permittivity of free space

FOR CAPACITOR 1:

Q = Q

d = d

A = A

V = V

Therefore,

V = Qd/A∈₀   --------------- equation (1)

FOR CAPACITOR 2:

V' = ?

Q' = Q

d' = d

A' = 2A

Therefore,

V' = Q'd'/A'∈₀

V' = Qd/2A∈₀

V' = (1/2)(Qd/A∈₀)

using equation (1):

<u>V' = V/2</u>

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The resistance of a wire depends upon the material's resistivity and the length and cross‑sectional area of the wire. What will
andriy [413]

Answer:

R' = 4R

The resistance will become 4 times the initial value.

Explanation:

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R = ρL/A   ----------- equation 1

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R = Resistance of wire

ρ = resistivity of the material

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A = Cross-sectional area of wire

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<u>The resistance will become 4 times the initial value.</u>

6 0
4 years ago
A horizontal 953 N merry-go-round of radius 1.68 m is started from rest by a constant horizontal force of 73.9 N applied tangent
solniwko [45]

Answer:

K.E=365.2 J

Explanation:

Given data

Weight w =953 N

radius r=1.68 m

F=73.9 N

t=2.55 s

g=9.8 m/s²

To find

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Solution

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I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}

The angular acceleration is given as

a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}

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3 0
3 years ago
How does free energy challenge the scientific definition of energy?
gtnhenbr [62]
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7 0
3 years ago
Helium–neon laser light (λ = 632.8 nm) is sent through a 0.350-mm-wide single slit. What is the width of the central maximum on
Cloud [144]

Answer:

The width of the central bright fringe is 7.24 mm.

Explanation:

Given that,

Wavelength = 632.8 nm

Width d= 0.350 mm

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Using formula of distance

y_{m}=\dfrac{\lambda D}{d}

Put the value into the formula

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Using formula of width

width = 2\times|y_{m}|

Put the value into the formula

width=2\times3.62

width = 7.24\ mm

Hence, The width of the central bright fringe is 7.24 mm.

5 0
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