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Svetlanka [38]
3 years ago
13

Two parallel plates that are initially uncharged are separated by 1.2 mm. What charge must be transferred from one plate to the

other if 11.0 KJ of energy are to be stored in the plates, if area of each plate is 19.0 mm
Physics
1 answer:
Gre4nikov [31]3 years ago
8 0

Answer:

Q=55.5 \mu C

Explanation:

We start by writing the formula of the energy stored between two charged parallel plates, which is:

U=\frac{Q^2}{2C}

Where <em>Q</em> is the charge on one of the plates and <em>C</em> its capacitance, which for two parallel plates is:

C=\frac{\epsilon A}{d}

Where A is the area of each plate, d the distance between them and \epsilon the permittivity of the medium, in our case the vacuum, \epsilon_0.

Putting all together:

U=\frac{Q^2d}{2\epsilon_0 A}

Which means:

Q=\sqrt{\frac{U2\epsilon_0 A}{d}}

Which for our values is:

Q=\sqrt{\frac{(11\times10^3J)2(8.85\times10^{-12}F/m)(0.000019m^2)}{(0.0012m)}}=55.5\times10^{-6}C

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telo118 [61]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The uncertainty in inverse frequency is  \Delta  [\frac{1}{w} ]=  \frac{3}{2000} \ s

Explanation:

From the question we are told that

   The value of the proportionality constant is  k  = 5  \frac{Hz }{T}

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The magnetic field is given as

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Where w is frequency

The uncertainty or error of the field is given as

         \Delta  B  =  \frac{k }{[\frac{1}{w}^]^2 }  \Delta [\frac{1}{w} ]

The uncertainty in inverse frequency is given  as

           \Delta  [\frac{1}{w} ]  = \frac{\Delta B}{k [\frac{1}{w^2} ]}

                    \Delta  [\frac{1}{w} ]=  \frac{\Delta B}{k (B)^2 }

substituting values

                  \Delta  [\frac{1}{w} ]=  \frac{3}{5 (20)^2 }

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6 0
3 years ago
A 0.23 kg mass at the end of a spring oscillates 2.0 times per second with an amplitude of 0.15 m
charle [14.2K]

Answer:

A) v = 1.885 m/s

B) v = 0.39 m/s

C) E = 0.03 J

D) x(t) = (0.15m)\cos(2\pi (2.0Hz)t)

Explanation:

Part A

We will use the conservation of energy to find the speed at equilibrium.

K_{eq} + U_{eq} = K_A + U_A\\\frac{1}{2}mv^2 + 0 = 0 + \frac{1}{2}kA^2\\v = \sqrt{\frac{k}{m}}A

where \omega = \sqrt{k/m} and \omega = 2\pi f

Therefore,

v = 2\pi f A = 2(3.14)(2)(0.15) = 1.885~m/s

Part B

The conservation of energy will be used again.

K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\\mv^2 + kx^2 = kA^2\\(0.23)v^2 + k(0.10)^2 = k(0.15)^2\\v^2 = \frac{k(0.15)^2-(0.10)^2}{0.23}\\v = \sqrt{0.054k}

where k = \omega^2 m = (2\pi f)^2 m = 2(3.14)(2)(0.23) = 2.89

Therefore, v = 0.39 m/s.

Part C

Total energy of the system is equal to the potential energy at amplitude.

E = \frac{1}{2}kA^2 = \frac{1}{2}(2.89)(0.15)^2 = 0.03~J

Part D

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x(t) = A\cos(\omega t + \phi)\\x(t) = (0.15m)\cos(2\pi (2.0Hz)t + \phi)

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5 0
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The right formula to use is: T^2= 2d/g
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6 0
4 years ago
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Answer:

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9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2

I=17.32\times 10^{-3}A

So current in the toroid will be I=17.32\times 10^{-3}A

5 0
4 years ago
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