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Marina86 [1]
3 years ago
5

A parallel-plate capacitor in air has a plate separation d = 4 mm and square plates with area A = 25.0 cm^2. The plates are main

tained to a constant potential difference of 255 V. Determine the charge on the plates and the energy stored by the capacitor.
Physics
1 answer:
Nikolay [14]3 years ago
3 0

Answer:179.79 \times 10^{-9} J

Explanation:

Given

Separation (d)=4 mm

Area of cross-section=25 cm^2

Potential difference=255 V

C=\frac{\epsilon A}{d}=5.53 \times 10^{-12}F

charge (Q)=CV=5.53\times 225=1411.10\times 10^{-12} C

Energy stored=\frac{cv^2}{2}

E=\frac{5.53\times 10^{-12} 255^2}{2}=179.79 \times 10^{-9} J

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Acceptame para ser amigos y luego te ayudo
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Rudy plans to conduct an experiment using three rosebushes of the same variety and size. His hypothesis is that the plant
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3 years ago
Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write
Eva8 [605]

Answer:

  • R = ( 4.831 m , 1.469 m )
  • Magnitude of R = 5.049 m
  • Direction of R relative to the x axis= 16°54'33'

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | its the magnitude and θ.

So, for our vectors, we will have:

\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )

\vec{D}=  ( 2.121 m , -2.121 m )

and

\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )

\vec{E}= ( 2.71 m , 3.59 m )

Now, we can take the sum of the vectors

\vec{R} = \vec{D} + \vec{E}

\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )

\vec{R} = ( 2.121 \ m  + 2.71 \ m , -2.121 \ m + 3.59 \ m )

\vec{R} = ( 4.831 \ m , 1.469 \ m )

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}

|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}

|\vec{R}| = \sqrt{25.496 m^2}

|\vec{R}| = 5.049 m

To find the direction, we can use

\theta = arctan(\frac{R_y}{R_x})

\theta = arctan(\frac{1.469 \ m}{4.831 \ m})

\theta = arctan(0.304)

\theta = 16\°54'33''

As we are in the first quadrant, this is relative to the x axis.

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4 years ago
Need help transforming one formula to another!
Sindrei [870]

Hello! I don't know if this will help you any but I worked it out, got nothing on it. So, I just had my twin help me and he had nothing either. But, this is just how to find the transformation. Seems like this is already complete to me.

To find the transformation, compare the equation to the parent function and check to see if there is a horizontal or vertical shift, reflection about the x-axis or y-axis, and if there is a vertical stretch.

Parent function : y=√a

Horizontal shift : None

Vertical shift : none

Reflection about the x-axis : none

Vertical stretch : Stretched.

I am so sorry if this doesn't help you but to me, in both my eyes and my brother's eyes, this already looks complete. I hope this helps you out. Again, so sorry if it doesn't.

-Karleif ☺

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An atom is probably less than 1 nano-meter in size
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