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2 years ago

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what is the capacitance of two square parallel plates 6.7 cm on a side that are separated by 1.8 mm of paraffin? the dielectric

constant of paraffin is 2.2.

1 answer:

almond37 [142]2 years ago

7 0

The capacitance of two square parallel plates will be 1.777 farads.

<h3>What is the capacitance of two parallel plates?</h3>

The body's capacity to hold electric charge is constrained by capacitance. There is a capacitance for each capacitor. Two metallic sheets of area A, spaced by d, make up the standard parallel-plate capacitor.

The parallel plate capacitor formula is given by:

C = ε Ad

Two square parallel plates 6.7 cm on a side are separated by 1.8 mm of paraffin. The dielectric constant of paraffin is 2.2. Then the capacitance of two square parallel plates will be given as,

C = ε Ad

C = 2.2 x 0.67 x 0.67 x 1.8

C = 1.777 farad

Two parallel square plates will have a capacitance of 1.777 farads.

More about the capacitance of two parallel plates link is given below.

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Answer:

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Explanation:

The resolution of the telescope is given by diffraction

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Where a is the separation of the linear slits, λ the wavelength, m an integer that determines the order of diffraction

In this case, suppose that the premieres meet Rayleigh's criteria, which establishes that the central maximum of the diffraction of an object coincides with the first minimum of diffraction of the second object. In this case m = 1

sin θ = λ / a

In the case of circular openings, polar coordinates must be used, so the equation is

sin θ = 1.22 λ / D

Where D is the diameter of the lens or tightness. Since the distances are very large and the small angles we can approximate the sine to the radian angle value

θ = 1.22 λ / D

Let's use trigonometry to find the angle. We have the separation of the premieres y = 3.7 10¹¹ m and

tan θ = y / L

θ = y / L

Let's replace

y / L = 1.22 λ / D

L = y D / 1.22 λ

calculate

L = 3.7 10¹¹ 1.03 / (1.22 550 10⁻⁹)

L = 5.68 10¹⁷ m

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3 years ago

What is the water potential of pure water?

DanielleElmas [232]

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3 years ago

A biologist looking through a microscope sees a bacterium at r⃗ 1=2.2i^+3.7j^−1.2k^μm(1μm=10−6m). After 6.2 s , it's at r⃗ 2=4.6

stiv31 [10]

The Average velocity for the bacterium is 0.75 unit/sec.

<u>Explanation:</u>

The given values are in the vector form

Where,

dS = distance covered

dT = time interval

Now, to calculate distance covered, we have

$|d S|=\sqrt{d S^{2}}$

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$d S=r_{2}-r_{1}$

d S=(4.6 i+1.9 k)-(2.2 i+3.7 j - 1.2 k)

d S=(4.6-2.2) i+(0-3.7) j+(1.9+1.2) k

d S=2.4 i-3.7 j+3.1 k

Now, putting these values in the standard formula to evaluate the average velocity, we get;

$v_{a v g}=\frac{|\mathrm{d} S|}{d T}$

$v(a v g)=\frac{|\sqrt{\left\{\left(2.4^{2}\right)+\left(3.7^{2}\right)+\left(3.1^{2}\right)\right\}}|}{7.2}$

As dT=7.2 sec

Now,

Solving the equation, we get;

$v(a v g)=\frac{5.390732789}{7.2}$

$\begin{aligned}&v(a v g)=\frac{5.39}{7.2}\\&v(a v g)=0.748611111\\&v(a v g)=0.75 \text { units / sec }\end{aligned}$

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4 years ago

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Gre4nikov [31]

Answer:

The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame

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From the question we are told that

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The velocity of the spacecraft is mathematically evaluated as

$v_s = \frac{d }{t}$

substituting values

$v_s = \frac{72000 * 3.0*10^{8} }{90000}$

$v_s = 2.40*10^{8} \ m/s$

The time elapsed in the spacecraft’s frame is mathematically evaluated as

$T = t * \sqrt{ 1 - \frac{v^2}{c^2} }$

substituting value

$T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }$

$T = 54000 \ s$

=> $T = 15 \ hours$

So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame

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3 years ago

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liberstina [14]

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$h = \frac{1}{2} gt^2$

Here,

$h = 8 m$

$g = 9.8 m/s^2$

Replacing,

$8 = 0.5 * 9.8 * t^2$

$t = 1.277 sec$

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$a = \sqrt{a_x^2 + a_y^2}$

$a_x = \frac{1.9}{0.5} = 3.8 m/s^2$

$a_y = g = 9.8 m/s^2$

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3 years ago