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Rudiy27
3 years ago
9

A capaitor has two parallel plates sepearted by 2mm and is connected across a 50V battery. i. What is the electric field between

the plates? ii. What is the surface charge density? iii. How much charge is stored on each plate if the area is 0.1m. iv. Calculate the capacitance. v. How much energy is stored in this capacitor? vi. What is the energy density of the electric field in the capacitor?
Physics
1 answer:
Oduvanchick [21]3 years ago
8 0

Answer:

i E=V/d=50/2*10^-3=25*10^3 N/C

ii It is a (+) and (-)

iii C=εA/d

C=12.56*10^-8 * 0.1/2*10^-4

C=62.83 μF

Q=CV=50*6.283*10^-6

Q=314 μC

iv E=0.5 QV

=0.5(50*314*10^-6)

=7850 μJoule

Explanation:

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Answer:

d=1.29*10^{-6}m

Explanation:

From the question we are told that:

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Second bright fringe y_2= 0.803 m

Let

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Generally the equation for Interference is mathematically given by

y=frac{n*\lambda*D}{d}

Where

d=\frac{n*\lambda*D}{y}

d=\frac{2*431 *10^{-9}m*1.4}{0.803}

d=1.29*10^{-6}m

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A plane traveled west for 4.0 hours and covered a distance of 4,400 meters what’s the velocity
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0.31m/s

Explanation:

Given parameters:

Time of travel = 4hrs = 4 x 60 x 60 = 14400s

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Unknown:

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Solution:

Velocity is defined as the displacement  per unit of time. It is expressed in m/s or km/hr:

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3 years ago
A spherical shell of radius 9.0 cm carries a uniform surface charge density σ= 9.0 nC/m2. The electric field at r= 9.1 cm is app
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Answer:

995.12 N/C

Explanation:

R = 9 cm = 0.09 m

σ = 9 nC/m^2 = 9 x 10^-9 C/m^2

r = 9.1 cm = 0.091 m

q = σ x 4π R² = 9 x 10^-9 x 4 x 3.14 x 0.09 x 0.09 = 9.156 x 10^-10 C

E = kq / r^2

E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)

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8 0
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10.
myrzilka [38]

Answer:

<em>The new period of oscillation is D) 3.0 T</em>

Explanation:

<u>Simple Pendulum</u>

A simple pendulum is a mechanical arrangement that describes periodic motion. The simple pendulum is made of a small bob of mass 'm' suspended by a thin inextensible string.

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

Where L is its length and g is the local acceleration of gravity.

If the length of the pendulum was increased to 9 times (L'=9L), the new period of oscillation will be:

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Taking out the square root of 9 (3):

T'=3*2\pi \sqrt{\frac{L}{g}}

Substituting the original T:

T'=3*T

The new period of oscillation is D) 3.0 T

4 0
3 years ago
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