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lara [203]
3 years ago
14

A capacitor consists of two square plates, 8.7 cm on a side, separated by a 2.0 mm air gap. How much energy would be stored in t

he capacitor if a mica dielectric is placed between the plates? Assume the mica is 2.0 mm thick (and therefore fills the space between the plates).The charges on the plates are equal and opposite and of magnitude 240 μC .
Physics
1 answer:
slavikrds [6]3 years ago
5 0

Answer:

122.84 J

Explanation:

Since plate is square, area, A is given by (8.7/100)^{2}=0.007569m^{2}

The distance between plates, d, is given in the question as 2mm=0.002m

Charge on plate, Q, as given in the question is 240 \mu c

Assuming mica dielectric constant, k of 7

Capacitance, C is given by

C=\frac {k\epsilon_{o}A}{d}=\frac {(7)(8.85*10^{-12})(0.007569)}{0.002}=2.34*10*^{-10}F

Stored energy, E is given by

E=\frac {Q^{2}}{2C}=\frac {(240*10^{-6})^{2}}{2*(2.34*10^{-10})}=122.84J

Therefore, the stored energy is 122.84 J

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According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this
Andrew [12]

Answer:

Part a)

\rho = 0.55 g/cm^3

Part b)

\rho_L = 1.49 g/cm^3

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

Explanation:

Part a)

As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force

So here we know

buoyancy force is given as

F_b = \rho_w V_{sub} g

F_b = (1 g/cm^3) (30 - 13.5) Ag

F_b = 16.5 Ag

Now we know that the weight of the cylinder is given as

W = \rho (30 cm)A g

now we have

\rho (30 cm) A g = 16.5 A g

\rho = 0.55 g/cm^3

Part b)

When the same cylinder is floating in other liquid then we will have

F_b = \rho_L (30 - 18.9 )A g

so we have

\rho_L (11.1) Ag = 0.55(30) Ag

\rho_L = 1.49 g/cm^3

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

3 0
3 years ago
A light ray in air enters and passes through a block of glass. What can be stated with regard to its speed after it emerges from
Alik [6]

Answer:

Speed is same as that before it entered glass.

Explanation:

Given:

A light ray enters and passes through the glass as shown in the diagram.

We have to analyze its speed.

Speed of light in air is 3\times 10^8\ ms^-^1 and speed of light in glass is 2.25\times 10^8\ ms^-^1

Whenever a light ray enters a glass block or slab there is bending of light at the interface of the two media.

So speed of light will decrease in glass medium and again it passes to the air.

So

Speed of light in air will again increase or will be equivalent to the earlier speed when it was entering the glass block.

Finally

Speed is same as that before it entered glass as it in the same medium (air).

6 0
3 years ago
Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.
olga2289 [7]

Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

     E = k q / r²

Point (0,0,0)

We calculate for the charge -q which is at a distance R = a

   E1 = k (-q) / a²

   E1 = - kq / a²

As the test charge is positive in the field it goes to the left, attractive force

We calculate for the charge that is also at R = a

    E2 = k q / a²

This field goes to the left, repulsive force

We find the total electric field

    Et = E1 + E2

    Et = kq / a² + kq / a²

    Et = 2 k q / a²

Point (0,0, R)

We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

     E1 = k q / (R + a)²

     E2 = kq / (R-a)²

     Et = kq [1 / (R + a)² + 1 / (R-a)²]

     Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

     Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that  R> a we can despise in the patents "a"

     (R² + a²) = R² (1+ a² / R²) ≈ R²

     (R + a)² = R² (1 + a / R)² ≈ R²

     (R- a)²  = R² (1-a / R)² ≈ R²

Substituting in the total electric field

     Et = kq {2 R²) / [R²R²]}

     Et =kq 2 / R²

7 0
3 years ago
Why is the motion of an athlete moving along the circular path with Constant speed considered to be an accelerated motion?
Strike441 [17]

The speed is changing its direction all the time. There is an acceleration which changes the direction of the speed – that is called centripetal acceleration. Only uniform linear motions are considered to have no acceleration.

This is the general formula for acceleration

a = dv/dt

When calculating dv, you should keep in mind the change in the velocity vector’s direction. You can easily see in a graph that with dt tending to 0 (so the length of the arc covered is also tending to 0), the difference between vectors Vf and V0 has a direction which is perpendicular to velocity (the shorter the arc, the closest the angle is to 90 degrees).

There is a formula (which can be deducted from the previous formula) which allows you to calculate the acceleration:

a = v^2/r

Let’s talk about the units:

v is in m/s

r is in m

so v^2/r

is in (m/s)^2/m = (m^2/s^2)/m = m/s^2

which is the same unit as dv/dt:

dv/dt = (m/s)/s= m/s^2

5 0
3 years ago
The kinetic energy of the molecules inside the balloon _______ which
11111nata11111 [884]

Answer:

Increase,.faster

Explanation:

The kinetic energy of the molecules inside the balloon

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I hope this helps you :)

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3 years ago
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