A parallel plate vacuum capacitor has 8.40 J of energy stored. The separation between plates is 2.30 mm. If the separation is decreased to 1.15 mm what is the energy stored if (a) the charge Q on the plates is held constant, and (b) the voltage V across the plates is held constant

Answers

Answer 1
Answer:

(a) 4.20 J

(b) 16.74 J

Explanation:

For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;

C = A∈₀ / d              --------------------(i)

Where;

∈₀ = constant called permittivity of vacuum.

The energy U stored in such capacitor is given by;

U = [tex]\frac{1}{2}[/tex]CV²             ----------------------(ii)

or

U =  [tex]\frac{1}{2}[/tex](Q²/C)        -------------------(**)

Where;

V = potential difference or voltage across the plates.

Q = charge on the plates.

(a) If the charge is held constant

Combine equations (i) and (**) to give;

U =  [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)     -----------------------(iii)

From the question;

The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e

U = 8.40J

d = 2.30mm = 0.023m

Substitute these values into equation (iii)

8.40 =  [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)

8.40 =  [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)

Multiply through by 2

2 x 8.40 = Q² x (0.023 / A∈₀)

16.80 = Q² x (0.023 / A∈₀)

Divide through by 0.023

16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023

730.4 = Q² / (A∈₀)

Make Q² subject of the formula

Q² = 730.4(A∈₀)

Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e

d = 1.15mm = 0.0115m

Q = constant [this means that Q² still remains 730.4(A∈₀) ]

The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;

U =   [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)

U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)

U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)

U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)

U = 4.20J

Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J

(b) If the voltage is held constant

Combine equations (i) and (ii) to give;

U =  [tex]\frac{1}{2}[/tex](A∈₀ / d)V²     -----------------------(iv)

From the question;

The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e

U = 8.40J

d = 2.30mm = 0.023m

Substitute these values into equation (iv)

8.40 =  [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²

Multiply through by 2 x 0.023

2 x 0.023 x 8.40 = (A∈₀)V²

2 x 0.023 x 8.40 = (A∈₀)V²

0.385 = (A∈₀)V²

Make V² subject of the formula

V² = 0.385/(A∈₀)

Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e

d = 1.15mm = 0.0115m

V = constant [this means that V² still remains 0.385/(A∈₀) ]

The energy stored is found by substituting these values of d and V² into equation (iv) as follows;

U =  [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]  

U = [tex]\frac{1}{2}[/tex](0.385/0.0115)

U = 16.74

Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J


Related Questions

Once a disk forms around a star, the process of planetary formation can begin. Rank the evolutionary stages for the formation of planets from earliest to latest.

a. Small clumps of matter stick together via the process of accrection to form plantesimals a few hundred kilometers in diameter
b. Dust keeps matter inside the disk cool long enough for planet formation to start
c. Planetisimals begin to accrete, forming protoplanets
d. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter
e. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star

Answers

Answer: See explanation

Explanation:

The evolutionary stages for the formation of planets from earliest to latest will be:

1. Dust keeps matter inside the disk cool enough for planet formation to start

2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.

3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.

4. Planetesimals begin to accrete, forming protoplanets.

5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star

What is the momentum of a lab cart with a mass of 0.60 [kg] and a speed of 2.2 [m/s]?

-3.67 [kg m/s)
-0.27 [kg m/s]
-1.32 [kg m/s)
-0.82 [kg m/s)

Answers

Answer:

1.32kgm/s

Explanation:

use the formula: p=mv

The momentum of a lab cart  = -1.32 kg m/s

What is momentum?

It is measure of the inertia of a body/ object .It can be calculated by multiplying mass with velocity .

General formula for momentum = M = m * v

given

mass = 0.60  kg

speed = 2.2 m/s

velocity = - 2.2 m/s ( answer is in negative , since mass is a scaler quantity  but velocity is a vector quantity  hence ,  velocity can be negative )

momentum = mass * velocity

                   = 0.60 * (-2.2 ) = -1.32 kg m/s

The momentum of a lab cart  =c) -1.32 kg m/s

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A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is fixed in place a distance d above the first plate. Assume that d is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V.

Required:
a. What is the tension in the cable?
b. Compute the energy stored in the electric field after the top plate was raised.

Answers

Answer:

A) F = V²E_o•πr²/2d²

B) U = E_o•Aπr²V²/2d

Explanation:

A) Since we have two circular plates, the formula for the electric field is expressed as;

E = V/d

Where;

V is voltage

d is distance

However, the net electric field produced is given by;

E' = V/2d

The tension in the cable can then be expressed as;

F = qE'

Where q is charge

Thus;

F = qV/2d - - - (eq 1)

We also know that;

C = q/V = E_o•A/d

A is area = πr²

Thus;

q/V = E_o•πr²/d

q = VE_o•πr²/d

Let's put VE_o•πr²/d for q in eq 1 to get;

F = V²E_o•πr²/2d²

B) formula for the energy stored in the electric field is;

U = ½CV²

From earlier, we saw that; C = E_o•A/d

Thus;

U = ½E_o•AV²/d

A = πr²

Thus;

U = E_o•Aπr²V²/2d

People who do very detailed work close up, such as jewellers, often can see objects clearly at much closer distance than the normal 25 cm. a. What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.5 cm

Answers

This question is incomplete, the complete question is;

People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm.

a) What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.50 cm? (Assume the lens-to-retina distance is 2.00 cm.)

b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)  __ mm

Answer:

1) the power in D of the eyes of a woman is 61.7647 D

2) the size in mm of an image of a 8.00 mm object is -1.882 mm

Explanation:

Given the data in the question;

a) power in D of the eyes of woman who can see an object clearly at a distance of only 8.5 cm and the lens-to-retina distance is 2.00 cm,

so

u = 8.5 cm = ( 8.5 / 100 )m = 0.085 m

v = 2.00 cm = ( 2 / 100 )m =  0.02 m

Now, we know that power of lens p = 1 / u + 1 / v

so we substitute

p = ( 1 / 0.085 ) + ( 1 / 0.02 )

p = 11.7647 + 50

p = 61.7647 D

Therefore,  the power in D of the eyes of a woman is 61.7647 D

b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)

we know that;

m = -v / u

we substitute

m = -0.02 / 0.085

m = -0.2353

since H₀ = 8.0 mm

H[tex]_i[/tex] = m × H₀

H[tex]_i[/tex] = -0.2353 × 8.0

H[tex]_i[/tex] = -1.882 mm

the size in mm of an image of a 8.00 mm object is -1.882 mm

Two people that have identical weight are holding onto a massless pole while standing on horizontal frictionless ice. 1)If the guy on the left starts to pull on the pole, where do they meet

Answers

Answer:

Explanation:

From the missing image attached below, it is obvious that there no external force. This implies that they cannot change their position by merely just pulling the ropes. As a result, there will be no movement and no net force will exist.

So, if there is no external force;

The center of mass of the two people is:

[tex]X_{cm}= \dfrac{m_1x_1+m_2x_2}{m_1+m_2} \\ \\ X_{cm}= \dfrac{m(-3m)+m(+3m)} {m+m}\\ \\ X_{cm}= \dfrac{0}{2m} \\ \\ X_{cm} =0[/tex]

Thus, In the system, no movement occurs and all forces remain the same.

Please helppppppp!!!!!!!!!!!!!!

Answers

Answer:

circuit breaker

Explanation:

A circuit breaker is a device used for electrical safety. It consists of a switch designed to protect an electrical circuit from damage that may result from heating due to overload in the circuit.

Its basic function is to interrupt current flow through its switch that consists of metal stripe which bends when it gets hot.

Fuse has similar action with circuit breaker, the only difference is that fuse can only be used once because it melts when it gets hot.

Therefore, the correct answer is "circuit breaker"

A 4.9 A current is set up in a circuit for 4.7 min by a rechargeable battery with a 12 V emf. By how much is the chemical energy of the battery reduced

Answers

Answer:

E = 16581.6 J

Explanation:

Given that,

Current, I = 4.9 A

Time for which the current is set up, I = 4.7 min = 282 s

The voltage of the battery, V = 12 V

We need to find how much chemical energy of the battery reduced. Let It is E. We know that,

E = P t

Where

P is power of battery, P = VI

So,

[tex]E=VIt[/tex]

Put all the values,

[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]

So, 16581.6 J of chemical energy of the battery is reduced.

Engineers are working on a design for a cylindrical space habitation with a diameter of 7.50 km and length of 29.0 km. The habitation will simulate gravity by rotating along its axis. With what speed (in rad/s) should the habitation rotate so that the acceleration on its inner curved walls equals 8 times Earth's gravity

Answers

Answer:

The speed will be "0.144 rad/s".

Explanation:

Given that,

Diameter,

d = 7.50 km

Radius,

R = [tex]\frac{7.5}{2} \ Km[/tex]

Acceleration on inner curve,

= 8 times

Now,

As we know,

⇒ [tex]\omega^2R=8g[/tex]

or,

⇒ [tex]\omega=\sqrt{\frac{8g}{R} }[/tex]

On substituting the values, we get

⇒     [tex]=\sqrt{\frac{8\times 9.8}{\frac{7.5}{2}\times 10^3 } }[/tex]

⇒     [tex]=\sqrt{\frac{78.4}{3750} }[/tex]

⇒     [tex]=\sqrt{0.0209}[/tex]

⇒     [tex]=0.144 \ rad/s[/tex]

A voltage of 75 V is placed across a 150 Ω resistor. What is the current through the resistor?

Answers

Answer:

0.5 A

Explanation:

Applying,

V = IR.................. Equation 1

Where V = Voltage, I = current, R = Resistance.

make I the subject of the equation

I = V/R............... Equation 2

From the question,

Given: V = 75 V, R = 150 Ω

Substitute these values into equation 2

I = 75/150

I = 0.5 A.

Hence the cuurent through the resistor is 0.5 A

Who is a socio-economically disadvantaged child? Explain any four ways forhelping such a child

Answers

Explanation:

A socio-economically disadvantaged child is the one who is disadvantaged in terms of social position and economic position. Such children have limited resources in terms of education, money and future options. Four ways of helping such children are as follows:

1) Help them in education - You can help such children by giving them free tuition. If you belong to a well off family, you can get them admitted in schools as well. Provide them with books and uniform.

2) Encourage them to do well in school and pursue their passions.

3) Provide them with meals, if they do not have access to regular meals.

4) encourage them to go school regularly.

Answer:

sorry i dont know the answer bit mark me as BRAINLIST

Air in a thundercloud expands as it rises. If its initial temperature is 292 K and no energy is lost by thermal conduction on expansion, what is its temperature when the initial volume has tripled

Answers

Answer:

Explanation:

It is a case of adiabatic expansion .

[tex]T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}[/tex]

T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.

Given ,

V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .

[tex]( 3 )^{\gamma-1}= \frac{292}{ T_2}[/tex]

[tex]( 3 )^{1.4-1}= \frac{292}{ T_2}[/tex]

1.552 = 292 / T₂

T₂ = 188 K .

the boiling point of F2 much lower than the boiling point of NH3​

Answers

Answer:yeah it A

Explanation:

Dos cargas puntuales iguales y negativas, q1=q2=-24micro C se localizan en x=0 y y=38m y x=0 y y=-7m, respectivamente. Calcula la magnitud de la fuerza electrica total en N que ejercen estas dos cargas sobre una tercera, tambien puntual, Q=26micro C en y=0 y x=16m

Answers

Answer:

 F_net = 9.87 10⁻⁴ N

Explanation:

Let's use that force is a vector magnitude

         ∑ F = F₁₃ + F₂₃

De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract

 

the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m

       ∑ F = F₁₃ - F₂₃

       F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]

in this case q₁ = q₂ = q

       F_net = k q q₃  (  )

 

      let's look for the distance

      r₂₃ = y₂ - y₃

      r₂₃ = -7 -16

      r₂₃ = - 23 m

       

      r₁₃ = 38 - 16

      r₁₃ = 22 m

let's calculate

      F_net = 9 10⁹ 24 26 10⁻¹² ( )

      F_net = 5.616 ( 1.758 10⁻⁴ )

      F_net = 9.87 10⁻⁴ N

In some situations, matter demonstrates wave behavior rather than particle behavior. This is best illustrated by which phenomenon? A. Emission spectra of atoms B. Blackbody radiation C. Interference patterns of electrons D. Photoelectric effect​

Answers

Answer:

In some situations, matter demonstrates wave behavior rather than particle behavior. This is best illustrated by which phenomenon is:

C. Interference patterns of electrons.

Answer:

C. Interference patterns of electrons

By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect the earth's rotational effects

Answers

Answer:

[tex]Weight\ loss=1.6321N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=85.9kg[/tex]

Altitude [tex]h= 6.33 km[/tex]

Let

Radius of Earth [tex]r=6380km[/tex]

Gravity [tex]g=9.8m/s^2[/tex]

Generally the equation for Gravity at altitude is mathematically given by

 [tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]

 [tex]g_s=9.781m/s^2[/tex]

Therefore

Weight at sea level

 [tex]W_s=9.8*85.9[/tex]

 [tex]W_s=841.82N[/tex]

Weight at 6.33 altitude

 [tex]W_a=9.781*85.9[/tex]

 [tex]W_a=840.2N[/tex]

Therefore

 [tex]Weight loss=W_s-W_b[/tex]

 [tex]Weight loss=841.82-840.2[/tex]

 [tex]Weight loss=1.6321N[/tex]

Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Treat this problem as one-dimensional, and use the relativistic relationbetweenEandp. Give your answer to 2 significant figures. (The large value you will find is a strongargument against the presence of electrons inside nuclei, since no known mechanism could contain anelectron with this much energy.)

Answers

Answer:

39.40 MeV

Explanation:

Determine the minimum possible Kinetic energy

width of region = 5 fm

From Heisenberg's uncertainty relation below

ΔxΔp ≥ h/2 , where : 2Δx = 5fm ,  Δpc = hc/2Δx = 39.4 MeV

when we apply this values using the relativistic energy-momentum relation

E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,

Also in a nuclear confinement ( E, P >> mc )

while The large value will portray a Non-relativistic limit  as calculated below

K = h^2 / 2ma^2 = 1.52 GeV

A FBD of a rocket launching into space should include:

Answers

Answer:

Explanation:

the force of the rocket engine pushing it up,  the force of gravity pulling it down,    maybe some force of air resistance as the rocket goes fast,   hmmm    Free Body Diagrams  (FBD)  should have any and all forces on the model,  unless they are negligible . or so slight they really make little difference in the total  outcome.  

A 9.2 resistor and a 5.2 resistor are connected in series with a battery. The potential difference across the 5.2 resistor is measured as 12 V. Find the potential difference across the battery.

Answers

Answer:

V = 33.12 V

Explanation:

Given that,

A 9.2 resistor and a 5.2 resistor are connected in series with a battery.

The potential difference across the 5.2 resistors is measured as 12 V.

In series combination of resistors, the current is same throughout the circuit. Let I is the current. Using Ohm's law,

[tex]I=\dfrac{V}{R}[/tex]

Put all the values,

[tex]I=\dfrac{12}{5.2}\\\\I=2.30\ A[/tex]

The equivalent resistance is :

R = 9.2 + 5.2

= 14.4 ohms

Let V os the potential difference across the battery. So,

[tex]V=2.3\times 14.4\\\\V=33.12\ V[/tex]

So, the potential difference across the battery is equal to 33.12 V.

Many adventures like to go rafting on the Colorado River through Grand Canyon National Park. There are many locations where the river becomes more narrow, both the distance between the canyon walls as well as the depth changes due to debris like boulders on the bottom of the river; this leads to changes in the water speed. In the park, the Colorado River has an average width of 100m and an average depth of 8m, and an average speed of 3 m/s. At the Lava Falls Rapids, the river has an average width of about 25m and an average depth of about 15m. What is the approximate speed of the water in this location

Answers

Answer:

6.4 m/s

Explanation:

Given that :

The average width of the Colorado river = 100 m

Average depth of the river is = 8 m

Therefore, area = [tex]$A_1= 100 \ m \times 8 \ m$[/tex]

Speed of the river, [tex]$v_1 = 3 \ m/s$[/tex]

After the lava falls on the river,

Width of the river becomes = 25 m

Depth of the river became = 15 m

Therefore, area = [tex]$A_2= 25 \ m \times 15 \ m$[/tex]

Now, since the volume flow rate of the Colorado river is same, then from the Continuity equation,

[tex]$Q_1=Q_2$[/tex]

[tex]$A_1v_1=A_2v_2$[/tex]

∴ [tex]$100 \times 8 \times3 = 25 \times 15 \times v_2$[/tex]

[tex]$v_2=\frac{100 \times 8 \times 3}{25 \times 15}$[/tex]

    = 6.4 m/s

Therefore, the speed of the river in this location is 6.4 m/s

Compared to its weight on Earth, a 5kg object on the moon will weigh
The same amount
Less
More

Answers

Answer:

Less

Explanation:

Weight is a force measurement. The object's mass is 5kg not its weight. To find its weight you have to take the mass of an object and multiply it by the acceleration of gravity. The acceleration of gravity is greater on earth than on the moon so therefore the object will weigh less on the moon.

The maximum amount of pulling force a truck can apply when driving on
concrete is 8760 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
O A. 8760 N
O B. 12,680 N
O C. 10,950 N
O D. 7240 N

Answers

Answer:

8760 N

Explanation:

think this is the right answer :)

Keesha is looking at a beetle with a magnifying glass. She wants to see an upright, enlarged image at a distance of 25 cm. The focal length of the magnifying glass is +5.0 cm. Assume that Keesha's eye is close to the magnifying glass.
(a) What should be the distance between the magnifying glass and the beetle?
(b) What is the angular magnification?

Answers

Answer:

a)   p = 4.167 cm, b)   m = + 6

Explanation:

a) For this exercise we must use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

In this case the distance to the image q = 25 cm and the focal length is f = 5.0 cm

Since the object and its image are on the same side of the lens, the distance to the image by the sign convention must be negative.

       [tex]\frac{1}{p } = \frac{1}{f} - \frac{1}{q}[/tex]

      [tex]\frac{1}{p} = \frac{1}{5} - \frac{1}{-25}[/tex]

      [tex]\frac{1}{ p}[/tex] = 024

      p = 4.167 cm

 

b) angular magnification

     m = h ’/ h = - q / p

     m = - (-25) /4.167

     m = + 6

the positive sign indicates that the image is straight and enlarged

PLEASE HELPPPPPP ME PLS PLS​

Answers

power = work / time --> time = work / power = 3600 J / 275 watts = 13.1 seconds.

effieiency of simple machine is always less than 100% why​

Answers

Answer:

efficiency of a machine is less than 100% because some part is energy is utilized to overcome some opposing forces like friction which is wasted as heat ,sound energy etc

Explanation:

Answer this
a) which ink is likely to be pure? Why?
b) What does the chromatography tell us about ink Y
c) Why are the three different spots separated out from ink Y found at different heights?​

Answers

Answer:

a) Ink X is likely to be pure because it only contain 1 spot.

b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.

c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.

The different spots from Y are found at various heights because they represent different compounds.

What is chromatography?

The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.

We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.

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The index of refraction of n-propyl alcohol is 1.39. Find the angle of refraction of light in that medium if light comes from air with an angle of incidence of 55 degrees.

Answers

Answer:

36.11 degrees

Explanation:

index of refraction n = sin i/sinr

i is the angle of incidence

r is the angle of refraction

Substitute into the expression

1.39 = sin55/sin(r)

1.39 = 0.8191/sin(r)

sin(r) = 0.8191/1.39

sin(r) = 0.5893

r = arcsin(0.5893)

r = 36.11

hence the angle of refraction of light is 36.11 degrees

One coulomb represents how many electrons?
a. 1 electron
b. 100 electrons
C. 6.25 quintillion electrons
d. 6.25 million-million electrons
e, none of the above

Answers

Answer:

6.24 x 1018 electrons.

Explanation:

So I think C

If two dogs are pulling a bone with force-20Newtons in opposite direction, then the resultant force is​

Answers

Answer:

Newtons third law of motion: Balanced forces

Every action has a corresponding and opposing response, according to Newton's third law of motion. As a result, forces always work in pairs.   Once more, tug-of-war is a prime illustration.

What force in opposite direction follow newton law?

The third law of motion by Newton states that equal, but diametrically opposed forces always act in pairs. There is an equal but opposite reaction to every action, to put it another way.

The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope. There is hence no motion.

Although equal and opposite in nature, action and reaction forces cannot be balanced since they act on separate things and do not cancel one another out.

Therefore, This means that when you push against a wall, the wall pushes back against you with an equal amount of force.

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Last stage of a medium mass star

Answers

White Dwarf Stage

This ring is called a planetary nebula. when the last of the helium atoms in the core are fused into carbon atoms, the medium size star begins to die. Gravity causes the last of the star's matter to collapse inward and compact. This is the white dwarf stage.

the last stage of a medium mass star is the white dwarf stage

Parallel rays from a distant object are traveling in air and then are incident on the concave end of a glass rod with a radius of curvature of 15.0 cm. The refractive index of the glass is 1.50. What is the distance between the vertex of the glass surface and the image formed by the refraction at the concave surface of the rod? Is the image in the air or in the glass?

Answers

Answer:

the distance of image from the vertex is 45 cm and the image formed is in the glass.

Explanation:

distance of object, u = - infinity

radius of curvature, R = - 15 cm

refractive index, n = 1.5

Let the distance of image is v.

Use the formula

[tex]-\frac{n1}{u}+\frac{n2}{v}=\frac{n2- n1}{R}\\\\-\frac{1}{\infty }+\frac{1.5}{v}=\frac{1.5-1}{-15}\\\\v=45 cm[/tex]

The image is in the glass.  

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