Light of intensity I0 and polarized horizontally passes through three polarizes. The first and third polarizing axes are horizontal, but the second one is oriented 20.0� to the horizontal. In terms of I0, what is the intensity of the light that passes through the set of polarizers?
A) 0.442 I0
B) 0.180 I0
C) 0.780 I0
D) 0.883 I0

Answers

Answer 1

Answer:

Option C.

Explanation:

Suppose that we have light polarized in some given direction with an intensity I0, and it passes through a polarizer that has an angle θ with respect to the polarization of the light, the intensity that comes out of the polarizer will be:

I(θ) = I0*cos^2(θ)

Ok, we know that the light is polarized horizontally and comes with an intensity I0

The first polarizer axis is horizontal, then the intensity after this polarizer is:

then θ = 0°

I(0°) = I0*cos^2(0°) = I0

The intensity does not change. The axis of polarization does not change.

The second polarizer is oriented at 20° from the horizontal, then the intensity that comes out of this polarizer is:

I(20°) =  I0*cos^2(20°) = I0*0.88

And the axis of polarization of the light that comes out is now 20° from the horizontal

Now the light passes through the last polarizer, which has an axis oriented horizontally, so the final intensity of the light will be:

note that here the initial polarization is  I0*0.88

and the angle between the axis is 20° again.

Then the final intensity is:

I(20°) =  I0*0.88*cos^2(20°) = I0*0.78

Then the correct option is C.


Related Questions

A 55 g soapstone cube--a whisky stone--is used to chill a glass of whisky. Soapstone has a density of 3000 kg/m3, whisky a density of 940 kg/m3. What is the approximate normal force of the bottom of the glass on a single stone?

Answers

Answer:

[tex]N=0.37N[/tex]

Explanation:

Mass [tex]m=55g=>0.055kg[/tex]

Soapstone Density [tex]\rho_s=3000kg/m^2[/tex]

Whisky Density [tex]\rho_w=940kg/m^2[/tex]

Generally the equation for Force is mathematically given by

 [tex]F=U+N[/tex]

Therefore

 [tex]N=m*g-(\frac{m}{\rho_s})*\rho_w*g[/tex]

 [tex]N=0.055*9.81 - {(\frac{0.055}{3000})*940*9.81}[/tex]

 [tex]N=0.37N[/tex]

A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140 N. Fp is parallel to the displacement of the block. The final speed of the block is 2.35 m/s.
a) How much work was converted to thermal energy? What work did friction do on the box?
b) What is the coefficient of friction?

Answers

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

[tex]u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\[/tex]

Using formula:

[tex]v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\[/tex]

   [tex]= 0.55 \ \frac{m}{sec^2}\\\\[/tex]

[tex]F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\[/tex]

Calculating the Work by net force

[tex]W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\[/tex]

The above work is converted into thermal energy.

Now,

[tex]F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J[/tex]

The gravitational field strength due to its planet is 5N/kg What does it mean?

Answers

Answer:

The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.

Weight can be calculated using the equation:

weight = mass × gravitational field strength

This is when:

weight (W) is measured in newtons (N)

mass (m) is measured in kilograms (kg)

gravitational field strength (g) is measured in newtons per kilogram (N/kg)

What is significant about the primary colors of pigments?
They can be mixed together to make almost any other color.
Any two primary colors of pigments combine to make white pigment.
Each primary color of pigment absorbs all other colors.
Any two primary colors of pigments combine to make black pigment.

Answers

Answer:

They can be mixed together to make almost any other color.

Explanation:

All the three primary colors can mix to form white color.

Blue and red mix to form a black color.

A loop of wire is in a magnetic field such that its axis is parallel with the field direction. Which of the following would result in an induced emf in the loop?
A. Moving the loop outside of the magnetic field region.
B. Change the diameter of the loop.
C. Change the magnitude of the magnetic field.
D. Spin the loop such that its axis does not consistently line up with the magnetic field direction.

Answers

Answer:

All the given options will result in an induced emf in the loop.

Explanation:

The induced emf in a conductor is directly proportional to the rate of change of flux.

[tex]emf = -\frac{d \phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux\\\\\phi = BA\ cos \theta[/tex]

where;

A is the area of the loop

B is the strength of the magnetic field

θ is the angle between the loop and the magnetic field

Considering option A, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.

Considering option B, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.

Option C has a similar effect with option A, thus both will result in an induced emf.

Finally, considering option D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will change the angle between the loop and the magnetic field. This effect will also result in an induced emf.

Therefore, all the given options will result in an induced emf in the loop.

The north pole of magnet A will __?____ the south pole of magnet B

Answers

Answer:

A will attract

B will repare

A cannon and a supply of cannonballs are inside a sealed railroad car of length L, as in Fig. 7-33. The cannon fires to the right; the car recoils to the left. The cannonballs remain in the car after hitting the far wall. (a) After all the cannonballs have been fired, what is the greatest distance the car can have moved from its original position

Answers

Answer:

Initially let n cannonballs with a total mass of m be to the left of the center of mass at L /2 and the mass of the car at L/2

x1 =  [-m / (m + M)] * L / 2   is the original position of the CM

x2 = (m (x + L/2) + M x) / (m + M) * L/2 final position of CM with all cannon balls to the right

[-m x - m L / 2 + m x - M x] / (M + m) * L/2

= - ( m L / 2 + M x) / (m + M) * L/2 = Xcm

Check the math, but maximum distance occurs when the cannonballs of mass m move from -L/2 to L/2 and the car of mass M moves from zero to -x

A 64-ka base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.

Required:
a. How much mechanical energy is tout due to friction acting on the runner?
b, How far does he slide?

Answers

Answer:

Explanation:

From the given information:

mass = 64 kg

speed = 3.2 m/s

coefficient of friction [tex]\mu =[/tex] 0.70

The mechanical energy touted relates to the loss of energy in the system as a result of friction and this can be computed as:

[tex]W = \Delta K.E[/tex]

[tex]\implies \dfrac{1}{2}m(v^2 -u^2)[/tex]

[tex]= \dfrac{1}{2}(64.0 \kg) (0 - (3.2 \ m/s^2))[/tex]

Thus, the mechanical energy touted = 327.68 J

According to the formula used in calculating the frictional force

[tex]F_r = \mu mg[/tex]

= 0.70 × 64  kg× 9.8 m/s²

= 439.04 N

The distance covered now can be determined as follows:

d = W/F

d = 327.68 J/  439.04 N

d = 0.746 m

Momentum is defined as mass ___ velocity

Answers

Answer:
p=mv
Momentum = mass x velocity

momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.

I hope this helped :)
p=mv momentum is the answer

Find the starting pressure of CCl4 at this temperature that produces a total pressure of 1.1 atm at equilibrium. Express the pressure in atmospheres to three significant figures.

Answers

The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.

Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.

Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Explanation:

The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.

[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.

                   [tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Initial:            a                0          0

Equilibrium:  (a - x)          0          2x

Total pressure = (a - x) + 2x = a + x

As it is given that the total pressure is 1.1 atm.

So, a + x = 1.1

a = 1.1 - x

Now, expression for equilibrium constant for this equation is as follows.

[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]

Hence, the value of 'a' is calculated as follows.

a + x = 1.1 atm

a = 1.1 atm - x

  = 1.1 atm - 0.31 atm

  = 0.79 atm

Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

What are stepdown transformers used for

Answers

Answer:

Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.

Explanation:

pls mark me as brainlist

Thanks a lot

PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)

Answers

Answer:

multiply mp and c^2

Explanation:

e=mc^2

A system is acted on by its surroundings in such a way that it receives 50 J of heat while simultaneously doing 20 J of work. What is its net change in internal energy

Answers

Answer:

30J

Explanation:

Given data

The total quantity of heat recieved= 50J

Quantity of heat used to do work= 20J

Hence the net change is

ΔU= Total Heat - Net work

ΔU= 50-20

ΔU= 30J

Hence the change in the internal energy is 30J

a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?​

Answers

Answer:

Explanation:

Speed= distance/time

Or time = distance/speed

According to your question

Speed=15m/s

and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s

D= 1.2km = 1.2×1000m =1200meter

Time = distance/ speed

1200/15 =80second

Or. 1min and 20 sec will be your answer.

need help pleaseee,question is in the pic​

Answers

Explanation:

For engine 1,

Energy removed = 239 J

Energy added = 567 J

[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]

For engine 2,

Energy removed = 457 J

Energy added = 789 J

[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]

For engine 3,

Energy removed = 422 J

Energy added = 1038 J

[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]

So, the engine 2 has the highest thermal efficiency.

Part C – RC Circuits in AC Mode 1. Derive Equation 5-6 from Equation 5-5. 2. Using the τ’s you calculated and your measured resistance: a. Calculate the capacitances of the capacitors. b. Compare your calculated and measured values via percent error.

Answers

Answer: hi your question is incomplete attached below is the complete question

1) attached below

2) a)  31 Ω,   302.9 Ω

   b)  17.3 Ω ,  26.4 Ω

Explanation:

1) Deriving Eqn 5-6 from Eqn 5-5

attached below

2) using τ’s calculated and measured resistance

use given data

Tm1 = Rm1 * Cm1

       = 329.3 * 333 * 10^-9 = 109.65 μs

Tm2 = Rm2 * Cm2

       = 329.3 * 200 * 10^-9 = 658.6 μs

a) Capacitance of capacitors

For Cm1

   t₁₂ = 72 μs = R₁' Cm1 log²

 ∴ R₁' = ( 72 * 10^-6 ) / ( 333 * 10^-9 log² )

         = 31 Ω

For Cm2

 t₁₂ = 40 μs  , ∴  R₂' = 302.9 Ω

b) comparing calculated and measured values via percent error

errors

Rm1 - R₁' = 17.3 Ω

Rm2 - R₂' = 26.4 Ω

g A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 3.00 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do

Answers

The desk is in equilbrium, so Newton's second law gives

F (horizontal) = p - f = 0

F (vertical) = n - mg = 0

==>   n = mg

==>   p = f = µn = µmg = 0.400 (80.0 kg) g = 313.6 N

The student pushes the desk 3.00 m, so she performs

W = (313.6 N) (3.00 m) = 940.8 Nm ≈ 941 J

of work.

NEED HELP ASAP- Please show work

The angular position of an object is given by θ = 4t3 +10t −40 , where θ is in radians and t is in seconds what is:

(a) (5 points) The angular velocity at t = 2 s?

(b) (5 points) The angular acceleration at t = 2 s?

Answers

Answer:

Look at work

Explanation:

Θ= 4t^3+10t-40

a) In order to find ω, we need to find displacement so plug in t=2 to find Θ.

Θ= 4*8+20-40=12

use ω=Θ/t

Plug in values

ω=6 rad/s

b) In order to find α we use ω/t.

Plug in values

α=6/2= 3 rad/s^2

When you hammer a nail into wood, the nail heats up. 30 Joules of energy was absorbed by a 5-g nail as it was hammered into place. How much does the nail's temperature increase (in °C) during this process? (The specific heat capacity of the nail is 450 J/kg-°C, and round to 3 significant digits.

Answers

Answer:

13.33 K

Explanation:

Given that,

Heat absorbed, Q = 30 J

Mass of nail, m = 5 g = 0.005 kg

The specific heat capacity of the nail is 450 J/kg-°C.

We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:

[tex]Q=mc\Delta T\\\\\Delta T=\dfrac{Q}{mc}\\\\\Delta T=\dfrac{30}{0.005\times 450}\\\\=13.33\ K[/tex]

So, the increase in temperature is 13.33 K.

g How much buoyancy force, in N, a person with a mass of 70 kg experiences by just standing in air

Answers

Answer:

686.7N

Explanation:

Given data

Mass= 70kg

We know that the buoyant force experienced by the person is equal to the weight of the person

Hence the weight is

Weight = mass* Acceleration

Weight= 70*9.81

Weight= 686.7N

Therefore the weight is 686.7N

The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distance between them is 5000 miles as measured along the earth's surface.
a. Through what angle do you turn, relative to the earth, if you fly from Kampala to Singapore? Give your answer in both radians and degrees.
b. The flight from Kampala to Singapore take 9 hours. What is the plane's angular speed relative to the earth?

Answers

Answer:

a) the required angle in both radian and degree is  1.25 rad and 71.6°

b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

Explanation:

Given the data in the question;

a)

we know that The expression for the angle subtended by an arc of circle at the center of the circle is,

θ = Length / radius

given that Length is 5000 miles and radius is 4000 miles

we substitute

θ = 5000 miles / 4000 miles

θ = 1.25 rad

Radian to Degree

θ = 1.25 rad × ( 180° / π rad )

θ =  71.6°

Therefore, required angle in both radian and degree is  1.25 rad and 71.6°

b)

The flight from Kampala to Singapore take 9 hours.

the plane's angular speed relative to the earth = ?

we know that, the relation between angular velocity and angular displacement is;

ω = θ / t

given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec

we substitute

ω = 1.25 rad / 32400 sec

ω = 3.86 × 10⁻⁵ rad/sec

Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

A body of mass 4kg is moving with a velocity of 108km/h . find the kenetic energy of the body.​

Answers

Answer:

KE = 2800 J

Explanation:

Usually a velocity is expressed as m/s. Then the energy units are joules.

[tex]\frac{108 km}{hr} * \frac{1000m}{1 km} * \frac{1 hour}{3600 seconds} =\frac{108*1000 m}{3600sec}[/tex]

v = 30 m / sec

KE = 1/2 * 4 * (30)^2

KE =2800 kg m^2/sec^2

KE = 2800 Joules

A student measure the length of a laboratory bench with a meter ruler. Which of the following values is the most approbriate way to record the result ? a.4.022m b.4.02m c.4.0m d.4m​

Answers

Answer:

Well a meter stick has increments of a centimeter, and since 1 cm=0.01m he should record it as 4.02m(b)

Explanation:

You place an 8 kg ball on the top of your 2 cm^2 finger tip. Calculate the
PRESSURE. Show MATH, answer and unit.

Answers

Answer:

the pressure exerted by the object is 392,000 N/m²

Explanation:

Given;

mass of the object, m = 8 kg

area of your finger, A = 2 cm² = 2.0 x 10⁻⁴ m²

acceleration due to gravity, g = 9.8 m/s²

The pressure exerted by the object is calculated as;

[tex]Pressure = \frac{F}{A} = \frac{mg}{A} = \frac{8 \times 9.8}{2\times 10^{-4}} = 392,000 \ N/m^2[/tex]

Therefore, the pressure exerted by the object is 392,000 N/m²

Light energy is part of a larger form of energy known as __________.

Answers

Light energy is part of a larger form of energy known as electromagnetic energy. Details about electromagnetic energy can be found below.

What is electromagnetic radiation?

Electromagnetic spectrum is the entire range of wavelengths of all known electromagnetic radiations extending from gamma rays through visible light, infrared, and radio waves, to X-rays.

Visible light is the part of the electromagnetic spectrum, between infrared and ultraviolet, that is visible to the human eye.

Therefore, Light energy is part of a larger form of energy known as electromagnetic energy.

Learn more about electromagnetic spectrum at: https://brainly.com/question/23727978

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Two sinusoidal waves have the same frequency and wavelength. The wavelength is 20 cm. The two waves travel from their respective sources and reach the same point in space at the same time, resulting in interference. One wave travels a larger distance than the other. For each of the possible values of that extra distance listed below, identify whether the extra distance results in maximum constructive interference, maximum destructive interference, or something in-between.
a. 10 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
b. 15 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
c. 20 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
d. 30 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
e. 35 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
f. 40 cm - (A) in-between (2) maximum destructive (3) maximum constructive.

Answers

Answer:

Explanation:

When the path difference is equal to wave length or its integral multiple, constructive interference occurs . If it is odd multiple of half wave length , then destructive interference occurs.

For constructive interference , path diff = n λ

For destructive interference path diff = ( 2n+ 1 ) λ /2

where λ is wave length of wave , n is an integer.

a )

path diff = 10 cm which is half the wavelength , so maximum destructive interference will occur.

b )

path diff = 15 cm which is neither  half the wavelength nor full wavelength , so in between is the right option.

c )

path diff = 20 cm which is equal to  the wavelength , so maximum constructive  interference will occur.

d)

path diff = 30 cm which is 3 times half the wavelength , so maximum destructive interference will occur.

e)

path diff = 35 cm which is neither integral multiple of half the wavelength , nor integral multiple of wavelength so in between is th eright answer.

f )

path diff = 40 cm which is 2 times the wavelength , so maximum constructive  interference will occur

A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an electric field through an electrical potential difference of 330 V. Determine the magnitude of the magnetic field, if the proton travels in a circular path with a radius of 23 cm.

Answers

Answer:

 B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

        Em₀ = U = q V

final point. Right out of the electric field

        em_f = K = ½ m v²

energy is conserved

       Em₀ = Em_f

       q V = ½ m v²

       v = [tex]\sqrt{2qV/m}[/tex]

we calculate

       v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]

       v = [tex]\sqrt{632.3353 \ 10^8}[/tex]

       v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

        F = m a

the force is

       F = q v x B

as the velocity is perpendicular to the magnetic field

       F = q v B

acceleration is centripetal

       a = v² / r

we substitute

       qvB =1/2  m v² / r

       B =  v[tex]\frac{m v}{2 q r}[/tex]

we calculate

       B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]

       B = 1.1413 10⁻² T

A ball is thrown from ground level with an initial speed of 24.5 m/s at an angle of 35.5 degrees above the horizontal. The ball hits a wall that is 25.8 meters horizontally from where it started. How high (meters) does the ball hit on the wall?

Answers

6.07 m

Explanation:

Given:

[tex]v_0=24.5\:\text{m/s}[/tex]

[tex]\theta_0 = 35.5°[/tex]

First, we need to find the amount of time it takes to travel a horizontal distance of 25.8 m. We know that

[tex]x = v_{0x}t \Rightarrow t = \dfrac{x}{v_0 \cos \theta_0}[/tex]

or

[tex]t = 1.29\:\text{s}[/tex]

To find the vertical height where the ball hit the wall, we use

[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]

[tex]\:\:\:\:=(24.5\:\text{m/s})\sin 35.5(1.29\:\text{s}) \\ - \frac{1}{2}(9.8\:\text{m/s}^2)(1.29\:\text{s})^2[/tex]

[tex]\:\:\:\:=6.07\:\text{m}[/tex]

what is the average velocity if the initial velocity is at rest and the final velocity is 16 m/s

Answers

Answer:

8m/s

Explanation:

Vavg= 16-0/2=8m/s

a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.00 x10^7 m/s in a circular path 2.00m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.b) Is this field strength obtainable with today's technology or is it a futuristic possibility?

Answers

Charge me and do I name for meters
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