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spayn [35]
11 months ago
10

What is the maximum resultant intensity in terms of the intensity of the initial unpolarized incoming light io?

Physics
1 answer:
Dennis_Churaev [7]11 months ago
6 0

The unpolarized light becomes less intense. When unpolarized light is run through a polarizer, its intensity is diminished by a factor of ½.

<h3>Unpolarized light:</h3>

Unpolarized light is diminished by a factor of ½ when it travels through a polarizer. Cos 2θ averages out across all angles to be ½ in size. I transmitted = I₀ and ½ I₀ at all angles The unpolarized light's intensity decreases to zero.

The light-producing atoms on the heated filament's surface behave independently of one another. Each of these emissions can be roughly represented as a brief "wave train" lasting between approximately 109 and 108 seconds. These wave trains, each with its unique polarization direction, are superimposed to form the electromagnetic wave that is coming from the filament.

Unpolarized light is produced by all common light sources, including the Sun, incandescent and fluorescent lighting, and flames.

Learn more about unpolarized light here:

brainly.com/question/15562548

#SPJ4

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What are the applications of pascal's principle​
Murrr4er [49]

Explanation:

  • The applications are, hydraulic lift- to transmit equal pressure throughout a fluid.
  • Hydraulic jack- used in the braking system of cars.
  • use of a straw- to suck fluids, which goes because of air pressure.
<h3>The question simply asks, where pressure can be applied. There are many others, such as <em><u>l</u></em><em><u>i</u></em><em><u>f</u></em><em><u>t</u></em><em><u> </u></em><em><u>p</u></em><em><u>u</u></em><em><u>m</u></em><em><u>p</u></em><em><u>.</u></em></h3>
5 0
3 years ago
Which one of the following statements is true concerning the magnitude of the electric field at a point in space? It is a measur
IceJOKER [234]

Answer:

It is a measure of the electric force per unit charge on a test charge.

Explanation:

The magnitude of the electric field is defined as the force per charge on the test charge.

Since we define electric field as the force per charge, it will have the units of  force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).

5 0
2 years ago
Which equation relates charge, time, and current?
Ierofanga [76]

Answer:

I = Δq / t

Explanation:

The quantity of electricity i.e charge is related to current and time according to the equation equation:

Q = It

Δq = It

Where:

Q => is the quantity of electricity i.e charge

I => is the current.

t => is the time.

Thus, we can rearrange the above expression to make 'I' the subject. This is illustrated below:

Δq = It

Divide both side by t

I = Δq / t

6 0
2 years ago
J.J. Thomson discovered the electron by noticing that: A. molecules with the same atoms exhibited the same chemical properties.
GenaCL600 [577]

J.J. Thomson discovered the electron by noticing that a beam of particles could be influenced by an electric or magnetic force.. That is option B.

<h3>What is an electron?</h3>

An electron can be defined as the part of an atom that is negatively charged and is found revolving round the nucleus of an atom.

J.J. Thomson was the scientist that discovered electrons through subjecting two oppositely-charged electric plates around the cathode ray.

He noticed that the particles where deflected by both the magnetic and electric fields.

Learn more about cathode rays here:

brainly.com/question/4441361

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4 0
2 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
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