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spayn [35]
1 year 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]1 year 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

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What is the tendency of an object to resist a change in motion
liubo4ka [24]
The tendency of an object to resist change in its motion is known as inertia
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Given the initial wavefunction Ψ (x, 0) = Axexp (-k x) withx&gt; 0 andk&gt; 0, and Ψ (x, 0) = 0 forx &lt;0, what value must A ta
madam [21]

Answer with explanation:

The Normalization Principle states that

\int_{-\infty }^{+\infty }f(x)dx=1

Given

f(x)=xe^{-kx}(x>0\\\\0(x

Thus solving the integral we get

\int_{0 }^{+\infty }A\cdot xe^{-kx}dx=1\\\\A\int_{0 }^{+\infty }\cdot xe^{-kx}dx=1

The integral shall be solved using chain rule initially and finally we shall apply the limits as shown below

I=\int xe^{-kx}dx\\\\x\int e^{-kx}dx-\int \frac{d(x)}{dx}\int e^{-kx}dx\\\\-\frac{xe^{-kx}}{k}-\int 1\cdot \frac{-e^{-kx}}{k}\\\\\therefore I=\frac{e^{-kx}}{k}-\frac{xe^{-kx}}{k}

Applying the limits and solving for A we get

I=\frac{1}{k}[\frac{1}{e^{kx}}-\frac{x}{e^{kx}}]_{0}^{+\infty }\\\\I=-\frac{1}{k}\\\\\therefore A=-k

3 0
3 years ago
The breaking of ________ bonds i food releases energy for your body to use.
erik [133]
Chemical because when you eat, the food give off a chemical bond
6 0
4 years ago
Read 2 more answers
A fugitive tries to hop on a freight train traveling at a constant speed of 5.5 m/s. Just as an empty box car passes him, the fu
stiks02 [169]
Th equations to be used here are the following:

a = (v - v₀)/t
x = v₀t + 0.5at²

The speed of the fugitive is the sum of his own speed plus the speed of the train. Thus, 
v₀ = 0 + 5.5 m/s = 5.5 m/s
v = 8 m/s + 5.5 m/s = 13.5 m/s

a.) We use the first equation to determine time
2.5 m/s² = (13.5 m/s - 5.5 m/s)t
Solving for t,
t = 3.2 seconds

b.) We use the answer in a) and the 2nd equation:

x = (5.5 m/s)(3.2 s) + 0.5(2.5 m/s²)(3.2 s)²
x = 30.4 meters
3 0
3 years ago
8. What variables shift both the long-run and short-run aggregate-supply curves?
vfiekz [6]

Variables shift both the long-run and short-run aggregate-supply curves are productivity, labor Wage Costs and taxes and other costs.

<h3>What is aggregate-supply?</h3>

The entire quantity of goods and services that enterprises in a national economy intend to sell during a certain time period is known as aggregate supply

Variables shift both the long-run and short-run aggregate-supply curves are ;

1. Productivity

2. Labor Wage Costs.

3. Taxes and other costs -

4. Material Prices.

5. Size of labor force.

6. Stock of Capital.

Hence, variables shift both the long-run and short-run aggregate-supply curves are productivity, labor Wage Costs and taxes and other costs.

To learn more about the aggregate-supply, refer:

brainly.com/question/16419230

#SPJ1

4 0
2 years ago
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