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1 year ago

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

1 answer:

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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A 6 kg weight is lifted off the ground to a height that gives it 70.56 j of gravitational potential energy. what is its height?

Marina CMI [18]

GPE= 70.56 J -------------------> GPE= mgh-------------> X= height
70.56 = 6(kg) * 9.8(m/s/s) * X
70.56 = 58.8X
70.56/58.8= 58.8X/58.8
X= 1.2
The height is 1.2 feet or meters (whatever unit you are using in this problem)

5 0

3 years ago

Read 2 more answers

Which force would result in a balanced force for the

ohaa [14]

Answer:

F= 45 N

Explanation:

4 0

3 years ago

A shoe and a shirt are released from the same height. They take different amounts of time to fall to the ground. How can this be

Law Incorporation [45]

The best explanation for the difference in time is: A. The difference in weight doesn't affect the time, but they are affected differently by air resistance.

<h3>What is weight?</h3>

Weight can be defined as the force acting on an object or a physical body due to the effect of gravity. Also, the weight of an object (body) is typically measured in Newton.

<h3>The factors that affect weight.</h3>

Some of the factors that affect the weight that is possessed by an object or a physical body include the following:

Mass

Distance

Air resistance

In conclusion, the weight possessed by the shoe and shirt has no effect on time but would be affected differently by air resistance.

Read more on weight here: brainly.com/question/13833323

4 0

2 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

Help is requested. Will give brainliest to anyone who answers correctly.

sp2606 [1]

The answer should be d because they are constantly rotating

4 0

4 years ago