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Mars2501 [29]
3 years ago
7

Can anyone help me learn Malus's law in optics?​

Physics
1 answer:
Harlamova29_29 [7]3 years ago
8 0

Answer:

It is also called law of Malus,the law states that the intensity of a beam of plane-polarized light after passing through a rotatable polarizer varies as square of the cosine of the angle through which the polarizer is rotated from the position that gives maximum intensity expand.

Before u can know it u have to understand it, so go through it over and over again and u will get it.

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Suppose you are standing such that a 35-foot tree is directly between you and the sun. If you are 7 feet tall and the tree casts
Juli2301 [7.4K]

Answer:

60 ft

Explanation:

The tree man and their shadow system forms two similar right angle triangles.

the figure is in the attachment.

let ∠BAC= θ

Therefore, in triangle ABC

tanθ = BC/AC= 35/75= 7/15

now in ΔAEF also

tanθ = EF/AE = 7/75-x= 7/15

solving we get x=60 ft

Now AC and FA are the shadows of the tree and the man respectively.

now FA =75-x=75-60= 15 ft

Therefore, the man must stand at a distance of 60 ft from the tree can you stand and still be completely in the shadow of the tree

3 0
4 years ago
An electron that has a velocity with x component 1.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic fi
Sergio039 [100]

Answer:

(a) 5.056 x 10^-14 N

(b) 5.056 x 10^-14 N

Explanation:

X component of velocity of electron is 1.6 × 10^6 m/s

Y component of velocity of electron is 2.4 × 10^6 m/s

X component of magnetic field is 0.025 T

Y component of magnetic field is  -0.16 T

charge on electron, q = - 1.6 x 10^-19 C

Write the velocity and magnetic field in the vector forms.

\overrightarrow{v}=1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j}

\overrightarrow{B}=0.025\widehat{i}-0.16\widehat{j}

The force on the charge particle when it is moving in the magnetic field is given by

\overrightarrow{F}=q\left ( \overrightarrow{v}\times \overrightarrow{B} \right )

(a) Force on electron is given by

\overrightarrow{F}=-1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )

\overrightarrow{F}=5.056\times 10^{-14}\widehat{k}

Magnitude of force is 5.056 x 10^-14 N.

(b) Force on a proton is given by

\overrightarrow{F}=1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )

\overrightarrow{F}=-5.056\times 10^{-14}\widehat{k}

Magnitude of force is 5.056 x 10^-14 N.

Thus, the magnitude of force remains same but the direction of force is opposite to each other.

Explanation:

4 0
3 years ago
Because acceleration is a quantity that has both magnitude and direction, it is a(n)___________
swat32

Answer:

Vector Quantity

Explanation:

A Vector quantity is a quantity has both magnitude and direction while scalar quantity has only magnitude.

4 0
4 years ago
Read 2 more answers
An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m
Tomtit [17]

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

Find: a and t

(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

40100 = ½ (47.5 + 34.3)t

t = 980 s

7 0
4 years ago
Determine the minimum angle at which a roadbedshould be banked
poizon [28]

To solve this problem, apply the concepts related to the relationship given between the centripetal Force and the Weight.

The horizontal force component is equivalent to the weight of the car, while the vertical component is linked to the centripetal force exerted on the car, therefore,

T cos\theta = mg \rightarrow T = \frac{mg}{cos\theta}

Tsin\theta = \frac{mv^2}{r} \rightarrow T = \frac{mv^2/r}{sin\theta}

Equating both equation we have that,

\frac{mv^2}{r} = mgtan\theta

tan\theta = \frac{v^2}{rg}

Rearranging to find the angle we have that,

\theta = tan^{-1} (\frac{v^2}{rg})

Our values are given as,

r = 2.00*10^2m

v = 20m/s

\theta = tan^{-1}(\frac{20^2}{(2*10^{2})(9.8)})

\theta = 11.53 \°

Therefore the minimum angle will be 11.53°

5 0
3 years ago
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