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Irina-Kira [14]
2 years ago
12

light that is polarized along the vertical direction is incident on a sheet of polarizing material. only 87% of the intensity of

the light passes through the sheet. what angle is the axis of the polarizer oriented, with respect to the vertical?
Physics
1 answer:
nekit [7.7K]2 years ago
8 0

Polarized light waves are those that have only one plane of vibration. Polarization is the process by which non-polarized light is converted into polarised light. There are numerous ways to polarise light.

A characteristic of transverse waves called polarisation identifies the geometric angle of the oscillations. A transverse wave's oscillation direction is perpendicular to the wave's motion direction.

given

Only 72.0% of the light's intensity makes it through the sheet and hits another sheet of polarising material. The second sheet lets no light flow through it.

Given that no light passes through the second sheet because both sheets are crossed, the angle the transmission axis of the second sheet makes with the vertical is

θ₂ = θ₁ + 90°

   = 31.94° + 90°

   = 1 21.94°

for the first sheet and

Icos²θ₁ = I

I/I₀ = 0.72

θ₁ = cos⁻¹√(0.72)

    = 31.94°

for the second sheet, respectively. This is because the malus law states that I = Iocos²θ₁ for the first sheet.

To know more about light that is polarized, click on the link below:

brainly.com/question/3092611

#SPJ4

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Radio waves are a type of______
Verizon [17]

Answer:

I already know the first one is electromagnetic but I don't know the second one sorry

Explanation:

5 0
3 years ago
What is the value of work done on an object when a 50-newton force moves it 15 meters in the same direction as the force?
Volgvan

Answer:

W = 750 Joules.

Explanation:

Given that,

Force acting on the object, F = 50 N

Distance moved by the object, d = 15 m

To find,

Work done by the object

Solution,

Let W is the work done by the object. It is given by the product of force and distance. Its expression is given by :

W=F\times d\ cos\theta

Here, \theta=0 (force and distance are in same direction)

W=F\times d

W=50\ N\times 15\ m

W = 750 Joules

So, the work done by the force is 750 Joules.

4 0
3 years ago
A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes angle of 28.1 with the tensil
NISA [10]

Answer:

we have to find out the critical resolved shear stress. As it it given in the question

Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.

a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.

cos(62.4°) = 0.46

cos(72.0°) = 0.31

cos(81.1°) = 0.15

Thus, the slip direction is at the angle of 62.4° along the tensile axis.

b) now the critical resolved shear stress can be find out by the following equation.

τ_{crss} = σ_{Y} ( cosФ cosλ)_{max}

now by putting values,

     = (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23

3 0
4 years ago
PLEASE SOLVE QUICKLY!!!<br> Solve for A, B, and C from graph
hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

<u>Explanation:</u>

We know,

x = \frac{L}{\frac{W}{F} +1}

and L = x+y

1.

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm

Therefore, position weight placed from fulcrum is 59.35cm

2.

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8  = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W  = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g

Therefore, weight of the beam is 196.56g

3.

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8  = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

4 0
3 years ago
It is proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. W
Anvisha [2.4K]

Answer:

962291.57928 m²

Explanation:

P_r = Pressure = 2\dfrac{I}{c}  (full reflection)

I = Intensity = \dfrac{P}{A}=\dfrac{P}{4\pi r^2}

P = Power = 3.9\times 10^{26}\ W

c = Speed of light = 3\times 10^8\ m/s

M = Mass of Sun = 1.99\times 10^{30}\ kg

m = Mass of ship = 1500 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Force of radiation is given by

F_r=P_rA\\\Rightarrow F_r=2\dfrac{I}{c}\times A\\\Rightarrow F_r=2\dfrac{P}{4\pi r^2c} A

This force will balance the gravitational force as stated in the question

\dfrac{GMm}{r^2}=2\dfrac{P}{4\pi r^2c} A\\\Rightarrow A=\dfrac{4\pi cGMm}{2P}\\\Rightarrow A=\dfrac{4\times \pi\times 3\times 10^8\times 6.67\times 10^{-11}\times 1.99\times 10^{30}\times 1500}{2\times 3.9\times 10^{26}}\\\Rightarrow A=962291.57928\ m^2

The area of the must be 962291.57928 m²

3 0
4 years ago
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