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12345 [234]
3 years ago
6

Unpolarized light with an intensity of (25.0 A) units is passed through two successive polarizing filters, the first with its po

larization axis aligned with the vertical and the second with its polarization axis rotated (55.0 B) from the vertical. Find the intensity of the light after passing through the two polarizing filters. Give your answer in same unit as the original light and with 3 significant figures.
Physics
1 answer:
coldgirl [10]3 years ago
6 0

Answer:

the intensity of the light after passing through the two polarizing filters is 4.11 units  

 

Explanation:

Given the data in the question;

the intensity of an unpolarized light; I₀ = 25.0 units

when the unpolarized light passes through the first polarizer, its intensity reduces to half of its initial value;

⇒ I₁ = I₀/2 = 25/2 = 12.5 units

the angle between the transmission axes of two polarizers is;

∅ = 55° - 0° = 55°

The intensity of the light after passing through two polarizing filters will be;

I₂ = I₁cos²∅      

we substitute

I₂ = 12.5 × cos²(55)

I₂ = 12.5 × 0.3289899

I₂ = 4.11 units

Therefore, the intensity of the light after passing through the two polarizing filters is 4.11 units

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JulijaS [17]

Explanation:

→ Volume of cone = πr² × h/3

Here,

  • Radius (r) = 13 cm
  • Height (h) = 27 cm

→ Volume of cone = π(13)² × 27/3 cm³

→ Volume of cone = 169π × 9 cm³

→ Volume of cone = 1521π cm³

→ Volume of cone = 1521 × 22/7 cm³

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4 0
3 years ago
An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part
guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

E=\dfrac{kq}{r^2}

k is electrostatic constant

if r = 2 m, electric field is given by :

E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)

If r = 1 m, electric field is given by :

E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)

Dividing equation (1) and (2) we get :

\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0
3 years ago
A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s
TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

vf^{2} =vo^{2}+2*g*h\\

We clear the vo (initial speed)

vo=\sqrt{vf^{2}-2*g*h }

v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}

vo=5.87m/s

7 0
3 years ago
Sandra wanted to see how fast she could kick a soccer ball. She recorded her results below. Unfortunately, she forgot to record
alexira [117]
Kick | Distance (M) | Time (s) | Average Speed

1. 55. 5.0. 11

2. 50. 5.0. 10

3. 20. 3.0. 10
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3 years ago
The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can acceler
Kay [80]

Answer:

Part a)

v_f = 4 m/s

Part b)

t = 0.001 s

Part c)

d = 0.815 m

Explanation:

Part a)

As we know that initially the grass hopper is at rest at the ground position

Now the acceleration is given as

a = 4000 m/s^2

distance of the legs that it stretched is given as

s = 2.0 mm

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(4000)(0.002)

v_f = 4 m/s

Part b)

time taken to reach this speed is given as

v_f - v_i = at

4 - 0 = 4000 t

t = 0.001 s

Part c)

as the grass hopper reach the maximum height its final speed would be zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 4^2 = 2(-9.81) d

d = 0.815 m

5 0
3 years ago
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