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sergejj [24]
2 years ago
11

Two polarizers are oriented at 68 ∘ to one another. unpolarized light falls on them. part a what fraction of the light intensity

is transmitted?
Physics
2 answers:
kykrilka [37]2 years ago
7 0

Answer:

Fraction=0.0701

or

Fraction=7.01%

Explanation:

Apply Malus Law

I₁=I₀/2

I₂=I₁cos²θ

I₂=(I₀/2)cos²(68)

I₂=I₀(0.1403/2)

Fraction of light is given as

I₂/I₀=0.1403/2=0.0701

To get in %

Fraction=0.0701×100

Fraction=7.01%

Radda [10]2 years ago
4 0
Your answer is.07 hope this helped 
You might be interested in
what is the kinetic energy of an object that has mass of 30 kilograms and move with a velocity of 20 m/s
kotegsom [21]
Kinetic Energy = 1/2 * m * v² 1/2 * 30 * 20² 1/2 * 30 * 400 12000/2 6000 J.
7 0
2 years ago
In Example 2.7, we investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on sol
igor_vitrenko [27]

Answer:

a) 20 seconds

b) No.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s (given in the question)

v = Final velocity = 0 (because the jet will stop at the end)

s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)

a = Acceleration = -5.00 m/s² (slowing down, so it is negative)

a) Equation of motion

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.

5 0
2 years ago
PLEASE HELP AND HURRY!!!!!!!!!
never [62]

Answer:

I believe the answer would be A. point x

4 0
3 years ago
INSTI
max2010maxim [7]

Answer:

B

Explanation:

8 0
2 years ago
An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
3 years ago
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