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pav-90 [236]
2 years ago
12

Light of intensity I0 and polarized horizontally passes through three polarizes. The first and third polarizing axes are horizon

tal, but the second one is oriented 20.0� to the horizontal. In terms of I0, what is the intensity of the light that passes through the set of polarizers?
A) 0.442 I0
B) 0.180 I0
C) 0.780 I0
D) 0.883 I0
Physics
1 answer:
Alex17521 [72]2 years ago
3 0

Answer:

Option C.

Explanation:

Suppose that we have light polarized in some given direction with an intensity I0, and it passes through a polarizer that has an angle θ with respect to the polarization of the light, the intensity that comes out of the polarizer will be:

I(θ) = I0*cos^2(θ)

Ok, we know that the light is polarized horizontally and comes with an intensity I0

The first polarizer axis is horizontal, then the intensity after this polarizer is:

then θ = 0°

I(0°) = I0*cos^2(0°) = I0

The intensity does not change. The axis of polarization does not change.

The second polarizer is oriented at 20° from the horizontal, then the intensity that comes out of this polarizer is:

I(20°) =  I0*cos^2(20°) = I0*0.88

And the axis of polarization of the light that comes out is now 20° from the horizontal

Now the light passes through the last polarizer, which has an axis oriented horizontally, so the final intensity of the light will be:

note that here the initial polarization is  I0*0.88

and the angle between the axis is 20° again.

Then the final intensity is:

I(20°) =  I0*0.88*cos^2(20°) = I0*0.78

Then the correct option is C.

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(A) 140 j/sec (b) 1.26 K

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We have given the heat heat flowing into the refrigerator = 40 J/sec

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(b) Total heat absorbed =140 j/sec =140\times 3600=504000j/hour

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A skater slides across the ice with an initial velocity of 5.0 m/s. She slows 10 points
zvonat [6]

Explanation:

Given that,

The initial velocity of a skater is, u = 5 m/s

She slows to a velocity of 2 m/s over a distance of 20 m.

We can find the acceleration of skater. It is equal to the rate of change of velocity. So, it can be calculated using third equation of motion as follows :

v^2-u^2=2as

a = acceleration

a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{(2)^2-(5)^2}{2\times 20}\\\\a=-0.525\ m/s^2

So, her acceleration is 0.525\ m/s^2 and she is deaccelerating. Also, her initial velocity is given i.e. 5 m/s.

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What crop is least likely to do well when the temperatures are very hot?
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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

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543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
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