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jolli1 [7]
4 years ago
10

Vertically polarized light with an average intensity of is incident on a polarizer whose transmission axis makes an angle of 30o

with respect to the horizontal. What is the average intensity of the light that is transmitted through the polarizer?
Physics
1 answer:
ladessa [460]4 years ago
3 0

Answer:

Explanation:

Average intensity of incident light I₀ .

intensity of transmitted light

I = I₀ cos²θ

if angle between transmission axis and angle of polarization of incident light be θ

In the given case

θ = 90 - angle the transmission axis makes with horizontal

= 90- 30

= 60°

I = I₀ cos²θ

= I₀ cos²60

=   I₀ /  4

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3.00 m^3 of water is at 20.0°C.
krok68 [10]

Answer:

9m^3

Explanation:

Given data

volume  v1=  3m^3

volume  v2=  ???

Temperature T1= 20.0°C.

Temperature T2= 60.0°C.

Applying the relation for temperature and volume

V1/T1= V2/T2

substitute

3/20= V2/60

3*60= V2*20

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3 years ago
Why is the picture above not an accurate representation of charges and their field lines?
pochemuha
Because field lines don’t all go in to the pole. The bottom half go outwards not inwards.
7 0
3 years ago
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Which describes the changes in visible light moving from red to violet?
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The wavelength decreases to roughly half.
(The frequency roughly doubles.)
3 0
4 years ago
A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600
3241004551 [841]

Answer:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}

Where:

v_{o} - Initial speed of the rocket, in m/s.

v_{ex} - Exhaust gas speed, in m/s.

m_{o} - Initial total mass of the rocket, in kg.

m - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

m(t) = m_{o} + r\cdot t

The initial total mass of the rocket is:

m_{o} = 750\,kg

The fuel consumption rate is:

r = -\frac{600\,kg}{30\,s}

r = -20\,\frac{kg}{s}

The function for the current total mass of the rocket is:

m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t

The speed function of the rocket is:

v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}

The speed of the rocket at given instants are:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

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What are some characteristics of soil? ​
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