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Anettt [7]
3 years ago
5

Unpolarized light of intensity i0 is incident on three polarizing filters. the axis of the first is vertical, that of the second

is 43° from vertical, and that of the third is horizontal. what light intensity emerges from the third filter?
Physics
1 answer:
Eva8 [605]3 years ago
7 0
To determine the intensity, the key idea is to work through the system filter-by-filter and applying either the one-half rule for unpolarized light or the cosine-squared rule for an already polarized light.

When the light reach the first filter, it is unpolarized, so the intensity is :
            I1 = 1/2 I0                     (One-half rule)

Because the light reaching the second filter is polarized, given 43° angle difference, the intensity is : 
            I2 = I0 cos^2 (angle difference)
                = I0 cos^2 43°
                = 0.535 I0

Lastly, when the light reached the third filter, it is polarized and has angle diference equals to the angle between the second and last filter 47°. Using the same rule, we get :
           I3 = I2 cos^2 47°
               = (0.535 I0) cos^2 47°
               = (0.535 I0) (<span>0.465)
               = 0.249 I0

So, the intensity which emerges is 0.249 I0.</span>
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When you run, your nose is moving in a straight line relative to a stone on the ground, and it's also moving back and forth relative to your left foot. But it's at rest relative to your ear.
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If an astronaut travels to different planets, which of the following planets will the astronaut’s weight be the same as on Earth
Nataly_w [17]
<h2>Astronaut travels to different planets - Option 4 </h2>

If an astronaut travels to different planets, none of the planets will the astronaut’s weight be the same as on Earth. On jupiter, weight will be more than the weight on earth. For instance if an astronaut has 100kg on earth then he will have 252 kg on jupiter.

On Mars, weight will be less than the weight on the earth. For instance, if an astronaut has 68 kg on earth then he will has 26 kg on mars. On Mercury, weight of an astronaut will be less than the weight on earth. Example if he has 68 kg on earth then he will have 25.7kg on mercury.

Hence, none of these planets the weight of astronaut will be same as on earth.

3 0
3 years ago
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How many neutrons does element X have if its atomic number is 48 and its mass number is 167?
Llana [10]
If its atomic number is 48, then it has 48 protons in the nucleus
of each atom.  Any more mass than that is supplied by the neutrons
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If the mass is 167, and 48 of those are protons, then there are

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8 0
3 years ago
Consider a rectangular ice floe 5.00 m high, 4.00 m long, and 3.00 m wide. a) What percentage of the ice floe is below the water
artcher [175]

Answer:

(a) 92 %

(b) 6.76 %

Explanation:

length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3

density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3

(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

                                                                      of ice x g

v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x  

                                                                      density of ice x g

v x 13600 x g = V x 920 x g

v / V = 0.0676

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4 0
3 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

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We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
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