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Andrews [41]
3 years ago
10

Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po

larizing filter whose axis is at 45.0 ∘∘ to that of the first. Part A Determine the intensity of the beam after it has passed through the second polarizer. Enter the factor only. For example, enter 0.2500.250, that means your answer is I=0.250I0I=0.250I0. II = nothing I0I0 SubmitRequest Answer Part B Determine its state of polarization. Determine its state of polarization. The light is linearly polarized along the axis of the first filter. The light is linearly polarized along the axis of the second filter. The light is linearly polarized perpendicular to the axis of the second filter. The light is linearly polarized perpendicular to the axis of the first filter. SubmitRequest Answer Provide Feedback
Physics
1 answer:
Montano1993 [528]3 years ago
3 0

Answer:

A) I = Io 0.578,   B) he light that leaves the polarized is completely polarized, being perpendicular to the axis of the second filter

Explanation:

A) Light passing through a polarizer must comply with the / bad law

          I = Io cos2 tea

Where is at the angle of the polarizer and incident light

          I = Io cos2 45

         I = Io 0.578

Therefore the beam intensity is 0.578 of the incident intensity

.B) the light that leaves the polarized is completely polarized, being perpendicular to the axis of the second filter

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The answer to this question is "Buffeting". This is an unusual strong wind condition that resulted in the loss of vehicle control. This condition occurs in roads and bridges, across and along mountains that affected vehicles control. The drivers must be alert and put their full attention to overcome this condition.
4 0
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Which particles in an atom are acted upon by the strong force? A. Electrons and neutrons B. Neutrons and protons O C. Protons an
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\large \sf \pmb{B) \: Neutrons \:  and  \: Protons}

  • <em>Neutrons And Protons</em> is a particles in an atom and are acted upon by the strong force.
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2 years ago
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit
Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

d_1=4.7\times 10^{10} m

v_i=9.5\times 10^4 m/s

d_2=6\times 10^{12} m

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

K_i+U_i=K_f+U_f

\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}

\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f

v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})

v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}

Using the formula

v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}

v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}

Where mass of sun=M=1.98\times 10^{30} kg

G=6.7\times 10^{-11}

v_f=58515.9 m/s

4 0
3 years ago
For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia
Oliga [24]

Answer with Explanation:

We are given that

m_1=0.3 kg

m_2=0.4 kg

Spring constant,k=200 N/m

h=0.250 m

a.Speed of ham before collision

u=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.25}=2.2m/s

Collision is inelastic

According to law of conservation of momentum

m_1u_1=(m_1+m_2)v

0.3\times 2.2=(0.3+0.4)v

v=\frac{0.3\times 2.2}{0.7}=0.94 m/s

Kinetic energy of masses=Spring potential energy at maximum distance

\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}kA^2

A^2=\frac{(m_1+m_2)v^2}{k}

A=\sqrt{\frac{(m_1+m_2)v^2}{k}}

Substitute the values

A=\sqrt{\frac{(0.3+0.4)\times (0.94)^2}{200}}=0.056 m=0.056\times 100=5.6 cm

1 m=100 cm

b.Time period,T=2\pi\sqrt{\frac{(m_1+m_2)}{k}}

T=2\pi\sqrt{\frac{(0.3+0.4)}{200}}=0.37 s

7 0
3 years ago
The star Antares has an apparent magnitude of 1.0, whereas the star Procyon has an apparent magnitude of 0.4. Which star appears
Anika [276]

Answer:

Procyon appears brighter in the sky

Explanation:

Apparent magnitude of star Antares = 1.0

Apparent magnitude of star Procyon = 0.4

Pogson's Ratio

m₂-m₁ =  -2.50 log(B₂/B₁)

where, m is the apparent magnitude

B = Brightness of star or flux coming towards us (W/m²)

∴Larger magnitudes correspond to fainter stars so here 1>0.4 which means Antares appears dimmer and Procyon appears brighter.

6 0
3 years ago
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