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AleksandrR [38]

3 years ago

12

Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the

light passes through this combination, what is the angle between the transmission axes of the two filters?

1 answer:

Zepler [3.9K]3 years ago

5 0

Answer:

51.94°

Explanation:

$I_0$ = Unpolarized light

$I_2$ = Light after passing though second filter = $0.19I_0$

Polarized light passing through first filter

$I_1=\frac{I_0}{2}$

Polarized light passing through second filter

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}$

The angle between the two filters is 51.94°

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Explanation:

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1. The net force's magnitude is 4N and it's direction is to the right.

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calculate the gravitational force in between two objects of mass 25 kg and 20 kg if distance between them is 5m

sladkih [1.3K]

Answer:

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Explanation:

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Nata [24]

Answer:

$Nu = 30.311$

Explanation:

Let consider that pipe is a horizontal cylinder. The Nusselt number is equal to:

$Nu = \left\{0.6+\frac{0.387\cdot Ra_{D}^{\frac{1}{6} }}{[1+\left(\frac{0.559}{Pr} \right)^{\frac{9}{16}} ]^{\frac{8}{27} }} \right\}^{2}$ , for $Ra_{D} \le 10^{12}$ .

Where $Ra_{D}$ is the Rayleigh number associated with the cylinder.

The Rayleigh number is:

$Ra_{D} = \frac{g\cdot \beta\cdot (T_{pipe}-T_{air})\cdot D^{3}}{\nu^{2}}\cdot Pr$

By assuming that air behaves ideally, the coefficient of volume expansion is:

$\beta = \frac{1}{T}$

$\beta = \frac{1}{283.15\,K}$

$\beta = 3.532\times 10^{-3}\,\frac{1}{K}$

The cinematic and dynamic viscosities, thermal conductivity and isobaric specific heat of air at 10 °C and 1 atm are:

$\nu = 1.426\times 10^{-5}\,\frac{m^{2}}{s}$

$\mu = 1.778\times 10^{-5}\,\frac{kg}{m\cdot s}$

$k = 0.02439\,\frac{W}{m\cdot ^{\textdegree}C}$

$c_{p} = 1006\,\frac{J}{kg\cdot ^{\textdegree}C}$

The Prandtl number is:

$Pr = \frac{\mu\cdot c_{p}}{k}$

$Pr = \frac{(1.778\times 10^{-5}\,\frac{kg}{m\cdot s} )\cdot (1006\,\frac{J}{kg\cdot ^{\textdegree}C} )}{0.02439\,\frac{W}{m\cdot ^{\textdegree}C} }$

$Pr = 0.733$

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$Ra_{D} = 12.486\times 10^{6}$

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