Supermarkeoloma

The average intensity of light emerging from a polarizing sheet is 0.775 W/m2, and that of the horizontally polarized light inci

rewona [7]

Answer:

the transmission axis of polarizing sheet makes an angle of $\Theta =26.74^{\circ}$ with the horizontal

Explanation:

We have given that intensity of light incident on the sheet $I_0=0.970W/m^2$

Average intensity of light emerging from a polarizing sheet $I=0.775W/m^2$

We have to find the angle between transmission axis with the horizontal

Intensity of light polarizing from sheet is equal to $I=I_0cos^2\Theta$

So $0.775=0.970cos^2\Theta$

$cos^2\Theta =0.798$

$cos\Theta =0.893$

$\Theta =26.74^{\circ}$

So the transmission axis of polarizing sheet makes an angle of $\Theta =26.74^{\circ}$ with the horizontal

5 0

3 years ago

Which of the following effects results from the interference of two sound

Natali5045456 [20]

Answer:beats

Explanation:l got it right gang

7 0

3 years ago

Make a VIR chart.<br> All voltage, resistance, and current.

KiRa [710]

$\fbox{Hope This Helps You .}$

5 0

3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago

A law of motion that states that an object at rest stays at rest and one in motion stays

adoni [48]

Answer:

that is the first law of motion or newtons first law of motion

6 0

3 years ago