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3 years ago

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A horizontal beam of light of intensity 25 W/m2 is sent through two polarizing sheets. The polarizing direction of the first mak

es a 15° angle with that of the second. Suppose that the initial beam is unpolarized. What then is the intensity of the transmitted light? A. 7.2 W/m2 B. 11.7 W/m2 C. 12.1 W/m2 D. 23.3 W/m E. 24.1 W/m

1 answer:

Zina [86]3 years ago

7 0

Answer:

option (B)

Explanation:

Intensity of unpolarised light, I = 25 W/m^2

When it passes from first polarisr, the intensity of light becomes

$I'=\frac{I_{0}}{2}=\frac{25}{2}=12.5 W/m^{2}$

Let the intensity of light as it passes from second polariser is I''.

According to the law of Malus

$I'' = I' Cos^{2}\theta$

Where, θ be the angle between the axis first polariser and the second polariser.

$I'' = 12.5\times Cos^{2}15$

I'' = 11.66 W/m^2

I'' = 11.7 W/m^2

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2 years ago

5) What is the net force caused by the moon acting on earth when the moon is 3.86x10^8 m away? The moon has a mass of 7.46x10^23

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Answers:

5) $1.99(10)^{21} N$

6) $1.37(10)^{18} N$

7) $1.64(10)^{21} N$

8) $4.29(10)^{10} N$ more than Venus force of gravity on Pluto

Explanation:

According to Newton's law of Universal Gravitation, the force $F$ exerted between two bodies of masses $M$ and $m$ and separated by a distance $R$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{Mm}{R^{2}}$ (1)

Where $G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

This is the equation we will use to solve each question in this problem.

<h3>5) Gravitational force between Earth and Moon</h3>

In this case we have:

$F_{earth-moon}$ is the gravitational force between Earth and Moon

$M=5.97(10)^{24} kg$ is the mass of the Earth

$m=7.46(10)^{23} kg$ is the mass of the Moon

$R=3.86(10)^{8} m$ is the distance between Earth and Moon

Solving:

$F_{earth-moon}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(5.97(10)^{24} kg)(7.46(10)^{23} kg)}{(3.86(10)^{8} m)^{2}}$ (2)

$F_{earth-moon}=1.99(10)^{21} N$ (3)

<h3>6) Gravitational force between Jupiter and Venus</h3>

Assuming for a moment that the planets are perfectly aligned and all are in the same orbital period, we can make a rough estimation of the distance between Jupiter and Venus, knowing the distance of each to the Sun:

distance between Sun and Jupiter - distance between Sun and Venus=distance between Jupiter and Venus= $R_{jupiter-venus}$ (4)

$R_{jupiter-venus}=778.3(10)^{9} m - 108(10)^{9} m=6.703(10)^{11} m$ (5)

Using this value in the Law of Universal Gravitation equation:

$F_{jupiter-venus}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.90(10)^{27} kg)(4.87(10)^{24} kg)}{(6.703(10)^{11} m)^{2}}$ (6)

$F_{jupiter-venus}=1.37(10)^{18} N$ (7)

<h3>7) Gravitational force between Saturn and Mars</h3>

Using the same assumption we made in the prior question:

distance between Sun and Saturn - distance between Sun and Mars=distance between Saturn and Mars= $R_{saturn-mars}$ (8)

$R_{saturn-mars}=1427(10)^{9} m - 227.9(10)^{9} m=227.9(10)^{9} m$ (9)

Using this value in the Law of Universal Gravitation equation:

$F_{saturn-mars}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.989(10)^{30} kg)(6.42(10)^{23} kg)}{(227.9(10)^{9} m)^{2}}$ (10)

$F_{saturn-mars}=1.64(10)^{21} N$ (11)

<h3>8) How much more is earths force of gravity on Pluto than Venus force of gravity on Pluto?</h3>

Firstly, we need to find $F_{earth-pluto}$ and then find $F_{venus-pluto}$ in order to find the difference.

<u>For $F_{earth-pluto}$ :</u>

$M=5.97(10)^{24} kg$ is the mass of the Earth

$m=1.46(10)^{22} kg$ is the mass of Pluto

$R_{earth-pluto}=5.7504(10)^{12} m$ is the distance between Earth and Pluto

$F_{earth-pluto}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(5.97(10)^{24} kg)(1.46(10)^{22} kg)}{(5.7504(10)^{12} m)^{2}}$ (12)

$F_{earth-pluto}=1.759(10)^{11} N$ (13) Force between Earth and Pluto

<u></u>

<u>For $F_{venus-pluto}$ :</u>

$M=4.87(10)^{24} kg$ is the mass of Venus

$m=1.46(10)^{22} kg$ is the mass of Pluto

$R_{venus-pluto}=5.792(10)^{12} m$ is the distance between Venus and Pluto

$F_{venus-pluto}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(4.87(10)^{24} kg)1.46(10)^{22} kg)}{(5.792(10)^{12} m)^{2}}$ (14)

$F_{venus-pluto}=1.33(10)^{11} N$ (15) Force between Venus and Pluto

Calculating the difference:

$F_{earth-pluto}-F_{venus-pluto}=1.759(10)^{11} N-1.33(10)^{11} N$

Finally:

$F_{earth-pluto}-F_{venus-pluto}=4.29(10)^{10} N$ (16)

Hence:

Earths force of gravity on Pluto is $4.29(10)^{10} N$ than Venus force of gravity on Pluto.

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