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Lina20 [59]
3 years ago
12

nitially, scientists described atoms as the smallest particles of matter. However, smaller particles within atoms were discovere

d. How did the scientific world respond to this new information?
Physics
1 answer:
Kisachek [45]3 years ago
7 0
<h2>Answer</h2>

They get involved in this new information

<h2>Expalantion</h2>

In the ancient time, the scientists believed that there is the smallest particle which can control the nature of reaction in the environment which they called them. Then a lot of experiment was conducted in which new coming scientists discovered the protons, electrons, neutrons, and other smallest particles. Firstly, scientists performed experiments for this new discoveries to know better about them and then include them in their next experiment for better exposures.

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At a beach the light is generallypartially polarized owing to reflections off sand and water. At a particular beach on a particu
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a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = E_{h}

Let the vertical electric field component = E_{v}

The formula for light intensity is given by:

I = \frac{E_{m} ^{2} }{2c \mu}..............................(1)

E_{m} is the resolution of the vertical and horizontal components, E_{h} and    E_{v}

E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}..................(2)

Light intensity before the glasses were put on:

I_{1}  = \frac{E_{m} ^{2} }{2c \mu_{1} }.............................(3)

Put equation (2) into equation (3)

I_{1}  = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }.............................(4)

After the glasses were put on the horizontal component vanishes, i.e. E_{h} = 0

I_{2}  = \frac{ E_{v} ^{2}}{2c \mu_{2} }...................................(5)

Divide equation (5) by equation (4)

\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}...............................(6)

But E_{h} = 2.3E_{v}......................(7)

Insert equation (7) into (6)

\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2}  }{(2.3E_{v})^{2}   + E_{v} ^{2}   } \\\frac{I_{2} }{I_{1} } =  \frac{E_{v}^{2}  }{5.29E_{v}^{2}   + E_{v} ^{2}   }\\\frac{I_{2} }{I_{1} } =  \frac{E_{v}^{2}  }{6.29E_{v}^{2}  }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\

\frac{I_{2} }{I_{1} }= 0.159

b) When the sunbather lies on his side, the vertical component vanishes, i.e E_{v} = 0

\frac{I_{2} }{I_{1} } = \frac{E_{h} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}

\frac{I_{2} }{I_{1} } = \frac{(2.3E_{v} )^{2}  }{E_{v} ^{2} +(2.3E_{v} )^{2}}

\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2}  }{E_{v} ^{2} +5.29E_{v}^{2} }

\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2}  }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } = \frac{5.29}{6.29} \\\frac{I_{2} }{I_{1} } = 0.84

8 0
3 years ago
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