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kondaur [170]
3 years ago
8

Which layer contains solid rock and magma

Physics
1 answer:
BlackZzzverrR [31]3 years ago
3 0
The semi surface of earth
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2. Explain brightness of light using the wave model of light.
Dafna11 [192]
The amplitude of a wave tells us about the intensity or brightness of the light relative to other light waves of the same wavelength.
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2 years ago
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Unpolarized light passes through a combination of two ideal polarizers. The transmission axes of the first polarizer and the sec
Yuliya22 [10]

Answer:

62.5 %

Explanation:

Let the initial intensity of unpolarized light is Io.

After first polariser the intensity of light becomes I'.

So, I' = \frac{I_{0}}{2}

Now it passes through another polariser. The angle between the first polariser and the second polariser is given by Ф. The intensity is I''.

According to the law of Malus

I'' = I' Cos^{2}\phi

Here, Ф = 30 degree

I'' = \frac{I_{0}}{2} Cos^{2}30=0.375I_{0}

The percentage change in the intensity is given by

\frac{I_{0}-I''}{I_{0}}\times 100=\frac{I_{0}-0.375I_{0}}{I_{0}}\times100

= 62.5 %

7 0
3 years ago
What qualities did Galilea, Sir Isaac Newton, and Albert Einstein have?
MA_775_DIABLO [31]
The qualities that Galileo , Sir Isaac Newton, and Albert Einstein is that they all created models of nature. Galileo discovered Jupiter's four moons and declared that the earth revolves around the sun.Newton developed the three laws of motion, which formed the basic principles of modern physics.Albert Einstein discovered the general theory of relativity, but best known for his mass-energy equivalence formula E=mc².
4 0
3 years ago
Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c
Travka [436]

Answer:

22.1 V

Explanation:

We are given that

A_1=0.7 cm^2=0.7\times 10^{-4} m^2

A_2=2.5 cm^2=2.5\times 10^{-4} m^2

A_3=2.2 cm^2=2.2\times 10^{-4} m^2

A_4=3 cm^2=3\times 10^{-4} m^2

Using 1cm^2=10^{-4} m^2

We know that

R=\frac{\rho l}{A}

In series

R=R_1+R_2+R_3+R_4

R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}

R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}

Substitute the values

R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})

R=\rho l(2.62\times 10^4)

V=145 V

I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}

Voltage across the 2.5 square cm wire=IR=I\times \frac{\rho l}{A_2}

Voltage across the 2.5 square cm wire=\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V

Voltage across the 2.5 square cm wire=22.1 V

6 0
3 years ago
ILL MARK BRAINIEST IF YOU DO THIS CORRECTLY!!!
IRINA_888 [86]

Explanation:

1)5.8m/s

2)5.15m/s^2

3)12.69m/s

4)

8 0
3 years ago
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