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Gekata [30.6K]
3 years ago
6

What four elements composed everything in nature according to the beliefs of the ancient greeks? air, earth, fire, and water air

, ashes, fire, and water air, earth, salt, and water smoke, earth, fire, and water none of the above?
Physics
2 answers:
Vesna [10]3 years ago
5 0
According to the beliefs of the ancient Greeks the four elements composing everything in nature are; the air, earth, fire and water. 
For example according to the Greek philosopher Aristotle in 350 BC, the five basic elements that composed everything in nature is air, earth, fire, water and ether. Additionally according to Greek philosopher Empedocles in 450 BC, the four basic elements that composed everything in nature are the air, fire water and earth. 
Nimfa-mama [501]3 years ago
3 0
The ancient Greeks, in a theory that was suggested around 450 BC, believed that everything was from from four elements. These elements included earth, water, air and fire. Later, Aristotle supported this theory.

Therefore, the first option is correct.
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How did each scientist contributed to cell theory?
LekaFEV [45]

Answer:

Matthias Schleidan

Explanation:

because Matthias Schleiden found that all plants are composed of cells, and communicated the finding to Schwann, who had found similar structures in the cells. Other researchers confirmed the similarity, as explained in his book, where he concluded, "All living things are composed of cells and cell products.

This became the cell theory.

I learn that in my old school.

6 0
3 years ago
Which picture correctly shows the path of the reflected light rays given an object outside the focal point?Select one:a. Ab. Bc.
nignag [31]

We will have that the graph that describes the scenario is given by graph B.

6 0
1 year ago
The sound intensity of a certain type of food processor in normally distributed with standard deviation of 2.9 decibels. If the
Maru [420]

Answer: (48.41,\ 52.19)

Explanation:

The confidence interval for population mean is given by :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}

Given : Sample size : n=9

Sample mean : \ovreline{x}=50.3\text{ decibels}

Standard deviation : \sigma=2.9\text{ decibels }

Significance level : \alpha=1-0.95=0.05

Critical value : z_{\alpha/2}=z_{0.025}=1.96

Now, the 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type :-

50.3\pm (1.96)\dfrac{2.9}{\sqrt{9}}\\\\\approx50.3\pm1.89\\\\=(50.3-1.89,\ 50.3+1.89)=(48.41,\ 52.19)

6 0
3 years ago
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree
kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is    

                    I_f_3 = 27.57 W/cm^2

2 When third  polarizer is removed the intensity after it passes through the stack is    

                I_f_2 = 102.24 W/cm^2

Explanation:

  From the question we are told that

       The angle of the second polarizing to the first is  \theta_2 = 21^o  

        The angle of the third  polarizing to the first is     \theta_3 = 61^o

        The unpolarized light after it pass through the polarizing stack   I_u = 60 W/cm^2

Let the initial intensity of the beam of light before polarization be I_p

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

                     I_1 = \frac{I_p}{2}

Now according to Malus’ law the  intensity of light that would emerge from the second polarizing filter is mathematically represented as

                    I_2 = I_1 cos^2 \theta_1

                       = \frac{I_p}{2} cos ^2 \theta_1

The intensity of light that will emerge from the third filter is mathematically represented as

                  I_3 = I_2 cos^2(\theta_2 - \theta_1 )

                          I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]

making I_p the subject of the formula

                  I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}

    Note that I_u = I_3 as I_3 is the last emerging intensity of light after it has pass through the polarizing stack

         Substituting values

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}

                           =234.622W/cm^2

When the second    is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

                      I_f_3 = \frac{I_p}{2} cos ^2 \theta_2

I_f_3 is the intensity of the light emerging from the stack

                     

substituting values

                     I_f_3 = \frac{234.622}{2} * cos^2(61)

                       I_f_3 = 27.57 W/cm^2

  When the third polarizer is removed  the  second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be  

                  I_f_2 = \frac{I_p}{2} cos ^2 \theta_1

I_f_2 is the intensity of the light emerging from the stack

Substituting values

                  I_f_2 =  \frac{234.622}{2} cos^2 (21)

                     I_f_2 = 102.24 W/cm^2

   

7 0
3 years ago
What kind of relationship exists between sound and the temperature of material
KATRIN_1 [288]

speed increases with temp maybe

8 0
4 years ago
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