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Katen [24]
9 months ago
14

1. What did observations between 1912 and 1917 show?_____

Engineering
1 answer:
valentinak56 [21]9 months ago
7 0

The following set of prompts is astronomy related.

1) It is to be noted that the observations between 1912 and 1917  revealed that the earth was millions of years old and cooled down from a once molten state.

2) Lemaitres argued that the physical world began as a single particle—the "primeval atom," as he termed it—that disintegrated in an explosion, resulting in space and time and the cosmos' ongoing expansion.

3) It is TRUE to state that between 1912 and 1922, astronomer Vesto Slipher at the Lowell Observatory in Arizona uncovered that the spectra of light from many of these celestial entities were systematically shifted to longer wavelengths, or redshifted. The objects referenced here were galaxies in the distance.

4)  The important discovery that Hubble and his assistant discovered about galaxies was that some nebulae were galaxies <u>expanding </u>beyond our own galaxies.

5) It means that the universe has been growing or expanding ever since the Big Bang.

6) By observing very distant objects, scientists can tell whether or not the universe is moving, growing, or shrinking. This concept is referred to as redshift.

7) In 1965, the American Radio Astronomers Arno Penzias and Robert Wilson inadvertently discovered Cosmic Microwave Background Radiation.

8) According to all recent observations and studies, the Universe has no center.

9) It is estimated the Universe is 13.8 Billion Years old.

10) According to the Big Bang Theory, the whole Universe existed within a bubble millions of times smaller than the size of a pinhead. It was hotter and denser than we could have imagined. Then it burst into flames. According to this belief, this explosion produced life as we know it. It should be highlighted that, while the hypothesis has been utilized to explain numerous scientific facts, it is still an unproven theory.

<h3>When did the study of astronomy begin?</h3>

The Assyro-Babylonians made the earliest written records of regular astronomical observations approximately 1000 BCE. Astronomers had built an extensive knowledge of the celestial bodies and documented their periodic movements in Mesopotamia, which is located in the southern section of modern-day Iraq.

Astronomy is the oldest natural science, dating back to antiquity, with roots in prehistoric religious, mythical, cosmological, calendrical, and astrological beliefs and practices: vestiges of these can still be found in astrology, a discipline long intertwined with public and governmental astronomy.

Learn more about astronomy:
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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
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Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

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The elongation is =21.29mm

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as \sigma_T.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

              \sigma _T =K\epsilon^n_T

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        K = \frac{\sigma_T}{(\epsilon_T)^n}

No substituting  345MPa \ for  \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and  \ 0.22 \ for  \ n from the question we have

                     K = \frac{345}{(0.02)^{0.22}}

                          = 815.82MPa

Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K  \ and  \  0.22 \ for \ n

       \epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }

            =0.0443

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

Where

       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

            Elongated \ Length =l_i - l_o

Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i

          Elongated \ Length = 491.28 - 470

                                       =21.29mm

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