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1 year ago

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

Engineering

1 answer:

valentinak56 [21]1 year 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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1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by

zvonat [6]

Answer:

299.36 feet

Explanation:

$To \ find \ the \ distance \ of \ the \ ball \ from \ the \ home \ plate. \\ \\ From \ the \ given \ information:$

$Height \ h = 4 \ ft$

$Initial \ speed \ V_o = 98 \ ft/s ec$

$The \ angle \ \theta = 45^0$

$Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s$

$U_x = V_o \ cos 45 = \dfrac{98}{\sqrt{2}}$

$U_y = V_o \ sin 45 = \dfrac{98}{\sqrt{2}}$

So;

$S_y = u_y t - \dfrac{1}{2}gt^2$

$-1 =\dfrac{98}{\sqrt{2}}t - \dfrac{1}{2}*32*1.85t^2$

By solving:

$t_1 = 4.32 \ sec$

Thus;

$horizontal \ distance = U_x t$

$= \dfrac{98}{\sqrt{2}}\times 4.32$

$\mathbf{=299.36 \ feet}$

$\mathbf{Thus \ , the \ distance \ from \ the \ home \ plate \ = \ 299.36 \ feet}$

5 0

3 years ago

For a fluid flowing through a pipe assuming that pressure drop per unit length of pipe (P/L) depends on the diameter of the pipe

Lyrx [107]

Answer:

Explanation:

La vaca

El pato

7 0

3 years ago

A pipe produces successive harmonics at 300 Hz and 350 Hz. Calculate the length of the pipe and state whether it is closed at on

MAXImum [283]

Answer:

The pipe is open ended and the length of pipe is 3.4 m.

Explanation:

For identification of the type of pipe checking the successive frequencies in both the open pipe and closed pipe as below

Equation for nth frequency for open end pipe is given as

$f_n=\frac{nv}{2L}$

For (n+1)th value the frequency is

$f_{n+1}=\frac{(n+1)v}{2L}$

Taking a ratio of both equation and solving for n such that the value of n is a whole number

$\frac{f_{n+1}}{f_n}=\frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}}\\\frac{350}{300}=\frac{(n+1)}{n}\\350n =300n+300\\50n =300\\n =6\\$

So n is a whole number this means that the pipe is open ended.

For confirmation the nth frequency for a closed ended pipe is given as

$f_n=\frac{(2n+1)v}{4L}$

For (n+1)th value the frequency is

$f_{n+1}=\frac{(2n+3)v}{4L}$

Taking a ratio of both equation and solving for n such that the value of n is a whole number

$\frac{f_{n+1}}{f_n}=\frac{\frac{(2n+3)v}{2L}}{\frac{(2n+1)v}{2L}}\\\frac{350}{300}=\frac{(2n+3)}{(2n+1)}\\700n+350 =600n+900\\100n =550\\n =5.5\\$

As n is not a whole number so this is further confirmed that the pipe is open ended.

Now from the equation of, with n=6, v=340 m/s and f=300 Hz

$f_n=\frac{nv}{2L}\\300=\frac{6 \times 340}{2L}\\L=\frac{2040}{600}\\L=3.4 m$

The value of length is 3.4m.

5 0

3 years ago

A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip

JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

$DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})$

Above equation can be written as:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})$

First Consider no wire i.e d/D=0

Above expression will become:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}$

Part A: (d/D=0.1)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574$

$DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.574}{1}*100$

DeltaV percent=42.6%

Part B:(d/D=0.01)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783$

$DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.783}{1}*100$

DeltaV percent=21.7%

5 0

3 years ago

What properties should the head of a carpenter’s hammer possess? How would you manufacture a hammer head?

BabaBlast [244]

Properties of Carpenter's hammer possess

Explanation:

1.The head of a carpenter's hammer should possess the impact resistance, so that the chips do not peel off the striking face while working.

2.The hammer head should also be very hard, so that it does not deform while driving or eradicate any nails in wood.

3.Carpenter's hammer is used to impact smaller areas of an object.It can drive nails in the wood,can crush the rock and shape the metal.It is not suitable for heavy work.

How hammer head is manufactured :

1.Hammer head is produced by metal forging process.

2.In this process metal is heated and this molten metal is placed in the cavities said to be dies.

3.One die is fixed and another die is movable.Ram forces the two dies under the forces which gives the metal desired shape.

4.The third process is repeated for several times.

5 0

3 years ago