Field of NoThingness - spl BiNal

"All other mysterious lies downstream of this question. It matters to me because I am human and do not like not knowing." ~ Ann Druyan Roen Kelly According to the standard Big Bang model, the universe was born during a period of inflation that began about 13.8 billion years ago. Like a rapidly expanding balloon, it swelled from a tiny, indeed - far, far smaller than the electron to nearly its current size within a tiny fraction of a second. This is the generally agreed principle and powerful model that explains many of the things scientists see when they look up in the sky. But there is the idea about the existence of the universe, that makes uneasy to the scientist. The idea is that the universe underwent a period of rapid inflation early in its history cannot be directly tested and it relies on the existence of a mysterious form of energy in the universe's beginning that has long since disappeared. Roen Kelly According to Eric Agol, "Infla...

The isotopes $Ra-226$ undergoes $\alpha$ decay with a half-life of $1620$ years. What is the activity of $1\;g$ of $Ra-226$? Avogadro number = $6.023 * 10^{23}\; /mole$.

Given, Half life $T_{1/2} = 1620$ years = $5.1 * 10^{10}$ Sec Mass of $Ra-226$ = 1 gm Activity of $Ra-226\; ; $ $(\frac{dN}{dt}) = \; ?$ Now, $\lambda = $ $\frac{0.693}{5.1 \; * \; 10^{10}}$ $ = 1.36 * 10^{-11}\; Sec^{-1}$ Again, 226 gm of $Ra-226$ contains $6.023 * 10^{23}$ atoms 1 gm of $Ra-226$ contains $\frac{6.023 \; * \; 10^{23}}{226}$ $ = 2.66 * 10^{21}$ atoms $\therefore \;\;\; N= 2.66 * 10^{21}$ atoms Now, $\frac{dN}{dt}$ $= \lambda \,N = 1.36 * 10^{-11} * 2.66 * 10^{10}\; dis/Sec.$ Return to Main Menu

The Origin of Universe from Nothingness | Quantum Fluctuation

Metastable false Vacuum has "neither matter nor space or time", but is a form of wave function referred to as "quantum potential." In order to understand the Origin of the universe, we need to combine the General Theory of Relativity with quantum theory. The favored interpretation by most scientists, is that it indicates that the General Theory of Relativity breaks down in the very strong gravitational fields in the early universe. It has to be replaced by a more complete theory. One would expect this anyway, because General Relativity does not take account of the small scale structure of matter, which is governed by quantum theory. This does not matter normally, because the scale of the universe is enormous compared to the microscopic scales of quantum theory. But when the universe is the Planck size, a billion trillion trillionth of a centimetre, the two scales are the same, and quantum theory has to be taken into account. According to Heisenberg’s uncertainty p...

20 Things to know about Relativity

Galileo invented it, Einstein understood it, and Eddington saw it. Credit: vchal/Shutterstock 1. Who invented relativity? Bzzzt - wrong. Galileo hit on the idea in 1639, when he showed that a falling object behaves the same way on a moving ship as it does in a motionless building. 2. And Einstein didn’t call it relativity. The word never appears in his original 1905 paper, “On the Electrodynamics of Moving Bodies,” and he hated the term, preferring “invariance theory” (because the laws of physics look the same to all observers - nothing “relative” about it). 3. Space-time continuum? Nope, that’s not Einstein either. The idea of time as the fourth dimension came from Hermann Minkowski , one of Einstein’s professors, who once called him a “lazy dog.” 4. But Einstein did reformulate Galileo’s relativity to deal with the bizarre things that happen at near-light speed, where time slows down and space gets compressed. That counts for something. 5. Austrian physicist Friedrich Hasenöhrl pub...

What force is required to stretch a steel wire of cross-sectional area $1\;cm^2$ to double its length?

Given, Area of the cross-section $(A) = 1\;cm^2 = 1 * 10^{-4}\,m$ Young's modulus of steel $(Y) = 2 * 10^{11}\,N/m^2$ Let, length of the wire $(l) = x\;cm$ After extension its length $(L)$ becomes double i.e $2x$ So extension $(e) = \Delta l = L - l = 2x - x = x\;cm$ Force $(F) = ?$ Now we have, $Y = \frac{F/A}{e/l} = \frac{F/A}{x/x} = \frac{F}{A}$ ⇒ $F = Y\,*\,A = 2 * 10^{11} \; * \; 1 \; * \; 10^{-4} = 2 * 10^{7}\;N$ Return to Main Menu

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