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Field of NoThingness - spl BiNal

Let's see the invisible: "Yato Vancho Nivirtante (From where the speech returns)"!!

'... the journey to the unknown field called NoThingness. And from such field, all things are created.' ~ spl BiNal.

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Physics in a second language | Symmetry Magazine

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English Laungage Physics Physics Laungage Quantum Physics Symmetry Magazine

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Study theoretical physics is to take risks. | By Evelyn Lamb

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A theory of Everything Quantum Physics Theoretical Physics

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Big Bounce Simulations Challenge the Big Bang

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Big Bang Quantum Physics Universe Universe from Nothing

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A thundercloud and the earth can be regarded as a parallel plate capacitor. Taking the area of the thundercloud to be $50 \;km^2$, its height above the earth as $1\; km$ and it's potential $100\; KV$, calculate the energy stored.

Given, Area of the thundercloud $(A) = 50 km^2 = 50\;*\;10^{6}\;m^2$ Distance of cloud from the earth $(d) = 1\;km = 1 \; * \; 10^{3}\;m$ Potential $(V) = 100\;kV = 100 \; * \; 10^{3}\;V$ Energy Stored $(E) = ?$ Since the capacitance of parallel plates capacitor is $C = $ $\frac{1}{2}$ $\;C\;V^2 $ Energy stored $(E)= $ $\frac{1}{2}\;*\; \frac{\epsilon_0\;A}{d}$ $V^2 = $ $\frac{1}{2}\;*\; \frac{8.85\;*10^{-12}\;50\;*\;10^6\;*\;(10^5)^2}{10^3}$ ∴ $E = 2.2 \; * \; 10^3\;J$ Return to Main Menu

A 300V battery is connected across capacitors of 3μF and 6μF in parallel. Calculate the energy stored in each capacitor.

Given, Potential $V = 300\;V$ Capacitor $C_1 = 3\mu F$ Capacitor $C_2 = 6\mu F$ Energy stored in each capacitor $E = ?$ When the capacitors are connected in parallel, then the equivalent capacitance $C = C_1 + C_2 = (3 + 6)\;\mu F = 9 \; \mu F$ We have, Energy stored in a capacitor ($C_1$); $E_1 = \frac{1}{2}C\;V^2 = \frac{1}{2} * 3 * 10^{-6} * 300^2 = 0.135\;J$ Energy stored in a capacitor ($C_2$); $E_2 = \frac{1}{2}C\;V^2 = \frac{1}{2} * 6 * 10^{-6} * 300^2 = 0.27\;J$ Energy stored in the equivalence capacitance ($C$); $E = \frac{1}{2}C\;V^2 = \frac{1}{2} * 9 * 10^{-6} * 300^2 = 0.405\;J$ Return to Main Menu

Making Sense of Quantum Physics

"The situation cannot declare itself until you've asked your question. But the asking of one question prevents and excludes the asking of another.” ~ John Archibald Wheeler Is quantum mechanics weird if no one is paying attention to it? Lots of people are paying attention to it now, and they seem to agree it’s weird. Credit: gettyimages The most widely accepted explanation of the “meaning” of quantum mechanics was the so-called orthodox, interpretation (although “orthodox” seems an odd descriptor for such a radical worldview). Also called the Copenhagen interpretation, because Bohr lectured on it in Copenhagen in the late 1920s, it holds that we cannot specify the nature of fundamental reality. Subatomic entities exist in a probabilistic limbo of many possible “superposed” states until they are brought into focus by the act of measurement. "Quantum mechanics is everywhere." Quantum Mechanics is the cornerstone of physical theories dealing wi

What is the terminal velocity of a glass ball falling through a tall jar containing glycerine? The densities of the glass ball and glycerine are $2.6\;g/cc$ and $1.32 \;g/cc$ respectively and the viscosity of the glycerine is $0.85\;poise$ and radius of the glass ball is $2\;mm$.

Given, Density of glass $(\rho) = 2.6\;g/cc = 2600\;kg/m^3$ Density of the glycerin $(\sigma) = 1.32\;g/cc = 1320\;kg/m^3$ Viscosity of glycerin $(\eta) = 0.85\;poise = 0.085 \; decapoise$ Radius of the ball $(r) = 2\;mm = 2 * 10^{-3}\;m$ Terminal Velocity $(v) = ?$ We know, $v = \frac{2\; r^2 \; (\rho \; - \; \sigma)\;g}{9\; \eta} = \frac{2 \; * \; (2 \; * \; 10^ {-3})^2 \; * \; (2600 \; - \; 1320) \; 10}{9 \; * \; 0.085}$ $= 0.132 \; m/s$ Return to Main Menu

A steel cable with a cross-sectional area $3\;cm^2$ has an elastic limit of $2.4\;*\;10^8\;pa$. Find the maximum upward acceleration that can be given a $1200\;kg$ elevator supported by the cable if the stress does not exceed one-third of the elastic limit.

Given, Area of the cross-section of cable $(A) = 3\;cm^2 = 3\,*\,10^{-4}\;m^2$ Elastic limit $ = 2.4 * 10^8\;pa$ Stress $ = \frac{1}{3} * 2.4 * 10^8 \;pa$ Mass $(M) = 1200\;kg$ Upward Acceleration $(a) = ?$ We have, $F = Stress * A = \frac{1}{3}* 2.4 * 10^8 * 3 * 10^{-4} = 2.4 * 10^4\;N$ ⇒ $a = $ $\frac{F}{M} = \frac{2.4 \; * \; 10^4}{1200}$ $ = 20\;m/s^2$ Return to Main Menu

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