Field of NoThingness - spl BiNal

Given, Terminal velocity of the each small raindrop $(v) = 0.15 \; m/s$ Terminal velocity of the big raindrop $(V) = ?$ Let $r$ be the radius of the small rain droplets. Then the terminal velocity of each small raindrop is, $v = \frac{2\,r^2 (\rho \; - \; \sigma)\,g}{9\; \eta}$ .......... (i) Let $R$ be the radius of the combined drop. Then the terminal velocity of the large drop is, $V = \frac{2\,R^2 (\rho \; - \; \sigma)\,g}{9\; \eta}$ .......... (ii) Since the volume of big raindrop = volume of eight small raindrops. Then $\frac{4}{3}\; \pi\; R^3 = 3 *\;\frac{4}{3}\; \pi \; r^3$ ⇒ $R = 3^{1/3}\;r$ .......... (iii) Now, Dividing equation (ii) by (i), we get $\frac{V}{v} = \frac{R^2}{r^2}$ .......... (iv) Using equation (iii) in equation (iv), then we get or, $V = v * \frac{R^2}{r^2} = v * \frac{(3^{1/3} \; r)^2}{r^2} = 0.15 * 2.08 = 0.312 \;m/s$ $\therefore$ Terminal velocity of the big raindrop $(V) = 0.312\;m/s$ Return to Main M...

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

Three capacitors of 1μF , 2μF and 3μF are connected first in series. The same capacitors are again connected in parallel. Compare their equivalent capacitance. Which one of these combinations gives a larger value of capacitance.

Given, Capacitance of there capacitors are, $C_1 = 1 \mu F$; $C_2 = 2 \mu F$; $C_3 = 3 \mu F$ When they are connected in series, their equivalent capacitance $C_s$ is, $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6}{11}\mu F$ ⇒ $C_s = \frac{11}{6}\mu \; F$ When they are parallel, their equivalent capacitance $C_p$ is $C_p = C_1 + C_2 + C_3 = 1 + 2 + 3 = 6 \; \mu F$ Now, to compare the capacitance of series and parallel combination, $\frac{C_p}{C_s} = \frac{6} {\frac{6}{11}}$ $= 11$ $C_p = 11\;C_s$ The capacitance of the parallel combination is 11 times greater than that of the series combination. Return to Main Menu

Quantum Computer | How does it Work?

By Michael Tabb , Andrea Gawrylewski , Jeffery DelViscio Adapted from Scientific America , July 7, 2021 If you understand how these systems operate, then you understand why they could change everything. If someone asked you to picture a quantum computer, what would you see in your mind? Maybe you see a normal computer-- just bigger, with some mysterious physics magic going on inside? Forget laptops or desktops. Forget computer server farms. A quantum computer is fundamentally different in both the way it looks, and ,more importantly, in the way it processes information. There are currently several ways to build a quantum computer. But let’s start by describing one of the leading designs to help explain how it works. Imagine a lightbulb filament, hanging upside down, but it’s the most complicated light you’ve ever seen. Instead of one slender twist of wire, it has organized silvery swarms of them, neatly braided around a core. They ar...

How much force is required to punch a hole 1 cm in diameter in a steel sheet $5\;mm$ thick whose shearing strength is $2.76 \; * \; 10^8\;N/m^2$.

Given, Diameter of the hole $(d) = 1\;cm = 0.01\;m$ Thickness of the sheet $(t) = 5\;mm = 5 * 10^{-3}\;m$ Shear of strength $(\eta) = 2.76 * 10^8 \;N/m^2$ Force required $(F) = ?$ We have, Shear strength $= \frac{F}{A}$ .......... (i) But $A = l * t = 2 . \pi . r * t$ [∵ Length of the circumference $l = 2 . \pi . r$] From the equation $1^{st}$, $F = Shear \;\; strength * A = 2.76 * 10^8 * 2 . \pi . r * t $ $ \;\;\;\; = 2.76 * 10^8 * 2 . \pi . 0.01 * 5 * 10^{-3} = 43353\;N$ Return to Main Menu

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