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Author: Tinku Tara

What-is-the-acceleration-due-to-gravity-g-on-the-moon-if-g-is-10m-s-2-on-the-earth-

Question Number 12316 by tawa last updated on 18/Apr/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity}\:,\:\mathrm{g},\:\mathrm{on}\:\mathrm{the}\:\mathrm{moon}\:,\:\mathrm{if}\:\mathrm{g}\:\mathrm{is}\:\mathrm{10m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{earth}. \\ $$ Answered by chux last updated on 18/Apr/17 $$\mathrm{since}\:\mathrm{gravity}\:\mathrm{in}\:\mathrm{moon}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{6}}\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{earth}'\mathrm{s}…

proof-tg-2-36-tg-2-72-5-

Question Number 143384 by mathdanisur last updated on 13/Jun/21 $${proof}:\:\:{tg}^{\mathrm{2}} \left(\mathrm{36}°\right)\:\centerdot\:{tg}^{\mathrm{2}} \left(\mathrm{72}°\right)\:=\:\mathrm{5} \\ $$ Answered by Dwaipayan Shikari last updated on 13/Jun/21 $${tan}\mathrm{36}°=\frac{{sin}\left(\mathrm{36}°\right)}{{cos}\left(\mathrm{36}°\right)}=\frac{\mathrm{1}−{cos}\mathrm{72}°}{{sin}\mathrm{72}°}=\frac{\mathrm{1}−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}}{\:\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}=\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}} \\ $$$${tan}\mathrm{72}°=\frac{{cos}\mathrm{18}°}{{sin}\mathrm{18}°}=\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}…

let-f-x-arctan-2-x-2-1-calculate-f-n-x-and-f-n-0-2-if-f-x-a-n-x-n-find-the-sequence-a-n-

Question Number 143381 by Mathspace last updated on 13/Jun/21 $${let}\:{f}\left({x}\right)={arctan}\left(\sqrt{\mathrm{2}}{x}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right){and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){if}\:{f}\left({x}\right)=\Sigma{a}_{{n}} {x}^{{n}} \:\:{find}\:{the}\: \\ $$$${sequence}\:{a}_{{n}} \\ $$ Answered by…

In-the-equation-B-0-H-0-M-why-is-the-polarization-of-the-vacuum-accounted-for-by-constant-0-if-the-vacuum-is-absolutely-empty-

Question Number 77845 by OTC-X1 last updated on 11/Jan/20 $$\boldsymbol{\mathrm{I}}\mathrm{n}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\boldsymbol{\mathrm{B}}=\mu_{\mathrm{0}} \boldsymbol{\mathrm{H}}×\mu_{\mathrm{0}} \boldsymbol{\mathrm{M}}\: \\ $$$$\mathrm{why}\:\mathrm{is}\:\mathrm{the}\:\mathrm{polarization}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vacuum}\: \\ $$$$\mathrm{accounted}\:\mathrm{for}\:\mathrm{by}\: \\ $$$$\mathrm{constant}\:\mu_{\mathrm{0}} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{vacuum}\:\mathrm{is}\:\mathrm{absolutely}\:\mathrm{empty}? \\ $$$$ \\…

Two-similar-boxes-B-i-i-1-2-contain-i-1-red-and-5-i-1-black-balls-One-box-is-chosen-at-random-and-two-balls-are-drawn-randomly-what-is-the-probability-that-both-balls-are-of-different-colours-

Question Number 12306 by Gaurav3651 last updated on 18/Apr/17 $${Two}\:{similar}\:{boxes}\:{B}_{{i}} \:\left({i}=\mathrm{1},\mathrm{2}\right){contain} \\ $$$$\left({i}+\mathrm{1}\right){red}\:{and}\:\left(\mathrm{5}−{i}−\mathrm{1}\right)\:{black}\:{balls}. \\ $$$${One}\:{box}\:{is}\:{chosen}\:{at}\:{random}\:{and} \\ $$$${two}\:{balls}\:{are}\:{drawn}\:{randomly}. \\ $$$${what}\:{is}\:{the}\:{probability}\:{that}\:{both} \\ $$$${balls}\:{are}\:{of}\:{different}\:{colours}? \\ $$$$\left({a}\right)\:\:\mathrm{1}/\mathrm{2} \\ $$$$\left({b}\right)\:\:\mathrm{3}/\mathrm{10}…