Question Number 18107 by Tinkutara last updated on 15/Jul/17 $$\mathrm{The}\:\mathrm{first}\:\mathrm{and}\:\mathrm{second}\:\mathrm{ionization} \\ $$$$\mathrm{potentials}\:\mathrm{of}\:\mathrm{helium}\:\mathrm{atoms}\:\mathrm{are}\:\mathrm{24}.\mathrm{58}\:\mathrm{eV} \\ $$$$\mathrm{and}\:\mathrm{54}.\mathrm{4}\:\mathrm{eV}\:\mathrm{per}\:\mathrm{mole}\:\mathrm{respectively}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{energy}\:\mathrm{in}\:\mathrm{kJ}\:\mathrm{required}\:\mathrm{to} \\ $$$$\mathrm{produce}\:\mathrm{1}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{He}^{\mathrm{2}+} \:\mathrm{ions}. \\ $$ Terms of Service Privacy…
Question Number 18106 by Tinkutara last updated on 15/Jul/17 $$\mathrm{The}\:\mathrm{ionization}\:\mathrm{potential}\:\mathrm{of}\:\mathrm{hydrogen}\:\mathrm{is} \\ $$$$\mathrm{13}.\mathrm{60}\:\mathrm{eV}/\mathrm{mole}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{energy}\:\mathrm{in} \\ $$$$\mathrm{kJ}\:\mathrm{required}\:\mathrm{to}\:\mathrm{produce}\:\mathrm{0}.\mathrm{1}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{H}^{+} \\ $$$$\left.\mathrm{ions}.\:\mathrm{Given},\:\mathrm{1}\:\mathrm{eV}\:=\:\mathrm{96}.\mathrm{49}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} \right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 83638 by bshahid010@gmail.com last updated on 04/Mar/20 Answered by TANMAY PANACEA last updated on 05/Mar/20 $${S}=\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}}{{x}^{\mathrm{4}} +\mathrm{1}}+…+\frac{\mathrm{2}^{{n}} }{{x}^{\mathrm{2}^{{n}} } +\mathrm{1}}\:\left({i}\:{think}\right) \\ $$$${S}=\underset{{n}=\mathrm{0}}…
Question Number 83639 by niroj last updated on 04/Mar/20 $$ \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equations}}: \\ $$$$\:\:\:\left(\mathrm{i}\right)\:\mathrm{log}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{ax}+\mathrm{by} \\ $$$$\:\:\:\left(\mathrm{ii}\right)\:\mathrm{x}\:\mathrm{cos}\:\mathrm{y}\:\mathrm{dy}=\left(\mathrm{x}\:\mathrm{e}^{\mathrm{x}} \mathrm{log}\:\mathrm{x}\:+\mathrm{e}^{\mathrm{x}} \right)\mathrm{dx} \\ $$$$ \\ $$ Answered…
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Question Number 149171 by liberty last updated on 03/Aug/21 $$\:\mathrm{2}^{\mathrm{2x}\:} \:+\:\mathrm{4}^{\mathrm{3x}} \:=\:\mathrm{128}\:\Rightarrow\mathrm{x}=? \\ $$ Answered by Ar Brandon last updated on 03/Aug/21 $$\mathrm{2}^{\mathrm{2}{x}} +\mathrm{4}^{\mathrm{3}{x}} =\mathrm{128}\:\:\Rightarrow\:\:\mathrm{4}^{{x}}…
Question Number 18095 by Tinkutara last updated on 15/Jul/17 $$\mathrm{A}\:\mathrm{boy}\:\mathrm{travelling}\:\mathrm{in}\:\mathrm{an}\:\mathrm{open}\:\mathrm{car}\:\mathrm{moving} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{levelled}\:\mathrm{road}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{speed} \\ $$$$\mathrm{tosses}\:\mathrm{a}\:\mathrm{ball}\:\mathrm{vertically}\:\mathrm{up}\:\mathrm{in}\:\mathrm{the}\:\mathrm{air}\:\mathrm{and} \\ $$$$\mathrm{catches}\:\mathrm{it}\:\mathrm{back}.\:\mathrm{Sketch}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{ball}\:\mathrm{as}\:\mathrm{observed}\:\mathrm{by}\:\mathrm{a}\:\mathrm{boy}\:\mathrm{standing} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{footpath}.\:\mathrm{Give}\:\mathrm{explanation}\:\mathrm{to} \\ $$$$\mathrm{support}\:\mathrm{your}\:\mathrm{diagram}. \\ $$ Answered…
Question Number 18094 by Tinkutara last updated on 15/Jul/17 $$\overset{\rightarrow} {{A}},\:\overset{\rightarrow} {{B}}\:\mathrm{and}\:\overset{\rightarrow} {{C}}\:\mathrm{are}\:\mathrm{three}\:\mathrm{non}-\mathrm{collinear}, \\ $$$$\mathrm{non}\:\mathrm{co}-\mathrm{planar}\:\mathrm{vectors}.\:\mathrm{What}\:\mathrm{can}\:\mathrm{you} \\ $$$$\mathrm{say}\:\mathrm{about}\:\mathrm{direction}\:\mathrm{of}\:\overset{\rightarrow} {{A}}×\left(\overset{\rightarrow} {{B}}×\overset{\rightarrow} {{C}}\right)? \\ $$ Commented by ajfour…
Question Number 18093 by Tinkutara last updated on 15/Jul/17 $$\mathrm{The}\:\mathrm{equation}\:\mathrm{sin}{x}\:+\:\mathrm{sin2}{x}\:+\:\mathrm{2sin}{x}\mathrm{sin2}{x} \\ $$$$=\:\mathrm{2cos}{x}\:+\:\mathrm{cos2}{x}\:\mathrm{is}\:\mathrm{satisfied}\:\mathrm{by}\:\mathrm{values} \\ $$$$\mathrm{of}\:{x}\:\mathrm{for}\:\mathrm{which} \\ $$$$\left(\mathrm{1}\right)\:{x}\:=\:{n}\pi\:+\:\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{6}}\:,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{2}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:+\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{3}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:−\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{4}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:−\:\frac{\pi}{\mathrm{2}}\:,\:{n}\:\in\:{I} \\ $$…
Question Number 83629 by M±th+et£s last updated on 04/Mar/20 $${show}\:{that} \\ $$$$\underset{{n},{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}!\:{k}!}{\left({n}+{k}+\mathrm{2}\right)!}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$ Answered by Kamel Kamel last updated on 04/Mar/20…