Question Number 13886 by tawa tawa last updated on 24/May/17 Answered by ajfour last updated on 24/May/17 $$\mathrm{2}. \\ $$$${F}=\frac{{e}^{\mathrm{2}} }{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}^{\mathrm{2}} } \\ $$$$\:{r}^{\mathrm{2}}…
Question Number 79423 by Rio Michael last updated on 25/Jan/20 $${solve}\:{for}\:{x}\:{and}\:{y}\: \\ $$$$\:\:\:{sinh}\:{x}\:−\:\mathrm{2}{cosh}\:{y}\:=\:\mathrm{0} \\ $$$$\:\:\:\mathrm{3}{cosh}\:{x}\:+\:\mathrm{6}\:{sihn}\:{y}\:=\:\mathrm{5} \\ $$ Answered by mr W last updated on 25/Jan/20…
Question Number 13875 by tawa tawa last updated on 24/May/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 13872 by tawa tawa last updated on 24/May/17 $$\mathrm{An}\:\mathrm{alpha}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\:\mathrm{6}.\mathrm{68}\:×\:\mathrm{10}^{−\mathrm{27}} \:\mathrm{kg}\:\mathrm{and}\:\mathrm{charge}\:\:\mathrm{q}\:=\:+\mathrm{2},\:\mathrm{are}\: \\ $$$$\mathrm{accelerated}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{through}\:\mathrm{the}\:\mathrm{potential}\:\mathrm{difference}\:\mathrm{of}\:\:\mathrm{1kV}.\:\mathrm{it}\:\mathrm{then}\:\mathrm{enters} \\ $$$$\mathrm{a}\:\mathrm{magnetic}\:\mathrm{field}\:\mathrm{B}\:=\:\mathrm{0}.\mathrm{2}\:\mathrm{T}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{their}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{motion}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{their}\:\mathrm{path}. \\ $$ Answered by ajfour last updated…
Question Number 79394 by TawaTawa last updated on 24/Jan/20 Answered by john santu last updated on 25/Jan/20 $${v}_{{x}} ={v}_{\mathrm{1}} \mathrm{cos}\:\mathrm{30}^{{o}} =\mathrm{20}.\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{10}\sqrt{\mathrm{3}}\: \\ $$$${v}_{{y}} =\:{v}_{\mathrm{2}} \mathrm{sin}\:\mathrm{30}^{{o}}…
Question Number 13849 by Tinkutara last updated on 24/May/17 $$\mathrm{Given}\:\mathrm{below}\:\mathrm{is}\:\mathrm{a}\:\mathrm{graph}\:\mathrm{between}\:\mathrm{speed} \\ $$$$\mathrm{and}\:\mathrm{time}\:\mathrm{for}\:\mathrm{a}\:\mathrm{particle}.\:\mathrm{Is}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{undergoing}\:\mathrm{positive}\:\mathrm{displacement}\:\mathrm{or} \\ $$$$\mathrm{negative}\:\mathrm{displacement}? \\ $$ Commented by Tinkutara last updated on 24/May/17…
Question Number 13832 by tawa tawa last updated on 24/May/17 Answered by ajfour last updated on 24/May/17 Commented by ajfour last updated on 24/May/17 $${F}_{\mathrm{1}}…
Question Number 13830 by tawa tawa last updated on 24/May/17 Answered by ajfour last updated on 24/May/17 $${V}=\pi{R}^{\mathrm{2}} {H}+\frac{\mathrm{2}}{\mathrm{3}}\pi{R}^{\mathrm{2}} {h}\:\:\: \\ $$$$\:\:\:=\pi{R}^{\mathrm{2}} \left({H}+\frac{\mathrm{2}{R}}{\mathrm{3}}\right)\:\:\:\:\:\:\:\:\:\left(\because\:{as}\:{h}={R}\right) \\ $$$$\:\:\:\approx\frac{\mathrm{22}}{\mathrm{7}}×\mathrm{25}\left(\mathrm{10}+\frac{\mathrm{10}}{\mathrm{3}}\right)…
Question Number 13831 by tawa tawa last updated on 24/May/17 Answered by ajfour last updated on 24/May/17 $$\left(\mathrm{5}\right). \\ $$$${F}_{{e}} =\frac{{e}^{\mathrm{2}} }{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}^{\mathrm{2}} }\:\:\:;\:\:{F}_{{g}} ={G}\frac{{m}_{{e}}…
Question Number 144849 by Dwaipayan Shikari last updated on 29/Jun/21 $$\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{1}}{{e}^{\left\{{x}\right\}^{\mathrm{2}} } +\mathrm{1}}{dx}\:\:\:\left\{{x}\right\}\:\:{is}\:{fractional}\:{part}\:{of}\:{x} \\ $$ Answered by mathmax by abdo last updated on…