Question Number 25103 by Tinkutara last updated on 03/Dec/17 Answered by prakash jain last updated on 05/Dec/17 $$\int_{\mathrm{0}} ^{\mathrm{5}} \mid\mathrm{2}{t}−\mathrm{3}\mid{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{3}/\mathrm{2}} \left(\mathrm{3}−\mathrm{2}{t}\right){dt}+\int_{\mathrm{3}/\mathrm{2}} ^{\mathrm{5}}…
Question Number 25091 by Tinkutara last updated on 03/Dec/17 $${A}\:{particle}\:{of}\:{mass}\:{m}\:{moving}\:{with} \\ $$$${speed}\:{u}\:{collides}\:{perfectly}\:{inelastically} \\ $$$${with}\:{a}\:{sphere}\:{of}\:{radius}\:{R}\:{and}\:{same} \\ $$$${mass},\:{at}\:{rest},\:{at}\:{an}\:{impact}\:{parameter} \\ $$$${d}.\:{Find} \\ $$$$\left({a}\right)\:{Angle}\:{between}\:{their}\:{final}\:{velocities} \\ $$$$\left({b}\right)\:{Magnitude}\:{of}\:{their}\:{final} \\ $$$${velocities} \\…
Question Number 25086 by Tinkutara last updated on 03/Dec/17 Commented by Tinkutara last updated on 03/Dec/17 Commented by ajfour last updated on 04/Dec/17 $${I}_{{z}} ={I}_{\mathrm{1}}…
Question Number 25058 by Tinkutara last updated on 02/Dec/17 $${Two}\:{objects}\:{slide}\:{over}\:{a}\:{frictionless} \\ $$$${horizontal}\:{surface}.\:{The}\:{first}\:{object}, \\ $$$${mass}\:{m}_{\mathrm{1}} \:=\:\mathrm{5}\:{kg},\:{is}\:{propelled}\:{with}\:{a} \\ $$$${speed}\:{u}\:=\:\mathrm{4}.\mathrm{5}\:{m}/{s}\:{towards}\:{the}\:{second} \\ $$$${object},\:{mass}\:{m}_{\mathrm{2}} \:=\:\mathrm{5}\:{kg},\:{which}\:{is} \\ $$$${initially}\:{at}\:{rest}.\:{After}\:{the}\:{collision}, \\ $$$${both}\:{objects}\:{have}\:{velocities}\:{which}\:{are} \\…
Question Number 90581 by hardylanes last updated on 24/Apr/20 $${given}\:{that}\:\alpha\:{and}\:\beta\:{are}\:{roots}\:{of}\:\:{the}\:{equation}\: \\ $$$${a}\chi^{\mathrm{2}} +{b}\chi+{c}=\mathrm{0}.\:{show}\:{that}\:\lambda\mu{b}^{\mathrm{2}} ={ac}\left(\lambda+\mu\right)^{\mathrm{2}\:} \\ $$$${where}\:\frac{\alpha}{\beta}=\frac{\lambda}{\mu} \\ $$ Answered by maths mind last updated on…
Question Number 25038 by Tinkutara last updated on 02/Dec/17 $$\mathrm{For}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rotating}\:\mathrm{rigid}\:\mathrm{body}, \\ $$$${v}\:=\:{r}\omega.\:\mathrm{So} \\ $$$$\left(\mathrm{1}\right)\:\omega\:\propto\:\left(\mathrm{1}/{r}\right) \\ $$$$\left(\mathrm{2}\right)\:\omega\:\propto\:{v} \\ $$$$\left(\mathrm{3}\right)\:{v}\:\propto\:{r} \\ $$$$\left(\mathrm{4}\right)\:\omega\:\mathrm{is}\:\mathrm{independent}\:\mathrm{of}\:{r} \\ $$ Answered by ajfour…
Question Number 25013 by Tinkutara last updated on 01/Dec/17 $$\mathrm{With}\:\mathrm{reference}\:\mathrm{to}\:\mathrm{figure}\:\mathrm{of}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{of} \\ $$$$\mathrm{edge}\:{a}\:\mathrm{and}\:\mathrm{mass}\:{m},\:\mathrm{state}\:\mathrm{whether}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{are}\:\mathrm{true}\:\mathrm{or}\:\mathrm{false}.\:\left(\mathrm{O}\:\mathrm{is}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}.\right) \\ $$$$\left(\mathrm{1}\right)\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{cube} \\ $$$$\mathrm{about}\:{z}-\mathrm{axis}\:\mathrm{is},\:{I}_{{z}} \:=\:{I}_{{x}} \:+\:{I}_{{y}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{cube} \\…
Question Number 25000 by ErickDN last updated on 30/Nov/17 $${find}\:{x},{y}\:{from}\:{the}\:{equation}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}−{yi}+\frac{\mathrm{1}}{\mathrm{1}+{i}}=\left(\sqrt{\mathrm{1}+\omega^{\mathrm{8}} }+\sqrt{\mathrm{1}+\omega^{\mathrm{10}} }\right)^{\mathrm{4}} \\ $$$$ \\ $$ Answered by ajfour last updated on 01/Dec/17…
Question Number 156065 by VIDDD last updated on 07/Oct/21 $$\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:\mathrm{expression}\:\mathrm{M}=\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{B}}}{\mathrm{2}}\mathrm{sin}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\mathrm{sin}\frac{\boldsymbol{\mathrm{B}}}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 24997 by Tinkutara last updated on 30/Nov/17 Terms of Service Privacy Policy Contact: info@tinkutara.com