Question Number 224570 by Abdulazim last updated on 19/Sep/25 Answered by fantastic last updated on 19/Sep/25 $$\mathrm{30}^{\mathrm{0}} \\ $$$${AB}=\mathrm{2}{R}\mathrm{sin}\:\alpha \\ $$$${CH}=\mathrm{2}{R}\mathrm{cos}\:\alpha\left[{R}={circumradius}\right] \\ $$$${CH}=\sqrt{\mathrm{3}}{AB} \\ $$$$\mathrm{2}{R}\mathrm{cos}\:\alpha=\sqrt{\mathrm{3}}×\mathrm{2}{R}\mathrm{sin}\:\alpha…
Question Number 224562 by gregori last updated on 19/Sep/25 $$\:\:\: \\ $$ Commented by Frix last updated on 19/Sep/25 $${r}=\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }=\frac{{a}_{\mathrm{3}} }{{a}_{\mathrm{2}} }\:\Leftrightarrow\:\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}\:\Leftrightarrow\:\frac{\mathrm{cos}^{\mathrm{3}}…
Question Number 224556 by mr W last updated on 18/Sep/25 Commented by mr W last updated on 18/Sep/25 $${if}\:{the}\:{friction}\:{coefficient}\:{between} \\ $$$${a}\:{small}\:{ball}\:{with}\:{mass}\:\boldsymbol{{m}}\:{and}\:{a} \\ $$$${thin}\:{tube}\:{ring}\:{with}\:{radius}\:\boldsymbol{{R}}\:{and} \\ $$$${mass}\:\boldsymbol{{M}}\:{is}\:\boldsymbol{\mu},\:{find}\:{the}\:{angle}\:\boldsymbol{\theta}\:{which}…
Question Number 224538 by MirHasibulHossain last updated on 17/Sep/25 $$\mathrm{32}^{\mathrm{4r}^{\mathrm{2}} −\mathrm{8}} =\mathrm{1}\:\:\:\:\mathrm{then}\:\mathrm{find}\:\mathrm{r}=? \\ $$ Answered by fantastic last updated on 17/Sep/25 $$\left(\mathrm{4}{r}^{\mathrm{2}} −\mathrm{8}\right)=\mathrm{log}\:_{\mathrm{32}} \mathrm{1}=\mathrm{0} \\…
Question Number 224539 by fantastic last updated on 17/Sep/25 $${a}^{{x}} ={m},\:{a}^{{y}} ={n}\:,{a}^{\mathrm{2}} =\left({m}^{{y}} {n}^{{x}} \right)^{{z}} \\ $$$${prove}\:{xyz}=\mathrm{1} \\ $$ Answered by som(math1967) last updated on…
Question Number 224529 by ajfour last updated on 17/Sep/25 Commented by ajfour last updated on 17/Sep/25 $${Find}\:{r}/{R}\:{if}\:{r}_{\mathrm{1}} ={r}_{\mathrm{2}} ={r}_{\mathrm{3}} \:{and}\:{R}_{\mathrm{2}} ={R}_{\mathrm{1}} . \\ $$ Answered…
Question Number 224531 by ajfour last updated on 17/Sep/25 Commented by ajfour last updated on 17/Sep/25 A plank inclined at angle phi to the level ground has coefficient of friction such that it just permits enough friction for a ring of mass M and radius R to purely roll down the fixed incline. Now our plan is to weld a bead at the periphery of the ring so that this new ring plus bead combination when olaced upright on the same incline only slides down not rotating at all. we hsve to find the minimum bead mass. assume coefficient of static and kinetic friction to be same. Commented by mahdipoor last updated on 17/Sep/25 $${i}\:{get}\:…
Question Number 224510 by Rogerkeyter last updated on 15/Sep/25 Answered by maths2 last updated on 16/Sep/25 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}+\frac{{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}}…
Question Number 224505 by EmGent last updated on 15/Sep/25 $$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{ax}} {J}_{\mathrm{0}} \left(\jmath_{\mathrm{0}{m}} {x}\right){dx} \\ $$$$\mathrm{Where}\:{J}_{\mathrm{0}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{and}\:\jmath_{\mathrm{0}{m}} \:\mathrm{its}\:{m}-\mathrm{th}\:\mathrm{zero} \\ $$ Terms of Service Privacy…
Question Number 224516 by Rogerkeyter last updated on 15/Sep/25 Answered by Mathswiz last updated on 17/Sep/25 Terms of Service Privacy Policy Contact: info@tinkutara.com