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Question-226292

Question Number 226292 by Spillover last updated on 25/Nov/25 Answered by Frix last updated on 25/Nov/25 $$=\int\frac{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}\:\overset{\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right]} {=} \\ $$$$=\mathrm{4}\int\frac{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)}{dt}\:\overset{\left[\mathrm{decompose}\:\mathrm{etc}.\right]} {=}…

A-hemispherical-bowl-of-radius-R-with-maimum-water-in-it-without-needing-to-spill-is-spinning-with-the-content-at-constant-Find-volume-of-water-in-bowl-

Question Number 226293 by ajfour last updated on 25/Nov/25 $${A}\:{hemispherical}\:{bowl}\:{of}\:{radius}\:{R} \\ $$$$\:{with}\:{maimum}\:{water}\:{in}\:{it}\:{without} \\ $$$${needing}\:{to}\:{spill}\:{is}\:{spinning}\:{with}\:{the} \\ $$$${content}\:{at}\:{constant}\:\omega.\:{Find}\:{volume} \\ $$$${of}\:{water}\:{in}\:{bowl}. \\ $$$$ \\ $$ Commented by fantastic2…

Question-226290

Question Number 226290 by Spillover last updated on 25/Nov/25 Answered by Frix last updated on 25/Nov/25 $$=\int\sqrt{\frac{\mathrm{cos}\:{x}\:\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x}\right)}{\mathrm{1}−\mathrm{cos}^{\mathrm{3}} \:{x}}}\:{dx}= \\ $$$$=\int\mathrm{sin}\:{x}\sqrt{\frac{\mathrm{cos}\:{x}}{\mathrm{1}−\mathrm{cos}^{\mathrm{3}} \:{x}}}\:{dx}\:\overset{\left[{t}=\mathrm{cos}^{\frac{\mathrm{3}}{\mathrm{2}}} \:{x}\right]} {=} \\…

Question-226291

Question Number 226291 by fantastic2 last updated on 25/Nov/25 Answered by mr W last updated on 25/Nov/25 $${m}\omega^{\mathrm{2}} \left({r}+{l}\:\mathrm{sin}\:\theta\right)={mg}\:\mathrm{tan}\:\theta \\ $$$$\Rightarrow\omega=\sqrt{\frac{{g}\:\mathrm{tan}\:\theta}{{r}+{l}\:\mathrm{sin}\:\theta}} \\ $$ Commented by…