Question Number 199181 by mr W last updated on 29/Oct/23 Commented by mr W last updated on 29/Oct/23 $${an}\:{uniform}\:{rope}\:{with}\:{length}\:\boldsymbol{{L}}\:{and} \\ $$$${unit}\:{mass}\:\boldsymbol{\rho}\:{is}\:{fixed}\:{at}\:{one}\:{end}\:{as} \\ $$$${shown}.\:{its}\:{other}\:{end}\:{is}\:{connected}\:{with} \\ $$$${a}\:{mass}\:{such}\:{that}\:{the}\:{rope}\:{and}\:{the}…
Question Number 198832 by ajfour last updated on 24/Oct/23 Commented by ajfour last updated on 24/Oct/23 $${The}\:{radii}\:{are}\:{a}\left({green}\right),\:{b}\left({brown}\right), \\ $$$$\:{c}\left({blue}\right). \\ $$$${Find}\:{v}\:{and}\:{V}\:\:{when}\:{green}\:{ball}\:{comes} \\ $$$${sliding}\:{at}\:{speed}\:{u}\:{its}\:{centre}\:{is}\:{at} \\ $$$$\left({a},{a},{z}\right)\:\:\:\:{z}>{a}.\:{This}\:{pushes}\:{other}…
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Question Number 198395 by BHOOPENDRA last updated on 19/Oct/23 Commented by mr W last updated on 22/Oct/23 $${what}\:{is}\:{the}\:{question}? \\ $$ Commented by mr W last…
Question Number 198363 by BHOOPENDRA last updated on 18/Oct/23 Answered by mahdipoor last updated on 18/Oct/23 $${F}_{{A}} ={A}\left({cos}\mathrm{30}\boldsymbol{\mathrm{j}}+{sin}\mathrm{30}\boldsymbol{\mathrm{i}}\right) \\ $$$${F}_{{B}} ={B}_{{x}} \boldsymbol{\mathrm{i}}+{B}_{{y}} \boldsymbol{\mathrm{j}} \\ $$$$\Sigma{F}=\mathrm{0\begin{cases}{\boldsymbol{\mathrm{j}}\Rightarrow\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{A}−\frac{\mathrm{12}}{\mathrm{13}}\left(\mathrm{350}\right)+{B}_{{y}}…
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Question Number 197619 by mr W last updated on 24/Sep/23 Commented by mr W last updated on 24/Sep/23 $${assume}\:{the}\:{hill}\:{has}\:{the}\:{shape}\:{of}\:{a}\: \\ $$$${parabola}.\:{find}\:{the}\:{minimum}\:{speed} \\ $$$${with}\:{which}\:{a}\:{projectile}\:{should}\:{be} \\ $$$${launched}\:{from}\:{point}\:{C}\:{such}\:{that}\:{it}…
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Question Number 196785 by Spillover last updated on 31/Aug/23 Answered by aleks041103 last updated on 03/Oct/23 $${Separation}\:{of}\:{variables} \\ $$$$\Psi\left({x},{t}\right)={X}\left({x}\right){T}\left({t}\right) \\ $$$$\Rightarrow{X}''{T}=−\frac{\mathrm{1}}{{c}^{\mathrm{2}} }{XT}'' \\ $$$$\Rightarrow\frac{{X}''}{{X}}=−\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\:\frac{{T}\:''}{{T}}=−{a}^{\mathrm{2}}…
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