Question Number 20761 by Tinkutara last updated on 02/Sep/17 $${A}\:\mathrm{5}\:{kg}\:{block}\:{B}\:{is}\:{suspended}\:{from}\:{a} \\ $$$${cord}\:{attached}\:{to}\:{a}\:\mathrm{40}\:{kg}\:{cart}\:{A}.\:{Find} \\ $$$${the}\:{accelerations}\:{of}\:{both}\:{the}\:{block}\:{and} \\ $$$${cart}.\:\left({All}\:{surfaces}\:{are}\:{frictionless}\right) \\ $$$$\left({g}\:=\:\mathrm{10}\:{m}/{s}^{\mathrm{2}} \right) \\ $$ Commented by Tinkutara last…
Question Number 20745 by Tinkutara last updated on 02/Sep/17 $$\mathrm{Consider}\:\mathrm{a}\:\mathrm{disc}\:\mathrm{rotating}\:\mathrm{in}\:\mathrm{the}\:\mathrm{horizontal} \\ $$$$\mathrm{plane}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{angular}\:\mathrm{speed}\:\omega \\ $$$$\mathrm{about}\:\mathrm{its}\:\mathrm{centre}\:{O}.\:\mathrm{The}\:\mathrm{disc}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{shaded}\:\mathrm{region}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diameter} \\ $$$$\mathrm{and}\:\mathrm{an}\:\mathrm{unshaded}\:\mathrm{region}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{side}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{Figure}.\:\mathrm{When}\:\mathrm{the} \\ $$$$\mathrm{disc}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{orientation}\:\mathrm{as}\:\mathrm{shown},\:\mathrm{two} \\ $$$$\mathrm{pebbles}\:{P}\:\mathrm{and}\:{Q}\:\mathrm{are}\:\mathrm{simultaneously} \\…
Question Number 20724 by Tinkutara last updated on 01/Sep/17 $$\mathrm{A}\:\mathrm{sphere}\:\mathrm{is}\:\mathrm{rolling}\:\mathrm{without}\:\mathrm{slipping}\:\mathrm{on} \\ $$$$\mathrm{a}\:\mathrm{fixed}\:\mathrm{horizontal}\:\mathrm{plane}\:\mathrm{surface}.\:\mathrm{In}\:\mathrm{the} \\ $$$$\mathrm{Figure},\:\mathrm{A}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{of}\:\mathrm{contact},\:{B}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sphere}\:\mathrm{and}\:{C}\:\mathrm{is}\:\mathrm{its}\:\mathrm{topmost} \\ $$$$\mathrm{point}.\:\mathrm{Then} \\ $$$$\left({a}\right)\:\overset{\rightarrow} {{v}}_{\mathrm{C}} \:−\:\overset{\rightarrow} {{v}}_{\mathrm{A}} \:=\:\mathrm{2}\left(\overset{\rightarrow} {{v}}_{\mathrm{B}}…
Question Number 86198 by Rio Michael last updated on 27/Mar/20 $$\mathrm{any}\:\mathrm{methods}\:\mathrm{to}\:\mathrm{sketch}\:\mathrm{these}\:\mathrm{curves} \\ $$$$\:\mathrm{r}\:=\:\mathrm{a}\left(\mathrm{1}−\mathrm{cos}\theta\right) \\ $$$$\:\mathrm{r}=\:\mathrm{a}\:+\:\mathrm{b}\:\mathrm{cos}\theta\:\:{a}>{b} \\ $$$$\:\mathrm{r}=\:\mathrm{a}\:+\:\mathrm{bcos}\theta\:\:\:{a}<{b} \\ $$ Commented by Prithwish Sen 1 last…
Question Number 86140 by Wepa last updated on 27/Mar/20 Commented by Prithwish Sen 1 last updated on 27/Mar/20 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2020}} \underset{\boldsymbol{\mathrm{k}}=\mathrm{2}} {\overset{\mathrm{2020}} {\boldsymbol{\sum}}\mathrm{k}}.\mathrm{2}^{\boldsymbol{\mathrm{k}}} \:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$…
Question Number 151677 by Tawa11 last updated on 22/Aug/21 Answered by Olaf_Thorendsen last updated on 22/Aug/21 $$\mathrm{R}\:=\:\mathrm{1}\Omega+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{3}\Omega+\mathrm{3}\Omega}+\frac{\mathrm{1}}{\mathrm{3}\Omega+\mathrm{3}\Omega}} \\ $$$$\mathrm{R}\:=\:\mathrm{1}\Omega+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{6}\Omega}+\frac{\mathrm{1}}{\mathrm{6}\Omega}} \\ $$$$\mathrm{R}\:=\:\mathrm{1}\Omega+\mathrm{3}\Omega\:=\:\mathrm{4}\Omega \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{V}}{\mathrm{R}}\:=\:\frac{\mathrm{12V}}{\mathrm{4}\Omega}\:=\:\mathrm{3A} \\ $$…
Question Number 86128 by M±th+et£s last updated on 27/Mar/20 $$\lfloor\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\rfloor=\lfloor\mid{x}\mid−\frac{\mathrm{1}}{\mathrm{2}}\rfloor=\mathrm{2}{x}−\mathrm{2} \\ $$ Answered by mr W last updated on 27/Mar/20 $$\lfloor\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\rfloor−\lfloor\mid{x}\mid−\frac{\mathrm{1}}{\mathrm{2}}\rfloor=\mathrm{2}{x}−\mathrm{2} \\ $$$$\lfloor\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\rfloor−\lfloor\mid{x}\mid−\frac{\mathrm{1}}{\mathrm{2}}\rfloor=\mathrm{2}{x}−\mathrm{2}={n}=\begin{cases}{\mathrm{2}{k}}\\{\mathrm{2}{k}+\mathrm{1}}\end{cases} \\ $$$$\Rightarrow{x}=\frac{{n}}{\mathrm{2}}+\mathrm{1}=\begin{cases}{{k}+\mathrm{1}}\\{{k}+\frac{\mathrm{3}}{\mathrm{2}}}\end{cases}…
Question Number 86132 by M±th+et£s last updated on 27/Mar/20 $$\int{x}^{\mathrm{3}} \:{sin}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}\right)^{\mathrm{5}} \:{dx} \\ $$ Answered by Kunal12588 last updated on 27/Mar/20 $${I}=\int{x}^{\mathrm{3}} \:{sin}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}\right)^{\mathrm{5}}…
Question Number 20579 by Tinkutara last updated on 28/Aug/17 $$\mathrm{A}\:\mathrm{1}\:\mathrm{kg}\:\mathrm{block}\:\mathrm{is}\:\mathrm{being}\:\mathrm{pushed}\:\mathrm{against}\:\mathrm{a} \\ $$$$\mathrm{wall}\:\mathrm{by}\:\mathrm{a}\:\mathrm{force}\:{F}\:=\:\mathrm{75}\:\mathrm{N}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{figure}.\:\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{is} \\ $$$$\mathrm{0}.\mathrm{25}.\:\mathrm{The}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{acceleration}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{block}\:\mathrm{is} \\ $$ Commented by Tinkutara last updated…
Question Number 20581 by Tinkutara last updated on 28/Aug/17 $${The}\:{surface}\:{between}\:{wedge}\:{and}\:{block} \\ $$$${is}\:{rough}\:\left({Coefficient}\:{of}\:{friction}\:\mu\right). \\ $$$${Find}\:{out}\:{the}\:{range}\:{of}\:{F}\:{such}\:{that}, \\ $$$${there}\:{is}\:{no}\:{relative}\:{motion}\:{between} \\ $$$${wedge}\:{and}\:{block}.\:{The}\:{wedge}\:{can}\:{move} \\ $$$${freely}\:{on}\:{smooth}\:{ground}. \\ $$ Commented by Tinkutara…