Question Number 171908 by ajfour last updated on 21/Jun/22 Commented by ajfour last updated on 21/Jun/22 $${Man}\:{has}\:{to}\:{water}\:{the}\:{tree} \\ $$$${once}\:{with}\:{volume}\:{V}_{\mathrm{0}} \:.\:{His} \\ $$$${bucket}\:{leaks}\:{at}\:{a}\:{rate}\:\frac{{dV}}{{dt}}=−{kV}. \\ $$$${his}\:{speed}={u}\left(\mathrm{2}−\frac{{m}}{{m}+\rho{V}}\right) \\…
Question Number 106256 by mr W last updated on 03/Aug/20 Commented by mr W last updated on 03/Aug/20 $$\left({an}\:{unsolved}\:{old}\:{question}\right) \\ $$ Commented by mr W…
Question Number 106094 by Dwaipayan Shikari last updated on 02/Aug/20 Commented by Dwaipayan Shikari last updated on 02/Aug/20 Dwaipayan Shikari: The wedge's Surface is friction less (where the ball of mass 'm' will roll) and the friction coefficient between wedge and the ground is also negligible.( An oily surface ��) Calculate the velocity of the wedge when the ball will fly off. Additional information is on the picture Commented by Dwaipayan Shikari last updated…
Question Number 40551 by ajfour last updated on 24/Jul/18 Commented by ajfour last updated on 24/Jul/18 $${A}\:{chain}\:{of}\:{mass}\:\boldsymbol{{m}},\:{length}\:\boldsymbol{{l}} \\ $$$${hangs}\:{the}\:{table}\:{edge}\:{only}\:{as}\:{much} \\ $$$${so}\:{that}\:{it}\:{just}\:{starts}\:{slipping} \\ $$$${down}.\:{Friction}\:{coefficient}\: \\ $$$${between}\:{chain}\:{and}\:{table}\:{being}\:\boldsymbol{\mu}.…
Question Number 40396 by LXZ last updated on 21/Jul/18 $${A}\:{block}\:{lying}\:{on}\:{a}\:{horizontal}\:{conv}− \\ $$$${eyor}\:{belt}\:{moving}\:{at}\:\:{a}\:{constant}\: \\ $$$${velocity}\:{receives}\:{a}\:{velocity}\:\mathrm{5}{m}/{s} \\ $$$${at}\:{t}=\mathrm{0}\:{sec}.\:{relative}\:{to}\:{the}\:{ground}\: \\ $$$${in}\:{the}\:{direction}\:{opposite}\:{to}\:{the}\:{dir}− \\ $$$${ction}\:{of}\:{motion}\:{of}\:{the}\:{conveyor}. \\ $$$${Aftert}=\mathrm{4}{sec},{the}\:{velocity}\:{of}\:{the}\: \\ $$$${block}\:{becomes}\:{equal}\:{to}\:{the}\:{velocity} \\…
Question Number 105925 by ajfour last updated on 01/Aug/20 Commented by ajfour last updated on 02/Aug/20 $${Find}\:{u}_{{min}} \:{such}\:{that}\:{the}\:{solid}\:{ball} \\ $$$${can}\:{climb}\:{up}\:{the}\:{plank}.\:{Mass}\:{of} \\ $$$${ball}\:{m},\:{and}\:{that}\:{of}\:{plank}\:{is}\:{M}. \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\…
Question Number 105842 by aurpeyz last updated on 01/Aug/20 $$\mathrm{A}\:\mathrm{wheel}\:\mathrm{revolving}\:\mathrm{at}\:\mathrm{6rev}/\mathrm{s}\:\mathrm{has}\:\mathrm{an}\:\mathrm{angular}\: \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{4rad}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\: \\ $$$$\mathrm{turns}\:\mathrm{the}\:\mathrm{wheel}\:\mathrm{must}\:\mathrm{take}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{26rev}/\mathrm{s} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{time}\:\mathrm{required}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 105838 by aurpeyz last updated on 01/Aug/20 $${A}\:{mass}\:{of}\:\mathrm{1}.\mathrm{5}{kg}\:{out}\:{in}\:{space}\:{moves}\:{in}\: \\ $$$${a}\:{circle}\:{of}\:{radius}\:\mathrm{25}{cm}\:{at}\:{a}\:{constant}\:\mathrm{2}{rev}/{s}. \\ $$$${Calculate}\:\left({a}\right)\:{the}\:{tangential}\:{speed}\: \\ $$$$\left({b}\right)\:{acceleration}\:{and}\:\left({c}\right)\:{the}\:{required}\: \\ $$$${centripetal}\:{force}\:{for}\:{the}\:{motion}. \\ $$ Terms of Service Privacy Policy…
Question Number 105840 by aurpeyz last updated on 01/Aug/20 $${A}\:{Car}\:{wheel}\:\mathrm{30}{cm}\:{in}\:{radius}\:{is}\:{turning}\:{at}\:{a} \\ $$$${rate}\:{of}\:\mathrm{8}{rev}/{s}\:{when}\:{the}\:{car}\:{begins}\:{to}\:{slow}\:\: \\ $$$${uniformly}\:{to}\:{rest}\:{in}\:{a}\:{time}\:\mathrm{14}{s}.\:{find}\:{the}\: \\ $$$${number}\:{of}\:{revolutions}\:{made}\:{by}\:{the}\:{wheel} \\ $$$${and}\:{the}\:{distance}\:{the}\:{car}\:{goes}\:{in}\:{the}\:\mathrm{14}{s}. \\ $$ Terms of Service Privacy Policy…
Question Number 105839 by aurpeyz last updated on 01/Aug/20 $${A}\:{wheel}\:\mathrm{25}{cm}\:{in}\:{a}\:{radius}\:{turning}\:{at}\:\mathrm{120}{rpm} \\ $$$${un}\mathrm{i}{formly}\:{increases}\:{its}\:{frequency}\:{to}\:\mathrm{660}{rpm} \\ $$$${in}\:\mathrm{9}{s}.\:{Find}\:\left({a}\right)\:{the}\:{constant}\:{angular}\: \\ $$$${acceleration}\:{in}\:{rad}/{s}^{\mathrm{2}} \:{and}\:\left({b}\right)\:{tangential} \\ $$$${acceleration}\:{of}\:{a}\:{point}\:{on}\:{its}\:{rim}. \\ $$ Terms of Service Privacy…