Question Number 47005 by ajfour last updated on 03/Nov/18 Commented by ajfour last updated on 03/Nov/18 $${A}\:{stick}\:{of}\:{length}\:{L}\:{is}\:{held}\:{balanced} \\ $$$${against}\:{two}\:{mutually}\:{perpendicular} \\ $$$${frictionless}\:{walls},\:{with}\:{the}\:{help} \\ $$$${of}\:{two}\:{strings}\:{AC}\left({length}\:{b}\right)\:{and} \\ $$$${BC}\left({length}\:{a}\right).\:{Find}\:{Tensions}…
Question Number 178037 by mr W last updated on 12/Oct/22 Commented by mr W last updated on 12/Oct/22 $${a}\:{rope}\:{with}\:{mass}\:{m}\:{and}\:{length}\:{b}\:{is} \\ $$$${connected}\:{on}\:{one}\:{end}\:{with}\:{a}\:{rod}\:{and} \\ $$$${its}\:{other}\:{end}\:{is}\:{fixed}\:{on}\:{the}\:{ceiling} \\ $$$${in}\:{a}\:{height}\:{h}\:{above}\:{the}\:{ground}.\:{the}\:…
Question Number 112403 by ajfour last updated on 07/Sep/20 Commented by ajfour last updated on 07/Sep/20 $${The}\:{elliptical}\:{frictionless}\:{track}\:{has} \\ $$$$\:{axes}\:{lengths}\:\boldsymbol{{a}},\:\:\boldsymbol{{b}}. \\ $$$${Find}\:{initial}\:{speed}\:\boldsymbol{{u}},\:{that}\:{must}\:{be} \\ $$$${given}\:{to}\:{a}\:{block}\:{such}\:{that}\:{its} \\ $$$${trajectory}\:{is}\:{as}\:{depicted}.…
Question Number 46629 by Necxx last updated on 29/Oct/18 $${An}\:{object}\:{is}\:{projected}\:{from}\:{a} \\ $$$${height}\:{of}\:\mathrm{80}{m}\:{above}\:{the}\:{ground} \\ $$$${with}\:{a}\:{velocity}\:{of}\:\mathrm{40}{m}/{s}\:{at}\:{an} \\ $$$${angle}\:{of}\:\mathrm{30}\:{degree}\:{to}\:{the} \\ $$$${horizontal}.{What}\:{is}\:{the}\:{tume}\:{of} \\ $$$${flight}? \\ $$ Answered by tanmay.chaudhury50@gmail.com…
Question Number 46503 by malwaan last updated on 27/Oct/18 $$\mathrm{true}\:\mathrm{or}\:\mathrm{false}\:? \\ $$$$\theta_{\mathrm{1}} +\theta_{\mathrm{2}} =\mathrm{90}^{°} \:\mathrm{and}\:\mathrm{m}_{\mathrm{1}} \neq\mathrm{m}_{\mathrm{2}} \\ $$$$\mathrm{in}\:\mathrm{elastic}\:\mathrm{collision} \\ $$ Terms of Service Privacy Policy…
Question Number 177548 by peter frank last updated on 07/Oct/22 $$\mathrm{A}\:\mathrm{projectile}\:\mathrm{is}\:\mathrm{fired}\:\mathrm{with}\:\mathrm{velocity}\left(\mathrm{v}_{\mathrm{o}} \right) \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{two} \\ $$$$\mathrm{points}\:\mathrm{both}\:\mathrm{a}\:\mathrm{distance}\left(\mathrm{h}\right)\:\mathrm{above}\:\mathrm{the} \\ $$$$\mathrm{horizontal}.\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{the}\:\mathrm{gun}\:\mathrm{is}\:\mathrm{adjusted} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{separation}\:\mathrm{of}\:\mathrm{two}\:\mathrm{position}\:\mathrm{is} \\ $$$$\boldsymbol{\mathrm{d}}=\frac{\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{o}}} \sqrt{\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{o}}}…
Question Number 177546 by peter frank last updated on 07/Oct/22 $$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{vertically} \\ $$$$\mathrm{upward}\:\mathrm{in}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{gravitation} \\ $$$$\mathrm{field}\:\mathrm{with}\:\mathrm{initial}\:\mathrm{speed}\:\left(\mathrm{v}_{\mathrm{0}} \right).\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{there}\:\mathrm{is}\:\mathrm{retarding}\:\mathrm{proportional} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{the}\:\mathrm{instantaneous} \\ $$$$\mathrm{speed},\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{partical} \\ $$$$\mathrm{when}\:\mathrm{it}\:\mathrm{returns}\:\mathrm{on}\:\mathrm{the}\:\mathrm{initial} \\…
Question Number 177446 by mr W last updated on 05/Oct/22 Commented by mr W last updated on 05/Oct/22 $${a}\:{rope}\:{with}\:{length}\:{L}\:{has}\:{uniform} \\ $$$${mass}.\:{one}\:{end}\:{of}\:{the}\:{rope}\:{is}\:{fixed}\:{on} \\ $$$${the}\:{top}\:{of}\:{a}\:{pole}\:{with}\:{heigth}\:{h}\:{above}\:{a} \\ $$$${inclined}\:{plane}\:\left({h}<{L}\right).\:{the}\:{other}\:{end}…
Question Number 46326 by ajfour last updated on 24/Oct/18 Commented by ajfour last updated on 25/Oct/18 $${small}\:{black}\:{circles}\:{are}\:{rollers}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 46299 by rahul 19 last updated on 23/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18 $${area}\:{remaining}\:{portion}=\pi{R}^{\mathrm{2}} −\pi\left(\frac{{R}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$=\pi{R}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\right)=\frac{\mathrm{8}\pi{R}^{\mathrm{2}} }{\mathrm{9}} \\ $$$${mass}\:{per}\:{unit}\:{area}\left(\rho\right)=\frac{{M}}{\frac{\mathrm{8}\pi{R}^{\mathrm{2}}…