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Question Number 82084 by oyemi kemewari last updated on 18/Feb/20 Commented by john santu last updated on 18/Feb/20 $$\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{4}^{\mathrm{2}} −\mathrm{1}\right)}+… \\ $$$$\underset{{k}=\mathrm{2}} {\overset{\infty}…
Question Number 147623 by mari last updated on 22/Jul/21 Commented by tabata last updated on 22/Jul/21 $$\left(\mathrm{2}\right)\:{it}\:{is}\:{clearley}\:\:{that}\:\:{f}\left({z}\right)=\frac{\mathrm{1}}{{z}−\mathrm{1}}−\frac{\mathrm{1}}{{z}−\mathrm{2}} \\ $$$$ \\ $$$$\left({a}\right)\:\mid{z}\mid<\mathrm{1} \\ $$$$ \\ $$$${f}\left({z}\right)=−\frac{\mathrm{1}}{\mathrm{1}−{z}}+\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{{z}}{\mathrm{2}}\right)}=−\underset{{n}=\mathrm{0}}…
Question Number 147622 by mari last updated on 22/Jul/21 Commented by Sozan last updated on 22/Jul/21 $$??????? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 16505 by Tinkutara last updated on 23/Jun/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{magnitude}\:\mathrm{of} \\ $$$$\mathrm{change}\:\mathrm{in}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{motorcycle} \\ $$$$\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{speed}\:{v}_{\mathrm{0}} \:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{circular}\:\mathrm{path}\:\mathrm{of}\:\mathrm{length}\:{l}\:=\:\frac{\pi}{\mathrm{3}}{R}\:\mathrm{and} \\ $$$$\mathrm{radius}\:\mathrm{R}?\:\mathrm{Treat}\:\mathrm{motorcycle}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{particle} \\ $$$$\left(\mathrm{1}\right)\:\mid\Delta\overset{\rightarrow} {{v}}\mid\:=\:\frac{{v}_{\mathrm{0}} }{\mathrm{2}}…
Question Number 16506 by Tinkutara last updated on 23/Jun/17 $$\mathrm{A}\:\mathrm{train}\:\mathrm{1}\:\mathrm{moves}\:\mathrm{from}\:\mathrm{east}\:\mathrm{to}\:\mathrm{west} \\ $$$$\left(\mathrm{clockwise}\right)\:\mathrm{and}\:\mathrm{another}\:\mathrm{train}\:\mathrm{2}\:\mathrm{moves} \\ $$$$\mathrm{from}\:\mathrm{west}\:\mathrm{to}\:\mathrm{east}\:\left(\mathrm{anticlockwise}\right)\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{equator}\:\mathrm{with}\:\mathrm{equal}\:\mathrm{speeds}\:\mathrm{relative} \\ $$$$\mathrm{to}\:\mathrm{ground}.\:\mathrm{The}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{their} \\ $$$$\mathrm{centripetal}\:\mathrm{acceleration}\:\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{2}} }\:\mathrm{relative}\:\mathrm{to} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{earth}\:\mathrm{is} \\…
Question Number 16504 by Tinkutara last updated on 23/Jun/17 $$\mathrm{An}\:\mathrm{object}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{xy}\:\mathrm{plane}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{at}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{When}\:\mathrm{the} \\ $$$$\mathrm{object}\:\mathrm{is}\:\mathrm{at}\:{x}\:=\:−\mathrm{2}\:\mathrm{m},\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{is} \\ $$$$−\left(\mathrm{4}\:\mathrm{m}/\mathrm{s}\right)\overset{\wedge} {{j}}.\:\mathrm{Then}\:\mathrm{objects}\:\mathrm{velocity}\:\mathrm{at} \\ $$$${y}\:=\:\mathrm{2}\:\mathrm{m}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{4}\:\mathrm{m}/\mathrm{s}\:\overset{\wedge} {{i}} \\…
Question Number 16499 by Tinkutara last updated on 23/Jun/17 $$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{parabolic}\:\mathrm{path} \\ $$$${x}^{\mathrm{2}} \:=\:{y},\:\mathrm{with}\:\mathrm{constant}\:\mathrm{speed}\:{u}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{when}\:\mathrm{it} \\ $$$$\mathrm{crossess}\:\mathrm{origin}.\:\mathrm{Also}\:\mathrm{find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of} \\ $$$$\mathrm{curvature}\:\mathrm{at}\:\mathrm{origin}. \\ $$ Answered by ajfour last…
Question Number 82034 by Dah Solu Tion last updated on 17/Feb/20 $$\boldsymbol{{P}}{rove}\:\:{by}\:\:{maths}\:\:{induction}\:\:{tbat} \\ $$$$\boldsymbol{{n}}^{\mathrm{5}} \:−\:\boldsymbol{{n}}^{\mathrm{3}} \:\:\boldsymbol{{is}}\:\boldsymbol{{divisible}}\:\boldsymbol{{by}}\:\mathrm{24}. \\ $$ Commented by MJS last updated on 17/Feb/20…
Question Number 16497 by Tinkutara last updated on 23/Jun/17 $$\mathrm{Two}\:\mathrm{particles}\:\mathrm{are}\:\mathrm{revolving}\:\mathrm{on}\:\mathrm{two} \\ $$$$\mathrm{coplanar}\:\mathrm{circles}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{angular} \\ $$$$\mathrm{velocities}\:\omega_{\mathrm{1}} \:\mathrm{and}\:\omega_{\mathrm{2}} \:\mathrm{respectively}.\:\mathrm{Their} \\ $$$$\mathrm{time}\:\mathrm{periods}\:\mathrm{are}\:{T}_{\mathrm{1}} \:\mathrm{and}\:{T}_{\mathrm{2}} \:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{second}\:\mathrm{particle} \\ $$$$\mathrm{to}\:\mathrm{complete}\:\mathrm{one}\:\mathrm{revolution}\:\mathrm{more}\:\mathrm{than} \\…