Question Number 16504 by Tinkutara last updated on 23/Jun/17 $$\mathrm{An}\:\mathrm{object}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{xy}\:\mathrm{plane}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{at}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{When}\:\mathrm{the} \\ $$$$\mathrm{object}\:\mathrm{is}\:\mathrm{at}\:{x}\:=\:−\mathrm{2}\:\mathrm{m},\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{is} \\ $$$$−\left(\mathrm{4}\:\mathrm{m}/\mathrm{s}\right)\overset{\wedge} {{j}}.\:\mathrm{Then}\:\mathrm{objects}\:\mathrm{velocity}\:\mathrm{at} \\ $$$${y}\:=\:\mathrm{2}\:\mathrm{m}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{4}\:\mathrm{m}/\mathrm{s}\:\overset{\wedge} {{i}} \\…
Question Number 147573 by phally last updated on 22/Jul/21 Commented by phally last updated on 22/Jul/21 $$\:\mathrm{help}\:\mathrm{me}? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 147572 by liberty last updated on 22/Jul/21 $$\:\:\:\underset{{n}\geqslant\mathrm{1}} {\sum}\:\frac{\mathrm{4}{n}−\mathrm{3}}{\left({n}^{\mathrm{2}} +\mathrm{2}{n}\right)\left({n}+\mathrm{3}\right)}\:=? \\ $$ Answered by mathmax by abdo last updated on 22/Jul/21 $$\mathrm{let}\:\mathrm{S}=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty}…
Question Number 16502 by Tinkutara last updated on 23/Jun/17 $$\mathrm{The}\:\mathrm{acceleration}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{in} \\ $$$$\mathrm{uniform}\:\mathrm{circular}\:\mathrm{motion}\:\mathrm{averaged}\:\mathrm{over} \\ $$$$\mathrm{one}\:\mathrm{cycle}\:\mathrm{is}\:\mathrm{a}\:\mathrm{null}\:\mathrm{vector}.\:\left(\mathrm{True}/\mathrm{False}\right) \\ $$ Answered by ajfour last updated on 23/Jun/17 $$\bar {{a}}=\:−\frac{{v}_{\mathrm{0}}…
Question Number 16501 by gokux2123 last updated on 23/Jun/17 $$\mathrm{3}+\mathrm{3} \\ $$ Answered by Tinkutara last updated on 23/Jun/17 $$\mathrm{3}\:+\:\mathrm{3}\:=\:\mathrm{6} \\ $$ Terms of Service…
Question Number 147569 by Sozan last updated on 21/Jul/21 $${find}\:{the}\:{taylor}\:{series}\:{of}\:{f}\left({z}\right)={sinz}\:,{z}=\frac{\pi}{\mathrm{4}}\:{in}\:{complex}\:{number} \\ $$ Answered by mathmax by abdo last updated on 22/Jul/21 $$\mathrm{f}\left(\mathrm{z}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{f}^{\left(\mathrm{n}\right)} \left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{n}!}\left(\mathrm{z}−\frac{\pi}{\mathrm{4}}\right)^{\mathrm{n}}…
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 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 16497 by Tinkutara last updated on 23/Jun/17 $$\mathrm{Two}\:\mathrm{particles}\:\mathrm{are}\:\mathrm{revolving}\:\mathrm{on}\:\mathrm{two} \\ $$$$\mathrm{coplanar}\:\mathrm{circles}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{angular} \\ $$$$\mathrm{velocities}\:\omega_{\mathrm{1}} \:\mathrm{and}\:\omega_{\mathrm{2}} \:\mathrm{respectively}.\:\mathrm{Their} \\ $$$$\mathrm{time}\:\mathrm{periods}\:\mathrm{are}\:{T}_{\mathrm{1}} \:\mathrm{and}\:{T}_{\mathrm{2}} \:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{second}\:\mathrm{particle} \\ $$$$\mathrm{to}\:\mathrm{complete}\:\mathrm{one}\:\mathrm{revolution}\:\mathrm{more}\:\mathrm{than} \\…
Question Number 82030 by naka3546 last updated on 17/Feb/20 $${a}\:−\:{b}\:+\:{c}\:−\:{d}\:\:=\:\:\mathrm{2} \\ $$$${a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:−\:{d}^{\mathrm{2}} \:\:=\:\:\mathrm{6} \\ $$$${a}^{\mathrm{3}} \:−\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} \:−\:{d}^{\mathrm{3}} \:\:=\:\:\mathrm{20} \\ $$$${a}^{\mathrm{4}} \:−\:{b}^{\mathrm{4}}…