Question Number 16634 by Tinkutara last updated on 24/Jun/17 $$\mathrm{Consider}\:\mathrm{a}\:\mathrm{rubber}\:\mathrm{ball}\:\mathrm{freely}\:\mathrm{falling} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{height}\:{h}\:=\:\mathrm{4}.\mathrm{9}\:\mathrm{m}\:\mathrm{onto}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{elastic}\:\mathrm{plate}.\:\mathrm{Assume}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{duration}\:\mathrm{of}\:\mathrm{collision}\:\mathrm{is}\:\mathrm{negligible} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{collision}\:\mathrm{with}\:\mathrm{the}\:\mathrm{plate}\:\mathrm{is} \\ $$$$\mathrm{totally}\:\mathrm{elastic}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:\mathrm{and}\:\mathrm{the}\:\mathrm{height}\:\mathrm{as} \\ $$$$\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:\mathrm{will}\:\mathrm{be} \\…
Question Number 16633 by tawa tawa last updated on 24/Jun/17 $$\sqrt{\mathrm{1}}\:.\:\sqrt{\mathrm{2}}\:.\:\sqrt{\mathrm{3}}.\:…\:\sqrt{\mathrm{n}}\:\:=\:\mathrm{S}_{\mathrm{n}} \\ $$$$\mathrm{what}\:\mathrm{is}\:\:\mathrm{S}_{\mathrm{n}} \:? \\ $$ Commented by Arnab Maiti last updated on 02/Jul/17 $$\mathrm{S}_{\mathrm{n}}…
Question Number 147706 by aliibrahim1 last updated on 22/Jul/21 Answered by mr W last updated on 22/Jul/21 Commented by aliibrahim1 last updated on 22/Jul/21 $${sorry}\:{sir}\:{i}\:{tried}\:{to}\:{build}\:{on}\:{that}\:{before}\:{sending}\:{it}\:{didnt}\:{work}\:{with}\:{me}…
Question Number 16632 by tawa tawa last updated on 24/Jun/17 $$\mathrm{e}^{\mathrm{xy}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{sin}\left(\mathrm{x}\:+\:\mathrm{y}\right) \\ $$$$\mathrm{Express}\:\mathrm{the}\:\mathrm{variable}\:\mathrm{y}\:\mathrm{interms}\:\mathrm{of}\:\mathrm{x}\:\mathrm{only}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147697 by mathdanisur last updated on 22/Jul/21 $${f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+…+{f}\left({n}\right)={n}!−{n}\centerdot{a} \\ $$$${f}\left(\mathrm{16}\right)=\mathrm{15}\centerdot\left(\mathrm{15}!−\mathrm{1}\right) \\ $$$${find}\:\:\boldsymbol{{a}}=? \\ $$ Answered by Olaf_Thorendsen last updated on 22/Jul/21 $$\underset{{k}=\mathrm{0}} {\overset{{n}}…
Question Number 147699 by mathdanisur last updated on 22/Jul/21 $$\boldsymbol{{a}}^{\mathrm{2}} \:=\:\mathrm{36}^{\mathrm{7}} \:+\:\mathrm{9}^{\boldsymbol{{b}}} \:+\:\mathrm{6}^{\mathrm{8}} \:\:;\:\:\boldsymbol{{a}}\in\mathbb{Z} \\ $$$${find}\:\:\boldsymbol{{b}}=? \\ $$ Answered by Olaf_Thorendsen last updated on 22/Jul/21…
Question Number 82160 by M±th+et£s last updated on 18/Feb/20 $${prove}\:{that} \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\:{n}^{\mathrm{2}} \:\sqrt{\left(\mathrm{1}−{cos}\left(\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)…}}\right.}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by MJS last updated on 18/Feb/20 $${w}=\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{1}}{{n}}\right)\sqrt{…}}} \\…
Question Number 16623 by Tinkutara last updated on 24/Jun/17 $$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{up}\:\mathrm{in}\:\mathrm{a}\:\mathrm{lift}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:{u}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lift}.\:\mathrm{If}\:\mathrm{it} \\ $$$$\mathrm{returns}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{in}\:\mathrm{time}\:{t},\:\mathrm{then} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{is}\:\left[\boldsymbol{\mathrm{Answer}}:\right. \\ $$$$\left.\frac{\mathrm{2}{u}\:−\:{gt}}{{t}}\:\mathrm{upwards}\right] \\ $$ Answered by ajfour last updated…
Question Number 16621 by Tinkutara last updated on 24/Jun/17 $$\mathrm{A}\:\mathrm{body}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:\mathrm{whose}\:\mathrm{square}\:\mathrm{decreases} \\ $$$$\mathrm{linearly}\:\mathrm{with}\:\mathrm{displacement}\:\mathrm{between} \\ $$$$\mathrm{two}\:\mathrm{points}\:{A}\:\mathrm{and}\:{B}\:\mathrm{as}\:\mathrm{shown}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{particle}.\:\left[\boldsymbol{\mathrm{Answer}}:\:\frac{\mathrm{8}}{\mathrm{3}}\:\mathrm{ms}^{−\mathrm{2}} \right] \\ $$ Commented by…
Question Number 147689 by mathdanisur last updated on 22/Jul/21 $$\left(\mathrm{234}\right)_{\mathrm{5}} \:\centerdot\:\left(\mathrm{23}\right)_{\mathrm{5}} \:=\:\left({x}\right)_{\mathrm{5}} \:\:\Rightarrow\:\:\boldsymbol{{x}}=? \\ $$ Answered by Olaf_Thorendsen last updated on 22/Jul/21 $$\left(\mathrm{234}\right)_{\mathrm{5}} .\left(\mathrm{23}\right)_{\mathrm{5}} \\…