Question Number 13478 by Tinkutara last updated on 20/May/17 $$\mathrm{The}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{is}\:{u} \\ $$$$\left(\mathrm{at}\:{t}\:=\:\mathrm{0}\right)\:\mathrm{and}\:\mathrm{acceleration}\:{f}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${f}\:=\:{at}\:\mathrm{where}\:{t}\:\mathrm{is}\:\mathrm{time}\:\mathrm{and}\:'{a}'\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}. \\ $$$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{relations}\:\mathrm{is} \\ $$$$\mathrm{valid}? \\ $$$$\left(\mathrm{1}\right)\:{v}\:=\:{u}\:+\:{at}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{v}\:=\:{u}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:{at}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:{v}\:=\:{u}\:+\:{at} \\…
Question Number 79015 by ajfour last updated on 22/Jan/20 $${Rigorously}\:{over}\:{one}\:{month}'{s} \\ $$$${time},\:{I}\:{developed}\:{a}\:{formula}\:{for} \\ $$$${general}\:{cubic}. \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${let}\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}} \\ $$$$\boldsymbol{{pq}}=\boldsymbol{{m}},\:\boldsymbol{{p}}+\boldsymbol{{q}}=\boldsymbol{{s}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\…
Question Number 13475 by Tinkutara last updated on 20/May/17 $$\mathrm{An}\:\mathrm{object}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{acceleration}\:\mathrm{4}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} ,\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distance} \\ $$$$\mathrm{travelled}\:\mathrm{by}\:\mathrm{object}\:\mathrm{in}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{half}\:\mathrm{second}. \\ $$ Answered by ajfour last updated on 20/May/17…
Question Number 13449 by Tinkutara last updated on 20/May/17 $$\mathrm{Four}\:\mathrm{particles}\:{A},\:{B},\:{C}\:\mathrm{and}\:{D}\:\mathrm{are}\:\mathrm{situated} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{corners}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:{ABCD}\:\mathrm{of}\:\mathrm{side} \\ $$$${a}\:\mathrm{at}\:{t}\:=\:\mathrm{0}.\:\mathrm{Each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particles}\:\mathrm{moves} \\ $$$$\mathrm{with}\:\mathrm{constant}\:\mathrm{speed}\:{v}.\:{A}\:\mathrm{always}\:\mathrm{has}\:\mathrm{its} \\ $$$$\mathrm{velocity}\:\mathrm{along}\:{AB},\:{B}\:\mathrm{along}\:{BC},\:{C}\:\mathrm{along} \\ $$$${CD}\:\mathrm{and}\:{D}\:\mathrm{along}\:{DA}.\:\mathrm{At}\:\mathrm{what}\:\mathrm{time}\:\mathrm{will} \\ $$$$\mathrm{these}\:\mathrm{particles}\:\mathrm{meet}\:\mathrm{each}\:\mathrm{other}? \\ $$ Answered…
Question Number 13373 by Stargazer last updated on 19/May/17 Commented by Stargazer last updated on 19/May/17 $${find}\:{the}\:{area}\:{of}\:{the}\:{ring}? \\ $$ Commented by prakash jain last updated…
Question Number 78880 by naka3546 last updated on 21/Jan/20 $${x}\:+\:\frac{\mathrm{1}}{{x}}\:\:=\:\:\mathrm{3}\:\:\:,\:\:\:\:{x}\:\in\:\mathbb{R} \\ $$$$\left({x}^{\mathrm{2020}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2020}} }\right)\:\:{mod}\:\left(\mathrm{10}\right)\:\:=\:\:{y} \\ $$$${y}^{\mathrm{2}} \:−\:\mathrm{1}\:\:=\:\:? \\ $$ Answered by mind is power last…
Question Number 78851 by ketto255 last updated on 21/Jan/20 Commented by MJS last updated on 21/Jan/20 $$\mathrm{we}\:\mathrm{cannot}\:\mathrm{find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{without}\:\mathrm{any}\:\mathrm{length} \\ $$$$\mathrm{given} \\ $$ Terms of Service Privacy…
Question Number 13302 by tawa tawa last updated on 17/May/17 $$\left(\mathrm{a}\right) \\ $$$$\mathrm{A}\:\mathrm{body}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}\:\mathrm{initially}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{at}\:\mathrm{a}\:\mathrm{point}\:\mathrm{O}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{surface}. \\ $$$$\mathrm{A}\:\mathrm{horizontal}\:\mathrm{force}\:\mathrm{F}\:\mathrm{is}\:\mathrm{applied}\:\mathrm{to}\:\mathrm{the}\:\mathrm{body}\:\mathrm{and}\:\mathrm{caused}\:\mathrm{it}\:\mathrm{to}\:\mathrm{move}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{line}\:\mathrm{accross}\:\mathrm{the}\:\mathrm{surface}.\:\mathrm{The}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{F}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{F}\:=\:\frac{\mathrm{1}}{\mathrm{s}\:+\:\alpha},\:\mathrm{Where}\:\mathrm{S}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{from}\:\mathrm{O}\:\mathrm{and}\:\alpha\:\mathrm{is}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{constant}.\:\mathrm{If}\:\mathrm{v}\:\mathrm{is}\:\mathrm{the}\:\mathrm{speed} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{at}\:\mathrm{any}\:\mathrm{moment},\:\mathrm{Show}\:\mathrm{that}\:\mathrm{S}\:=\:\alpha\:\left(\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{mv}^{\mathrm{2}} } −\:\mathrm{1}\right).\: \\ $$$$\left(\mathrm{b}\right)…
Question Number 144367 by Gbenga last updated on 25/Jun/21 Commented by Gbenga last updated on 25/Jun/21 $$\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{some}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{help}}?????? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 13261 by Tinkutara last updated on 17/May/17 $$\mathrm{Balance}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{K}_{\mathrm{2}} \mathrm{CrO}_{\mathrm{4}} \:+\:\mathrm{HCl}\:\rightarrow\:\mathrm{K}_{\mathrm{2}} \mathrm{Cr}_{\mathrm{2}} \mathrm{O}_{\mathrm{7}} \:+\:\mathrm{KCl}\:+\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$ Answered by Joel577 last updated…