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Author: Tinku Tara

A-spring-of-force-constant-k-is-cut-into-two-pieces-such-that-one-piece-is-twice-as-long-as-the-other-Then-the-longer-piece-will-have-a-force-constant-of-

Question Number 18924 by Tinkutara last updated on 01/Aug/17 $$\mathrm{A}\:\mathrm{spring}\:\mathrm{of}\:\mathrm{force}\:\mathrm{constant}\:{k}\:\mathrm{is}\:\mathrm{cut}\:\mathrm{into} \\ $$$$\mathrm{two}\:\mathrm{pieces}\:\mathrm{such}\:\mathrm{that}\:\mathrm{one}\:\mathrm{piece}\:\mathrm{is}\:\mathrm{twice} \\ $$$$\mathrm{as}\:\mathrm{long}\:\mathrm{as}\:\mathrm{the}\:\mathrm{other}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{longer} \\ $$$$\mathrm{piece}\:\mathrm{will}\:\mathrm{have}\:\mathrm{a}\:\mathrm{force}\:\mathrm{constant}\:\mathrm{of} \\ $$ Answered by ajfour last updated on 01/Aug/17…

log-10-x-log-10-x-8log-10-y-log-10-2-x-log-10-2-y-3-log-10-y-8log-10-x-log-10-y-log-10-2-x-log-10-2-y-0-find-x-amp-y-

Question Number 84459 by jagoll last updated on 13/Mar/20 $$\begin{cases}{\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}\right)+\frac{\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}\right)+\mathrm{8log}_{\mathrm{10}} \left(\mathrm{y}\right)}{\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{y}\right)}=\mathrm{3}}\\{\mathrm{log}_{\mathrm{10}} \left(\mathrm{y}\right)+\frac{\mathrm{8log}_{\mathrm{10}} \left(\mathrm{x}\right)−\mathrm{log}_{\mathrm{10}} \left(\mathrm{y}\right)}{\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{y}\right)}=\mathrm{0}}\end{cases} \\…

Consider-a-block-sliding-over-a-smooth-inclined-surface-of-inclination-Relating-to-Newton-s-second-law-applied-on-the-block-select-the-incorrect-alternative-1-F-y-0-2-F-y-0-3-F-

Question Number 18922 by Tinkutara last updated on 01/Aug/17 $$\mathrm{Consider}\:\mathrm{a}\:\mathrm{block}\:\mathrm{sliding}\:\mathrm{over}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{inclined}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{inclination}\:\theta.\:\mathrm{Relating} \\ $$$$\mathrm{to}\:\mathrm{Newton}'\mathrm{s}\:\mathrm{second}\:\mathrm{law}\:\mathrm{applied}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{block},\:\mathrm{select}\:\mathrm{the}\:\mathrm{incorrect}\:\mathrm{alternative}. \\ $$$$\left(\mathrm{1}\right)\:\Sigma{F}_{{y}'} \:\neq\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\Sigma{F}_{{y}} \:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\Sigma{F}_{{x}} \:=\:−{mg}\:\mathrm{sin}\:\theta…

By-subs-u-2-4-x-evaluate-4-x-x-dx-

Question Number 149989 by ZiYangLee last updated on 08/Aug/21 $$\mathrm{By}\:\mathrm{subs}\:{u}^{\mathrm{2}} =\mathrm{4}+{x},\:\mathrm{evaluate}\:\int\:\frac{\sqrt{\mathrm{4}+{x}}}{{x}}\:{dx} \\ $$ Answered by puissant last updated on 08/Aug/21 $$\int\frac{\sqrt{\mathrm{4}+{x}}}{{x}}{dx}={Q} \\ $$$${u}=\sqrt{\mathrm{4}+{x}}\:\rightarrow\:{u}^{\mathrm{2}} =\mathrm{4}+{x}\:\rightarrow\:{x}={u}^{\mathrm{2}} −\mathrm{4}…

A-bird-sits-on-a-stretched-telegraph-wire-The-additional-tension-produced-in-the-wire-is-1-Zero-2-Greater-than-weight-of-the-bird-3-Less-than-the-weight-of-the-bird-4-Equal-to-the-weight-of-t

Question Number 18918 by Tinkutara last updated on 01/Aug/17 $$\mathrm{A}\:\mathrm{bird}\:\mathrm{sits}\:\mathrm{on}\:\mathrm{a}\:\mathrm{stretched}\:\mathrm{telegraph} \\ $$$$\mathrm{wire}.\:\mathrm{The}\:\mathrm{additional}\:\mathrm{tension}\:\mathrm{produced} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Zero} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Greater}\:\mathrm{than}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bird} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Less}\:\mathrm{than}\:\mathrm{the}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bird} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bird} \\ $$ Commented…

Question-149986

Question Number 149986 by mnjuly1970 last updated on 08/Aug/21 Answered by mindispower last updated on 08/Aug/21 $${ln}\left(\mathrm{1}+{e}^{{x}} \right)={x}+{ln}\left({e}^{−{x}} +\mathrm{1}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}}{\mathrm{1}+{e}^{\mathrm{2}{x}} }{dx}+\int_{\mathrm{0}} ^{\infty}…