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Author: Tinku Tara

A-uniform-flexible-chain-of-length-3-2-m-rests-on-a-fixed-smooth-sphere-of-radius-R-2-pi-m-such-that-one-end-A-of-chain-is-on-the-top-of-the-sphere-while-the-other-end-B-is-hanging-freely-Chain-

Question Number 22370 by Tinkutara last updated on 16/Oct/17 $$\mathrm{A}\:\mathrm{uniform}\:\mathrm{flexible}\:\mathrm{chain}\:\mathrm{of}\:\mathrm{length}\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{m} \\ $$$$\mathrm{rests}\:\mathrm{on}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{smooth}\:\mathrm{sphere}\:\mathrm{of} \\ $$$$\mathrm{radius}\:{R}\:=\:\frac{\mathrm{2}}{\pi}\:\mathrm{m}\:\mathrm{such}\:\mathrm{that}\:\mathrm{one}\:\mathrm{end}\:{A}\:\mathrm{of} \\ $$$$\mathrm{chain}\:\mathrm{is}\:\mathrm{on}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sphere}\:\mathrm{while} \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{end}\:{B}\:\mathrm{is}\:\mathrm{hanging}\:\mathrm{freely}.\:\mathrm{Chain} \\ $$$$\mathrm{is}\:\mathrm{held}\:\mathrm{stationary}\:\mathrm{by}\:\mathrm{a}\:\mathrm{horizontal} \\ $$$$\mathrm{thread}\:{PA}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{chain}\:\mathrm{when}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{string}\:{PA} \\…

y-3y-2y-10sin-x-2cos-2x-

Question Number 87904 by john santu last updated on 07/Apr/20 $$\mathrm{y}\:''\:−\mathrm{3y}'\:+\mathrm{2y}\:=\:\mathrm{10sin}\:\mathrm{x}\:+\:\mathrm{2cos}\:\mathrm{2x} \\ $$ Commented by niroj last updated on 07/Apr/20 $$\:\:\mathrm{y}^{''} −\mathrm{3y}^{'} +\mathrm{2y}=\mathrm{10sin}\:\mathrm{x}+\mathrm{2cos}\:\mathrm{2x} \\ $$$$\:\:\:\left(\mathrm{D}^{\mathrm{2}}…

The-first-and-second-ionization-potentials-of-helium-atoms-are-24-58-eV-and-54-4-eV-per-mole-respectively-Calculate-the-energy-in-kJ-required-to-produce-1-mole-of-He-2-ions-

Question Number 22368 by Tinkutara last updated on 16/Oct/17 $$\mathrm{The}\:\mathrm{first}\:\mathrm{and}\:\mathrm{second}\:\mathrm{ionization} \\ $$$$\mathrm{potentials}\:\mathrm{of}\:\mathrm{helium}\:\mathrm{atoms}\:\mathrm{are}\:\mathrm{24}.\mathrm{58}\:\mathrm{eV} \\ $$$$\mathrm{and}\:\mathrm{54}.\mathrm{4}\:\mathrm{eV}\:\mathrm{per}\:\mathrm{mole}\:\mathrm{respectively}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{energy}\:\mathrm{in}\:\mathrm{kJ}\:\mathrm{required}\:\mathrm{to} \\ $$$$\mathrm{produce}\:\mathrm{1}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{He}^{\mathrm{2}+} \:\mathrm{ions}. \\ $$ Terms of Service Privacy…

Question-22365

Question Number 22365 by A1B1C1D1 last updated on 16/Oct/17 Answered by sma3l2996 last updated on 16/Oct/17 $$=\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }}{\mathrm{1}+\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} {x}}}=\frac{\mathrm{1}−\mathrm{0}+\mathrm{0}}{\mathrm{1}+\mathrm{0}+\mathrm{0}}=\mathrm{1}\:\: \\ $$ Commented…

Question-153438

Question Number 153438 by BHOOPENDRA last updated on 07/Sep/21 Commented by MJS_new last updated on 07/Sep/21 $$\mathrm{to}\:\mathrm{reach}\:\mathrm{maximum}\:\mathrm{height}\:\mathrm{putting}\:\mathrm{a}\:\mathrm{hemisphere} \\ $$$$\mathrm{on}\:\mathrm{one}\:\mathrm{with}\:{r}_{\mathrm{1}} =\mathrm{1}\:\mathrm{you}\:\mathrm{need}\:{r}_{\mathrm{2}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\:\mathrm{this}\:\mathrm{should} \\ $$$$\mathrm{help}… \\ $$…