Question Number 24434 by Tinkutara last updated on 17/Nov/17 $$\mathrm{Two}\:\mathrm{cylindrical}\:\mathrm{hollow}\:\mathrm{drums}\:\mathrm{of}\:\mathrm{radii} \\ $$$${R}\:\mathrm{and}\:\mathrm{2}{R},\:\mathrm{and}\:\mathrm{of}\:\mathrm{a}\:\mathrm{common}\:\mathrm{height}\:{h}, \\ $$$$\mathrm{are}\:\mathrm{rotating}\:\mathrm{with}\:\mathrm{angular}\:\mathrm{velocities}\:\omega \\ $$$$\left(\mathrm{anti}-\mathrm{clockwise}\right)\:\mathrm{and}\:\omega\:\left(\mathrm{clockwise}\right), \\ $$$$\mathrm{respectively}.\:\mathrm{Their}\:\mathrm{axes},\:\mathrm{fixed}\:\mathrm{are} \\ $$$$\mathrm{parallel}\:\mathrm{and}\:\mathrm{in}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{plane} \\ $$$$\mathrm{separated}\:\mathrm{by}\:\left(\mathrm{3}{R}\:+\:\delta\right).\:\mathrm{They}\:\mathrm{are}\:\mathrm{now} \\ $$$$\mathrm{brought}\:\mathrm{in}\:\mathrm{contact}\:\left(\delta\:\rightarrow\:\mathrm{0}\right). \\…
Question Number 24436 by Tinkutara last updated on 17/Nov/17 $$\mathrm{For}\:\mathrm{a}\:\mathrm{reversible}\:\mathrm{reaction}\:\mathrm{A}\:\rightleftharpoons\:\mathrm{B}.\:\mathrm{Find} \\ $$$$\mathrm{K}_{\mathrm{eq}} \:\mathrm{at}\:\mathrm{2727}°\mathrm{C}\:\mathrm{temperature}. \\ $$$$\mathrm{Given}\::\:\Delta_{\mathrm{r}} \mathrm{H}°\:=\:−\mathrm{30}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} \:\left(\mathrm{at}\:\mathrm{2727}°\mathrm{C}\right) \\ $$$$\Delta_{\mathrm{r}} \mathrm{S}°\:=\:\mathrm{10}\:\mathrm{JK}^{−\mathrm{1}} \:\left(\mathrm{at}\:\mathrm{2727}°\mathrm{C}\right) \\ $$$$\mathrm{R}\:=\:\mathrm{8}.\mathrm{314}\:\mathrm{JK}^{−\mathrm{1}} \:\mathrm{mol}^{−\mathrm{1}} \\…
Question Number 89956 by swizanjere@gmail.com last updated on 20/Apr/20 $$\mathrm{simplify}\kappa\mathrm{giving}\kappa\mathrm{your}\kappa\mathrm{answer}\kappa\mathrm{in}\kappa\mathrm{index}\kappa\mathrm{form} \\ $$$$\sqrt{\frac{\mathrm{ac}^{\mathrm{2}} }{\mathrm{9a}^{\mathrm{2}} \mathrm{c}^{\mathrm{4}} }} \\ $$ Commented by john santu last updated on 20/Apr/20…
Question Number 89953 by swizanjere@gmail.com last updated on 20/Apr/20 $$\mathrm{solvethefollowingequation} \\ $$$$\mathrm{5}^{\mathrm{2x}+\mathrm{y}} =\mathrm{625and2}^{\mathrm{4x}\nmid\mathrm{2y}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$ Commented by jagoll last updated on 20/Apr/20 $$\mathrm{2x}+\mathrm{y}\:=\:\mathrm{4} \\…
Question Number 89936 by ~blr237~ last updated on 20/Apr/20 $${Prove}\:{that}\:\underset{{p}\geqslant\mathrm{1},{q}\geqslant\mathrm{1}} {\sum}\:\:\frac{\mathrm{1}}{{pq}\left({p}+{q}−\mathrm{1}\right)}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\: \\ $$ Answered by maths mind last updated on 20/Apr/20 $$\underset{{p},{q}\geqslant\mathrm{2}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 89934 by ~blr237~ last updated on 20/Apr/20 $$\left.{Let}\:{x}\in\right]\mathrm{0};\mathrm{1}\left[\:\:{Prove}\:{that}\right. \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{\mathrm{1}+{x}^{{n}} }\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−{x}\right)^{{n}} }{\mathrm{1}−{x}^{{n}} }\:=\:\mathrm{0} \\ $$ Terms of Service…
Question Number 89937 by ~blr237~ last updated on 20/Apr/20 $${Prove}\:{that}\:{for}\:{all}\:{complex}\:{such}\:{as}\:\mid{z}\mid<\mathrm{1}= \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{z}^{{n}} }{\left({z}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} }\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{nz}^{{n}} }{{z}^{{n}} −\mathrm{1}}\:=\:\mathrm{0}\: \\ $$ Commented by…
Question Number 24387 by chernoaguero@gmail.com last updated on 17/Nov/17 $$\boldsymbol{\mathrm{Two}}\:\boldsymbol{\mathrm{particle}}\:\boldsymbol{\mathrm{move}}\:\boldsymbol{\mathrm{along}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{axis}}. \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{position}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{particle}}\:\mathrm{1}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}}; \\ $$$$\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{x}}=\mathrm{6}.\mathrm{00}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{3}.\mathrm{00}\boldsymbol{\mathrm{t}}+\mathrm{2}.\mathrm{00}\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}}\right) \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{acceleration}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{particle}}\:\mathrm{2}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}} \\ $$$$\boldsymbol{\mathrm{by}}\:\:\boldsymbol{\mathrm{a}}=−\mathrm{8}.\mathrm{00}\boldsymbol{\mathrm{t}}\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}^{\mathrm{2}} }\right)\:\boldsymbol{\mathrm{and}},\mathrm{at}\:\mathrm{t}=\mathrm{0},\boldsymbol{\mathrm{its}} \\ $$$$\boldsymbol{\mathrm{velocity}}\:\boldsymbol{\mathrm{is}}\:\mathrm{20}\:\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}}\right).\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{velocities}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{particles}}\:\boldsymbol{\mathrm{match}}, \\…
Question Number 24371 by Tinkutara last updated on 16/Nov/17 $$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{are}\:\mathrm{moving}\:\mathrm{together}\:\mathrm{under} \\ $$$$\mathrm{the}\:\mathrm{action}\:\mathrm{of}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{horizontal} \\ $$$$\mathrm{external}\:\mathrm{force}\:{F}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{block}\:\mathrm{is} \\ $$$$\mathrm{at}\:\mathrm{rest}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bigger}\:\mathrm{block} \\ $$$$\mathrm{due}\:\mathrm{to}\:\mathrm{the}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{them},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{normal}\:\mathrm{reaction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{bigger} \\ $$$$\mathrm{block}\:\mathrm{and}\:\mathrm{floor}\:\mathrm{is} \\ $$ Commented…
Question Number 24369 by ajfour last updated on 16/Nov/17 Commented by ajfour last updated on 16/Nov/17 $${If}\:{a}\:{uniform}\:{rod}\:{of}\:{mass}\:\boldsymbol{{m}}, \\ $$$${length}\:\boldsymbol{{L}},\:{and}\:{hooked}\:{at}\:{end}\:{A} \\ $$$${to}\:{a}\:{string}\:{of}\:{length}\:\boldsymbol{{l}}\:{be} \\ $$$${released}\:{in}\:{horizontal}\:{position}, \\ $$$${then}\:{a}\:{little}\:{time}\:{later}\:{find}\:\boldsymbol{\phi}…