Question Number 23272 by ajfour last updated on 28/Oct/17 Commented by ajfour last updated on 28/Oct/17 $${If}\:{the}\:{mounted}\:{wheels}\:{are}\:{kept} \\ $$$${spinning}\:{at}\:{an}\:{angular}\:{velocity} \\ $$$$\boldsymbol{\omega}\:\left({as}\:{shown}\:{in}\:{diagram}\right),\:{find} \\ $$$${the}\:{time}\:{period}\:{of}\:{small}\:{oscillations} \\ $$$${of}\:{a}\:{rod}\:{of}\:{mass}\:\boldsymbol{{M}}\:{kept}\:{on}\:{them},\:…
Question Number 23270 by Tinkutara last updated on 28/Oct/17 $$\mathrm{The}\:\mathrm{reaction}\:\mathrm{CH}_{\mathrm{4}} \left(\mathrm{g}\right)\:+\:\mathrm{Cl}_{\mathrm{2}} \left(\mathrm{g}\right)\:\rightarrow\:\mathrm{CH}_{\mathrm{3}} \mathrm{Cl}\left(\mathrm{g}\right) \\ $$$$+\:\mathrm{HCl}\left(\mathrm{g}\right)\:\mathrm{has}\:\Delta\mathrm{H}\:=\:−\mathrm{25}\:\mathrm{kcal}\:\mathrm{and}\:\mathrm{bond} \\ $$$$\mathrm{dissociation}\:\mathrm{energy}\:\mathrm{of}\:\mathrm{C}\:−\:\mathrm{Cl},\:\mathrm{H}\:−\:\mathrm{Cl}, \\ $$$$\mathrm{C}\:−\:\mathrm{H}\:\mathrm{and}\:\mathrm{Cl}\:−\:\mathrm{Cl}\:\mathrm{is}\:\mathrm{given}\:\mathrm{as}\:\mathrm{84}\:\mathrm{kcal}/\mathrm{mol}, \\ $$$$\mathrm{103}\:\mathrm{kcal}/\mathrm{mol},\:\mathrm{x}\:\mathrm{kcal}/\mathrm{mol}\:\mathrm{and}\:\mathrm{y}\:\mathrm{kcal}/\mathrm{mol} \\ $$$$\mathrm{respectively}.\:\mathrm{Given}\:\mathrm{x}\::\:\mathrm{y}\:=\:\mathrm{9}\::\:\mathrm{5},\:\mathrm{then} \\ $$$$\mathrm{bond}\:\mathrm{energy}\:\mathrm{of}\:\mathrm{Cl}\:−\:\mathrm{Cl}\:\mathrm{bond}\:\mathrm{in}\:\mathrm{kcal}/\mathrm{mol}…
Question Number 154343 by Lekhraj last updated on 17/Sep/21 Answered by qaz last updated on 17/Sep/21 $$\frac{\mathrm{sin}\:\mathrm{x}}{\bullet\mathrm{O}}=\frac{\mathrm{sin}\:\angle\mathrm{M}\bullet\mathrm{C}}{\mathrm{OC}},\:\:\:\:\:\:\frac{\mathrm{sin}\:\mathrm{30}°}{\bullet\mathrm{O}}=\frac{\mathrm{sin}\:\mathrm{105}°}{\mathrm{OA}} \\ $$$$\because\:\:\mathrm{OA}=\mathrm{OC}\:\:\:\:\:\: \\ $$$$\therefore\:\:\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{105}°}\centerdot\mathrm{sin}\:\angle\mathrm{M}\bullet\mathrm{C}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}\mathrm{sin}\:\angle\mathrm{M}\bullet\mathrm{C} \\ $$$$\because\:\:\:\angle\mathrm{M}\bullet\mathrm{C}=\mathrm{180}°−\mathrm{135}°−\mathrm{x}=\mathrm{45}°−\mathrm{x} \\ $$$$\therefore\:\:\:\mathrm{sin}\:\mathrm{x}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}\mathrm{sin}\:\left(\mathrm{45}°−\mathrm{x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}\left(\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)…
Question Number 23269 by Rasheed.Sindhi last updated on 28/Oct/17 $$\mathbb{P}\mathrm{rove}\:\mathrm{that} \\ $$$$\:\mathrm{3k}+\mathrm{2}\:\mathrm{is}\:\mathrm{not}\:\mathrm{perfect}\:\mathrm{square}\:\mathrm{for} \\ $$$$\mathrm{all}\:\mathrm{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},…\right\}. \\ $$ Answered by Tinkutara last updated on 28/Oct/17 $${Any}\:{number}\:{can}\:{be}\:{written}\:{as}\:\mathrm{3}{p}, \\…
Question Number 154337 by SANOGO last updated on 17/Sep/21 $${etudier}\:{la}\:{continuite},\:{et}\:{derivabilite}: \\ $$$${x}^{\mathrm{2}} {sin}\left(\frac{\mathrm{1}}{{x}}\right)\:{si}\neq\mathrm{0}\:{et}\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$ Answered by puissant last updated on 17/Sep/21 $${posons}:\begin{cases}{{f}\left(\mathrm{0}\right)=\mathrm{0}\:{si}\:{x}=\mathrm{0}\:}\\{{f}\left({x}\right)={x}^{\mathrm{2}} {sin}\left(\frac{\mathrm{1}}{{x}}\right)\:,\:{sinon}..}\end{cases} \\…
Question Number 23262 by ajfour last updated on 28/Oct/17 Commented by ajfour last updated on 28/Oct/17 $$\frac{{CN}}{{AC}}=\frac{{x}}{{x}+{y}}\:\:\:;\:\:\:\frac{{AN}}{{AC}}=\frac{{y}}{{x}+{y}} \\ $$$$\frac{{MB}}{{AB}}\:=\frac{{y}}{{x}+{y}}\:;\:\:\frac{{AM}}{{AB}}=\frac{{x}}{{x}+{y}}\:. \\ $$$${based}\:{on}\:{similarity}\:\:{of}\:{triangles}. \\ $$$$\bigtriangleup{CNP}\:\sim\:\bigtriangleup{CAB} \\ $$$$\Rightarrow\:\:\frac{{CN}}{{AC}}=\frac{{NP}\left(={AM}\right)}{{AB}}=\frac{{CP}}{{BC}}…
Question Number 154334 by alcohol last updated on 17/Sep/21 $${the}\:{n}^{{th}} \:{term}\:{of} \\ $$$$\mathrm{1}\:,\:\mathrm{2},\:\mathrm{6},\:\mathrm{24},\:\mathrm{120}\:……..\:{is}? \\ $$ Commented by talminator2856791 last updated on 17/Sep/21 $$\:\mathrm{joke}\:\mathrm{question}. \\ $$…
Question Number 88794 by TawaTawa1 last updated on 12/Apr/20 Answered by john santu last updated on 12/Apr/20 $$\frac{\mathrm{2}{p}+\mathrm{1}−{p}−\mathrm{7}}{−\mathrm{1}−\mathrm{5}}\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\frac{{p}−\mathrm{6}}{−\mathrm{6}}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\:{p}−\mathrm{6}\:=\:−\mathrm{8} \\ $$$${p}\:=\:−\mathrm{2} \\ $$ Commented…
Question Number 154328 by liberty last updated on 17/Sep/21 $$\:{S}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{10}}+\frac{\mathrm{2}}{\mathrm{1}+\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{1}+\mathrm{10}^{\mathrm{4}} }+\frac{\mathrm{8}}{\mathrm{1}+\mathrm{10}^{\mathrm{8}} }+\ldots\: \\ $$$$ \\ $$ Answered by MJS_new last updated on 17/Sep/21 $$\mathrm{having}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{we}\:\mathrm{could}\:\mathrm{substract}\:\mathrm{the}…
Question Number 154330 by DELETED last updated on 17/Sep/21 Answered by DELETED last updated on 17/Sep/21 $$\mathrm{Jawab}: \\ $$$$\mathrm{Dik}=\mathrm{K}_{\mathrm{1}} =\mathrm{8},\mathrm{318}×\mathrm{10}^{−\mathrm{7}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{K}_{\mathrm{2}} =\mathrm{5},\mathrm{248}×\mathrm{10}^{−\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{H}^{+}…