Question Number 142359 by 777316 last updated on 30/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76819 by Rio Michael last updated on 30/Dec/19 $$\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{foci}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ellipse} \\ $$$$\:\:\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{9}}\:+\:\frac{{y}^{\mathrm{2}} }{\mathrm{4}}\:=\:\mathrm{1}\:\mathrm{is} \\ $$$$\mathrm{A}.\:\left(\mathrm{4},\mathrm{0}\right) \\ $$$$\mathrm{B}.\:\left(\mathrm{9},\mathrm{0}\right) \\ $$$$\mathrm{C}.\:\left(\mathrm{5},\mathrm{0}\right) \\ $$$$\mathrm{D}.\:\left(\sqrt{\mathrm{5}}\:,\:\mathrm{0}\right) \\ $$…
Question Number 11283 by suci last updated on 19/Mar/17 $$\mathrm{2}{a}+{b}+\mathrm{4}{c}….\left(\mathrm{1}\right) \\ $$$${a}+\mathrm{3}{c}….\left(\mathrm{2}\right) \\ $$$$−\mathrm{3}{a}−\mathrm{4}{b}−{c}….\left(\mathrm{4}\right) \\ $$$${a}=..?? \\ $$$${b}=..?? \\ $$$${c}=..?? \\ $$ Terms of Service…
Question Number 11284 by suci last updated on 19/Mar/17 $${a}+\mathrm{2}{b}+\mathrm{3}{c}….\left(\mathrm{1}\right) \\ $$$$−\mathrm{3}{a}−\mathrm{4}{b}+\mathrm{2}{c}….\left(\mathrm{2}\right) \\ $$$$\mathrm{2}{a}−{b}−{c}….\left(\mathrm{3}\right) \\ $$$${a}=…?? \\ $$$${b}=…?? \\ $$$${c}=…?? \\ $$ Terms of Service…
Question Number 76817 by Rio Michael last updated on 30/Dec/19 $$\mathrm{A}\:\mathrm{compound}\:\mathrm{pendulum}\:\mathrm{ocsillates} \\ $$$$\mathrm{through}\:\mathrm{angles}\:\theta\:\mathrm{about}\:\mathrm{its}\:\mathrm{equilibrium} \\ $$$$\mathrm{position}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{8}{a}\theta^{\mathrm{2}} \:=\:\mathrm{9}{g}\:{cos}\theta,\:{a}>\mathrm{0}.\:\mathrm{its}\:\mathrm{period}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\mathrm{2}\pi\sqrt{\frac{\mathrm{8}{a}}{\mathrm{9}{g}}} \\ $$$$\mathrm{B}.\:\frac{\mathrm{3}\pi}{\mathrm{8}}\sqrt{\frac{{a}}{{g}}} \\ $$$$\mathrm{C}.\:\mathrm{2}\pi\sqrt{\frac{\mathrm{9}{g}}{\mathrm{8}{a}}} \\…
Question Number 76813 by Rio Michael last updated on 30/Dec/19 $$\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \:=\: \\ $$$$\mathrm{A}.\:\infty \\ $$$$\mathrm{B}.\:\mathrm{1} \\ $$$$\mathrm{C}.\:−\mathrm{1} \\ $$$$\mathrm{D}.\:\mathrm{0} \\ $$ Commented…
Question Number 76811 by Rio Michael last updated on 30/Dec/19 $$\mathrm{The}\:\mathrm{eccentricity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hyperbola} \\ $$$$\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{64}}\:−\:\frac{{y}^{\mathrm{2}} }{\mathrm{36}}\:=\:\mathrm{1}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{B}.\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{C}.\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{D}.\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$…
Question Number 11272 by chux last updated on 18/Mar/17 Answered by mrW1 last updated on 18/Mar/17 $$\left(\mathrm{1}\right)\Rightarrow{y}+{z}=\mathrm{1}−{x} \\ $$$$\left(\mathrm{2}\right)\Rightarrow{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{35}−{x}^{\mathrm{2}} \\ $$$$\left({y}+{z}\right)^{\mathrm{2}} =\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \\…
Question Number 76809 by Rio Michael last updated on 30/Dec/19 Commented by Rio Michael last updated on 30/Dec/19 $$\mathrm{I}_{\mathrm{q}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{rod} \\ $$$$\mathrm{AB}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}\:\mathrm{about}\:\mathrm{its}\:\mathrm{centre}\:\mathrm{and}\:\mathrm{I}_{\mathrm{p}} \:\mathrm{is}\:\mathrm{the}\: \\ $$$$\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{about}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{shown}\:\mathrm{above}.…
Question Number 76802 by kavinila last updated on 30/Dec/19 $$\left.\mathrm{3}\right)\alpha\mathrm{particle}\:\mathrm{of}\:\mathrm{energy}\:\mathrm{5MeV}\:\mathrm{pass} \\ $$$$\mathrm{through}\:\mathrm{an}\:\mathrm{ionisation}\:\mathrm{chamber}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{rate}\:\mathrm{of}\:\mathrm{10}\:\mathrm{pwe}\:\mathrm{second}\:.\:\mathrm{Assum}\:\mathrm{that}\:\mathrm{all} \\ $$$$\mathrm{the}\:\mathrm{energy}\:\mathrm{is}\:\mathrm{used}\:\mathrm{in}\:\mathrm{producing}\:\mathrm{ion}\: \\ $$$$\mathrm{pairs},\mathrm{calculate}\:\mathrm{the}\:\mathrm{current}\:\mathrm{produced}. \\ $$$$\left(\mathrm{35MeV}\:\mathrm{is}\:\mathrm{required}\:\mathrm{for}\:\mathrm{producing}\:\:\mathrm{an}\:\right. \\ $$$$\left.\mathrm{ion}\:\mathrm{pair}\:\mathrm{and}\:\mathrm{e}=\mathrm{1}.\mathrm{6}×\mathrm{10}^{-\mathrm{19}} \:\mathrm{C}\right) \\ $$$$\boldsymbol{\mathrm{solution}}:…