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Question-216477

Question Number 216477 by Jubr last updated on 08/Feb/25 Answered by A5T last updated on 08/Feb/25 $$\left.\mathrm{2}\right)\:\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}=\mathrm{u}\:;\:\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}}=\mathrm{v} \\ $$$$\Rightarrow\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} =\mathrm{2}\:;\:\mathrm{uv}=\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\sqrt{−\mathrm{3}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}}=\mathrm{2} \\ $$$$\Rightarrow\left(\mathrm{u}+\mathrm{v}\right)^{\mathrm{2}}…

Question-216478

Question Number 216478 by Tawa11 last updated on 08/Feb/25 Commented by mr W last updated on 09/Feb/25 $${the}\:{vertex}\:{of}\:{the}\:{cone}\:{may}\:{touch}\: \\ $$$${any}\:{point}\:{of}\:{one}\:{end}\:{of}\:{the}\:{cylinder}, \\ $$$${we}\:{always}\:{have} \\ $$$${V}_{{S}} =\mathrm{2}{V}…

Question-216153

Question Number 216153 by Tawa11 last updated on 28/Jan/25 Answered by AntonCWX last updated on 29/Jan/25 $${i}={i}_{{L}} =\frac{\mathrm{12}{V}}{\mathrm{1}\Omega+\mathrm{5}\Omega}=\mathrm{2}{A} \\ $$$${v}_{{C}} =\left(\mathrm{1}\Omega+\mathrm{4}\Omega\right)\left(\mathrm{2}{A}\right)=\mathrm{10}{V} \\ $$$$ \\ $$$${W}_{{C}}…

Question-216016

Question Number 216016 by Tawa11 last updated on 25/Jan/25 Answered by som(math1967) last updated on 25/Jan/25 $${current}\:\mathrm{20}\Omega\:{when}\:{both}\:{switch} \\ $$$${oppen}=\frac{\mathrm{6}}{\mathrm{80}}=\frac{\mathrm{3}}{\mathrm{40}}{amp} \\ $$$${current}\:{flow}\:{cct}\:{when}\:{s}_{\mathrm{1}} \:{s}_{\mathrm{2}} \:{both} \\ $$$${close}\:\mathrm{6}\boldsymbol{\div}\left(\mathrm{50}+\frac{\mathrm{20}×{R}}{\mathrm{20}+{R}}\right)…