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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)…

Question-215975

Question Number 215975 by Tawa11 last updated on 23/Jan/25 Answered by Tawa11 last updated on 23/Jan/25 $$\mathrm{v}\:\:=\:\:\mathrm{1}.\mathrm{98}\:\mathrm{m}/\mathrm{s}\:\:\:\:\:\mathrm{at}\:\mathrm{the}\:\mathrm{back}\:\mathrm{of}\:\mathrm{the}\:\mathrm{book}. \\ $$ Answered by mr W last updated…

Question-215943

Question Number 215943 by sitholenonto last updated on 22/Jan/25 Answered by MathematicalUser2357 last updated on 23/Jan/25 $$\mathrm{Q}.\mathrm{2}.\mathrm{1} \\ $$$$\sqrt{\left(\mathrm{30}−\mathrm{10}×\mathrm{tan}\:\mathrm{60}°\right)^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} } \\ $$$$\mathrm{16}.\mathrm{14836}\checkmark \\ $$$$\mathrm{Q}.\mathrm{2}.\mathrm{2}…

t-33-ab-t-00-ab-Clearing-up-defintions-t-33-a-a-36-3-t-00-a-a-36-0-t-a-a-36-t-a-transformation-of-a-

Question Number 215527 by alephnull last updated on 09/Jan/25 $${t}_{\mathrm{33}} '{ab}+{t}_{\mathrm{00}} '{ab}=? \\ $$$$ \\ $$$$\mathrm{Clearing}\:\mathrm{up}\:\mathrm{defintions}: \\ $$$${t}_{\mathrm{33}} '{a}=\left({a}−\sqrt{\mathrm{36}}\right)×\mathrm{3} \\ $$$${t}_{\mathrm{00}} '{a}=\left({a}−\sqrt{\mathrm{36}}\right)×\mathrm{0} \\ $$$${t}'{a}={a}−\sqrt{\mathrm{36}} \\…