Question Number 209122 by Spillover last updated on 02/Jul/24 Answered by Spillover last updated on 02/Jul/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 209121 by Spillover last updated on 02/Jul/24 Answered by Spillover last updated on 04/Jul/24 $${r}={r}_{\mathrm{0}} {e}^{{k}\theta} \\ $$$${In}\:{central}\:{force}\:{field}, \\ $$$${the}\:{angular}\:{momentum}\:{L}\:{is}\:{conserved} \\ $$$${L}={mr}^{\mathrm{2}} \frac{{d}\theta}{{dt}}\:\:\:\:\:{where}\:{m}={mass}\:{of}\:{the}\:{particle}…
Question Number 209098 by Spillover last updated on 01/Jul/24 Answered by A5T last updated on 02/Jul/24 $$\mathrm{4}=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{\mathrm{2}{g}}\Rightarrow{sin}\theta=\frac{\sqrt{\mathrm{8}{g}}}{{v}_{\mathrm{0}} } \\ $$$${R}={v}_{\mathrm{0}} {cos}\theta×\mathrm{2}{t}={v}_{\mathrm{0}} \sqrt{\mathrm{1}−\frac{\mathrm{8}{g}}{{v}_{\mathrm{0}}…
Question Number 209099 by Spillover last updated on 01/Jul/24 Answered by mr W last updated on 02/Jul/24 $${say}\:{particle}\:\mathrm{2}\:{starts}\:{time}\:{T}\:\:{later} \\ $$$${than}\:{particle}\:\mathrm{1}.\: \\ $$$$ \\ $$$${particle}\:\mathrm{2}\:{at}\:{time}\:{t}: \\…
Question Number 209024 by Spillover last updated on 30/Jun/24 Commented by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 01/Jul/24 Answered by…
Question Number 208999 by Spillover last updated on 30/Jun/24 Answered by mr W last updated on 30/Jun/24 Commented by mr W last updated on 30/Jun/24…
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Question Number 206394 by 0202391905769 last updated on 13/Apr/24 Commented by mr W last updated on 13/Apr/24 $$\Rightarrow{Q}\mathrm{206321} \\ $$ Terms of Service Privacy Policy…
Question Number 206273 by EmGent last updated on 10/Apr/24 $$\mathrm{Does}\:\mathrm{anyone}\:\mathrm{know}\:\mathrm{how}\:\mathrm{this}\:\mathrm{works}\:? \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{have}\:{d}\psi\:=\:\left({x}^{\mathrm{2}} -{cy}^{\mathrm{2}} \right){dy} \\ $$$$ \\ $$$$\mathrm{And}\:\mathrm{my}\:\mathrm{physics}\:\mathrm{teacher}\:\mathrm{says}\:\mathrm{it}\:\mathrm{is}\:\left(\mathrm{or}\:\mathrm{can}\right. \\ $$$$\left.\mathrm{be}\right)\:\mathrm{a}\:\mathrm{harmonic}\:\mathrm{function}\:\left(\Delta\psi\:=\:\mathrm{0}\right) \\ $$$$\mathrm{Can}\:\mathrm{anyone}\:\mathrm{explain}\:? \\…