Question Number 155560 by peter frank last updated on 02/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 155381 by ajfour last updated on 29/Sep/21 Commented by mr W last updated on 30/Sep/21 Commented by mr W last updated on 30/Sep/21…
Question Number 155292 by peter frank last updated on 28/Sep/21 Answered by peter frank last updated on 30/Sep/21 $$\mathrm{R}=\mathrm{a}=\frac{\mathrm{v}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta}{\mathrm{g}}=\frac{\mathrm{2v}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{g}}…\left(\mathrm{i}\right) \\ $$$$\mathrm{H}=\mathrm{x}=\frac{\mathrm{v}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2g}}…
Question Number 155293 by peter frank last updated on 28/Sep/21 Answered by mr W last updated on 28/Sep/21 Commented by mr W last updated on…
Question Number 155294 by peter frank last updated on 28/Sep/21 Answered by peter frank last updated on 30/Sep/21 Commented by peter frank last updated on…
Question Number 155033 by peter frank last updated on 24/Sep/21 Answered by peter frank last updated on 25/Sep/21 $$\mathrm{work}\:\mathrm{done}=\gamma×\bigtriangleup\mathrm{A}=\gamma×\left(\mathrm{A}_{\mathrm{2}} −\mathrm{A}_{\mathrm{1}} \right) \\ $$$$\mathrm{A}_{\mathrm{1}} =\mathrm{4}\pi\mathrm{R}^{\mathrm{2}} \:\:=\mathrm{1}.\mathrm{256}×\mathrm{10}^{−\mathrm{3}}…
Question Number 155035 by peter frank last updated on 24/Sep/21 Answered by mr W last updated on 24/Sep/21 $${h}_{{max}} =\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}{g}} \\ $$$${p}.{e}.\:{at}\:{h}=\frac{{h}_{{max}}…
Question Number 155032 by peter frank last updated on 24/Sep/21 Answered by peter frank last updated on 25/Sep/21 Commented by peter frank last updated on…
Question Number 155034 by peter frank last updated on 24/Sep/21 Commented by peter frank last updated on 24/Sep/21 $$\mathrm{i}\:\mathrm{think}\:\mathrm{h}_{\mathrm{o}} ={l} \\ $$$${u}_{{o}} ={max}\:{speed} \\ $$…
Question Number 154888 by physicstutes last updated on 22/Sep/21 Answered by mr W last updated on 23/Sep/21 Commented by physicstutes last updated on 23/Sep/21 $$\mathrm{brilliant}\:\mathrm{sir}\:\mathrm{thanks}…