Question Number 6648 by Tawakalitu. last updated on 07/Jul/16 $${What}\:{mass}\:{of}\:{ice}\:{at}\:\:−\mathrm{14}\:{will}\:{be}\:{needed}\:{to}\:{cool}\:\mathrm{200}\:{cm}^{\mathrm{3}} \:{of} \\ $$$${an}\:{orange}\:{drink}\:\left({essentially}\:{water}\right)\:{from}\:\mathrm{25}°{C}\:\:{to}\:\:\mathrm{10}°{C} \\ $$$$\left({specific}\:{latent}\:{heat}\:{of}\:{fusion}\:{of}\:{ice}\:=\:\mathrm{3}.\mathrm{36}\:×\:\mathrm{10}^{\mathrm{5}} \:{jkg}^{−\mathrm{1}} \right. \\ $$$${specific}\:{heat}\:{capacity}\:{of}\:{ice}\:=\:\mathrm{2100}\:{jkg}^{−\mathrm{1}} {k}^{−\mathrm{1}} \\ $$$${specific}\:{heat}\:{capacity}\:{of}\:{water}\:=\:\mathrm{4200}\:{jkg}^{−\mathrm{1}} {k}^{−\mathrm{1}} . \\…
Question Number 137717 by mnjuly1970 last updated on 05/Apr/21 Commented by Dwaipayan Shikari last updated on 05/Apr/21 $${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}−{a}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−{a}}{\mathrm{2}}\right)\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}−{a}}{\mathrm{2}}\right)=\frac{\mathrm{1}−{a}}{\mathrm{4}}.\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}{a}\right)} \\ $$$$=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−{a}\right){sec}\left(\frac{\pi}{\mathrm{2}}{a}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}{f}'\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{12}} \\ $$ Terms…
Question Number 6644 by WAI LIN last updated on 07/Jul/16 $${Prove}\:\frac{{d}}{{dt}}\:{e}_{\rho} =\phi\:{e}_{\phi} \:,\:\frac{{d}}{{dt}}\:{e}_{\phi} =\:−\phi\:{e}_{\rho} \:{where}\:{dots}\:{denote}\: \\ $$$${differentiation}\:{with}\:{respect}\:{to}\:{time}\:{t}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 72178 by ajfour last updated on 25/Oct/19 Commented by ajfour last updated on 25/Oct/19 $${If}\:{released}\:{as}\:{shown},\:{in}\:{what} \\ $$$${time}\:{does}\:{the}\:{block}\:{separate} \\ $$$${from}\:{the}\:{hemisphere}? \\ $$ Commented by…
Question Number 6642 by WAI LIN last updated on 07/Jul/16 $${Represent}\:{the}\:{vector}\:{A}={zi}−\mathrm{2}{xj}+{yk}\:{in}\:{cylindrical}\:{coordinates}. \\ $$$${Thus}\:{determine}\:{A}_{\rho} ,\:{A}_{\phi} ,\:{A}_{{z}} . \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 6641 by WAI LIN last updated on 07/Jul/16 $${Q}\left(\mathrm{1}\right)\:{Express}\:{div}\:{A}\:=\:\bigtriangledown\:.\:{A}\:{in}\:{orthogonal}\:{coordinates}. \\ $$$${Q}\left(\mathrm{2}\right)\:{Express}\:\bigtriangledown^{\mathrm{2}} \:\phi\:{in}\:{orthogonal}\:{curvilinear}\:{coordinates}. \\ $$$${Q}\left(\mathrm{3}\right)\:{Express}\:\left({a}\right)\:\bigtriangledown\:×\:{Aand}\:\left({b}\right)\:\bigtriangledown^{\mathrm{2}} \phi\:{in}\:{spherical}\:{coordinates}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 6640 by Rasheed Soomro last updated on 07/Jul/16 $$\mathrm{ABC}\:{is}\:{a}\:{triangle},{whose}\:{one}\:{vertex}\: \\ $$$${is}\:{moving}\:{along}\:{a}\:{circular}\:{path}.\: \\ $$$${Discuss}\:{the}\:{nature}\:{of}\:{the}\:\:{path},\:{along} \\ $$$${which}\:{the}\:\boldsymbol{{centroid}}\:{of}\:{the}\:{triangle} \\ $$$${is}\:{moving}. \\ $$ Commented by Rasheed Soomro…
Question Number 137708 by mohammad17 last updated on 05/Apr/21 Commented by mohammad17 last updated on 05/Apr/21 $${prove} \\ $$ Answered by mr W last updated…
Question Number 72175 by mr W last updated on 26/Oct/19 $$\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} =\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{3}} \\ $$$$ \\ $$$${how}\:{large}\:\left({area}\right)\:{is}\:{your}\:{heart}\:\heartsuit\:? \\ $$ Commented by mr W…
Question Number 6638 by Tawakalitu. last updated on 07/Jul/16 $${The}\:{radius}\:{of}\:{curvature}\:{of}\:{a}\:{thin}\:{equiconvex}\:{lens}\:{is}\:\mathrm{40}.\mathrm{0}\:{cm} \\ $$$${the}\:{lens}\:{is}\:{made}\:{of}\:{glass}\:{of}\:{refractive}\:{index}\:\mathrm{1}.\mathrm{660}.\:\:{find}\:{the}\: \\ $$$${position}\:{of}\:{the}\:{image}\:{of}\:{an}\:{object}\:{placed}\:{on}\:{the}\:{axis}\:{of}\:{the}\: \\ $$$${axis}\:{of}\:{the}\:{lens}\:{at}\:{a}\:{distance}\:{of}\:\mathrm{50}.\mathrm{0}\:{mm}\:\:{from}\:{the}\:{pole}. \\ $$$$ \\ $$ Terms of Service Privacy Policy…