Question Number 22483 by Tinkutara last updated on 19/Oct/17 $$\mathrm{Predict}\:\mathrm{the}\:\mathrm{density}\:\mathrm{of}\:\mathrm{Cs}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{density}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{elements} \\ $$$$\mathrm{K}\:\mathrm{0}.\mathrm{86}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \:\:\:\:\:\:\:\:\mathrm{Ca}\:\mathrm{1}.\mathrm{548}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \\ $$$$\mathrm{Sc}\:\mathrm{2}.\mathrm{991}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \:\:\:\:\:\mathrm{Rb}\:\mathrm{1}.\mathrm{532}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \\ $$$$\mathrm{Sr}\:\mathrm{2}.\mathrm{68}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \:\:\:\:\:\:\:\:\mathrm{Y}\:\mathrm{4}.\mathrm{34}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \\ $$$$\mathrm{Cs}\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Ba}\:\mathrm{3}.\mathrm{51}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{La}\:\mathrm{6}.\mathrm{16}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}}…
Question Number 87952 by ajfour last updated on 07/Apr/20 Commented by ajfour last updated on 07/Apr/20 $${what}\:{is}\:{c}\:?\:\:{Sir}. \\ $$ Commented by ajfour last updated on…
Question Number 87561 by jagoll last updated on 05/Apr/20 $$\mathrm{for}\:\mathrm{x},\mathrm{y}\:\in\mathbb{R}\:,\: \\ $$$$\mathrm{x}\:,\mathrm{y}\:>\:\mathrm{0}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{3}} \mathrm{y}−\mathrm{xy}^{\mathrm{3}} \:=\:\mathrm{24}}\\{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{10}\:}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}? \\ $$$$\left(\mathrm{a}\right)\:\mathrm{6}\:\:\:\left(\mathrm{b}\right)\:\mathrm{5}\:\:\:\left(\mathrm{c}\right)\:\mathrm{4}\:\:\:\left(\mathrm{d}\right)\:\mathrm{3}\sqrt{\mathrm{2}}\:\:\left(\mathrm{e}\right)\:\mathrm{2} \\ $$ Commented…
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Question Number 21837 by math solver last updated on 05/Oct/17 Commented by math solver last updated on 05/Oct/17 $${anyone}?? \\ $$$${will}\:{anyone}\:{answer}\:{my}\:{question}???\:\:\:\:\:\:\:\: \\ $$$$;\left(\right. \\ $$…
Question Number 152682 by alcohol last updated on 31/Aug/21 Answered by Olaf_Thorendsen last updated on 31/Aug/21 $$\mathrm{Time}\:\mathrm{taken}\:\mathrm{1}\:\mathrm{second}. \\ $$$$\mathrm{Potential}\:\mathrm{energy}\:\mathrm{E}_{\mathrm{P}} \:: \\ $$$$\mathrm{E}_{\mathrm{P}} \:=\:{mgh}\:=\:\mathrm{50}×\mathrm{9},\mathrm{81}×\mathrm{40}\:=\:\mathrm{19620}\:\mathrm{J} \\ $$$$\mathrm{Kinetic}\:\mathrm{energy}\:\mathrm{E}_{\mathrm{K}}…
Question Number 151899 by liberty last updated on 24/Aug/21 Answered by john_santu last updated on 24/Aug/21 $$\mathrm{A}=\int_{\mathrm{0}} ^{\mathrm{6}} \left(\frac{−\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}=\left[−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{24}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{6}} \\…
Question Number 20819 by ajfour last updated on 03/Sep/17 Commented by ajfour last updated on 03/Sep/17 $${A}\:{particle}\:{moves}\:{along}\:{the}\:{spiral} \\ $$$${shown}.\:{Determine}\:{the}\:{magnitude} \\ $$$${of}\:{the}\:{velocity}\:{of}\:{the}\:{particle}\:{in} \\ $$$${terms}\:{of}\:\boldsymbol{{b}},\:\boldsymbol{\theta},\:\overset{.} {\boldsymbol{\theta}}\:. \\…
Question Number 86063 by john santu last updated on 26/Mar/20 Answered by mr W last updated on 26/Mar/20 $$\left(\frac{\mathrm{1}}{{R}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{R}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }\right) \\…
Question Number 20511 by dioph last updated on 27/Aug/17 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{sphere}\:\mathrm{of}\:\mathrm{unit}\:\mathrm{radius}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circular} \\ $$$$\mathrm{spot}\:\mathrm{on}\:\mathrm{the}\:\mathrm{sphere}'\mathrm{s}\:\mathrm{surface}\:\mathrm{given} \\ $$$$\mathrm{the}\:\mathrm{latitude}\:\beta\:\mathrm{and}\:\mathrm{the}\:\mathrm{longitude}\:\lambda \\ $$$$\mathrm{of}\:\mathrm{its}\:\mathrm{center}\:\mathrm{and}\:\mathrm{its}\:\mathrm{angular}\:\mathrm{radius}\:{r}. \\ $$ Terms of Service Privacy Policy…