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Question-193345

Question Number 193345 by yaslm last updated on 11/Jun/23 Answered by aleks041103 last updated on 13/Jun/23 $${The}\:{trajectry}\:{of}\:{the}\:{water}\:{is}\:{a}\:{freefall}\:{parabolla} \\ $$$$\Rightarrow{H}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \Rightarrow{t}=\sqrt{\frac{\mathrm{2}{H}}{{g}}} \\ $$$${D}={vt}={v}\sqrt{\frac{\mathrm{2}{H}}{{g}}} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{gD}^{\mathrm{2}} }{\mathrm{2}{H}}}…

Question-193268

Question Number 193268 by 073 last updated on 09/Jun/23 Answered by qaz last updated on 09/Jun/23 $$\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{x}\right)^{{n}} }{{n}!}\right)\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{x}\right)^{{n}} {x}}{{n}!\left({n}+\mathrm{1}\right)}\right)={xe}^{−{x}} \underset{{n}=\mathrm{0}} {\overset{\infty}…