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Question Number 45129 by Necxx last updated on 09/Oct/18

Commented by Necxx last updated on 09/Oct/18

$$12and13pls$$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18

$$13)v_0=\sqrt{2gh}=\sqrt{2ga}hereh=aheightofliquid$$$$levelaboveholeforfirstfigure$$$$v_1=\sqrt{2{gh}_1}{h}_1=\frac{a\sqrt{2}}{2}=halfofdiagonal$$$$\frac{v_1}{v_o}=\frac{\sqrt{2g}\times \sqrt{h_1}}{\sqrt{2g}\times \sqrt{h}}={\left(\frac{a}{\sqrt{2}}\times \frac{1}{a}\right)}^{\frac{1}{2}}={\left(\frac{1}{2}\right)}^{\frac{1}{4}}$$$$v_1=v_0\times {\left(\frac{1}{2}\right)}^{\frac{1}{4}}$$$$$$

Commented by Necxx last updated on 09/Oct/18

$$IammostgratefulmrTanmay$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18

$$thankyou...$$

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