Water-flows-out-of-a-tank-through-a-hole-of-diameter-2cm-above-the-hole-1-Determine-the-velocity-of-outflow-2-The-rate-of-outflow-when-the-level-of-the-water-in-the-tank-is-2cm-above-the-hole-

FacebookTweetPin

Question Number 12499 by tawa last updated on 23/Apr/17

$$\mathrm{Water}\mathrm{flows}\mathrm{out}\mathrm{of}\mathrm{a}\mathrm{tank}\mathrm{through}\mathrm{a}\mathrm{hole}\mathrm{of}\mathrm{diameter}2\mathrm{cm}\mathrm{above}\mathrm{the}\mathrm{hole}$$$$\left(1\right)\mathrm{Determine}\mathrm{the}\mathrm{velocity}\mathrm{of}\mathrm{outflow}$$$$\left(2\right)\mathrm{The}\mathrm{rate}\mathrm{of}\mathrm{outflow}\mathrm{when}\mathrm{the}\mathrm{level}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{in}\mathrm{the}\mathrm{tank}\mathrm{is}2\mathrm{cm}\mathrm{above}$$$$\mathrm{the}\mathrm{hole}.$$

Answered by sandy_suhendra last updated on 24/Apr/17

$$\mathrm{h}=2\mathrm{cm}=0.02\mathrm{m}$$$$\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$$$$\mathrm{D}=2\mathrm{cm}=0.02\mathrm{m}$$$$\mathrm{A}=\frac{1}{4}\pi {\mathrm{D}}^2=\frac{1}{4}\times 3.14\times 0.{02}^2=3.14\times {10}^{-4}{\mathrm{m}}^2$$$$\mathrm{a})\mathrm{v}=\sqrt{2\mathrm{gh}}=\sqrt{2\times 9.8\times 0.02}=0.626\mathrm{m}/\mathrm{s}$$$$\mathrm{b})\mathrm{Q}=\mathrm{A}.\mathrm{v}=3.14\times {10}^{-4}\times 0.626=1.97\times {10}^{-4}{\mathrm{m}}^3/\mathrm{s}$$

Commented by tawa last updated on 24/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin