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Question Number 138217 by otchereabdullai@gmail.com last updated on 11/Apr/21

$$\mathrm{A}\mathrm{rectangular}\mathrm{water}\mathrm{tank}\mathrm{is}\mathrm{being}\mathrm{filed}$$$$\mathrm{at}\mathrm{the}\mathrm{constant}\mathrm{rate}\mathrm{of}70\mathrm{lt}/\mathrm{s}.\mathrm{The}$$$$\mathrm{base}\mathrm{of}\mathrm{the}\mathrm{tank}\mathrm{has}\mathrm{width}\mathrm{w}=9\mathrm{m}$$$$\mathrm{and}\mathrm{length}\mathrm{length}\mathrm{l}=16\mathrm{m}\mathrm{if}\mathrm{the}\mathrm{volume}$$$$\mathrm{of}\mathrm{the}\mathrm{tank}\mathrm{is}\mathrm{v}=\mathrm{w}\times \mathrm{l}\times \mathrm{h}\mathrm{where}\mathrm{h}\mathrm{is}$$$$\mathrm{the}\mathrm{hight}\mathrm{of}\mathrm{the}\mathrm{tank}.\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{rate}$$$$\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{the}\mathrm{hight}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{the}$$$$\mathrm{tank}$$

Commented by Dwaipayan Shikari last updated on 11/Apr/21

$$V=wlh$$$$\Rightarrow \frac{dV}{dt}=wl\frac{dh}{dt}\Rightarrow 70\frac{litre}{s}=160dcm\times 90dcm\frac{dh}{dt}$$$$\frac{dh}{dt}=\frac{70}{160\times 90}.\frac{dm}{s}=\frac{7}{14400}.\frac{m}{s}$$

Commented by otchereabdullai@gmail.com last updated on 11/Apr/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}!$$

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