Water tank

A 288 hectoliter of water was poured into the tank with dimensions 12 m and 6 m bottom and 2 m depth. What part of the volume of the tank water occupied? Calculate the surface of tank wetted with water.

Correct result:

p =  20 %
A =  86.4 m²

Solution:

$$\begin{gathered} a=12\text{ m } \\ b=6\text{ m } \\ c=2\text{ m } \\ S=a\cdot b=12\cdot 6=72{\text{m}}^2 \\ V=S\cdot c=72\cdot 2=144{\text{m}}^3 \\ {V}_1=288\mathrm{/}10={\displaystyle \frac{144}{5}}=28.8{\text{m}}^3 \\ p=100\cdot V_1\mathrm{/}V=100\cdot 28.8\mathrm{/}144=20=20\mathrm{\%} \\ {c}_1=V_1\mathrm{/}S=28.8\mathrm{/}72={\displaystyle \frac{2}{5}}=0.4\text{ m} \end{gathered}$$

$A=S+c_1\cdot (2\cdot a+2\cdot b)=72+0.4\cdot (2\cdot 12+2\cdot 6)={\displaystyle \frac{432}{5}}={\displaystyle \frac{432}{5}}{\text{m}}^2=86.4{\text{m}}^2$

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Jimbob

This question wording make no sense in brainhole.

2 years ago  Like

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