Question #9be26

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Question #9be26

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Answer 1 · 2017-10-02T07:51:25

$0.42 m$ $0.42"m"$

Explanation:

$specific gravity of wood \to \frac{700 \frac{kgm}{m^{3}}}{1000 \frac{kgm}{m^{3}}} = 0.7 (no units!)$ $"specific gravity of wood" ->(700"kgm"/m^3)/(1000"kgm"/m^3) = 0.7 ("no units!")$

$% of wood immersed = 0.7 \cdot 100 = 70 %$ $% "of wood immersed" = 0.7*100 = 70%$

$volume of cube = {(0.6 m)}^{3} = 0.216 m^{3}$ $"volume of cube" = (0.6m)^3 = 0.216m^3$

$depth of wood immersed = 70 % \cdot 0.215 = 0.1512 m^{3}$ $"depth of wood immersed" = 70%*0.215 = 0.1512m^3$

Two of its sides will remain the same while immersed. So the depth $(d)$ $(d)$ immersed,

$0.6 \cdot 0.6 \cdot d = 0.1512$ $0.6*0.6*d = 0.1512$

$\Rightarrow d = \frac{0.1512 m^{3}}{0.6 m \cdot 0.6 m} = 0.42 m$ $=>d=(0.1512m^3)/(0.6m*0.6m)=0.42m$

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Question #9be26

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