Calculate 80793

The liquid has a specific gravity of γ = 9050 N. m-3. Calculate its specific gravity, the mass of 2 m³ of this liquid, and also of 3 m³ of this liquid.

Correct answer:

ρ =  922.528 kg/m³
m₂ =  1845.0561 kg
G₃ =  27150 N

Step-by-step explanation:

$$\begin{gathered} \gamma =9050{\text{N/m}}^3 \\ g=9.81{\text{m/s}}^2 \\ G=mg \\ \gamma =\rho \cdot g \\ \rho =\gamma /g=9050/9.81=922.528{\text{kg/m}}^3 \end{gathered}$$

$$\begin{gathered} V_2=2{\text{m}}^3 \\ {m}_2=\rho \cdot V_2=922.528\cdot 2=1845.0561\text{kg} \end{gathered}$$

$$\begin{gathered} V_3=3{\text{m}}^3 \\ {G}_3=\gamma \cdot V_3=9050\cdot 3=27150\text{N} \end{gathered}$$

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