Seawater has a density of 1025 kg/m3, ice 920 kg/m3. 8 liters of seawater froze and created a cube. Calculate the size of the cube edge.

Correct result:

x2 =  2.073 dm


V1=8 ldm3=8 1 dm3=8 dm3 x1=V13=83=2 dm  V2=1025920 V1=1025920 8205238.913 dm3  x2=V23=8.9133=2.073 dmV_{1}=8 \ l \rightarrow dm^3=8 \cdot \ 1 \ dm^3=8 \ dm^3 \ \\ x_{1}=\sqrt[3]{ V_{1}}=\sqrt[3]{ 8 }=2 \ \text{dm} \ \\ \ \\ V_{2}=\dfrac{ 1025 }{ 920 } \cdot \ V_{1}=\dfrac{ 1025 }{ 920 } \cdot \ 8 \doteq \dfrac{ 205 }{ 23 } \doteq 8.913 \ \text{dm}^3 \ \\ \ \\ x_{2}=\sqrt[3]{ V_{2}}=\sqrt[3]{ 8.913 }=2.073 \ \text{dm}

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