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Sagot :
The length and density of the aluminum cube is 15.46 cm and 1,081.1 kg/m³ respectively
Volume of the aluminum
The volume of the aluminum is equal to the weight of the kerosene displaced.
V = 0.05 kg / 13.5 kg/m³
V = 3.7 x 10⁻³ m³
Density of the aluminum
density = mass/volume
density = 4 kg / 3.7 x 10⁻³ m³
density = 1,081.1 kg/m³
Length of the aluminum
V = L³
L = ∛V
L = ∛(3.7 x 10⁻³)
L = 0.1546 m
L = 15.46 cm
Thus, the length and density of the aluminum cube is 15.46 cm and 1,081.1 kg/m³ respectively.
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