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A wooden cube (density 650 kg m-3; edge length 20,0 cm) is floating in water (Density = 103 kg m-3). The height (in cm) of the block that is protruding above the water is:

Answer :

Final answer:

The height of the wooden cube that is protruding above the water is 5 cm.

Explanation:

To find the height of the wooden cube that is protruding above the water, we can use Archimedes' principle. According to this principle, an object will float in a fluid if the weight of the fluid displaced by the object is equal to the weight of the object. In this case, the weight of the wooden cube is equal to the weight of the water it displaces.

The weight of the wooden cube can be calculated using its density and volume. The density of the wooden cube is given as 650 kg/m³, and its volume can be calculated by cubing its edge length. The volume of the cube is (0.2 m)³ = 0.008 m³.

The weight of the water displaced by the cube can be calculated using the density of water and the volume of the cube that is submerged. The density of water is given as 103 kg/m³, and the height of the cube submerged in water is the same as the length of its edge, which is 0.2 m. Therefore, the volume of water displaced is (0.2 m)³ = 0.008 m³.

Since the weight of the cube is equal to the weight of the water it displaces, we can set up the equation: 650 kg/m³ * 0.008 m³ = 103 kg/m³ * h * 0.008 m³, where h is the height of the cube above the water.

Simplifying the equation gives h = (650 kg/m³ * 0.008 m³) / (103 kg/m³) = 0.05 m = 5 cm.

Learn more about Archimedes' here:

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