Answer: 3. 500 kg/m3
Explanation:
The only external forces acting upon the piece of wood, are the gravity (which we call weight) and the buoyant force due to the water removed by the wood to be partially submerged.
Both forces oppose each other, due to the gravity is always downward, and the buoyant force is always upwards.
In order to be in equilibrium, both forces must be equal in magnitude, so we can write the following:
Fg = Fb
If the wood has only half of his volume just above the Surface, this means that the buoyant force is equal to the weight of this volumen in water, as follows:
Fb = δh2o. V/2. g
This value must be equal to the weight of the full volumen of wood:
Fg = = δw . V. g
So, in order to Fg = Fb, it must be δw = δh2o / 2 = 500 Kg/m3
The density of the piece wood floating on the water is 500 kg/m³.
The given parameters;
The specific gravity of the piece of wood is calculated as follows;
[tex]S.G = \frac{volume \ submerged}{total \ volume} = \frac{0.5V}{V}= 0.5[/tex]
The density of the piece wood is calculated as follows;
[tex]S.G = \frac{\rho _{wood}}{\rho _{water}} \\\\\rho _{wood} = S.G \times \rho _{water}\\\\\rho _{wood} = 0.5 \times 1000 \ kg/m^3\\\\\rho _{wood} = 500 \ kg/m^3[/tex]
Thus, the density of the piece wood floating on the water is 500 kg/m³.
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