One cm of a 10-cm-long rod is made of metal, and the rest is wood. The metal has a density of 5000kg/m^3 and the wood has a density of 500kg/m^3. When the rod is set into pure water, the metal part points downward. How much of the rod is underwater?
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9.5 cm
To find out how much of the rod is underwater, we need to compare the densities of the materials.
First, let's find the volume of the entire rod. Given that the length of the rod is 10 cm, we can convert it to meters by dividing by 100: 10 cm / 100 cm/m = 0.1 m.
The volume of the rod is then calculated by multiplying the length by the cross-sectional area. Since the rod is cylindrical, the cross-sectional area is given by πr^2, where r is the radius.
The radius of the entire rod can be found by dividing the diameter, which is 1 cm (the metal part), by 2: 1 cm / 2 = 0.5 cm.
Converting the radius to meters: 0.5 cm / 100 cm/m = 0.005 m.
The cross-sectional area is then given by: π(0.005 m)^2 = 0.00007854 m^2.
Substituting the volume and the density of the entire rod into the formula:
mass = density x volume
mass = 500 kg/m^3 x 0.00007854 m^2 = 0.03927 kg
Now, let's find the volume of the metal part of the rod, which is 1 cm long and has a radius of 0.005 m (converted from cm to m).
The volume of the metal part = π(0.005 m)^2 x 0.01 m = 0.000000785 m^3.
Now, let's find the mass of the metal part using the density:
mass of metal = density of metal x volume of metal
mass of metal = 5000 kg/m^3 x 0.000000785 m^3 = 0.003925 kg.
Since the metal part is denser than water, it will sink and its entire mass will be underwater.
Finally, let's find the volume and mass of the wood part of the rod:
The volume of the wood part = volume of the entire rod - volume of the metal part
volume of the wood part = 0.00007854 m^2 - 0.000000785 m^3 = 0.000077755 m^3
The mass of the wood part can be found using the density:
mass of wood = density of wood x volume of wood
mass of wood = 500 kg/m^3 x 0.000077755 m^3 = 0.0388775 kg
Since the wood part is less dense than water, it will float, therefore, only the metal part is underwater, which is 1 cm or 0.01 m long.