A cubical object of sides 5 cm is immersed completely in water. What is the force of buoyancy on the object? (Assume g = 10 m/s2)
.05 m on a side
volume = .05^3 = 125 * 10^-6 m^3
mass of water displaced = 10^3 kg/m^3 * 125*10^-6 m^3 = .125 kg
Archimedes says force up = weight of water displaced
g = 10 m/s^2
so
m g = 1.25 Newtons
Thanks!!
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To calculate the force of buoyancy on an object, we need to know the volume of the object and the density of the fluid it is immersed in. In this case, the object is immersed in water, so we need to know the density of water.
The density of water is approximately 1000 kg/m^3. However, the side length of the object is given in centimeters, so let's convert it to meters first.
The side length of the object is 5 cm, which is equivalent to 0.05 meters (1 cm = 0.01 m).
To find the volume of the object, we can use the formula for the volume of a cube:
Volume = (side length)^3
So, the volume of the object is:
Volume = (0.05 m)^3 = 0.000125 m^3
Now we can calculate the force of buoyancy using Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
Buoyant Force = Density of fluid × Volume of fluid displaced × g
where g is the acceleration due to gravity.
In this case, the volume of the fluid displaced is the same as the volume of the object, since the object is completely immersed in water.
Buoyant Force = Density of water × Volume of object × g
Plugging in the values, we have:
Buoyant Force = 1000 kg/m^3 × 0.000125 m^3 × 10 m/s^2
Simplifying the calculation, we find:
Buoyant Force = 1.25 N
Therefore, the force of buoyancy on the object is 1.25 Newtons.