A paperweight, when weighed in air, has a weight of W = 7.5 N. When completely immersed in water, however, it has a weight of Win water = 4.2 N. Find the volume of the paperweight.
To find the volume of the paperweight, we can use Archimedes' principle, which states that the buoyant force acting on a submerged object is equal to the weight of the fluid displaced by that object.
The weight of the paperweight in air is given as W = 7.5 N.
The weight of the paperweight in water is given as Win water = 4.2 N.
The difference between these two weights represents the buoyant force acting on the paperweight in water:
Buoyant force in water = W - Win water
= 7.5 N - 4.2 N
= 3.3 N
According to Archimedes' principle, this buoyant force is equal to the weight of the water displaced by the paperweight when it is fully submerged. Let's denote this weight of the water displaced as Wdisplaced water.
Wdisplaced water = Buoyant force in water
= 3.3 N
Now, we need to calculate the volume of the water displaced by the paperweight. We can do this by using the equation:
Density of water * Volume of water = Weight of water
The density of water is a known constant, which is approximately 1000 kg/m^3.
Weight of water displaced = Wdisplaced water = 3.3 N
Substituting these values into the equation, we can solve for the volume of water displaced:
1000 kg/m^3 * Volume of water = 3.3 N
Since 1 N = 1 kg⋅m/s^2, we can convert N to kg⋅m/s^2 to cancel out units:
1000 kg/m^3 * Volume of water = 3.3 kg⋅m/s^2
Dividing both sides of the equation by 1000 kg/m^3, we get:
Volume of water = (3.3 kg⋅m/s^2) / (1000 kg/m^3)
= 0.0033 m^3
Therefore, the volume of the paperweight is approximately 0.0033 cubic meters.