A girl is floating in a freshwater lake with her head just above the water. If she weighs 395 N, what is the volume of the submerged part of her body?
To find the volume of the submerged part of the girl's body, 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 the object. Given that she weighs 395 N, the buoyant force acting on her in water is also 395 N, since she is floating with her head just above the water.
The buoyant force can be calculated using the formula:
Buoyant force = Weight of the fluid displaced
We can rearrange this equation to find the displaced volume:
Displaced volume = Buoyant force / Density of the fluid
In this case, the fluid is water, and the density of water is approximately 1000 kg/m^3.
First, let's convert the girl's weight from newtons (N) to kilograms (kg) using the acceleration due to gravity (g ≈ 9.8 m/s^2):
Weight = Mass × Acceleration due to gravity
395 N = Mass × 9.8 m/s^2
Mass ≈ (395 N) / (9.8 m/s^2)
Mass ≈ 40.31 kg
Now, we can calculate the volume of the submerged part of her body:
Displaced volume = Buoyant force / Density of water
Displaced volume ≈ (395 N) / (1000 kg/m^3)
Displaced volume ≈ 0.395 m^3
Therefore, the volume of the submerged part of the girl's body is approximately 0.395 cubic meters.