A person with a density of 980 kg/m3 floats in seawater of density 1025 kg/m3. What fraction of the person is submerged?
weight=buoyant force
mg= F
d₁•V•g =k•d₂•V•g
k= d₁/d₂ =980/1025 =0.956
To find the fraction of the person that is submerged, we need to compare their density with the density of the seawater.
Let's start by calculating the buoyant force acting on the person. The buoyant force is equal to the weight of the displaced liquid.
First, we need to calculate the volume of water displaced by the person. We can use the Archimedes' principle, which states that the weight of the displaced liquid is equal to the buoyant force acting on the object.
Let's assume the person has a mass of 1 kg. We can calculate the volume of water displaced using the formula:
Volume = Mass / Density
The volume of the person is given by:
Volume = 1 kg / 980 kg/m^3
Volume = 0.00102 m^3
Now, we can calculate the buoyant force using the formula:
Buoyant force = Density of seawater * Volume of the person * Acceleration due to gravity
Buoyant force = 1025 kg/m^3 * 0.00102 m^3 * 9.8 m/s^2
Buoyant force = 10.0124 N (approximately)
Comparing this with the weight of the person, we can determine the fraction that is submerged.
Weight of the person = Mass of the person * Acceleration due to gravity
Weight of the person = 1 kg * 9.8 m/s^2
Weight of the person = 9.8 N
Fraction submerged = Buoyant force / Weight of the person
Fraction submerged = 10.0124 N / 9.8 N
Fraction submerged = 1.022 (approximately)
Therefore, the person is approximately 1.022 times submerged in seawater.