Final answer:
The area of the Styrofoam slab that allows it to float awash in freshwater with a swimmer on top is 0.1072 m², which is calculated using Archimedes' Principle and the densities of the materials involved.
Step-by-step explanation:
The question asks about calculating the area of a Styrofoam slab that floats on water's surface with a swimmer on top by applying Archimedes' Principle. According to this principle, the weight of the water displaced by the submerged part of the slab must equal the total weight of the slab and the swimmer. The density of the object (Styrofoam) is given as 500 kg/m3, and the thickness of the slab is 10.7 cm, which is 0.107 m.
Freshwater's density is approximately 1000 kg/m3. To find the area of the slab, we must first calculate the volume of water that the slab and the swimmer displace. Since the slab is just awash, it displaces its own volume plus the volume of water equivalent to the swimmer's weight. The volume of water displaced is equal to the volume of the slab (which we can denote as V), and for the swimmer, it is their mass divided by the density of water (53.6 kg / 1000 kg/m3).
The volume of the slab (V) can be found with the formula V = m / rho object, where m is the mass of the slab. Since the slab is awash, it must displace a weight equal to its weight plus that of the swimmer. So, if we let A represent the area and use 'g' for gravitational acceleration (approximately 9.81 m/s2):
A * h * rho object * g + swimmer mass * g = A * h * 1000 kg/m3 * g.
We can cancel out 'g' and 'h' to solve for the area (A):
A * rho object + swimmer mass = A * 1000 kg/m3,
A = swimmer mass / (1000 kg/m3 - rho object).
Substituting the swimmer's mass and the densities, we get:
A = 53.6 kg / (1000 kg/m3 - 500 kg/m3) = 53.6 kg / 500 kg/m3 = 0.1072 m2.
Therefore, the area of the slab is 0.1072 m2.