Answer:
Part A
The volume of water displaced is 4.1887902 × 10⁻³ m³
Part B
The buoyant force is approximately 40.93 N
Explanation:
From the question, we have;
The radius of the ball suspended (barely floating) in the water, r = 0.1 m
The density of the water, ρ = 997 kg/m³
Part A
The volume of the ball = The volume of a sphere = (4/3)·π·r³
∴ The volume of the ball = (4/3) × π × 0.1³ = 0.0041887902 m³ = 4.1887902 × 10⁻³ m³
Therefore;
The volume of water displaced, V = The volume of the ball = 4.1887902 × 10⁻³ m³
The volume of water displaced, V = 4.1887902 × 10⁻³ m³
Part B
The buoyant force = The weight of the water displaced = Mass of the water, m × The acceleration due to gravity, g
The buoyant force = m × g
Where;
g ≈ 9.8 m/s²
The mass of the water, m = ρ × V
∴ m = 997 kg/m³ × 4.1887902 × 10⁻³ m³ = 4.17622383 kg
The buoyant force = 4.17622383 kg × 9.8 m/s² ≈ 40.93 N.
