Final answer:
Using Archimedes' principle to find the buoyant force, the apparent weight of the iron anchor when submerged is calculated. With the cable breaking at 5000 N, it is found that the anchor's submerged weight surpasses the cable's limit before it is completely out of the water, meaning 0% of the anchor would be out of the water when the cable breaks.
Step-by-step explanation:
To determine what percentage of an iron anchor's weight will be supported by buoyant force when submerged in saltwater, we can use Archimedes' principle, which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. First, we calculate the weight of the anchor in air by using the formula Weight = Mass × Gravity, which gives us 560 kg × 9.8 m/s² = 5488 N.
Next, to find the buoyant force, we determine the volume of the anchor using its mass and the density of iron. The volume V is Mass / Density, which gives us Volume = 560 kg / 7800 kg/m³ = 0.07179 m³. The buoyant force is then the volume of the anchor times the density of seawater times gravity, or Buoyant Force = 0.07179 m³ × 1025 kg/m³ × 9.8 m/s² = 725.3 N.
The anchor will be fully submerged, and its apparent weight will be the actual weight minus the buoyant force, which is 5488 N - 725.3 N = 4762.7 N. If the cable breaks at 5000 N, that means the cable supports 5000 N / 4762.7 N × 100% ≈ 105% of the submerged weight. Since it cannot support more than 100%, the anchor is already at the point of breaking the cable before being completely out of the water. Therefore, the answer is that 0% of the anchor will be out of the water when the cable breaks; hence none of the options (a, b, c, d) are correct.