The absolute pressure experienced by porpoises at a depth of 500 meters, with a seawater density of 1025 kg/m³, is calculated using the pressure formula P = P0 + ρgh. The result is 6,123,825 N/m² or roughly 60.53 atmospheres.
Step-by-step explanation:
To calculate the absolute pressure experienced by porpoises at a depth of 500 meters in the ocean, where the density of seawater is 1025 kg/m³ and assuming standard atmospheric pressure at the ocean surface, we use the formula:
P = P0 + ρgh
where P is the absolute pressure, P0 is the atmospheric pressure at the surface (1 atmosphere or 1 atm, which is 101,325 N/m²), ρ (rho) is the density of seawater, g is the acceleration due to gravity (9.8 m/s²), and h is the depth (500 meters in this case).
First, we need to convert atmospheric pressure to pascals:
P0 = 1 atm = 101,325 N/m²
Then we can calculate the pressure due to the water column:
P = 101,325 N/m² + (1025 kg/m³)(9.8 m/s²)(500 m)
P = 101,325 N/m² + (1025 kg/m³)(9.8 m/s²)(500 m)
P = 101,325 N/m² + 5,022,500 N/m²
P = 6,123,825 N/m²
Therefore, the absolute pressure experienced by porpoises at this depth is 6,123,825 N/m², or roughly 60.53 atm.
The probable question can be: "Porpoises, relatives of dolphins and whales, exhibit remarkable diving capabilities, reaching depths of 500 meters in the ocean. Considering the density of seawater as 1025 kg/m³, calculate the absolute pressure experienced by the porpoises at this depth. Assume standard atmospheric pressure at the ocean surface."