Final answer:
The seawater density at the given pressure is 0 psi/ft.
Step-by-step explanation:
The density of seawater depends on various factors such as temperature, pressure, and salinity. However, for this question, we can assume a density of 1025 kg/m³ for seawater. To calculate the seawater density at the given pressure, we need to convert the specific weight to the equivalent pressure.
Specific weight (G) is defined as the weight per unit volume. In this case, it is given as 63.6 lbf/ft³. To convert this to a pressure (P), we can use the equation P = G/g, where g is the acceleration due to gravity (32.2 ft/s²).
Using the above conversion, we can calculate the pressure P = 63.6 lbf/ft³ / 32.2 ft/s² = 1.976 psi (pound per square inch).
Now, we can use the equation for pressure at a given depth in a fluid to find the density. The equation is: P = ρgh, where ρ is the density, g is the acceleration due to gravity, and h is the depth of the fluid.
In this case, we assume the depth is at the surface of the seawater, so h = 0. Using the given pressure P = 1.976 psi and rearranging the equation, we can solve for the density:
ρ = P / (g * h) = 1.976 psi / (32.2 ft/s² * 0 ft) = 0 psi/ft.
Therefore, at this specific point with the given pressure, the seawater density is 0 psi/ft.