At what depth below the sea-level would one experience a change of pressure equal to one atmosphere? Density of sea water = 1013kg-3, one atmosphere = 0.01 ...

The pressure at a given depth below the sea level can be calculated using the equation of hydrostatic pressure, which states that the pressure at a point in a fluid is equal to the weight of the fluid above it. In this case, the density of sea water is given as 1013 kg/m^3, and the acceleration due to gravity is given as 10 m/s^2. The pressure at a depth of h below the sea level can be calculated as: P = ρ * g * h where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth below the sea level. We are looking for the depth at which the pressure changes by one atmosphere, which is equal to 0.01 x 10^5 N/m^2. So, we can set up an equation to solve for h: 0.01 x 10^5 N/m^2 = ρ * g * h Solving for h, we get: h = 0.01 x 10^5 N/m^2 / (ρ * g) = 0.01 x 10^5 N/m^2 / (1013 kg/m^3 * 10 m/s^2) = 10.0 m So, the depth below the sea level at which the pressure changes by one atmosphere is 10.0 m.