SAMMM asked in Science & MathematicsPhysics · 2 months ago

# Calculate the depth in the ocean at which the?

pressure is three times atmospheric pressure.

Atmospheric pressure is 1.013 × 105 Pa. The

acceleration of gravity is 9.81 m/s

2

and the

density of sea water is 1025 kg/m3

.

Answer in units of m.

Relevance
• Anonymous
2 months ago

if such pressure is the hydraulic pressure :

3*1.013*10^5 = g*h*ρw = 9.81*h*1025

h = (3*1.013*10^5) / (9.81*1025) = 30.22 m

if such pressure is the absolute pressure :

2*1.013*10^5 = g*h'*ρw = 9.81*h'*1025

h' = h*2/3 = 20.15 m

• 2 months ago

Think of a column of water that measures 1cm by 1cm and has a height of x cm

Pressure = Force / Area

Force = Mass * acceleration

Density = Mass / Volume

Volume = Area_Base * Height

Density * Volume = Mass

Density * Volume * Acceleration = Force

Density * Volume * Acceleration / Area = Pressure

Density * Area_Base * height * Acceleration / Area_Base = Pressure

Density * Acceleration * height = P

Remember, you have the air pressing down on the water with a pressure of 1.013 * 10^5 Pascals, so when you're feel 3 atm of pressure, you're really feeling 1 atm from the air and 2 atm from the water

P = 2 * 1.013 * 10^5 Pascals

D = 1025 kg/m^3

A = 9.81 kg/s^2

1025 kg/m^3 * 9.81 kg/s^2 * h m = 2.026 * 10^5 Pa

h kg * m / (m^3 * s^2) = 2.026 * 10^5 Pa / (1025 * 9.81)

1 Pascal = kg/(m * s^2)

h kg / (m^2 * s^2)  =  220600 / (1025 * 9.81)  kg/(m * s^2)

h = ‭21.938788195221401755301956689292‬...

21.94 meters, roughly.