calculate the increase in pressure at a depth of 2 m below the surface of water of density 1000 kg/m³​

Answers 2

Answer:

increase in pressure

δP = P(2m below) - P(atm)

= ρhg + Patm - Patm

= ρhg

= 1000×2×9.8

= 19600 N\m² or Pa

Answer:

19600 Pa

Explanation:

- To Calculate increase in pressure at a depth or in other words at the bottom of a liquid, The Formulae used is: P = hdg OR P = hpg.

- In the formulae :

P = stands for Pressure.

h = Stands for Depth.

d Or p (rho) = Stands for density.

And lastly g = stands for gravitational field strength. (which is equal to 9.8 or 10 rounded up.)

- For calculation:-

P = hdg

P = 2 x 9.8 x 1000

P = 19600 Pa

-Why Pascals? Because when Depth (h) is in m and density (p or d) is in kg/m³ Then pressure will be in Pascals (Pa).

Note: I know I made the answer seem long and complicated But To me when I see answers like this, It helps me understand better and so I hope you do too.

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