What is the absolute pressure at a depth of 9.91 m below the surface of a deep lake? Assume atmospheric pressure is 1.01×105 Pa .

Respuesta :

Answer:

P = 198.118 kPa

Explanation:

Given:

Atmospheric pressure = P[tex]_{atm\\}[/tex] = 1.01×10⁵

depth = h = 9.91 m

To find:

Absolute pressure P[tex]_{abs}[/tex]

Solution:

Density of water = ρ = 1.000x10 ³kg/m ³

acceleration due to gravity = ρ = 9.8 m/s²

P[tex]_{abs}[/tex] = P[tex]_{atm\\}[/tex] + ρgh

      = 1.01×10⁵ + 1.000x10 ³x 9.8 x 9.91

      = 101000 + 1000(9.8)(9.91)

      = 101000 + 97118

      = 198118 Pa

      = 198.118 kPa

P[tex]_{abs}[/tex] = 198.118 kPa

Lanuel

The absolute pressure at a depth below the surface of this deep lake is 198.118 kPa.

Given the following data:

  • Atmospheric pressure = [tex]1.01 \times 10^5 \;Pa[/tex]
  • Height (depth) = 9.91 meters.

Scientific data:

  • Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]
  • Density of water = 1000 [tex]kg/m^3[/tex]

To calculate the absolute pressure at a depth below the surface of a deep lake:

Mathematically, absolute pressure is given by this formula:

[tex]P_{abs} = P + \rho gh[/tex]

Substituting the given parameters into the formula, we have;

[tex]P_{abs} = 1.01 \times 10^5 + 1000 \times 9.8 \times 9.91 \\\\ P_{abs} = 101000+ 97118 \\\\[/tex]

Absolute pressure = 198118 Pa

Note: 1 kPa = 1000 Pa

Absolute pressure = 198.118 kPa

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