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2m below the surface of water of density 1000kgm-3. If the atmospheric pressure is 10 Pa, calculate the pressure on the diver. (g = 10ms−2)

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2 parts
1

116960 Pa\text{116960 Pa}

58480 kg/m3\text{58480 kg/m}^3

2

152000 Pa\text{152000 Pa}

Step-by-step solution

8 steps

Here are the solutions to your physics problems.

Part 1· 4 steps

The question asks to calculate the pressure on a membrane and then, based on an interpretation of the unclear second part, to calculate the density of a liquid in tube B.

Given:

  • Depth of membrane in mercury, hm=10 cm=0.10 mh_m = 10 \text{ cm} = 0.10 \text{ m}
  • Atmospheric pressure height in mercury, hatm,Hg=76 cm=0.76 mh_{atm, Hg} = 76 \text{ cm} = 0.76 \text{ m}
  • Density of mercury, ρHg=13600 kg/m3\rho_{Hg} = 13600 \text{ kg/m}^3
  • Height of liquid in tube B, hB=20.0 cm=0.20 mh_B = 20.0 \text{ cm} = 0.20 \text{ m}
  • Acceleration due to gravity, g=10 m/s2g = 10 \text{ m/s}^2 (assumed from the second question)

  1. Calculate the atmospheric pressure (PatmP_{atm}).

    The atmospheric pressure is equivalent to the pressure exerted by a 76 cm column of mercury.

    Patm=ρHgghatm,HgP_{atm} = \rho_{Hg} g h_{atm, Hg} Patm=(13600 kg/m3)(10 m/s2)(0.76 m)P_{atm} = (13600 \text{ kg/m}^3)(10 \text{ m/s}^2)(0.76 \text{ m}) Patm=103360 PaP_{atm} = 103360 \text{ Pa}

  2. Calculate the pressure due to the 10 cm column of mercury above the membrane (Pdepth,HgP_{depth, Hg}).

    Pdepth,Hg=ρHgghmP_{depth, Hg} = \rho_{Hg} g h_m Pdepth,Hg=(13600 kg/m3)(10 m/s2)(0.10 m)P_{depth, Hg} = (13600 \text{ kg/m}^3)(10 \text{ m/s}^2)(0.10 \text{ m}) Pdepth,Hg=13600 PaP_{depth, Hg} = 13600 \text{ Pa}

  3. Calculate the total pressure on the membrane (PmembraneP_{membrane}).

    The total pressure on the membrane is the sum of the atmospheric pressure and the pressure due to the mercury column above it.

    Pmembrane=Patm+Pdepth,HgP_{membrane} = P_{atm} + P_{depth, Hg} Pmembrane=103360 Pa+13600 PaP_{membrane} = 103360 \text{ Pa} + 13600 \text{ Pa} Pmembrane=116960 PaP_{membrane} = 116960 \text{ Pa}

  4. Calculate the density of the liquid in tube B (ρB\rho_B).

    The question states that this pressure is balanced by a liquid in tube B with a height of 20.0 cm. Given the phrasing "calculate the height of the liquid in tube B" while also stating its height is 20.0 cm, we will assume the question intends to ask for the density of the liquid in tube B.

    Pmembrane=ρBghBP_{membrane} = \rho_B g h_B ρB=PmembraneghB\rho_B = \frac{P_{membrane}}{g h_B} ρB=116960 Pa(10 m/s2)(0.20 m)\rho_B = \frac{116960 \text{ Pa}}{(10 \text{ m/s}^2)(0.20 \text{ m})} ρB=1169602 kg/m3\rho_B = \frac{116960}{2} \text{ kg/m}^3 ρB=58480 kg/m3\rho_B = 58480 \text{ kg/m}^3

    The pressure on the membrane is 116960 Pa\boxed{\text{116960 Pa}}.
    The density of the liquid in tube B is 58480 kg/m3\boxed{\text{58480 kg/m}^3}.

Part 2

152000 Pa\text{152000 Pa}

The question asks to calculate the total pressure on a diver.

Given:

  • Depth of diver, h=5.2 mh = 5.2 \text{ m}
  • Density of water, ρw=1000 kg/m3\rho_w = 1000 \text{ kg/m}^3
  • Atmospheric pressure, Patm=105 PaP_{atm} = 10^5 \text{ Pa}
  • Acceleration due to gravity, g=10 m/s2g = 10 \text{ m/s}^2

  1. Identify the formula for total pressure.

    The total pressure on the diver is the sum of the atmospheric pressure and the pressure due to the column of water above the diver.

    Ptotal=Patm+ρwghP_{total} = P_{atm} + \rho_w g h

  2. Substitute the given values into the formula.

    Ptotal=105 Pa+(1000 kg/m3)(10 m/s2)(5.2 m)P_{total} = 10^5 \text{ Pa} + (1000 \text{ kg/m}^3)(10 \text{ m/s}^2)(5.2 \text{ m})

  3. Calculate the pressure due to the water column.

    Pwater=(1000)(10)(5.2)=52000 PaP_{water} = (1000)(10)(5.2) = 52000 \text{ Pa}

  4. Add the atmospheric pressure to find the total pressure.

    Ptotal=100000 Pa+52000 PaP_{total} = 100000 \text{ Pa} + 52000 \text{ Pa} Ptotal=152000 PaP_{total} = 152000 \text{ Pa}

    The pressure on the diver is 152000 Pa\boxed{\text{152000 Pa}}.

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