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Calculate the force exerted by oxygen at a regulator opening of 0.2 m2, given a safe use pressure of 1000 kPa.

Physics

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Answer

200,000 N\text{200,000 N}

Step-by-step solution

3 steps
  1. Identify the given values.

    The pressure of oxygen for safe use (PP) is 1000 kPa1000 \text{ kPa}.
    The area of the regulator opening (AA) is 0.2 m20.2 \text{ m}^2.

  2. Convert the pressure to Pascals (Pa) for consistency with SI units.

    1 kPa=1000 Pa1 \text{ kPa} = 1000 \text{ Pa} P=1000 kPa=1000×1000 Pa=1,000,000 PaP = 1000 \text{ kPa} = 1000 \times 1000 \text{ Pa} = 1,000,000 \text{ Pa}
  3. Calculate the force (FF) using the formula F=P×AF = P \times A.

    We assume the question implicitly asks for the force exerted by the oxygen at the regulator opening at the safe use pressure.

    F=P×AF = P \times A F=(1,000,000 Pa)×(0.2 m2)F = (1,000,000 \text{ Pa}) \times (0.2 \text{ m}^2) F=200,000 NF = 200,000 \text{ N}

    The force exerted by the oxygen at the regulator opening is 200,000 N\text{200,000 N}.

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