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anonymous

  • one year ago

A cylindrical disk of wood weighing 48.0 N and having a diameter of 30 cm floats on a cylinder of oil of density 0.850 g/cm^3 (the figure ). The cylinder of oil is 75.0 cm deep and has a diameter the same as that of the wood. What is the gauge pressure at the top of the oil column? Suppose now that someone puts a weight of 95.0 on top of the wood, but no oil seeps around the edge of the wood. What is the change in pressure at the bottom of the oil? What is the change in pressure at the halfway down in the oil?

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  1. IrishBoy123
    • one year ago
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    what are you learning? basic pressure/force/area or bernoulli?

  2. rajat97
    • one year ago
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    I think that the gauge pressure of a level of liquid is just the pressure difference between the pressure at that layer and the pressure at the topmost layer of the liquid (i.e. \[\rho g h\]) so the gauge pressure at the top of the liquid will be zero. Next, the pressure at the bottom of the liquid will be \[P=P _{atm} + Pressure due \to the wooden block and the weight+ \rho g h\] where P atm is the atmospheric pressure So, i get the answer 1.0688 x 10^5 Pa. I hope this helps you

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