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Pressure in a tornado and net force on a door

  1. Jan 13, 2005 #1
    In a tornado, the pressure can be 15% below normal atmospheric pressure. Seomtimes a tornado can move so quickly that this pressure drop can occur in one second. Suppose a tornado suddnely occured outside your front door, whic is 182cm high and 91cm wide. what net force would be exerted on the door? in what direction would the force be exerted?

    Answer: Fnet = Foutside-Finside

    = (Poutside-pinside ) x a
    = (0.85x10^5 Pa - 1x10^5 Pa ) x ( 1.82m)(0.91m)
    = - 2.5x10^4 N( toward the outside )

    The above answer is written at the back of the text book for the question typed above. However, i dont undersatnd where did they get the figures of .85x10^5 Pa and 1x10^5 Pa from..... can anyone please temme what those figures are?

    thanks much
     
  2. jcsd
  3. Jan 13, 2005 #2

    dextercioby

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    "Pa" apud "Pascal" (apud Blaise Pascal,French phylosopher & physicist in the XVII-th century) is the SI unit for pressure.
    In those units,atmospheric pressure is 101325Pa which can be approximated to 10^{5}.

    Daniel.
     
  4. Jan 13, 2005 #3
    so where did they get the figure 0.85x10^5 from ?? and how does tyhe textbook assume that the atmopsheric pressure inside the house and outside it is different without evidence from the question?

    thanks
     
  5. Jan 13, 2005 #4

    dextercioby

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    It says that the tornado exerts a pressure 15% less than the atmospheric pressure (presumably found inside the house),so there's how they got that number.

    Daniel.
     
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