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Homework Help: Floating Instrument:

  1. Jun 10, 2005 #1

    DDS

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    A hollow brass tube (diameter = 3.23 cm) is sealed at one end and loaded with lead shot to give a total mass of 0.219 kg. When the tube is floated in pure water, what is the depth, z, of its bottom end?

    through manipulation of some laws i have broken it down to this stage:

    z = m/(πr2ρ)

    where:

    m=0.219 kg
    r=0.0323/2=0.01615m
    p=1000 kg/m^3

    thus i have :

    z= 0.219/ n(0.01615)2(1000)

    but what the hell is n.... i know its not the incidencs for water because nothing is getting reflected and i jsut tried to plug it in to see what would happen and it doesnt work any help?
     
  2. jcsd
  3. Jun 10, 2005 #2

    Chi Meson

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    That's not"n", it's "pi" (3.1415926535...)

    It came from the formula for the area of a circle.
     
  4. Jun 10, 2005 #3

    DDS

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    thanx so mucn, silly mistake on my part
     
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