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Floating Instrument:

  • Thread starter DDS
  • Start date

DDS

171
0
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?
 

Chi Meson

Science Advisor
Homework Helper
1,767
10
That's not"n", it's "pi" (3.1415926535...)

It came from the formula for the area of a circle.
 

DDS

171
0
thanx so mucn, silly mistake on my part
 

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