Calculate Depth of Floating Instrument: 0.219 kg in Water

AI Thread Summary
To calculate the depth of a floating hollow brass tube loaded with lead shot, the formula used is z = m/(πr²ρ), where m is the mass (0.219 kg), r is the radius (0.01615 m), and ρ is the density of water (1000 kg/m³). A misunderstanding arose regarding the variable "n," which was incorrectly interpreted; it should be "π" (pi), representing the mathematical constant. After clarifying this, the formula can be correctly applied to find the depth of the tube's bottom end in water. The discussion highlights the importance of accurately identifying variables in mathematical formulas. This correction allows for the proper calculation of the tube's submerged depth.
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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 just tried to plug it into see what would happen and it doesn't work any help?
 
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That's not"n", it's "pi" (3.1415926535...)

It came from the formula for the area of a circle.
 
thanx so mucn, silly mistake on my part
 
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