Calculate Depth of Floating Instrument: 0.219 kg in Water

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SUMMARY

The discussion centers on calculating the depth of a floating hollow brass tube loaded with lead shot, with a total mass of 0.219 kg, in water. The formula derived for depth, z, 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 common mistake identified was confusing the symbol "π" with "n," which led to confusion in the calculations. The correct application of the formula yields the depth of the tube's bottom end when floated in water.

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  • Familiarity with the formula for the area of a circle (A = πr²).
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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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