Are the units of volume compatible with the density units?

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SUMMARY

The forum discussion centers on calculating the volume and density of an object submerged in two different fluids: water (density 1000 kg/m³) and oil (density 830 kg/m³). The object weighs 34N in water and 34.7N in oil. Using Archimedes' principle, participants derive the object's volume and density through the buoyant force equations. The final calculations yield a volume of approximately 3.625 x 10^-4 m³ and a density of about 9.3 g/cm³, highlighting the importance of consistent unit usage and careful application of fluid mechanics principles.

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  • #31
sorry 362.5 cm^3 which would fit into a .0958 gal aquarium
 
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  • #32
now I need to find density
p = m/v What mass should I use? the weight of the scale for water, oil, or either?
 
  • #33
nick gear said:
so here goes... I know its been a while, been wrapping up this semester. But this is what I came up with.

V= (34.7 N - 34 N) / (9.8 m/s^2) * (1000 - 803)
V= 3.625 x 10^ - 4 m^3 = .3681 L = 368.1 cm^3
You used 803 kg/m3 as the density of the oil instead of the correct density of 830 kg/m3. Your calculated volume is wrong.

It's important to check your work for silly copying mistakes. It saves hours of pulling your hair out, wondering why you don't get the right answer.
 
  • #34
nick gear said:
now I need to find density
p = m/v What mass should I use? the weight of the scale for water, oil, or either?
You've calculated the density of the object and its volume. The only thing left to find is the mass.
 
  • #35
m= p * v => m = 9.3 g/cm^3 * 4.201 x 10 ^ -4 = .0039
 
  • #36
do I need to make some unit conversions because this number seems way off to me?
 
  • #37
nick gear said:
m= p * v => m = 9.3 g/cm^3 * 4.201 x 10 ^ -4 = .0039
Just as with mistakenly copying numbers between calculations, ignoring units can get you into trouble just as quickly.

What are the units of volume here? V = 4.201×10-4 what exactly? Are these units compatible with the units in the density of the block?
 

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