TorMcOst
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I have a bolt (13 kg) of stainless steel (1650 cm``). How much will this weight in sea water?
Thank you in advance!
Thank you in advance!
The discussion revolves around calculating the weight of a stainless steel bolt when submerged in seawater, utilizing Archimedes' principle to understand buoyancy effects. The original poster presents a specific weight and volume for the bolt and seeks clarification on the calculation process.
Participants have provided guidance on the calculation method and engaged in clarifying the distinction between mass and weight. There is an ongoing exploration of the correct interpretation of the values presented, with some participants expressing agreement on the calculations while others raise questions about unit usage.
There is a noted confusion regarding the use of kilograms as a unit of weight versus mass, and participants emphasize the importance of precision in terminology within the context of physics discussions.
Does one mean 13 (13.000) kg, or perhaps 13,000 g? kg (g) is a unit of mass, N (Newton) or dyne (used in cgs system) is a measure of weight. Mass * acceleration of gravity would give weight (force) due to gravity.TorMcOst said:Thanks a lot for your respond Born2bwire!
This means that my calculation will become something like this:
Weight in Air: 13,000 kg
- Weight of fluid displaced (1650 cm3*1,025): 1,691 kg
= Weight in Sea Water: 11,309 kg
Correct?
TorMcOst said:Thanks Astronuc!
However I got a little confused. By 13,000kg I meant 13 kg (in air). And my following answer was that this bolt would weight 11,3kg in water. Do you agree or disagree with this calculation?
As Born2bwire indicated, I was referring to the use of mass as weight. Being a physics forum, we wish to be accurate in such a matter.TorMcOst said:Thanks Astronuc!
However I got a little confused. By 13,000kg I meant 13 kg (in air). And my following answer was that this bolt would weight 11,3kg in water. Do you agree or disagree with this calculation?