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Archimedes' law (10 points)?
A piece of aluminum is attached to a string and suspended in a pool of oil with density 840 kg/m^3. If the apparent weight of the aluminum is 670 N, what is the volume of the aluminum?
Answer = 0.0368 m^3
Please show how to get to this answer, thank you.
2 Answers
- Anonymous8 years agoFavorite Answer
There must be a equilibrium, draw the piece of aluminium.
There are three forces acting on your piece of aluminium, earth's gravity, the apparent weight, and the buoyant force.
forces acting downward: earth's gravity : mass* gravity constant = Volume*density_aluminium* gravity constant
forces acting upward: string: 670 N and buoyant force = Volume (under water) * density_oil* gravity constant
The density of the aluminium is not given, but i found it online: 2700 (water 1000)
So when we write this al together we get:
2700*9.81* volume - 670 - 840*9.81*volume = 0
26487*volume - 670 - 8240.4* volume =0
volume = 0.0367 m^3
close enough :)
Source(s): I study for engineer - FlexLv 58 years ago
Weight = Density x Volume x Acceleration due to Gravity
W = ÏVg
V = W / Ïg
V = 670 /(840)(9.8)
V = 0.08139 m^3