Buoyant force of an aluminum block

AI Thread Summary
The discussion focuses on calculating the buoyant force on a submerged aluminum block with a mass of 0.25 kg and a density of 2700 kg/m³. The volume of the block is calculated as 9.25 x 10^-5 m³, which is equivalent to 92.5 cm³ when converted. For the buoyant force, the user attempts to calculate it using the formula involving the volume and the density of water, but there is confusion regarding units. The correct buoyant force calculation should yield approximately 0.0925 N, considering the gravitational acceleration. The discrepancy in volume arises from differing unit interpretations, with some users expressing it in cm³ instead of m³.
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Homework Statement



An aluminum block has a mass of 0.25kg and a density of 2700 kg/m cubic.

a) calculate the volume of the aluminum block.
b) Determine the buoyant force exerted on this block when it´s completely submerged in water of density 1000kg/m cubic.

Homework Equations



v = m/d

The Attempt at a Solution



I already solved for a and I got:

v = 0.25/2700 = 9.25 x 10 to the power of -5

Now for b), I am not sure... i did something like:

9.25 x 10-5 x 1000 x 10
9.25 x 10-5 x 1.0 to the power of 3 x 10
9.25 x 10-1 ?
 
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domyy said:
I already solved for a and I got:

v = 0.25/2700 = 9.25 x 10 to the power of -5
OK. (But don't leave out the units.)

Now for b), I am not sure... i did something like:

9.25 x 10-5 x 1000 x 10
9.25 x 10-5 x 1.0 to the power of 3 x 10
9.25 x 10-1 ?
If the block were submerged in water, that would be about right. (Using g = 10 m/s2.) (Again, don't leave out the units.)
 
Why do i have someone saying the answer should be 92.5 for a) ?
 
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domyy said:
Why do i have someone saying the answer should be 92.5 for a) ?
They are expressing the volume in cm3 instead of m3.
 
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