Buoyant force of an aluminum block

[domyy]

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 ?

 

[Doc Al]

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.)

 

[domyy]

Why do i have someone saying the answer should be 92.5 for a) ?

 

[Doc Al]

Why do i have someone saying the answer should be 92.5 for a) ?

They are expressing the volume in cm³ instead of m³.

 

[asteriatic]

The correct answer for b) is 2.44N.

The correct equation to use for calculating the buoyant force is Fb = ρVg, where ρ is the density of the fluid (in this case, water), V is the volume of the object, and g is the acceleration due to gravity (9.8 m/s^2). So, for this problem, the buoyant force would be:

Fb = (1000 kg/m^3)(9.25 x 10^-5 m^3)(9.8 m/s^2) = 2.44 N

This means that when the aluminum block is submerged in water, it will experience a buoyant force of 2.44 N, pushing it upwards. This is because the density of the block (2700 kg/m^3) is greater than the density of water (1000 kg/m^3), causing it to displace a volume of water equal to its own volume. This allows the block to float in water, with a portion of it submerged and a portion above the surface.