I know Volume would equal .09x.09xh, but wouldn't this height be referring to be side of the 1.3m? aren't we looking for the part of the .09 that is submerged? the post seems to be laying horizontally in the water, not straight up.
Also, when I reduce and isolate the height, I obtain .531m or...
Greater than the gravitational force or m x g. But on the table. the force acting on the post is equal to that of the post acting on the table. In other words, the gravitational force is equal to the normal force. Am I wrong to say that the buoyant force must be greater than the gravitational...
Ok, Archimedes's principle states that the magnitude of the buoyant force=weight of the fluid displaced by object
Fb=density of fluid x volume of object submerged x g
=1029 x (.09 x .09 x 1.3) x 9.8
I don't see where to go next.
A 9 cm square wooden post (density=420kg/m3) is 1.3m long and floats in sweater (density=1029 kg/m3. How deep is the post submerged in the water? The picture shows a 9cm x 9cm cross section of the post.
Fb=density of liquid x volume of liquid submerged x...
The tuna's density must be the same as the surrounding water. Would we do this:
absolute pressure of water=density of water x gravity x depth and solve for density of water?
we could then equate density=m/v and we have the mass of the fish.
2.44atm=density x 9.8 x 15m
Since we are given the pressure at the depth, is it redundant to include it in the equation?
ok, I see what you mean about water not changing density with depth. But how do we solve for the fish volume without dealing with the tuna's density?
I guess I do not understand the equation well. In P=pgh, density would be referring to that of the object, Pressure would be referring to that of the incompressible fluid, and h the height of the column of fluid above location. This is what I have in my notes. If we multiply the tuna's density...