Recent content by tennisgirl92

Got it! THANK you! You have been so helpful

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...

So setting it equal... 1029 x Vdw x 9.8=420 x (.09 x .09 x 1.3) x 9.8 Vdw=.0043 m3 How do we find the height from this? Do we subtract it from the volume of the post?

Fb=densityw x Volumedw x g =1029 x Vdw x 9.8 would we set this equal to the density of the post x Volume of post x g?

Ok, it can either be densitywaterx Volumepost x g=masspostxg or densitywaterx Volumepost x g=densitypost x Volumewater x g does this look right?

I believe the buoyant force is equal to the gravitational force, as the object is at rest.

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...

If the post is floating, wouldn't the buoyant force have to be greater? Would we use the density of the wood on the water to find it instead of the density of the water?

ha! oh yes-I did mean water. Sorry. Ok, no the wood is not entirely submerged. But the problem does not given us the percent it is submerged-how do we find that?

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 =106.187 N I don't see where to go next.

Homework Statement 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. Homework Equations 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 density=.01698...

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...

Ok, I think that would mean the body would have the same density as the fluid. But why does the above mean the tuna is reaching all the way to the surface? Are we calling where the tuna is as h=0?