For #4 the answer key, when calculating the Buoyant Force on the iceberg, why is the density of the ice used instead of the sea water?

Christine,

Since the ice block is floating, and not moving, the buoyant force equals the gravitation force on the block. 

F sub mg - F sub buoyant = 0

The gravitation force is (density ice)(Volume ice)g.

John

Hi all, I am struggling with the solution to 5 part a. Surly the answer that the velocity in the pipe is directly related to the area only works for when the pipe has one inlet and one outlet. Imagine a situation where the two outlet pipes have areas slightly smaller than the inlet pipe. In this case the pipe with the largest area will have the largest velocity because mass flow rate needs to be conserved. I do not think it is this simple.

Christopher, we can add the two pipes so A1v1 = A2v2 + A3v3 does work. However, we need to modify the problem - the two exit pipes should be at the same level - if one pipe is higher than the other, we have to add in Bernoulli as it takes energy to raise the water level to the higher pipe. Taking an extreme - if the top pipe is 20 above the source pipe, then no water will get up there - or at the very least, it would be very slow.

John

Thanks John, I was worried about that too. A1v1=A2v2+A3V3 indeed does work however I still do not believe that the jump to smallest area, largest velocity can be taken unless you compare the total areas before and after. In this case the total area after is smaller so that the velocities will naturally be faster on the right however in a world where the total area after is larger, I believe the velocities will be slower even if the diameter of each pipe is smaller. If one pipe split into 2 smaller equally sized pipes but the total area before and after did not change the water should not change speeds. Even if the pipes on end were not the same area I am not convinced that one would be faster than the other if the total area before and after does not change. Too many unknown variables.

I just want to ensure I'm reading your comment correctly - for the problem as stated, the answer is correct, but the explanation isn't quite right for the case where the area of the supply pipe is equal to the area of the two exit pipes. If I got that right, I will edit the response.

John