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## Homework Statement

An ideal liquid flows horizontally through a pipe of cross sectional area 3cm

^{2}with a velocity 1ms

^{-1}. The pipe narrows to a cross sectional area 1cm

^{2}. 2 vertical pipes are connected to the pipe, one in either region.

Calculate the height difference of the liquid between the 2 vertical pipes.

## Homework Equations

p

_{1}+ 0.5ρv

_{1}

^{2}+ ρgh

_{1}= p

_{2}+ 0.5ρv

_{2}

^{2}+ ρgh

_{2}

A

_{1}v

_{1}= A

_{2}v

_{2}

## The Attempt at a Solution

I think i'm a bit confused on what my variables represent in bernoullis equation:

Using the continuity equation gives v

_{2}= 3ms

^{-1}.

Then I said that for the pipes p

_{1}= p

_{2}(because they're both open to the atmosphere), putting everything into bernoullis equation and rearranging gives

h

_{1}- h

_{2}= 0.4 m

The book says the answer is 4x10

^{-3}m.

Am I not allowed to consider the heights of the fluid in the vertical pipe because I calculated the speeds of the fluid in the horizontal pipe? I can't think of any other way to do the problem.