Water flows at a rate of 25L/min through a horizontal 8.0-cm-diameter pipe under a pressure of 5.7Pa . At one point, calcium deposits reduce the cross-sectional area of the pipe to 40cm2 .

Question

Part A
What is the pressure at this point? (Consider the water to be an ideal fluid.)

Answer 1

I get really irritated when a student types in bogus titles. this has nothing to do with buoyance, does it?

Specifically, it deals with Bernoulli's equation, density remains constant, and mass flow is conserved (mass flow in = mass flow out).
http://scienceworld.wolfram.com/physics/BernoullisLaw.html
P+1/2 rho*v^2+rho g h=constant

Not that is it always needed, but it is safe to change units to SI.
25liters=25l(1m^3/1000l)=.025m^2
1min= 60 sec

so 25l/min= .025/60 m^3/sec

Now, have two sides to work with:
Pi+1/2 rho*vi^2+rho g h=Pf+1/2 rho*vf^2+rho g h

the rho g h term can be lost frm both sides, you have Pi, vi, you know rho.

But vf: use the law of mass continuity
Vi*areainput=Vf*areaoutput solve for vf
Now, finally, solve for Pf