Question
I can't seem to figure out this problem...what equation would I use? Thanks!
An incompressible, nonviscous fluid is initially at rest in the vertical portion of the pipe, where L = 2.17 m. When the valve is opened, the fluid flows into the horizontal section of the pipe. What is the speed of the fluid when all of it is in the horizontal section? Assume that the cross-sectional area of the entire pipe is constant.
An incompressible, nonviscous fluid is initially at rest in the vertical portion of the pipe, where L = 2.17 m. When the valve is opened, the fluid flows into the horizontal section of the pipe. What is the speed of the fluid when all of it is in the horizontal section? Assume that the cross-sectional area of the entire pipe is constant.
Answers
So, V^2=g*L
V=sqrt 9.8 * 2.17
V=sqrt 9.8 * 2.17
v_2=√gL=√((9.8)(2))=4.43 m/s
Related Questions
I got stuck on this problem and I cannot figure out what to do or explain this problem. Please help...
Let F,G be vector fields on the plane such that both com-positions F dot G, G dot F : R^2 -> R^2...
Thanks Reiny,that is my only problem trying to figure out the equation. I just cannot figure out for...
Bernoulli's equation for steady motion to an incompressible non-viscous fluid is given by p+1/2pv^2+...