Boo Physics

johnjohn320

Diamond Member
Jan 9, 2001
7,572
2
76
Small Question:

A venturi meter has a cross section of 40.0cm^2 at its entry and exit ports and a cross section of 25.0 cm^2 in its constricted reigon. Water enters at 3.5m/s. (a) What is the speed of the water through the constricted reigon? (b) What is the difference in pressure between the two reigons?

part A was no problem, I found it to be 5.6 m/s and I was correct. Part B on the other hand...the book plugs and chugs into P2-P1=(1/2)pv1^2-(1/2)v2^2 to get 9555. Where in the heck did that formula come from? Or should I just memorize it and not fight it? Cause...P (pressure) = pgv, but i dont understand where (1/2)v^2 comes from. Any help?

Big Question:

A stream of water (initially horizontal) flows from a small hole in the side of a tank. The stream is 2.5m from the side of the tank after falling 0.70m. How far below the surface of the water in the tank is the hole?

:confused:
 

So

Lifer
Jul 2, 2001
25,923
17
81
Originally posted by: johnjohn320
Small Question:

A venturi meter has a cross section of 40.0cm^2 at its entry and exit ports and a cross section of 25.0 cm^2 in its constricted reigon. Water enters at 3.5m/s. (a) What is the speed of the water through the constricted reigon? (b) What is the difference in pressure between the two reigons?

part A was no problem, I found it to be 5.6 m/s and I was correct. Part B on the other hand...the book plugs and chugs into P2-P1=(1/2)pv1^2-(1/2)v2^2 to get 9555. Where in the heck did that formula come from? Or should I just memorize it and now fight it? Cause...P (pressure) = pgv, but i dont understand where (1/2)v^2 comes from. Any help?

Big Question:

A stream of water (initially horizontal) flows from a small hole in the side of a tank. The stream is 2.5m from the side of the tank after falling 0.70m. How far below the surface of the water in the tank is the hole?

:confused:

Don't know offhand the answer to the first question (part B) but the second question is easy because the water flowing out can be treated like a standard newtonian particle (x and y components) [google for 'Torcelli's result'] therefore you know the x velocity, which should help you to your answer.
 

raptor13

Golden Member
Oct 9, 1999
1,719
0
76
The equation in part A is a simplified version of something called "the Bernoulli equation." Perhaps you've heard of that? It relates fluid pressure to velocity. Hence, we have airplanes.


You use the same equation to solve your second problem. After the water hits the ground, you have x and y velocities. You're in physics so you better be able to figure out the y velocity. Then you also have the time the water's in the air. Figure out vx. Suddenly, you have velocities. Holy crap! You're alse given one of the pressures! Solve for the other. That's your answer.
 

Kadarin

Lifer
Nov 23, 2001
44,296
16
81
"Boo Physics"

And here I thought this thread was going to be about explaining ghosts..