Recent content by HP8

Oh sorry, my bad, I never thought of moving the square root of the equation on the right to the left, instead, I canceled out the square root and got confused because moving the V_1 to the left will make it equal to 0. Sorry I caused you a little confused about my work. Finally, I got the value...

Wouldn't the two V_1 cancel each other out because I'll transpose V_1 after simplifying the radical expression. This is what I'm saying: V_1=(A_1\sqrt{0.021+V_1^2})/(A_2)

Hi, BVU suggested to me that I should get the difference in velocity, which is 0.14 m/s. I am still confused on how should I plug this value into my original equation when my answer is stated as difference in velocity? Q = A_1 V_1 = A_2 V_2

Well, after what you suggested, here's what happened: 2(P_1-P_2) = p(V_2^2-V_1^2) after that I plugged 10.5 Pa into pressure difference, then I divided it by the density of water (1000kg/m^3), so I got an answer to the difference between two velocities. 0.021 m^2/s^2 = p(V_2^2-V_1^2)...

What do you mean by h1 = h2? Equating them? And I didn't understand much this phrase "write it out in terms of one of the v and the A's".

Well there are some other formulas mentioned like this one: P_1 + \frac{1}{2} pv_1^2 + pgh = P_ + \frac{1}{2} pv_2^2 + pgh That's it. The one I mentioned earlier was a theorem on continuity.

Oh I'm very sorry! I just realized it right after you replied! I was really wondering why my question was somewhat getting irrelevant to others, thanks and I'll edit my question. By the way, the 10.5 Pa you are mentioning is really 10.5, in its decimal form.

Homework Statement The pressure difference between the main pipeline and the throat of a venturri meter is 10.5 Pascals (Pa). The areas of the pipe and the constriction are 0.1 m^2 and 0.05 m^2 respectively. How many cubic meters per second are flowing in the pipe? The liquid in the pipe is...

Off-Topic: Can I still post one more question related to this type of question?

With your reply, I tried to do the problem and this is what happened; -at a. I applied torricelli theorem, plugging the value 10 m to h and I got 14 m/s -at b. I used dV/dt = which was A * v and I got 0.0043 (m^3/s)? Am I doing it right?

1. A circular hole 2 cm in diameter is cut in the side of a large stand pipe 10 m below the water level in the standpipe. Find: a. the velocity of the efflux b. the volume discharged per unit time[/b] 2. P + 1/2 pv^2 + pgh Q = A1 V1 = A2 V2 where = a = area v =...