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>> All right. Let's take a look at an example problem that might have relevance for your homework. And let's do the example of the water tower. And the water tower example is this. We have a big tub of water. Down at the bottom we're going to have a little outlet. And water is going to shoot out. And in here we have some water. Now it's open on the top which means it's open to the atmosphere. And the spigot at the bottom is also open to the atmosphere. And what we want to calculate is what is this V2? What is the speed of the water shooting out the bottom of the tower? Okay. We'll put some legs on our tower and we'll say that this is height h. How far is the top of the water from that spigot? So how do we calculate this? Who has a thought on how we can calculate this? Rich. Can you grab the mic? All right. Richie, we want to calculate V2. What is the exit speed of the water -- this is water floating around in here -- coming out of the water tower? >> Well, I think we're going to have to use the conservation of mass or something? The area of the spigot coming out and the mass. >> Okay. >> Of the water? >> Maybe. Let's take a look at that. So you're saying the continuity equation, right? >> Yes. >> A1V1 equals A2V2. That's the one you want to use? All right. Maybe, but the problem is we don't really know what V1 is. V1 would be the speed of the water up there. V1 would be the speed of the water up there. And we don't really know what that is, right. We didn't give that to you in the problem. So I don't think that equation is really going to help us. So what else should we try, Richie? >> Bernouilli's equation? >> Bernouilli's equation. >> I just didn't want to say that because it was really big. >> [Laughs]. It's Bernouilli. It's not that bad. Bernouilli's equation. Okay. Good idea. So let's write down what Bernouilli's equation is. P1 plus one-half rho V1 squared plus rho gy1 equals P2 plus one-half rho V2 squared plus rho gy2. Okay. It's not so bad. Like we said before, it's just conservation of energy. Work, kinetic energy, potential energy equals work, kinetic energy, potential energy. All right. If this is location two down here and this is location one up there, then maybe we can figure out what some of these terms are. What does P1 equal? P1 is the pressure at the top of the water. If it's open to the atmosphere, what does it become? It becomes atmospheric pressure, P naught equals 1 atmosphere. Okay. And you got to put it into the right units. You got to get into Pascal's. But that just becomes P naught. What is V1 up at the top? Well, the water tower is so big and wide that the velocity of the water at the top is negligible. So that's zero. What about this guy? Well, on our scale we said this thing is the height h, so this just becomes rho gh. What about the other side? Ah, we said the pipe is also open to the atmosphere and so that is also a pressure of P naught. It has speed, V2, which is what we're interested in. And then we are at height zero. So quite a few of these terms dropped out. And look what we're left with: P naught plus rho gh equals P naught plus one-half rho V2 squared. And now I can subtract P naught from both sides. If it's open to the atmosphere on the top and it's open to the atmosphere on the bottom, that doesn't contribute to the speed of the water. And look what I get. I get one-half rho V2 squared equals rho gh. I can cross out the rho on both sides and I get V2 equals the square root of 2gh. That's the speed of the water coming out the bottom of this pipe. Does that sound familiar? Square root of 2gh? Yeah? What was your name again? >> Brandon. >> Brandon? >> Yeah. >> Can somebody hand the mic to Brandon? Brandon made the mistake of nodding his head in agreement and now he gets the mic. Where, have we heard that before? Square root of 2gh? >> We have. I'm trying to figure out, remember where. >> It sounds familiar, right? It sounds very familiar. Let's try the following. Let's take a rock and let's drop it a distance h and let's calculate V2. How do I do that? Conservation of energy says one-half m V2 squared has to equal mgh. V2 is therefore the square root of 2gh. Isn't that wild? If I drop a rock from the top of the water tower, how fast is it going when it hits the bottom? It's going V2, square root of 2gh. If I open the pipe and let water shoot out, how fast is this going? Exactly the same speed, square root of 2gh. So it's kind of weird to think about but when you turn on your hose, it's like that little water molecule fell from the top of the water tower. And that's it. That's the speed coming out. And now you understand why they put water towers up on the top of hills or on the top of buildings. You want that water to fall through a good amount of h in order to come flying out the pipe at a good speed. It's really just free fall. Same as free fall. Which I think is pretty cool.

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