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Water flowing through a garden hose of diameter $2.74 mathrm{~cm}$ fills a 25.0-L bucket in $1.50$ min. (a) What is the speed of the water leaving the end of the ho...

Question

Water flowing through a garden hose of diameter $2.74 mathrm{~cm}$ fills a 25.0-L bucket in $1.50$ min. (a) What is the speed of the water leaving the end of the hose? (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?

Water flowing through a garden hose of diameter $2.74 mathrm{~cm}$ fills a 25.0-L bucket in $1.50$ min. (a) What is the speed of the water leaving the end of the hose? (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?



Answers

Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m$^3$/s. (a) How fast will it shoot out of a hole 4.50 cm in diameter? (b) At what speed will it shoot out if the diameter of the hole is three times as large?

So we know for part of the volumetric flow rate is equaling the cross sectional area times velocity and this is the volumetric flow rate is equal in 1.20 meters cubed per second and so we can simply solve for the velocity. This would be 1.20 meters cubes for a second, divided by the area of pi R squared or pi multiplied by the radius 0.150 meters quantity squared. So the volume through that section 17.0 meters per second. This would be your answer for part A for part B. Ah, we're going to say that here we're trying to find the radius. We know that again the following much of flow rate is equal and 1.20 meters cube for a second and this is equaling v times pi times r squared and so the radius would be equaling 1.20 meters cubed per second and then this would be divided by the velocity times pi. Remember, we found the velocity. Uh, neither the velocity is given to us 3.80 meters per second, so the radius would be the square root of 1.20 meters cubed per second, divided by pi multiplied by 3.80 meters per second and the radius would be 0.317 meters. This would be our answer for part B. That is the end of the solution. Thank you for

In this problem. We have water flowing in a pipe with varying cross section and were asked to find a couple of velocities and radius when we're looking at two points in the pipe. So for this problem, what we need to do is use the definition for volume flow rate and the continuity equation for an in compressible fluid. We are given that our volume flow rates. The chief for this pipe is 1.20 cubic meters per second are radius one is there a 0.150 m and our velocity at our second point is 3.8 meters per second. Okay, so those equations that we're going to need, uh, our volume flurry So the capital B E t is equal to area on velocity. That's our definition for volume flow. And for our continuity equation, we're gonna go. Are area a cross sectional area. Times velocity at different points are equal to each other. So it any a one B one equals two, a two b two and for in compressible fluids like we have here, okay. And we have heard and be part asks us what is our velocity at this 0.1 where we have radius 0.150 m. Eso we're going to use our volume flow rate definition for this to say that this V one equals to Devi tt of volume divided by that a one and so our DVD t again 1.2 cubic meters per second and our area is going to be a circular cross section. That's when we pi times are one squared. That's 0.15 m squared, right? And if we calculate that out, we will get to three significant figures uh, velocity at one. 17 m per second and part B asks us what is our radius at the second point? Given that velocity and so what we can do here is we can use our continuity and volume flow rate definitions, so we'll say that are a to equals to a one V one divided by V two. So that's our continuity equation. And we can say that our A one B one is equal to that volume for a DVD T three d v d t over me too. And a two is equal to the area i r. Two squared and from there we can solve for the Radius R two and we'll get our two equals spirit of DVD T over pie on velocity to And that will be have our 1.2 cubic meters per second numerator and high 3.8 meters per second denominator. And if we calculate that out we will get 23 significant figures 0.317 m for our radius r two.

Okay, so into question 51. To water the yard, use a hose with a diameter of 3.6 centimeters. Would have flows from the hose with speed of 1.3 meters per second. And if you partially book the end of the hose, the effective diameter is now nor 0.52 centimeters. With what speed does the water now spread from the Harris? And now we're going to use something called the Continuity. Of course, he's the area through which fear it flows multiply by the volume that the velocity that Richard flows in one is equal to the products of the area and the velocity in another part. So effectively, what we're looking for is a reduction in area may increase our velocity. If the input flower is the same, let's unpack this So we have full area. Bye. I don't want to One over two the radius squared sounds The first velocity is equal to pi. Don't have to with you also squared times a second velocity. Naturally, the pies cancer and we end up with D one squared over two square, which is four. Just one is equal to D two squared also every square. What's it for you too? And naturally, these fools also cancer. So their velocity in the second hers is equal to the ratio of the two dynamics is squared because something squared over something out squared is equal to the ratio of the two things squared everyone. And now, because we're working in ratios, we really don't mind if we're working in centimeters or meters or inches and feet or furlongs or any of the others sitting that scares that we can use because we're just talking about a race. You remembering to square it afterwards. So you know that our first I amateur was 3.6. Senator Mitch is not second diamonds and no 0.5 to you gonna take that ratio square. It multiplied by the original velocity, but for 1.3 meters per second. And we can very quickly work out the new velocity 62.3, which again, four significant figures. You can say we're working with two significant figures. 62 me just a second. And that's the speech that it flares up

So. Question 56 asks us water flows at a rate of 3.11 kilograms per second through a hose with a diameter 3.22 sentences. What is the speed of water homes? So we remember a rate of mass flair. It's simply equals the density of the fluid flowing multiplied by the area through which it flows and the velocity of a cheap clothes. The first question is what the velocity. Therefore we take our rate of flow. Rates of master divide that our identity on the area through which the liquid flows. So it's simply a matter of choosing those values. Question say is 3.1 direct master be going What more kilograms? A second and also the density of water. Very straightforward. Valued 1000 kilograms permits you. It's something that used often, and now we need the area, which is pi r squared. So if the house has a diameter a 3.22 centerpieces, having that is 1.61 sentences. So we put one quite 61 times, 10 to the minus two, remembering that we converted to meet us if you would stand it. Measurements on we square that. So we put all of that in. So I calculated. And from that we get an answer. 3.82 needs his per second. Great. So the second part of the question that asks if the hoses attached to another with a diameter of no 0.732 acceptances, what is the speed of water in the nozzle? We got to another page here and would just remember our continuity equation that the area of one tube on the multiplied by the velocity of flow in that too the product is the same on the other side. This product remains constant to the area in the second to the velocity in the second to most, but together mystical the area the first student of lost first to hold together. So we re arranged for the lost in the 2nd 2 get me one a one a two. And now, once again, he writes out these areas I'm most like Why do you want was the pies cancer from the top? And we have a ratio of radio I squared, which means that we could use the ratio off the radio and square that that also means we can use the ratio of the diameter squared, which does make things a little easier for us as well. We don't have to have things that I calculated before. So the speed in the two 3.82 which we've just determine his reward well. Multiply that by the diameter, the 1st 2 divided by the jams there a second not forgetting after we've done this square because our units cancel within the brackets we were getting on. So three 73 0.9 needs his pa's second still not faster. And then, finally, part see of this question say it is the number of kilograms per second floor it to the nose of greater that less than equal to 3.14 kilograms per second. And explain now, really, we should just consider this somewhat common sense. There's any amount of fluid that passes through the hose in the given time, must also pass through the nozzle in the same amount of time, so the same amount must be flowing through the onset is that it has to be the same, and that's because water is in compressor. If water could be compressed into a smaller volume, then a great a mess could flow in the same time. But that's not the case for Annan Compressible fluid. We've certainly not taking into account. The world can't be compressed some of the second simplicity on and it's true to physics. The as water is compressible is in compressing, so it's not compressible, I should say the number of kilograms per second flying. See the nozzle is equal to three point won't want kilograms a second.


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