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Jirote74hAn inverted pyramid-shaped tank with square Cross section shownthe figure Deloy10 ft tall with 81 Ft2 cross sectionathe veryLankmark with liquicweight dens...

Question

Jirote74hAn inverted pyramid-shaped tank with square Cross section shownthe figure Deloy10 ft tall with 81 Ft2 cross sectionathe veryLankmark with liquicweight densitymuch work (in #-Ib) requiredDumo Jlthe liquidabove the tOPthe container? (Round voun Jnswerncare;C ineeger )

Jirote 74h An inverted pyramid-shaped tank with square Cross section shown the figure Deloy 10 ft tall with 81 Ft2 cross sectiona the very Lank mark with liquic weight density much work (in #-Ib) required Dumo Jl the liquid above the tOP the container? (Round voun Jnswer ncare;C ineeger )



Answers

The tank is used to store a liquid having a density of $80 \mathrm{lb} / \mathrm{ft}^{3} .$ If it is filled to the top, determine the magnitude of force the liquid exerts on each of its two sides $A B D C$ and $B D F E$.

In this problem, we're given that there is a thing which is eight meters long, four meters wide and two meters deep and that has failed. What in fluid with 820 clever kilograms per meter, you density and it is filled with that flute tow. It leapt off 1.5 meters. In part, they were us to find the pressure at the bottom off the stank. We know that pressure is density times G times. That sedan city for this fluid is 820 g is 9.8, and that is given that is 1.5 a few months old. We see that then the pressure is 12 1000 future for Paschal's, and we can write this one us about 12. Kill Paschal's in part B. Where has to find a force acting at the bottom of the cylinder. It means that we would need to talk plate the area that this pressure's acting on, and that would be this shaded area. Force will be pressure times area where area in this case as eight times for that is 32 meter skirt, so force would then be 12 1054 times 32. That would be three times 86 10 to the fifth in evidence and in party where has to find the force acting on Ah, one end off this tank's. Imagine that we're looking at this time from this side and this is the side of this Think so He had this one. And, um, certain this is four meters, since that is for me, it is white and the oil that is two meters of everyday this field. What if would up to 1.5 meters? So we're interested in this tin strip right now. Let's assume that we have this sense stripped and the thickness as D X sort of totaling as four area off. This would be four times DX. All right, now, let's assume not be having origin located here. And the positive why is in the downwards directions as we go down? What is positive? You know, that pressure is ro times G times, uh, that hear that is the distance measure for Miss Surface. Unless said, I bet his ex Sadat would be X and pressure then would be, um a 20 times 9.8 times. Thanks force would then be pressured times area forever. This is a 10 strip and if he some all those 10 strips up, we would find the total force acting on one answer I need It means that mean it Intergrated starting from zero to a 1.5. Since it is filled with water up to 1.5 meters, you would have a 20 times 9.8 times sex. That is pressure multiplied by area. And we found area to be four times the X from the species that we have a 20 times 9.8 times. For this is a constant and form this integral be would got excrete over to Now we need to let this between zero and 1.5. If you do so, we find a force to be 36,000 and 162 engines.

So we're looking at a tank partially submerged in water and want to find out how much mass we should put in of concrete to the tank to keep it submerged at an even level at 10 meters below the sea level. So what we know is that the buoyant force that acts up on, uh, this tank or any object that submerged in a liquid actually the buoyant force is gonna be equal to the weight of the fluid it displaces. So if we look at this diagram right here, the flu that's been displaced is in this entire volume here. So we go ahead and write that out in terms of an equation. What we have is the density of water, since that's the fluid that has been displaced times, gravitational acceleration times the volume which in this case, we can write in terms of the height on the cross sectional area in order for this to be at rest. So to be staying 10 meters below the water level, it has to be equal to the total force pushing down on this tank due to its mass. So we're gonna have the mass of the tank plus the massive concrete times, the gravitational acceleration. So these are actually the only two forces that we have to look at and keep in balance in order to solve for the mass of concrete that we need. So if we go ahead and rewrite we have is the gravitational acceleration will actually cancel out. So then we're left with this expression here where the density of water times the height, which is 10 meters times cross sectional area minus the mass of the tank and when we plug in with all the values that we know we find is that we need a mass of concrete in the tank. 19,000 910 kilograms.

In this question we have given to large open tanks that is tank and tank F. And it contains the same liquid. We have given a horizontal pipe that is B. C. D. And we have given dead all the flow is streamlined flow and there's the viscosity ease zero. We have to find the relation between the height H. Two and the height H. One. So first of all I am writing the, assuming that pressure at bonN sees that is P. C. And we can see that from Banal equation that is at point the end outside parts. So we can say that this can be written as the note plus off road. Zh one is equal to this is peanut plus off after all we D square. It is the velocity eight day. So from here we get this is half of roe V. D. Square as it was too rosy as one. Like this is our equation First. This equation can be written as that is peanut plus Rosie H one. This will equals two energy eight point see that is P. C. Plus half of this is roe V. C square. No, we can say that continue to question that is a C. Into V. C. Is equal to this is Eddie into VD. We have given that area sections is the half of area sexually. So from here we get the debt we seize two times of VD. Now put here. So we get this is peanut plus of Rosie H one is equal to this is Pc plus half of row and this is two times of reedy Holy square from here we get the value of pc and this will comes out to be peanut Plus Rosie H one minus. This is two times of roe V. D square. Now put the value of ravi D square from equation one. So we can see it from equation one we get busy. That is peanut plus this is Rosie, it's one minus. This is too Into this is two times of Rosie, it's one. So finally we get the value of pressure at sea. This will come out to be pronounced -3 times of OG. As one. This is the pressure to see that this is a request in 2nd. No pressure it she can also be written as that is from tank to that is P C. Plus Rosie at two is equal to P note. Now from here we get the value of pC. That is not minus Rosie H. To let this is our equation T. No compared to question 2nd and 3rd we get that this is peanut minus three times of rosie H one is equal to, this is peanut minus rosie H two or I can say extra is given by that is three times of H one. So this is the value of height H. Two and this is our answer for this evolution in terms of H one. Thank you

Yeah. Alright. For this question, we have a cylinder filled with gasoline. Mhm. We know that the cylinder has a radius of 8 m the height of 10 m, and we want to know how much work it will take to pump all of the gas out of the cylinder. So in this context, we can think of work as being the second you can think of the work as being the weight of the of an object. Yeah, times the distance is lifted. So what we want to do here is start by imagining our tank as being filled with sort of discrete disks of gasoline. It doesn't really matter what the thickness of these imaginary disks is. So we'll just say that the thickness it's going to be Delta X. So the total work is going to be the sun of work for each disk. The work for each disk, which I'll label w sub I hear okay, is still going to be equal to the weight times the distance lifted. So we need to think about what the weight is going to be. Well, we have a liquid. The weight of our liquid is going to be the area which is filling essentially the sort of cross sectional area of the disc of liquid that we have area times density tree fell row times the gravitational acceleration on Earth G yeah, times our thickness shows our Delta X and then times our distance. I'm just going to write distance explicitly there. So since we have a cylindrical tank, we know that the area is going to be pie R squared. And then we have row G Delta X. Yeah, and the distance that the each individual disk of gasoline will need to be raised is going to be here. It will be 10 minus. But I'll label as you I where you I is the distance from the top. Or rather sorry the distance of the disk. I should say the distance of the disc from the bottom is going to be our ui. So mhm we know that our radius was 8 m. So we'll have that our our square there will give us 64 our density. The row for gasoline is 720. Yeah, Then we have our pie Delta x 10 minus you. I so the total work that will take for pumping all of the gasoline out of the tank will be the limit as an approaches infinity, where n is the number of discs of the some from I equals one up to N of our w eyes or equivalently. That's the limit as an approaches infinity, the some from I equals one up to N of 64 times 720 here is going to be second. So I'll do 64 10 7, 20 times pi here. 14159 We'll have about 1 44 764.5. So I'll just round that to 1 44 seven. So 144,764 times the thickness Times 10 Minus you. I where I'll also note that our thickness, the thickness of each disk of gasoline that we're imagining Well, that Delta X taken here, I'll move that over. Yeah, the Delta X is itself going to be equaling 10 over. And because, you know, we have this, uh, total 10 m high cylinder and we are splitting it apart into in different disks inside. So I'll bring that down here 10 over and but keeping in mind that that is the same thing as Delta X. And of course, this should be familiar. The limit as n approaches infinity of the some specifically when we have this Delta acts going on there when we take end towards Infinity for the sum that's basically just introducing, um, or that's basically just treating it as an integral. So we are integrating from 0 to 10, 144,000, 764 times 10 minus. Now, I used you I to, uh, to specify the height of our disk that we're pushing up. So once we've generalized to the continuous case that you I will become an X and our Delta X becomes a D. X. So we just need to do our integration now. Yeah, so we can bring out the constant that we have out front. We have 144,764 times the integral from zero up to 10 of 10 minus x the x or 144,764 times 10 x minus X squared over two, which we are evaluating from zero up to 10 means that we'll have 144,000, 764 times. Well, 10 times 10 is going to give us 100. Then we'd have minus 10 squared. So minus 100 divided by two. So minus 50 or 144,000, 764 times 50 which is going to equal. Yeah, Yeah. 72 R one moment here. Ah, slight mistake. I'll need to, uh, jump back and write this down here because I forgot to multiply by G for each of these things. It doesn't make too much of a mathematical difference overall at this point, because G is just going to be scaling everything. Yeah, so we should have 9.8 times. 144,764 times 50. And that will give us Yeah. That gives us 70 million 934,589. Approximately 70,000 or 70,934,589 jewels of work that will take to pump all the liquid out of the cylinder


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