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A tank is is half full of oil that has a density of 900 kg/m? Find the work W required to pump the oil out of the spout: (Use 9.8 m/s? for g. Assume r = m and h = 3...

Question

A tank is is half full of oil that has a density of 900 kg/m? Find the work W required to pump the oil out of the spout: (Use 9.8 m/s? for g. Assume r = m and h = 3 m.) W = 310413703.95 X ]

A tank is is half full of oil that has a density of 900 kg/m? Find the work W required to pump the oil out of the spout: (Use 9.8 m/s? for g. Assume r = m and h = 3 m.) W = 310413703.95 X ]



Answers

(a) The tank shown is full of water. Given that the water weighs 62.5 $\mathrm{lb} / \mathrm{ft}^{3}$ , find the work required to pump the water out of the spout.
(b) What if the tank is half full of oil that has a density of 900 $\mathrm{kg} / \mathrm{m}^{3} ?$

We are going to find the work in pumping heat oil from a cylindrical tank. So the cylindrical tank is actually lying on its side. Um it is a full cylinder right now, I just have the bottom piece but I'll go ahead and make that top because it's only filled to that kind of halfway mark. Now we're going to set up the integral to find the work done to pump out the oil over the top of the tank and asked us to choose X equals zero for the position of the base of the tank. So most of our problems, we've been told that we can kind of do it any way that we want to, we're looking at the answer but they do want us to use zero for the bottom and so that will mean kind of two for the top there. Right, okay. So before um sometimes I put a negative two for the bottom of that and put the zero for the top. And that was just a regular circular formula. But if we're gonna put a zero for the bottom and a two for the top, that's a transformation of the circles. That's really x squared plus y minus two squared Equals that radius of four. So we're going to solve for X. So that we can tell what that kind of length is really, our area is going to be a rectangle and the front of that rectangle is really to Radi I so if we solve for that, we can go ahead and take our four subtract are Y -2 squared and then take the square root. And we are going to be dealing with um the positive side, there's a positive side and a negative side. We'll deal with the positive side and then multiply it by two to get that full length of that red line because the radius would just be the right half because we're using positive. So we have our density to put in front. Um that is um that doesn't include the 9.8 um for gravity. And then we have two of those sides that we found now they like to use X. As their variable. Um And so I'll go ahead and change my wise to excess whatever variable you use it doesn't matter you're it's just a piece of your solving process so we'll use excess and de access instead of wise and d. Wise. Okay so that is the front and then that's the only part of the area because now it's running that length of six and then the D. X. Is the thickness of each one of those cross sectional areas. So now we just need to figure out how we're far we're pumping. So remember we're just going from 0 to 2 for our integral. And so to think about how far we have to pump we put a zero in. We need to pump it for we put it to in we need to pump it to so that's going to be a four minus X. And that will make that work out so all of our constants, we're going to move to the front and then we can consider our inner girl from 0 to 2. We do have a square root here and um we also have the four minus X. So I did just throw this into the calculator to find um the value. And so it ends up that the work to pump this heating oil out will be 860, jewels.

We need to calculate the work required to pump all of the water out of the full tank where the distances are in meters and the density of water is 1000 kg per meter. Cute. So our tank is a hemisphere and the water is going to exit through a spouts. So recall that the work is equal to the inner role from where we start toe where we end of the force with respect to the distance. And here we need to find the force. So the force is equal to the acceleration due to gravity times the density tens the volume. So this is 9.8 times 1000 times the volume. So remember the radius of a layer of this hemisphere is going to be so let's first just talk about the radius. So the radius is the square root of 100 minus y squared, since there the radius of the circle on top of 10. So the volume is equal to pie times 100 minus y squared because it's pi r squared time still toe. Why So the force is 9.8 times 1000 tens pie. It's ah 100 minus y squared Delta y. So now we have The work is the integral from zero to 10 of 9800 pie times 100 minus. Why squared now? We ain't delta. Why so delta y is going to be? Why plus two? Because we need to get to the top of the hemisphere and then up to more units to get out of the spout. So why plus two de y So when we saw this integral, This is equal to 11 to 7 five zeroes pie over three in units of jewels. So this is equal to 1.18 times 10 to the eighth. Jules, Yes.

Hello. In that problem we have a fuel pump that is pumping 600 liters of fuel In a tank that is 20 m high above the level of the pump. So here, given that one centimeter cube of that oil ways open to 0.82 g. That Point here is to calculate the work done against gravity pumping these 20 leaders A distance 20 m high. So first, let's calculate the mass of this amount or the 600 L. So to calculate the mass, we have six 100 liters, let's convert them to centimeter cube. Side note, we have one leader Is equal to 1000 centimeter cube. So multiplying by 1000. This gives us the amount In cm Cube, multiplied by 0.82 which is the mass of one cm Cube. So we have the mass which is around 492,000 g or 492 kg. So that's the first part. The second part here is to calculate the work now and we know from previous problems that the work done against gravity is equal to MG age. So now multiplying that mass using the units in kilograms Times the gravitational acceleration, which is 9.82, multiplied by the height, Which is 20 m in in that problem. And we have the answer which is 9,613.8 jewels Approximately. That's equal to 96 kill jules. And that's that work done against gravity and pumping those 600 L to the thank 20 m high.

We need to find the work required to pump all of the water out of the container when the density of water is 1000 kg per meter cubed. So here we want to pump all of the water from the spout. So first, let's recall that the work is equal to the inner girl, from where we start to where we end of the force with respect to the distance. So here, since we're talking about pumping water out of a container, the force is equal to the acceleration due to gravity times the density times the volume. Okay, so first, the acceleration due to gravity is equal to 9.8. The density is 1000. So let's find the volume. The volume here is equal to the bottom is the dimensions air eight and four. So the volume is 32 delta y. So to get a number, our force is equal to three and 13,600. I'm still toe Why forces and units of Newton's. So now we need to find the work. So here the change and why we're going up the whole distance of the container, plus one so each layer must be lifted Y plus one. So the work you're going from zero to the top is five of 313 600 times. Why? Plus one t y. So this is equal to 156 800 times. Why plus one squared. We're going from 0 to 5. This gives us 5.488 times tend to the six, and work is in units of jewels.


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