3

5 ()'Pachce Poblems (0 ' 83 (0.8sI0.ii10.13 Cdltutlake At Jou HaDtg) * 'a 0g) H,0+ty at 600 k wing Hne Rettousng data +bty? O.c9) 4,O,t92 ka %.3X ...

Question

5 ()'Pachce Poblems (0 ' 83 (0.8sI0.ii10.13 Cdltutlake At Jou HaDtg) * 'a 0g) H,0+ty at 600 k wing Hne Rettousng data +bty? O.c9) 4,O,t92 ka %.3X I06 & (O k 2/4, 49) * 0,0) ath0 19 k = 1.8 X 1d3F ad 6OkKntules AGQlnk

5 ()' Pachce Poblems (0 ' 83 (0.8s I0.ii 10.13 Cdltutlake At Jou HaDtg) * 'a 0g) H,0+ty at 600 k wing Hne Rettousng data +bty? O.c9) 4,O,t92 ka %.3X I06 & (O k 2/4, 49) * 0,0) ath0 19 k = 1.8 X 1d3F ad 6Ok Kntules AG Qlnk



Answers

\text { In a }\text { wind }\text { tunnel experiment the pressures on the }\text { upper }\text { andlowers }\text { surfaces of the wings are } 0.90 \times 10^{5} \mathrm{~Pa}\text { and0.91 }\times 10^{5} \mathrm{~Pa} \text { respectively. } 1\text { If the area of the wing is }40^{\circ}\mathrm{m}^{2}the net lifting force on the wing is \begin{aligned} &\text { (a) } 2 \times 10^{4} \mathrm{~N} \\(b) $4 \times 10^{4} \mathrm{~N}$ &\text { (c) } 6 \times 10^{4} \end{aligned} (d) $8 \times 10^{4} \mathrm{~N}$

The design of a new airplane requires a gasoline tank of constant cross section area in each wind. A scale drawing of the cross section shown here. The tank must hold £5,000 of gasoline, which has a density of £42 per cubic feet, estimate the length of the tank we have here the uh huh Drawing of the cross section, the horizontal spacing of the measures wise of I is one ft. So they are one fell apart And Y zero. The first vertical measure is 1.5 ft. Why one is 1.6 ft Y 2148 ft by 31.9 by 42 ft. And y five equal wife six 2.1 ft. So we have 67 tools of same length one ft. So we came for example, apply Simpson's rule to calculate the across the area of the cross section because he uh tank has a constant protection area. We can say that the volume of the tank, that's a V. Is equal to cross section area times length of the tank. And that essentially it is true because the gasoline tank has a constant cross section area. So, uh we know that the length. We want to estimate the length of the wing, sorry, of the tank. Mhm is equal to volume of the tank over cross section area. So we need to calculate the cross section area and the volume. Now we know that density equals, let's put it this way, density equals mess over volume. And so the volume. Yes, mess over density. We know the density of the gasoline Given here for £2 per cubic feet. And we know the mess of the gasoline containing the tank That must be contained. The tank is £5,000. So mass equal 5000 pants. And density of the gasoline £42 per cubic fit. Okay. Than the volume of the tank is equal to. There is a vote the volume of the in fact we are calculating the volume of the gasoline contains a tank as assuming that the gasoline is going to fulfill the whole tank. So that's equivalent to the volume of the tank. So the volume of the tank is of is equal to let's put it here to be clear volume of the castle lean contains contained in the tank. Yeah, it's a fool capacity and that's equal to mass. That is £5000 over £42 per cubic feet. And this is about yeah, Let's see here is about 100 19 0.5. This case it will be a cure fit. Mhm. So volume of the tank Is about 119 05 cubic feet. So that's the first calculation we did you're using the mass of the gasoline fulfilling the tank or that must be holding the tank and the density of the gasoline. Now, using the first observation we made here to estimate the length of the tank, we get to estimate the cross section area of the tank because we already have the volume, the volume of the tank. V Okay, so we need to calculate uh to estimate the cross section area of the cross section area of the tank. Mhm. And for that we use Simpson's rule in that we can use mhm. Because the number mhm Of serve intervals is even six. So we can use Simpson's rule and we know that area. It's pretty here cross section area. Mhm is approximately equal to Simpson's rule with six of intervals and that's equal To age 3rd times 10 plus four way one plus two, Y two Plus four, Y 3 Plus 2. 4 plus four way five plus why six? Okay, So where h you see a personal space in which is equal to unfit? Okay mm. So At six is equal to 1/3 times. So we put the values now why Syria is 1.5? Um Okay, let's put the units maybe to be complete here. So it's one ft On one ft. And now here we put 1.5 ft plus four times Why? one is 1.6 ft? Yeah Plus two times 1.8 feet Plus four times 1.9 ft plus two times two ft plus four times 2.1 ft Plus 2.1 ft. We know there is, you know the unit feet is a common factor inside the square bracket and we have a feat outside here. And so the units of the result is square feet. So S six is equal to one third times. So now let's do this calculation inside the square brackets here, We get 33 0.6. Mhm. And we know his square feet. And the division here is um 11 0.2 square feet. So cross section, the cross section area, Yeah. Of the tank. Mhm. Is about 11.2 few square feet. And that's the second measurement we have. And with this one here and mm hmm this one here, we use this equation here to estimate the length of the tank. Okay, okay then the lens of the tank is equal to volume of the tank. Yeah, over cross section area. Yeah. And that's about okay, in this case, uh the volume, It's about 119.05 cubic fit over 11.2 square feet. So the units, it's in fits so the length of the tank, it's about this vision here is about 10.63 yeah, fit. And that the estimation of the length of the tank. And remember we have used uh huh the idea that the cross section mhm of the gasoline tank is a constant cross section area. There is any cross section of the tank has the same area and for that reason we know that the volume of the tank is equal to the cross section area. Trying to land on the tank from there. We got the length of the tank is the volume of the tank over the cross section area. Then, using the density equation must over volume, we calculated the volume of the gasoline to fulfill in the tank, which is equivalent to the volume of the tank to the volume of the tank where was calculated and it is equal to or about 100 Uh 19.3 or 5 to fit. And then he is in the Simpsons rule, we estimate the cross section area since the rule six of intervals and we can use that because the numbers of intervals even. And we put the values and found that the cross section areas about 11.2 square feet. And then we put that back into the formula we discussed at the start of the resolution of the problem length of the tank, equal volume of the tank of a cross section area without we estimated the length of the tank.

All right, So for this question, we have a protein that's being stretched from one centimeter to four centimeters. Has an elastic module is off 1.7 times 10 to the sixth has across sexual area of one millimeter squared. Now what we're asked to do was refined. The four does to find the force that has stretched this protein from one centimeters 47 years and we're gonna do that is to say that stress is equal to strain and the question for stresses force over area. We're just gonna be equal to the Youngs. Module is times the equation for strain Delta L over L. Now, if we wanna minute plates equation to be an equation for the four supply, we just multiply a on both sides. And now we can start plugging in things that we know in order to solve for the force. We know the young's modu, Alice. We know the cross sectional area A. You know the change in length. It's three centimeters. We know the initial length. It is one centimeter, so you can plug all those things in. But first we have to convert from millimeters squared and two meters squared so the way we're gonna do that's going to take one millimeter squared, then multiply it by one meter her 1000 meters twice. You might know how to do this conversion, but I just want to show you anyways, 1000 millimeters. Excuse me twice. So in the middle of meters, cancel the millimeters squared we'll get is one times 10 of the negative sixth meters squared. So that's what we're gonna plug in for A Let's solve for F here. Split everything in. And if you calculate all this correctly, you should get a force equal to 5.1 news. Then the next thing we want to know is what is the potential energy held within this block resin on the way we're gonna do that is we're gonna use the potential energy equation for, uh, any elastic spring system. One of the potential energy held within the stress of this object is going to be equal to 1/2 que x squared. Where X is our displacement from equilibrium and K. It's going to be equal to why a over l. L is our initial length. So now we can solve for K or even simply plug in. Why over l for K and R potential energy equation would say this is equal to 1/2. Why a over l Times X squared. And now that we know that A is equal to one times 10 to the negative sixth meter squared, we can plug everything that we know into this equation or to solve potential energy. So let's do that. Let's say potential energy is going to be toe 1/2 times the Youngs model. It's 1.7 times 10 to 6 times cross sectional area, one times 10 to the negative. Six all over the initial length is gonna be one centimeter. Times are displacement, which is 37 years. Swears Now we only do three centimeters because our initial length are relaxed. Length of the resin a wrestle in was one centimeter and then we displaced it. Three centimeters take a total of four centimeters. So if we calculate all this correctly, we should get a final potential energy. 0.765 jewels. That's your answer


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