5

You nre IrYInGsolveIz 2 0,9(I1,*2) = f1 + I2 56 f(rT2) subjeet (0fi 2 0, MLI respeul both f Ad strictly incrensing with The objective function dliscover that the va...

Question

You nre IrYInGsolveIz 2 0,9(I1,*2) = f1 + I2 56 f(rT2) subjeet (0fi 2 0, MLI respeul both f Ad strictly incrensing with The objective function dliscover that the value of the gTfdlient (714) (3,3) ATc [ You eXamine the POHt function is Vf(3,3) = (1.2). vetor of the objective problem? Explain your reasoning: (3.3) # optimal solution for the Is (T.T2) 2 0. Label FOu T2)-space, for T1 2 0 ad p2 Carefully draw diagram in the (T.= with your textual accurately complements and integrates diagTam suffic

You nre IrYInG solve Iz 2 0,9(I1,*2) = f1 + I2 56 f(rT2) subjeet (0fi 2 0, MLI respeul both f Ad strictly incrensing with The objective function dliscover that the value of the gTfdlient (714) (3,3) ATc [ You eXamine the POHt function is Vf(3,3) = (1.2). vetor of the objective problem? Explain your reasoning: (3.3) # optimal solution for the Is (T.T2) 2 0. Label FOu T2)-space, for T1 2 0 ad p2 Carefully draw diagram in the (T.= with your textual accurately complements and integrates diagTam sufficiently that it explanation_



Answers

(a) Take first and second derivatives with respect to time of $q$ given in Eq. $(30.28),$ and show that it is a solution of Eq. $(30.27) .$ (b) At $t=0$ the switch shown in Fig. 30.17 is thrown so that it connects points $d$ and $a ;$ at this time, $q=Q$ and $i=$ $d q / d t=0 .$ Show that the constants $\phi$ and $A$ in Eq. $(30.28)$ are given by
$$\tan \phi=-\frac{R}{2 L \sqrt{(1 / L C)-\left(R^{2} / 4 L^{2}\right)}}$ and $A=\frac{Q}{\cos \phi}$$

Also, we have the question for the charges said and instead, off time as small musicals too. Hey, it is bar minus said, Are you wanted by 12 t on Dhere? We have co sign off. Square it off. One you wanted by L C minus are square divided by four L square, this is our on. Then we have your tea plus five. This can be simplified is so molecules equals toe Hey, here is part of minus are divided by two t because I know off. Oh my God! Prime T plus five here were defined It is a mega prima's mega prime musicals to square it off one divided by L C minus R squared, divided by for and square. Now let's take their time The video off this required so we can write Dick, you divided by D T is equals to a square brackets open here We have ears. Part minus are divided by 12 t on then here we have in tow minus are divided by 12. Andi here we have because I know Omega T plus five minus. Here we have here is part minus are divided by 12 t on. Then here we have signed off. Omega Prime T, this is Michael Prime plus five. Andi, here we have my God Prime. So you can't records close. This is the first time the video off this very question we called the cold this equation as equation number one. Now, let's take the second time their value off this question so we can Raidi square human divided by D T Square. And this is written as, uh, capital and cracks open here. We have minus. Are you wanted by too? Well, uh, in this way here is part minus are divided by two t in tow minus are divided by 12. On here we have co sign off into a mega prime D plus five minus. Uh, minus. Are you wanted by two hour? This way. Andi. Then here we have, uh, it s par minus are divided by 12 t dog to Omega Prime. Then here we have signed off Omega Prime T plus five. So this is t well minus omega Prime. It is part minus are divided by 20 T on Dhere. We have minus are divided by 12. Then here we have a sign off. Oh, my, The prime T plus five plus omega prime square. It is far minus are divided by two hour T Onda. Then here we have co sign off. Oh my God. Prime T plus five. So square because close. So this is the equation Number two Now using equation number one in this equation number two, we can write dis square Q divided by D T square. Plus, are you wanted by held de que divided by d t plus two divided by L. C So this is equal to two in tow. Square brackets open. He are divided by 12 all square minus omega crime square minus R squared, divided by two outs square and then plus one you wanted by healthy describe X plus is equal to zero. So from here we can write The Omega Prime is omega Prime is equals to square root off one divided by Elsie plus minus are square You wanted by four L Square since the above equation, which is routinized musicals to a par minus. Are you wanting to our t? Because I know off Oh my God. Prime plus five satisfies this equation so that one is the solution of this equation. Thank you

Let's start from the question now for the charge it and instant off time, which is Cuba's equals toe. And here is part of minus. Are you wanted by two t? Because I know off squared off one you wanted by l C minus are square. You are in by four. I will square and here we have d plus five. The screen can be written as a two is equals two years power minus. Are you wanted by to l t. Because I know. Uh, Omega Prime T plus five. Here we have defined this, uh, Omega Prime is mega prime is equals to square it off When you wanted by L C minus our square you wanted by for L Square. Now let's take the first time the video off this equation so we can Why did you You wanted by DT is equals toe square bags open here. We can write in this part. Minus are rewarded by 20 t on here Would be minus are rewarded by two hour. Then here we have chosen off Mega prime D plus five minus. It is bar minus are divided by two t on Dhere. We have sign off Omega Prime T plus five, then here we have Oh, my God! Prime square bags Close their school. The question number one. Now let's take the second time video off this question so we can write the square Q divided by D T Square. This is equals Thio Scattered groups open on Dhere we can write to minus are divided by 12 Onda Then here we have here is bar minus are divided by 12 t Onda should be minus are divided by two hours on then here we have co sign off Omega Prime T plus five minus are divided by minus into minus are divided by two hour. Here is far minus are divided by 20 T on board then here we have a daughter. Omega Prime, sign off. Oh my God! Prime T plus five minus omega prime Here is bar minus are divided by two of t Andi. Then here we have minus are rewarded by two hour on. Then here we have sign off or make a prime T plus five on board. Plus omega prime square. It is bar minus. Are you wanted by two t? Because I know off Omega Prime T plus five So a square box close. This is our question number two. Now using equate number one and equal number two, we can write r d square kun rewarded by DT square lust Are you wanted by hell D two divided by D t plus Q rewarded by L C. Is equal to going toe square box open here we can write Are you wanted by two of the whole square plus omega prime squared? Plus uh, sorry that it's minus sign with mega prime square minus are square divided by two old square plus one divided by l C square X closed This is equal to zero So from here we can, right? Oh my God! Prime is supposed to square it off one divided by l C minus r squared Divided by four l square eso This shows that the human equation is the solution off this question how in part of this problem, we have to show that this tangent off five is equals to minus are divided by to all squared off wonder Why did bile see minus r squared divided by four square? So the square is up to this point on then we have to show is equals to capital coun divided by because I know five So we can say that at T is equals to zero. The charge goes to this small becomes equals toe Captain que so we can write I is equals to do Q divided by d t is equal to zero. So from this case, we can write capital too is equals to a go sign on five. So from here we can write This is equals to tuned You added by because I know five hands proved Now we can write this d two divided by d t is equals to into minus are divided by two hour on Dhere We have echo Sign off I minus omega prime. Yes, I know why is equals to zero from here? We can write minus are divided by two hour Because I know if I is equals to Omega Prime Sign off. So from here we can write it This sign off five divided by casino five equals two minus are divided by too well. Oh, my God! Prime! So this will give us Daniel Door five is equals to minus are divided by two l squared off one divided by L C minus R squared, divided by four l Square and proved so I know the question. Thank you.

In order to answer this question, we're going to have to look at the Newman projections for two metal butin. So to method butane is going to look like this. And I'll just label the carbons right now. So this is carbon one to three and four. And we're going to be looking at two methyl butin from this angle so that we can see the confirmations that are made as it rotates about the bond between carbons two and three. It's gonna be this fund. We're going to be looking directly at it like this. So I'm just going to draw a sample of what? Um a Newman confirmation is gonna look like, and then we'll get into each of the confirmations that can be made and then rank them on to an energy diagram. So starting off, I'm going to color everything on carbon too. Read. So label that here. So here, I'm gonna have my carbon to, And when you draw Newman production, you draw a circle around it. This represents the carbon that we're looking at from the front. And remember, carbons have four bonds. So as we can see here, we have this method group here and here. So kind of based off of that perspective, that probably be here, the left and at the bottom. And then don't forget, even though I didn't draw in, you have a hydrogen here and now we have our front. So now we can just draw what we have in the back. So obviously, we have the method group up here. Then again, we don't want to forget that we have our hydrogen is attached, and I forgot to mention, but this is all carbon three, and carbon three isn't visible. Um, but I'm gonna draw all of its attachments in green is invisible cause for looking directly again directly at this angle, so it would be right behind carbon number two. So at this angle, it looks like these hydrogen czar staggered staggered means that they're just, like, not directly on top of each other. This is the first confirmation that we're gonna be working with. And what we're going to be doing is we're going to be rotating this group for I guess all the groups in the back, um, we're going to be rotating each 60 degrees. We're gonna keep the front the same until we get the full 3 60 degrees around the entire circle, and that's how we're going to get all of our confirmations. So I actually drew out the front already for all of our confirmations, cause again, we want to keep the front. Still, we don't want to be moving both of them. We just want to be moving the back. So starting with the first confirmation against the one, we just true, we're gonna have ch three each and h So this is the first confirmation. Now we're going to rotate 60 degrees, and that will push the methyl group to eclipse this hydrogen right here. And that will also push the hydrogen ins to eclipse these methyl groups. So when I say eclipse, that means that they're very close to each other. So there's going to be a lot of hindrance here, so we'll keep that in mind when we draw the energy diagram again. We want to rotate this method group 60 degrees, so that will put our methyl group here for hydrogen up here. No, over here. So if we take a look at this confirmation that we just drew, you can actually tell that it's very similar to this 1st 1 And they have method groups that are across from each other. So here also, and Carl methyl groups that are, um, gosh to each other's there. They're not like on top of each other, but they're next to each other. So we have that as well. So these who are basically the same thing, they're going to be very similar in the energy diagram. So I'm gonna put a green star, so we remember that. So for the grain store right here, we can no rotates a another 60 degrees. So that is going to make another eclipsed confirmation. And this time, the methyl group will be right by another method group. Now we can make our next confirmation, so this time it will be staggered again. But the methyl group will be between two other method groups. So it does not look like the other two staggered confirmations that we made. So I'm not going to say that these have the same energy, so we'll just leave it alone for now, will rotate it another 60 degrees, and that will bring us to our last eclipsed. It was a little hard to draw this one close, but these are all eclipsed. And then finally, we rotated the last 60 degrees, and that brings it to its original position. That will be with the CH the group on top, your riches on the bottom again. I'll put a great and start because they're all the same and I just want to go back for a second. Um so confirmations, um, confirmations six and four look very similar. So we have to eclipsing hydrogen zahn both to eclipsing methyl groups and to, uh, and then eclipsing hydrogen and methyl group. So I'll put a Red Star in Exodus just so we know that these are the same thing, so will be the same energy level. So now let's start to write everything onto our energy diagram. So I drew out. My ACSI is already so why access is going to be potential energy on the X axis is going to be the guy he'd roll angle. So I'm just going to start with the first confirmation that we made. And I'm just gonna right the Roman numerals. That's why I gave them some numbers just so I don't have to redraw everything. So starting with number one. If we take a look at number one, the again, we have a staggered confirmation. It's pretty low energy. Um, these method groups are as far apart as they can be, and the Method group is only next to one other method groups. This is a pretty good, pretty low energy, um, confirmation. So we'll put it towards the bottom. Now we can move on to number two. Number two is a clip, so we know that's high energy. But if you look at all of the eclipse in, it's, um, all methyl groups and hydrogen. So it's not terribly high energy, but it's still much higher energy than this first confirmation over here. So we'll put it about here. I would say again, this is an exact This is all just estimations. So let's take a look. Ah, Confirmation. Three. We have a green star, and we know that it's the same energy level as number one, cause they have the same. Um, they're basically the same confirmation. So we'll put three over here. No, for a number four we have. Let's compare it to the other Eclipse to confirmation that we have so far, Um, so we have. As I've said before, I believe we have eclipsing hide regions and eclipsing hydrogen and methyl group. But very importantly, we have an eclipsing pair of metal groups. This is going to be much higher energy than any of the other eclipsing groups. So this is gonna make this confirmation very high energy because it is gonna have a lot of serik hindrance. So I will draw that in now so we can draw even higher than the second confirmation and are grafts. I'll put it right here. That was before number five. Let's compare it to the other staggered confirmation that we have. So, as I mentioned before, we have the methyl group in between these other methyl groups, so they're not eclipsing, so they're not on top of each other. So the stark hindrance isn't as bad as any of the eclipse in groups. But there's going to be slightly more Starik endurance over here than there is going to be between a medical group and hydrogen. So slightly more Starik hinderance or a spirit clash. So we can draw this confirmation a little bit higher energy than the other staggered confirmations. But lower than the eclipsed. So it would between me between these groups over here, so I'll draw it right there. It's was number five. You move on to number six and number six, we actually already saw that. Number six is the same as number four. So let's draw them at the same level once about here. So we have six, then finally, we have number seven hour full 3 60 degree rotation. Um, and again, it's a green star. So we know it's that lowest energy level, lowest amount of energy. So this was number seven. So now we have our relative maxes and men's, and we can just connect them. Husing curve will use blue. So again, this isn't a very exact, um, graph. It's just a general graph to give you an idea of how these different confirmations compared to one another in terms of stability or energy. So there we go. That is our final curb

We are going to be drawing some, um, different ways that this molecule to three dimethyl butin can be arranged via rotation. So first, let's just draw out our flat molecule. You cannot ignore everything else that I drew for now. So dimethyl butane butane is going to be apparent. Jane made up of four carbons. And on carbons two and three, which are these middle carbons, we're gonna have methyl groups. I'm just going to decide that this carpet here is going to be carbon one. This will be to three and then four. It doesn't matter if you start with this carbon or this carbon on the right. Either way, you'll get the same name. Um, so now that we have our flat molecule, what we're going to be doing is we're going to be looking at this molecule from this perspective here, looking at the bond carbon two and three share looking right at it. So when we look at this molecule, we will see right in the front is going to be carbon number two, which all draw in green, actually. So this will be carbon to and it's attached to three methyl groups. So here I'll even highlighted. This is carbon too. You're flat molecule it is attached to I might have said three month groups and attached to two metal groups and a hydrogen. So we can label these three substitutes on carbon number two as this methyl group, this methyl group and the Haijun. And this is what we're going to be looking at as our front of our molecule. And as you could see, I drew this for all of our rotations. We're going to be keeping the front stationary as we rotate the back. So we're going to be adding the back molecule. So I'm just gonna draw one example. Um so directly behind this green carbon, which is carbon number two is going to be carbon number three. So we won't see carbon number three on these new mid projections. We will see what's bonded to mow carbon number three's, that's going to be two methyl groups and then ah, hydrogen. So I can choose to draw that. However, however I want because once we rotate it around the full 3 60 degrees, it will come back and go through all of the possible positions. So this is what I'm going to start with? I'm gonna have my first, um, human production looking like this. So I'm going to make this number one. And so again, this is just the same thing. I'm just redrawing it. And then what we're going to be doing, as I said previously, is we're gonna take the back groups and we're gonna be rotating them. But we keep the front group stationary. So the first thing we're going to dio we're going to rotate this Newman production, which is right now in a staggered confirmation into an eclipsed confirmations. So what's going to happen is that this hydrogen will be right behind the green metal group, and then our methyl group will be behind this hydrogen, and then our other method group will be behind this other method group. So this increases the energy because there's going to be a little bit of clashing between these groups that are right in front of and behind each other. So now we're going to rotate again, and what we're going to end up making is another staggered confirmation. So our substitute mints are going to be a little more spread out. But as you can see the staggered confirmation looks different than the one that we had in our first human production. In our 1st 1 we have our hydrogen ins both across from each other and then our, um, methyl groups are They're too on this half and two on this half, but, um, in our noon production number three or hydrogen zehr a lot closer together, and that puts all of our methyl groups basically on the same side. So they're not going to be the same amount of energy. So moving on we again, we're going to rotate or back molecules and again it's going to be eclipsed. So this time we'll have a metal group behind a method group on top, and then we will have ah, hydrogen behind a hydrogen. In another method group behind another method group. This eclipse to confirmation is different from the eclipse confirmation. We had a number two because here we have a method group two methyl groups eclipse to hide regions and one, um, eclipsed double metals, I guess, like their two methyl groups that are eclipsing each other. But in the Newman prediction that we just drew, we have two method groups, eclipsing each other twice, and then hydrogen is eclipsing each other. So it's not gonna be the same mobile energy that's gonna be helpful for when we're organizing everything. Um, so we're going to rotate Lee back subsitute hints once again in this time, we will have our methyl group here. Ah, hydrogen at the bottom and another methyl group over here. So if we take a look at our previous Newman projections, we would actually see that this Newman projection is a lot like this one. So five looks a lot like three. So for us to remember that since they look a lot like they're gonna probably have similar energy levels, I'm gonna put a little star. So moving on, we're going to do another eclipsed. So again we're going to have the method group behind this method group. We'll have a method group behind a hydrogen, and then we'll have a hydrogen behind a method group. So again, if we look at the previous human projections that we drew, you would see that this Newman projection number six is going to look a lot like the one that we drew in number two, where we have to metals eclipsed with Hye Jin's and one eclipsing of two miracles. So again I'm going to put a mark. I guess I'll put a green star that works. So we can remember that these are gonna be the same energy level because they look the same. Finally, we'll have our last rotation, and it's gonna bring us back to the original confirmation that we had in our first human projection So we can put another star here so we can remember that these two are also the same. So now that we have everything, um, drawn out, we have the ones that are the same match. Together, we can start to plot them onto a curve. Um, so are curve already drew. The axes are X axis is going to be the angle, and the Y axis is going to be the potential energy. So let's start with confirmation. The first Newman production that we did Number one. So, as I said before, this is staggered, so it's pretty low energy. Someone's gonna put it pretty low. Is number one next? Let's move on to number two. Number two is eclipsed, so it's going to be higher energy than the staggered one so I'll just put it. Let's say right here. Number three is also staggered like number one, so let's compare it to number one to see how they're going to rank on the curve. So Number three we have, Um, as I've said before, we have all of arm Ethel's on kind of one side of the molecule, whereas these metals in the Newman production one are kind of across from each other like you have, uh, I'll drive all race this. But you have these methods being across from each other in these metals, being across from each other. Whereas here we only have one set of methods being across from each other. Uh, and that's going to increase the energy because there's going to be a little bit more clashing between, like this methyl group, this methyl group in this method group with this metal group in this month, a group this method group, whereas in the first human production, there's really only going to be some questioning here, here and here and here. So there's more clashing here. It's gonna be a little higher energy, so we're going to put three in between one and two. That's gonna be about here. Let's say moving on to number four Number four is also eclipsed, but it's not the same as number two, so let's compare the two. So, as I've said before, here we have only one eclipsing of two methyl groups. Auras In number four. We're gonna have to eclipse sings of methyl groups. And that's going to increase the energy overall of this confirmation or of this rotation. So this is going to be higher energy compared to number two. So we'll put number two or sorry. Welcome Number four higher than number two. So will be about here and now. From here on out, our molecules, we're going to have the same energy level is something previous. So number five has the same energy as number three. So let's put that the same level on the graph. And then number six has the same energy as number two. So we'll put those at about the same level on the graph. Might be a little slanted. No, it's about here. It's number six and the number seven is going to be the same amount of energy as number one. So it'll be down here and now we disconnect our lines so we have our minimums maximums on the curve. We'll be able to see our final curves for all of these different positions. You're these different. Newman projections how their energy levels compare Nega. So this is part A. Now we can move on to part B. So again, for now, let's just ignore this and let's just draw out our flat molecule. So we're going to have a beauty in again, which is four carbons on the parent chain. The middle number them 12 three. Before then, we'll have two metals on Corbyn's two and three ich since is what are flat molecule is gonna look like. And when me draw our human projections, we will again be looking directly at carbon number two, which again, all drawing green. And then we'll see that its attachments. I'm gonna call her in green as well. We're going to be three metals like that. So I'll draw them with that and then at the back, we're going to have, um, we're gonna have carbon number three, which we won't see, but we'll see. It's three substitute since the metals as well, and we can choose to draw those, however again. So I already drew out a couple of the muscles, um, on these circles. So I'm just going to fill in this 1st 1 that I just drew. And then when we rotate for the first time, you will get our metals all eclipsing each other. And what you're going to see is that since all of the substitutes in the front and the back are the same, there are really only two types of, um, confirmations that we're going to have two types of human projections. Um, since the actual substitute Wint's aren't going to change, it's just gonna be whether they are staggered or eclipsed to Really, These were the only two confirmations, but in total, there are going to be seven because you want to rotate the back. We want to rotate like the back all the way around 3 60 degrees. So that's why we'll have, like, position one here and then to and then three and so on and so on. Um, so that's that's the same thing as what we did up here, where we rotated it seven to get seven different new and projections. Well, I guess last ones identical. But, um, you can do the same thing here. So even though there's only two types of positions, we can still have multiple, um, curves until me molecule has made a 3 60 rotation around the back. So again, as I said before, staggered confirmations are lower energy, since there's not as much of, ah, clash between the molecules on their in its close together. So one is going to be lower, and then to will be higher, because again, these are groups that are directly in front and behind each other, going to have some stark endurance. And as I said before, there's only going to be those two types of confirmations for a human productions. So I'm not going to draw out all of them. Three is going to be identical to one because it's going to be staggered as well. And then, after staggered is another eclipsed, which will be appear. The same is number two, been a meeting before, and five again will be down here and then six again will be eclipsed and then seven will be the full 3 60 Going back to the same identical position is number one and then we can just fill in our curve with a line. No, you can see the maximum minimums. And again, for this case, the substitue INTs are all the same to the maximum and minimum Zehr All going to be the same. There's no difference. And finally get to the last one and then you get your final curve, your part B.


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Find dataset related to your department which is suitable for one- and two-way ANOVA tests. Explain the dataset and the research question: For each test What do you need to check first? Why? Set up the null and alternative hypotheses Calculate the test statistic_ Make decision and interpret. Comment...
5 answers
Ccess nten1 Find a particular 3.5 A particular solulion Is Yp(*) solullon 0 ol tho {ullotnno
ccess nten 1 Find a particular 3.5 A particular solulion Is Yp(*) solullon 0 ol tho {ullotnno...
5 answers
An object is shot straight upward from sea evel with an initial velocity of 550 fUsec: Assuming that gravity is the only force acting on the object give an upper estimate for its velocity after 5 sec have elapsed. Use g = 32 fusec? for the gravitational acceleration. Using At = sec, find _ lower estimate for the height attained after 5 sec.
An object is shot straight upward from sea evel with an initial velocity of 550 fUsec: Assuming that gravity is the only force acting on the object give an upper estimate for its velocity after 5 sec have elapsed. Use g = 32 fusec? for the gravitational acceleration. Using At = sec, find _ lower est...
5 answers
In Exercises 9 - 16, match the polynomial function with its graph. [The graphs are labeled (a), (b), (c), (d , (e), (f), (g), and (h).]$ f(x) = - rac{1}{3} x^3 + x^2 - rac{4}{3} $
In Exercises 9 - 16, match the polynomial function with its graph. [The graphs are labeled (a), (b), (c), (d , (e), (f), (g), and (h).] $ f(x) = -\frac{1}{3} x^3 + x^2 - \frac{4}{3} $...
3 answers
NttpauntMAT121 Sp19Homework: HW 2 (4.4, 4.5) Bcom- 0oln01017 MJcomote4,5.41IhtuFndEmc OtantteLnttiFumAnMacBook Air80"888Unu"
nttpaunt MAT121 Sp19 Homework: HW 2 (4.4, 4.5) Bcom- 0oln 01017 MJcomote 4,5.41 Ihtu Fnd Emc Otantte Lntti FumAn MacBook Air 80 "888 Unu"...
5 answers
X + (x _y)ex dx + xexdy = 0
x + (x _y)ex dx + xexdy = 0...
5 answers
0 0 D D Selectone: Wjle) = 5844 1
0 0 D D Selectone: Wjle) = 5844 1...
5 answers
Question 3151ptsFor : particular population; sample ofn = 'scures has a standard populaticn, sample ofn = 4 scores wouldhave error of 4 For the same stzndard error ofOk24OM-2Question 321ptsFot ? normal population withu C0 Ande probability of being obtained?wnicho the follow ing sAmples has thehighestM<42toasampleotn-Ms44t05 Simo = Aame -Mr4ztor a sampleofn0z44fora samp #ofnQuestion 33
Question 315 1pts For : particular population; sample ofn = 'scures has a standard populaticn, sample ofn = 4 scores wouldhave error of 4 For the same stzndard error of Ok24 OM-2 Question 32 1pts Fot ? normal population withu C0 Ande probability of being obtained? wnicho the follow ing sAmples ...
5 answers
Drw Anhtnhbnuduort, Jnc Uertne Tlhpie1 cuato chance tulroy wilI nutomobilc You Lnow0ul40uautonaREmedeeTour-Coor NcdelsESallZednarASel 4 door rtha sabo}
Drw Anhtnhbnu duort, Jnc Uertne Tlhpie1 cuato chance tulroy wilI nutomobilc You Lnow0ul40u autonaRE medee Tour-Coor Ncdels ESallZednar ASel 4 door rtha sabo}...
5 answers
Uestion 2 ot yet answeredlarked out of 1 00Flag questionA charge of uniform volume density (80 nC/m?) fills a cube with 5 cm edges. What is the total electric flux through the top surface of this cube?A 0.001507 Nm? /CB. 0.3766 Nm? /Cc 0 Nm? /CD.0.1883 Nm? /CE.1.13 Nm? /C
uestion 2 ot yet answered larked out of 1 00 Flag question A charge of uniform volume density (80 nC/m?) fills a cube with 5 cm edges. What is the total electric flux through the top surface of this cube? A 0.001507 Nm? /C B. 0.3766 Nm? /C c 0 Nm? /C D.0.1883 Nm? /C E.1.13 Nm? /C...

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