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Olutions the niernuunu mkein Amnuaetraflic light huving AL Witn the 86 Kg is horizontul_ hupnnnat (4o mncluucd #ints One *ire moke other wire makes 220 Ceculet letr...

Question

Olutions the niernuunu mkein Amnuaetraflic light huving AL Witn the 86 Kg is horizontul_ hupnnnat (4o mncluucd #ints One *ire moke other wire makes 220 Ceculet letrJonl And T Tuacantt . An object wcighs 3920 ioalr. Whcn Itasimninenud Hner #elgh- 2350 Cuculete (A) the huujunt (urce acting on the object (B) the Enitss of Uhe objcct (C)the volume of the object (D) the density of the objcct (E) the uccelerution with which the object sinks i M releascd insude AlrMcrry-%o - Hanie rnlulinnil Inerin; M

olutions the nier nuunu mkein Amnuae traflic light huving AL Witn the 86 Kg is horizontul_ hupnnnat (4o mncluucd #ints One *ire moke other wire makes 220 Ceculet letrJonl And T Tuacantt . An object wcighs 3920 ioalr. Whcn Itasimninenud Hner #elgh- 2350 Cuculete (A) the huujunt (urce acting on the object (B) the Enitss of Uhe objcct (C)the volume of the object (D) the density of the objcct (E) the uccelerution with which the object sinks i M releascd insude Alr Mcrry-%o - Hanie rnlulinnil Inerin; Mel Ken' and rudlus Ke 10 uvint initially rotating Fude- = Aner child ot UNKNOWN LuaessiLs 0n the edge & the mctty-go uund rdls Cucuutethc Inull5 the chlld_ the angular Yelocity elcuans(la Action 2 dtnmn the diutucter o lhe pipe i Wntcr now horirontal pipr [rom seclion cucllon dhmneler ulthe 10 €T; (luid velacity is V,=20 ns and Oluid pressure # BATM: Fuscal ~rctin acliot (KP Quid prASuT pipe [ JUcu _ Calculate (AMluid velocily (1 ATM=L.OISXlo' Fa) 200 = amt Lnaam Trotl uniform shon wcighing 400 nt Calculate the hariontal lorot sipn supporting Mire ( Caleulate (A) the tension in thc cered the binge on ne Tigure hclow (C) Calculute the rertical force exerted by the hluge= spring constant 16 Kg oscillates vhile attached surin moring #Ith 4 speed o 40 An object with muss equilibrium position,it ks object [s 2 m from ~ncco attained by Wc toiecl: K=8O0 Nm: When - motion the maxinum Calculate (A) amplitude of the thc lower surtuce the wing Kg; und the air flows pt [ X10 1600 m', An uirplare has Mass of the wings huve total surfuce to stay in the air . density of air is 129 Kym"v Furincl wing ifthe plane ' 160 ns. The flow over the upper calculate pard of the air additional Trcuucnci 24 Hz When an end of spring oscillates with thc tuluc = (A)he mas5 mass M at the Hz Calculate ;added to I; the frequency 12000 g mass (B) the spring constant K



Answers

Additional Problems
Figure 44-14 shows part of the experimental arrangement in which antiprotons were discovered in the 1950s. A beam of 6.2 GeV protons emerged from a particle accelerator and collided with nuclei in a copper target. According to theoretical predictions at the time, collisions between protons in the beam and the protons and neutrons in those nuclei should produce antiprotons via the reactions
$$\begin{array}{l}{\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{p}+\mathrm{p}+\overline{\mathrm{p}}} \\ {\mathrm{p}+\mathrm{n} \rightarrow \mathrm{p}+\mathrm{n}+\mathrm{p}+\overline{\mathrm{p}}}\end{array}$$
and
However, even if these reactions did occur, they would be rare compared to the reactions
$$\begin{array}{l}{\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{p}+\pi^{+}+\pi^{-}} \\ {\mathrm{p}+\mathrm{n} \rightarrow \mathrm{p}+\mathrm{n}+\pi^{+}+\pi^{-}}\end{array}$$
Thus, most of the particles produced by the collisions between the 6.2 GeV protons and the copper target were pions.
To prove that antiprotons exist and were produced by some limited number of the collisions, particles leaving the target were sent into a series of magnetic fields and detectors as shown in Fig. $44-14 .$ The first magnetic field (M1) curved the path of any charged particle passing through it; moreover, the field was arranged so that the only particles that emerged from it to reach arranged so that the only particles thad to be negatively charged the second magnetic field (Q1) had to be negatively charged (either a $\overline{\text { p}}$ or a $\pi^{-} )$ and have a momentum of 1.19 GeV/c. Field Q1 was a special type of magnetic field (a quadrapole field) that focused the particles reaching it into a beam, allowing them to pass through a hole in thick shielding to a scintillation counter S1. The passage of a charged particle through the counter triggered a signal, with each signal indicating the passage of either a 1.19 GeV/c $\pi^{-}$ or (presumably) a 1.19 GeV/c $\overline{\mathrm{p}} .$
After being refocused by magnetic field $\mathrm{Q} 2,$ the particles were directed by magnetic field $\mathrm{M} 2$ through a second scintillation counter S2 and then through two Cerenkov counters C1 and C2. These latter detectors can be manufactured so that they send a signal only when the particle passing through them is moving with a speed that falls within a certain range. In the experiment, a particle with a speed greater than 0.79c would trigger C1 and a particle with a speed between 0.75c and 0.78c would trigger C2.
There were then two ways to distinguish the predicted rare antiprotons from the abundant negative pions. Both ways involved the fact that the speed of a 1.19 GeV/c $\overline{p}$ differs from that of a 1.19 GeV/c $\pi^{-} :(1)$ According to calculations, a $\overline{\mathrm{p}}$ would trigger one of the Cerenkov counters and a $\pi^{-}$ would trigger the other. (2) The time interval $\Delta t$ between signals from $\mathrm{S} 1$ and $\mathrm{S} 2,$ which were separated by $12 \mathrm{m},$ would have one value for a $\mathrm{p}$ and another value for a $\pi^{-} .$ Thus, if the correct Cerenkov counter were triggered and the time interval $\Delta t$ had the correct value, the experiment would prove the existence of antiprotons.
What is the speed of (a) an antiproton with a momentum of 1.19 $\mathrm{GeV} / \mathrm{c}$ and $(\mathrm{b})$ a negative pion with that same momentum? (The speed of an antiproton through the Cerenkov detectors would actually be slightly less than calculated here because the antiproton would lose a little energy within the detectors. Which Cerenkov detector was triggered by (c) an antiproton and (d) a negative pion? What time interval $\Delta t$ indicated the passage of (e) an antiproton and (f) a negative pion? [Problem adapted from O. Chamberlain, E. Segre, C. Wiegand, and T. Ypsilantis, "Observation of Antiprotons," Physical Review, Vol. $100,$ pp. $947-950(1955) . ]$

Okay this question has many parts but in the first part we have to approved that the stiffness of simple pendulum is equal to M. G. Out of. Yeah. And the angular treatment C. Is equal to the older. And so in order to show this to immigration, we have to draw the simple pendulum. Suppose we have a simple pendulum of mars and and land in hang from a support making an angle theater. And then this is the displacement through which it is moving. Then the restoring force F. Is equal to minus mg. A scientist restoring forces of force which brings uh the simple pendulum divorces mean position. So this is equal to a physical minus MG scientist to here scientists that he is equal to exhale dove. And so we put eggs out of balance data scientist to then the restoring force ever become minus MG. X. Out of. And and the supreme constant K. Which is also called stiffness is equal to minus if odo X. So here it is restoring for. So we will put the available the restoring force which we call palatable, which is equal to minus into minus MG. X. Of dove in and divide by X. So K. Is equal to K. Is equal to MG out of which was required. And the angular frequency omega. Oh my God scare is equal to pay out of. And as we have compared with the. Okay so maybe scare is ignorant. Do is equal to M. G. O. D. And out of them. Therefore we can write the angle of frequency S. G. Out of. And when the theater is very very small, which is approximately equal to scientist to. Then we can write data as it will to eggs out of in and X. Is equal to in Quetta. And we know that the restoring force F. Is equal to minus MG scientist to. So we can write designed to essex MG went to minus MG. Ah to to and this is a question number. Uh two. And according to Newton's 2nd Law of Motion, according to new dance, New done second. No F. Is it will do M. X. Double dot. This is a vision number two comparing equation two and three. Then we get minus MG. Tita is Ecuador and and theater double edge. So rearranging the situation, we get heated about dash plus the G. O. N. And theater isn't going to zero. So here he had a geode up and have same dimension as frequency dear for or maybe as scary as it went to a year out of hand. And in the second part of the question we have to show ah there yeah we have integration which is the clinical pendulum. And this is this is a chronicle pendulum and it is see is is stiffness and this is in England theater. So the restoring force in this case is equal to F. Is equal to minus a c theater. This is situation number one and the moment of inertia, moment of inertia is Mhm. Moment of and then she yeah is equal to. I see the revelation comparing one end to this equation number one. And this is a question number comparing this equation we get seat is equal to I hit a devilish rearranging this. We get he did the will days plus see out of I data is you want to zero. So here see and I have the same dimension Air street mincy. So therefore we can write home investigators are going to see out of bye. And In the 3rd part we have a day you know which is it was this sp masi. Um And These are the two sittings tension T. And it is attention T. This is the displacement through which it is moving. It is the land of the system. So here the restoring force is equal to F. Is it will do Everything will to -2. T. Exhale out of and and According to Newton's 2nd law of motion are visible to M. X. The world dot comparing this to envision we get em X. Double door is equal to -2 d. x. by L. And by rearranging this to this equation we get X doubled up as a duty eggs. Out of L. M. Is equal to zero. Yeah to T. O. Out of L. Have the same dimension as pregnancy. So therefore we can write on investigators. It will do investigators England to T. Oh, darn in there. And in the third part of this question we have to uh we have to prove that the angular frequency is equal to the angle frequency is equal to which is only the skier is equal to uh to G. O. Tough. And so and I did to show this. We have to try the Degra. So we have a diagram which is a visit in U. Shape. This is yeah I'm sorry I have to try it again. Okay we're here for we have a U. Shape. Isn't U shaped visit. Okay. It is filled with a liquid liquid and the distance from here from neck of the visible to the league with his ex. And the distance from the two Ages is even 1- two x. And this is X. And it is filled with what and the displacement of the liquid with a height of X. Here's uh ah week. Hi there X. Is equal to X out of two. So here the displacement is exalted. Any stiffness has given us. K. Is equal to G. O. Golf. Explain to expert to any here the G. Is given as role X. G. Old of X. So we can write the stiffness case equal to puro. Uh Yeah. And the Newton's second love motion is equal to and excellent or not. So, so we can write, it is minus G is equal to M. X double lot. And this can be written as minus to roll A. G. X. Is in winter. Mm X double Lord. So here this And this can be written as -2 row G. X. Is equal to rule and and acceptable not as much as it were to then still into area. So therefore we can write it? S acceptable dot plus two G out of L. X. Is equal to zero. So therefore the omega scare is equal to buongiorno off at and in the earth part of the question we have to again show that or maybe the scare is it will do. Biggest fear is equal to gamma B. A. Out of el ruby. So here we have a a container like this. This is a container and it's more luminous. We this is the land of the neck of the container and displacement is X. By by taking the local time of the invasion which is we have integration. PV gamma is equal to constant. So taking log on both side, vegan it? S lago B bless plus grandma log of we is equal to constant. So this can be written as D P L. P plus gamma devi of the we is equal to zero and we can write it as D. P. Is equal to minus gum up the D. V. Older gov as changing volume which is David, L. V is equal to A. X. So we can write Delpy is equal to is equal to minus gamma B X. Out of of me as as the plus next has the moss which is equal to end the mass of the flask of the ask is equal to rosa. And so according to Newton's second I'll motion we can write A D. P is equal to M. X double dot. So therefore weekend write it as as D. P is equal to minus gamma. The a scared eggs divide by V is ignorant row and X double dot. So we can write this excitable dot plus gamma be divided by and ruby is able to zero. So oh my investigator is equal to gamma B. A. Old off L roby.

Okay. So after reading that paragraph, what is the speed of an anti proton? With a momentum of 1.19 Giga electron volts per seat. So they have a relativistic relationships. So we know that the momentum is gamma and the so we have The oversee is equal to two square root of 1 -1 over pc over EMC squared squared plus one. So for an anti proton EMC squared is 938 0.3 mega electron volts. PC is 1.19 Giga electron volts Which is 1190. My collection bolts. So, plugging that in and we get a velocity 0.785 C. Part B a negative pi on with the same momentum. So for a negative pion we have EMC squared is a 193.6 mega electron volts. It has the same pc. So it gives us a velocity of 0.993. See So for part C, an anti proton. So this beat of an anti proton is about .78 C. But not Over .79 seed. So an anti proton will trigger C. Two a negative pi on. So the speed of the negative pi on is over .79 c. So a negative pylon will trigger see one what time interval adult ET indicated the passage of an electron or for part of a negative pi on. So we have that delta T. Is the over V. We know the distance is 12 m. We know those velocities. So for part E. We have adult it T. of 51 nanoseconds and for part F. Yes, We have adult itty of 40 nanoseconds.

Okay, so we have a very long problem here. Um, I took the liberty of drawing it, so I'm going to try to 0.2 things in blue as necessary as I'm reading. Um, also, I did use Microsoft Excel to solve this problem, and I can show this to you as well. It says Figure 10.32 which I tried to draw on the screen. She was an apparatus used to measure rotational inertia is of various objects in this case, spheres of varying masses, M and radi. I are The spheres are made of different materials and some are hollow, while others are solid. Um, now, one thing that I know is that inertia for ah, hollow sphere is Let me look for that hollow sphere is 2/3 ammar squared and inertia for a solid sphere is to fifth m r squared. So as it spoke, the question spoke about hollow spheres and solid spheres. It made me think of the different coefficients that come in front of m R squared. It's either 2/5 or 2/3. Okay, let me continue reading to perform the experiment. A sphere is mounted to a vertical. Axl held in a frame with essentially frictionless bearings. So up here is the sphere. Ah, spool of radius. Be so I wrote radius beyond there equals 2.5 centimeters. So I should write that down. But I'm going to write point zero to five meters. Okay, is also mounted to the axle and a string is wrapped around this school. The string runs horizontally over and essentially frictionless pulley and is tied to a mass m so we can see that over here. Tying to em and um is 77.8 grams. But I would like to write it as 0.778 kilograms. I generally like to write things in S I base units as the mass falls the string in parts of torque to the spool slash axle slash disc combination resulting in angular acceleration so that mass is going to accelerate downward. And at the same time the spool is going to spin. Ah is going to accelerate its spinning. Ah, and the mass is tied to the pool by that Axl. Okay, The mass of the string is negligible, but the combination of excellent spool has non negligible rotational inertia. I subzero. We're going to read that. We need to figure out what that I subzero is. Um that value is not known in advance. In each experimental run, the mass is suspended at a height off 1.0 meters. So I wrote that down here. Looks like I should have had 20 is Okay, so that site above the floor and the rotating system is initially at rest. So our initial angular velocity and our initial speed are both zero. The mass is released and experimenters measure the time to reach the floor. Results are given in the tables below. Determine an appropriate function of Time T which, when plotted against other quantities, including M and R, should yield two straight lines, one for the hollow spheres and one for the solid ones. Plot your data established best fit lines and use the resulting slopes to verify the numerical factors. 2/5 and 2/3. Um, and I've already written those down there in the expressions for the rotational and nurses of spears given in table 10.2. You should also find the value of the rotational inertia of the axle and the drum. Together they don't mean drum there. I think that's a typo. They mean the spool, the axle and the spool. So, um, the axle and the spool are rotating together, so they're going toe, Have, um, one total inertia. Okay, that seems like a lot. Nevertheless, the first thing that I'm going to dio is I'm going to write the some of the forces equals mass times acceleration. And I'm going to draw that. This are these air the forces one the mass element. So one that mass element there is wait, which is m times G. And there's also tension. Now we know that it's going to accelerate downward. So I'm going to set my axis so that a is downward. So what that's going to give me is when I write this down, the some of the forces are tee minus MGI or rather, MGI minus T. Because I'm taking downward as the positive direction. So let me redo that. Ah, going there. You stress again m g minus t. And that is going to equal the mass time to the acceleration. We can solve this for acceleration and we get a equals m g minus t over em, which again shows us that a equal simplifying it a little bit more g minus t over em, and we are going to use this later. Okay? Now I'm going to draw some of the torques. So if I draw the spool, I'm gonna draw it from above, and I'm gonna change my colors here, maybe read. Draw the spool from above the torque. Um, which is, which is the tension times the distance is pulling from above, and then our acceleration is defined to be in the same direction. So some of the torque equals I health. Ah, Twerk some of the torque his eye Alfa The torque is detention times B, which is the radius of the spool. And that equals I off she be equals. We know that I is some constant. I'm gonna call that constant K que is either going to be 2/3 or two fists times m r squared Now, um, good. However, our inertia is not just que times m r square. There's also the small inertia that we have to add on. Um so the k times m r squared is the inertia of the sphere. That small inertia I subzero is the inertia of the spool and maybe of the rod. Okay, so now we need to take that times Alfa the gay. Next thing that we need to know and I'll change colors here is that why equals this is for movement of the mass. Why equals y subzero? Um, plus V y subzero t plus 1/2 a t squared. However, I do find a to be downward. And so I need to change that to minus 80. Squared nips one of that to be green. Okay, Visa zero is zero. Um, why at the end is also zero zero equals? Why subzero? Why subzero was age, which is one, um, minus 1/2 A t squared. And we can solve that for a a equals to h over t squared. Really? Cutting the Children room here to h over t squared. So we're gonna need that later, and we're going to need this later. There's three things we're gonna need later. Um, we're also gonna need something else, So let me go to another page. A equals off, uh, are we don't have in our in this problem as you look back at how we draw on this problem. Um, there's no lower case r r is raised this fear, but this this are that I'm talking about now is the radius of this school which is actually be so a equals alfa be. And so if I solve that for Alfa, we see that olive ah is a over B. Okay, so now I'm going to go back to the equation that said a equals T over AM minus G r a g minus t over on my bed right here ache with G minus t over em. So let me write it over here. And that should be a lower case. M is the mess of the object. Okay? But we know what T is. So a equals. Let's go back and look what he is. I think I'm going back here. Okay, So from this equation, right here T is K m r squared plus ice. Subzero times. Alfa Overbey, K M r squared plus ice of zero already for gotten K M R squared. Okay, so we got that k m r squared ice subzero times, Alfa over. I m okay, but ah, we know what Alfa is from over. Oh, no, I have it on this page. Alfa is a over B. So that's going to give us G minus k m r squared. Plus I subzero over em a over B. And let's look at what else we got here. We know down here that a is to age over t squared. Okay? I'm not gonna use that quite yet. What I'm gonna do is I'm gonna add this to both sides, and I'm going to get that g equals a plus k m r squared. Plus I subzero over MB a Okay, m r squared. Plus I subzero. Okay, good. Can factor out the A So that's going to be a times one plus k m r squared. Plus I subzero over and be okay, but now we know a trying to go back to it. A is two h over t squared. She is to h over t squared times one lus Okay, m r squared. Plus I subzero over and be Now what I want to do is I want toe Ah, factor out the MBI the one over m b from the inside. So that's gonna give me two h over m b t squared times, MB. I'm just realizing way back, way back when I wrote this equation right here, I forgot to divide by be on both sides. So I should have written k m r squared. Ah, plus I subzero Alfa Overbey. So when you realize you have a mistake, just go back and fix it. That's going to put it. Was writing right in here K m r squared. So that's gonna put a B down here. It's just going to give me another be here. It's going to give me a B squared here or B squared here, B squared here, B squared here, b squared here. Okay, now we're now we're doing good. Um, plus, let me think about what we're doing here. Okay? Um que m r squared? Plus I subzero. Okay, um r squared. Plus I subzero. Okay, um, now I'm going to invert this and multiply it onto both sides. I'm also going to rearrange so that ah, these two are going to reverse their order. And let me write it again. Next page G I m b squared over to h t squared equals I subzero plus m b squared plus k m r squared. This is what I wanted to come up with. So this 1st 1 I am going to plot this. So I'm gonna take the t value that's given in the, um, tables. Take the T value, and I'm going to calculate this, Okay? I'm going to plot it. Verses. Um, m r squared. So m r squared is going to be I'm gonna write verses this tous okay, on the x access. So I'm going to calculate Elmar squared for every single one of them. Um, all right, that's going to give me that. The ah slope is gonna be K and the Y intercept is going to be I subzero plus and B squared. So again, I have to calculate this for every single row, every single column, actually in the table. I also have to calculate this for every single column and then plot them this one on the X axis, this one on the Y axis. And then I can read from the graph the slope on the Y intercept. So I did this using Microsoft Excel. So here we have it. I put in the first table up there on the top. I calculated in column D I calculated m r squared, um, that second tables on the bottom and I calculated I'm r squared there also. Then I also calculated t squared times g m B over to H. That is in column E. I'm all applied by 1000 because in excel, it's a little bit hard to see if you didn't ah, have reasonable, reasonably large numbers. And so I plotted the points and you can see the points there. There are 10 points. I then sorted out the points so you would see that the top table is not the top table that you saw in the book. And the bottom table is not the bottom table that you saw in the book. I sorted them out because, um, some of them are hollow and some of them are solid. Um, now we can look at the Orange Line and the blue line. You put a line through the orange ones. You can see how it says seven e negative. Five X with a slope is seven. Really. The slope is 70.6666667 So let's go back to the white board. The slope of one of them was 66667 which is the same as 2/3. And if we look back here, 2/3 goes with hollow, hollow orange. Okay, so the orange line on the spreadsheet represents all of the hollow spheres. And again, they were from different columns. They weren't all in the first table. They weren't all in the second table. Ah, some of the hollow ones air in the first table, and some of the hollow ones are in the second team. All right, you will notice that, um, the blue line 0.4. So the blue line slope is is 0.4 or a multiple of 0.4, an order of magnitude of point for so slope number two four. If we look back, 2/5 is the same as point for. So those are the solid spheres. And so that was the blue line. All right, so we got our slope. Now we've got to do our Why intercept Now, notice that our Y intercept is ice subzero plus M b squared. Next page I subzero plus and B squared equals. Let's go back to the spreadsheet. The Y intercept. Now the Y intercept is shown to be point 978.978 However, um, masses in grams. And, um actually, that doesn't matter. Hey, it's ah, point 09 70. So let me write this down. But I have to divide that by 1000. And the reason why I have to divide that by 1000 is on the spreadsheet In column E. I multiplied by 1000 so that I could see the number for the Y intercept. If I didn't do that, that Excel would not be showing me the number very well. So I need to divide by 1000. All right, if we solve this using the values of M and B. So again, M was 0.778 be was 0.0 to 50 Put those into an equation and solve for I subzero. You will get I subzero equals 4.9 times 10 to the negative fifth power kilogram meter squared. That's just off by a little bit. But from what it says on the book, there's there's got to be a rounding error. Um, perhaps in my use of Microsoft Excel, Um, perhaps in the makers of the book and what they did. Um, but nevertheless, that is how you would solve the problem. So what I want to do now is I want to make some changes here. Teoh, Um, my spreadsheet. Notice that Put mess Ingram's. I'm gonna change that to mass in kilograms just to show you what's gonna happen to the and she. But we are done, by the way. That was the answer. Um, 0.73 0.432 0.677 0.947 0.8 to 1. And so now, if we can see, we have here for the, um, slope again, it's a little hard to see. Um, exactly this this first number that that it's, uh, 6667 Whatever. So I just want to try. Let's try changing this two meters. You're a 6 to 5 when you're 386 Hey, now, what just happened? Looks didn't see what that said. 942 Still can't see it. Um, man, let's try. Tried to small applying this by 10 trombone playing this times. No, uh, back toe. Would we have 1000 and multiplying that by 10,000? I'm not even sure that I'm messing with the right one right now. Um, trying to make it so we could see that easier. Um, that's pretty funky. Try multiplying this times. 1000. All right. Um, nevertheless, I clearly just made a mistake in something that I was doing there. Um, it was right up until I started just messing with the spreadsheet. Let's go back, Teoh. Here. Um, we said that the orange one is the one that was 10.6667 and I think it was messing with the wrong one anyway. So let's see what I can do with this orange one. I'm going Teoh go here. And instead of multiplying by 1000 I'm gonna multiply by 10,000 then drag this down. You know, I'm gonna get rid of the blue because I don't know what I did to the blue one, but whatever I did was not good. Um, 0.7 All right, so rise overrun. So I need to make this a much bigger number. Let's just give it three more zeros. I just want to see that slope. There we go. 6674 Finally. So if you do see a trick of multiplying. Um, then you can see, um, the slope and the Y intercept. And that's what I wanted to show you at 667 All right, So, um, thank you for watching that. Um, took some time and we went through three pages, but we came up with the answers, so I thank you for watching.


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Hormone that promotes growth of lateral buds and has negative effect on apical dominance is (a) cytokinin(b) gibberellin(c) auxin(d) both (b) and (c).
Hormone that promotes growth of lateral buds and has negative effect on apical dominance is (a) cytokinin (b) gibberellin (c) auxin (d) both (b) and (c)....
5 answers
Based on the periodic trend, arrange the following sets of elements in order of increasing atomic size.(a) $mathrm{Rb}, mathrm{S}, mathrm{Sr}$(b) $mathrm{Ca}, mathrm{Li}, mathrm{Mg}$(c) $mathrm{K}, mathrm{Ca}, mathrm{Br}$
Based on the periodic trend, arrange the following sets of elements in order of increasing atomic size. (a) $mathrm{Rb}, mathrm{S}, mathrm{Sr}$ (b) $mathrm{Ca}, mathrm{Li}, mathrm{Mg}$ (c) $mathrm{K}, mathrm{Ca}, mathrm{Br}$...
5 answers
Be sure t0 answer all parts:Civen the thermochemical cquation for photosynthesis; GH,O() 6COn(g) CcHuzOs(s) 6Oz(g)AH =+2803 kImol Cakulate the solar energy required t0 produce 7152 g of CcHnOs:Enter your answer in scientific notation:
Be sure t0 answer all parts: Civen the thermochemical cquation for photosynthesis; GH,O() 6COn(g) CcHuzOs(s) 6Oz(g) AH =+2803 kImol Cakulate the solar energy required t0 produce 7152 g of CcHnOs: Enter your answer in scientific notation:...
5 answers
Your W ! Silgtore Question revious Pagc Left:1.29.24 = Answer 2 temperature were nitride (SizNa) Si reacts point) Dinkey Ut made 1 Gopalbhal units Zatculaoe bne Thakkar the fonobimngg percent nitrogen gas vield realtoont (Si) 3 € pue rcaction; and nitrogen 3 gas 1 2 Page: at a 3
Your W ! Silgtore Question revious Pagc Left:1.29.24 = Answer 2 temperature were nitride (SizNa) Si reacts point) Dinkey Ut made 1 Gopalbhal units Zatculaoe bne Thakkar the fonobimngg percent nitrogen gas vield realtoont (Si) 3 € pue rcaction; and nitrogen 3 gas 1 2 Page: at a 3...
5 answers
JrcesIncorrect: Only one E2 elimination is possibleintsThe structure of the compound "B' is:A)SkillnQ88 19.45 1144HOB)KoStudyOH
Jrces Incorrect: Only one E2 elimination is possible ints The structure of the compound "B' is: A) Skill nQ88 19.45 1144 HO B) Ko Study OH...
5 answers
The differential equation 1y 2cy - f(t,y) In |zl+C. +y? has a solution of the forI Find the constant € which satisfies the condition y(1) = 1 initial0 2In 2In 2()Ine
The differential equation 1y 2cy - f(t,y) In |zl+C. +y? has a solution of the forI Find the constant € which satisfies the condition y(1) = 1 initial 0 2 In 2 In 2 () Ine...
5 answers
A codon is best defined asthree nucleotides that code for an amino acid.a sequence of nucleotides in DNA that codes for a functionalproduct.any random segment of DNA.a sequence of nucleotides in RNA that codes for a functionalproduct.
A codon is best defined as three nucleotides that code for an amino acid. a sequence of nucleotides in DNA that codes for a functional product. any random segment of DNA. a sequence of nucleotides in RNA that codes for a functional product....
1 answers
Find the pressure of 3.40 $\mathrm{mol}$ of gas if the gas temperature is $40.0^{\circ} \mathrm{C}$ and the gas volume is 22.4 $\mathrm{L}$
Find the pressure of 3.40 $\mathrm{mol}$ of gas if the gas temperature is $40.0^{\circ} \mathrm{C}$ and the gas volume is 22.4 $\mathrm{L}$...
5 answers
What are the coordinates of the centroid of the triangle shown in the figure? Notice that it shows the lines between cuts whose equations are: x =4in; (line between vertical cut) y = 5 in; (horizontal broken line)6 in5 in A18 inX = 4 in(6 in; 2 in)(16 in, 7 in)(12 in, 2 in)(18 in; 6 in)
What are the coordinates of the centroid of the triangle shown in the figure? Notice that it shows the lines between cuts whose equations are: x =4in; (line between vertical cut) y = 5 in; (horizontal broken line) 6 in 5 in A 18 in X = 4 in (6 in; 2 in) (16 in, 7 in) (12 in, 2 in) (18 in; 6 in)...
5 answers
Question 9 (2 points) The = enzyme phenylalanine dehydrogenase is deficient in patients with phenylketonuriaTrueFalseQuestion 10 (2 points) The pancreatic lipase needs colipase to properly attach to the TAG in the micelle; micelle and digest theTrue FalseQuestion 11 (5 points) Name the 2 most essential fatty acids (both thesc fatty acids dietary essential? omega and chemical names). Why are
Question 9 (2 points) The = enzyme phenylalanine dehydrogenase is deficient in patients with phenylketonuria True False Question 10 (2 points) The pancreatic lipase needs colipase to properly attach to the TAG in the micelle; micelle and digest the True False Question 11 (5 points) Name the 2 most e...
5 answers
Doloring [70 D-value r.790 Ulnn Uie ( Lablo Mounarusui in you #w CnirternuTUsJt shaxntnoh scloc'ng chokue tonbeonp-valaAlfaAU CThuiulureCccemanconduon Made in Eurqaand hArsumdnn neudgomor lo purlcrirpfopodur? 9444Wei applyThu @innlIndnonndeniTnerpcoulutoneuneeucudenallnentaningnnandemcnuueon meangunkrorn und eJu40o HJctn 0liru Jamo U6 0re higiur Iran J0
Doloring [70 D-value r.790 Ulnn Uie ( Lablo Mouna rusui in you #w Cn irternu TUsJt shaxntnoh scloc'ng chokue tonbeon p-vala AlfaAU C Thuiulure Ccceman conduon Made in Eurqaand h Arsumdnn neudg omor lo purlcrir pfopodur? 9444 Wei apply Thu @innl Indnonndeni Tnerpcoulutoneuneeucudenallnentaningnn...
5 answers
Provide two cations that can be added to Cl that will cause precipitate?HTMLEdltora
Provide two cations that can be added to Cl that will cause precipitate? HTMLEdltora...
5 answers
2. Given the function f (x) =x 3bx- (c + 2), delermine the values of b and such that f (1) = 0 und f'(3) = 0.
2. Given the function f (x) =x 3bx- (c + 2), delermine the values of b and such that f (1) = 0 und f'(3) = 0....

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