4

2 2 66+6 E 0' -1sa = 1S 0 Xct 22 6656 + 5 c9eut (t:2onia') 6? () faxt 0 .0 22 66$6X Lo 6j) 1So 1fo '0226 f6 X 2osolv:~j LJa set ;Xkw) 29'32 jm 8...

Question

2 2 66+6 E 0' -1sa = 1S 0 Xct 22 6656 + 5 c9eut (t:2onia') 6? () faxt 0 .0 22 66$6X Lo 6j) 1So 1fo '0226 f6 X 2osolv:~j LJa set ;Xkw) 29'32 jm 8 L~tinj oourd of #o lort;m J 1 Pul- + -0 &' Ma 5 0 1Sox 2 Ln Ylt-Checi lal 4 > 6 Jemahiy xt 1or tmc A > Ao M (x) Y0 - }x6a 0 Jm MW 05 _ 9 , 1 Cv) 5 0 3 () 309m M+ND)T time 04uliieh( 'Y holhel kooned Rr) 903nesknt 4e0 22 6656 + Fifo3 0 = 1S0 0 ,0 21 6 6f6 5 0,0 226616 + ske' (1-{ 0'0226656 € % T t2 %o. J

2 2 66+6 E 0' -1sa = 1S 0 Xct 22 6656 + 5 c9 eut (t:2onia') 6? () faxt 0 .0 22 66$6X Lo 6j) 1So 1fo '0226 f6 X 2o solv:~j LJa set ; Xkw) 29'32 jm 8 L~tinj oourd of #o lort;m J 1 Pul- + -0 &' Ma 5 0 1Sox 2 Ln Ylt- Checi lal 4 > 6 Jemahiy xt 1or tmc A > Ao M (x) Y0 - }x6a 0 Jm MW 05 _ 9 , 1 Cv) 5 0 3 () 309m M+N D) T time 04 uliieh ( 'Y holhel kooned Rr) 903n eskn t 4 e 0 22 6656 + Fifo 3 0 = 1S0 0 ,0 21 6 6f6 5 0,0 226616 + ske' (1-{ 0'0226656 € % T t2 %o. Jau' 629m Put



Answers

1. Draw a graph of position y as a function of time t. 2. Assume a relationship of the form: y=at^n where a and n are constants. Do you think it is a reasonable assumption? 3. If, as conjectured, y and t are related by a power function, a graph of log(y) as a function of log(t) should be a straight line. Calculate of the values of the log(t) and log (y) and fill them in the table. 4. Draw a graph of log(y) vs. log(t). Use the graph to determine the constants a and n (see equation in step 2 above) and give the empirical relationship between y and t. Illustrate on your graph and show the calculations on the space below.

In the first of all the questions. Okay, I'm gonna find what is the total number of efficient that has occurred. Never given that at the start to start off the efficient chain reaction total of and not nuclei will be introduced to the uranium sample. So the sixties zero generation. And it will cost and not a number of fission reaction. Now from this fission reaction w reproduction constant that tells us how many new clones. Sorry, how many neutrons are actually uh how many neutrons are produced? These nutrients will then go on and create more fusion. Right so the reproduction constant is key. And this tells us a total of key times and not number of neutrons will be produced. And this key and not well produce care and not number of fish fishing reactions. And so this is the first generation of fission reaction then so on and so forth. The second efficient second generation. And it'll generation until the end generation should be as such. So I define what is the total number of fission reaction that has occurred until the end generation. My total fission reaction. Let's just say and right. This big end is the total number will be and not plus and not K. Plus and keep square etcetera until and not kid power and specter is this out and out. So we're going to ah small little trick over here. After the fact arising out tea and not we're going to is we're going to multiply wolf left hand side right inside by key. So one times K. You give a K. Kate. I'm scared give U K square plus Q. So on and so forth. And the U K plus what? And after this we're going to add 1, 2 both sites plus one. Close to one plus. Okay. Plus key square plus Q. Thus And the UKN Plus one. Now, do you see that? Yes. Until just before came plus one. Which is until okay. And Yeah. Plus one. So until kn All right, this is our and over and not right from here. So we can replace this as an over and not plus K plus what? Now? Trying to rearrange this equation a little bit. All right. And the way I'm not gonna bring it over to the other side. Get it over and not ki minus one. Close to Kate above N -1, Sorry. And plus one -1. Right? This -1 is from the left hand side. Finally we get and offer not cause too cheap. Plus 1 -1/-1. Let me get in. It goes to a not for profit by K. M plus one minus one over key. Witness what? And this is our final equation. Then we needed to prove now we're going to use this equation in a hypothetical situation, right? For a uranium weapon. So we're given that uranium to uranium 25. You want to find out what is the time required to completely efficient? 5.5 kg of uranium. So five or 5 kg. You first you find how many new client they are. So a number of Uranium 2 5 nuclei Take 5.5 kg defined by D mass off each individual UK. There is 235. You converted into kilograms. That is 1.41. I'm standing about 25 your place next. We're going to use the equation that we previously solved to find out how many efficient generations is required in order to use this number of UK. So we equip our end To be close to this 1.41. Don't set apart for 35 equals two and not times. Okay. And plus one this one or he might have sweat and were given that the and not 90 0 generation. Yes, 10 to about 20. Okay. Is given as 1.1. All right. So we just have to substitute the equation in and we arrange a little bit so and fired by a not close to multiplied by K -1 which is 2.1. 1. -1 is 2.1 because two K And plus 1 -1, I'm going to bring the -1 to the other side and take the natural law so long of K plus one because to 2.1 in fact don't plus one. So our end and plus one will be lawn of the entire thing .1 and over and not plus one to fight by long key. So Can bring the -1 over to the other side. And we should get 99 0.2. So they we gotta round up since an easy integer value. So the minimum number of generations required is be 100 generations in order to fully Uh fishing or the Ukraine. You know 235 Ukraine's. We are given that the time required for each of this between each generation is 10 nanoseconds. So the delta T. Is 10 nanoseconds. So the total time quite for efficient be 100 times 10. The power of -9 there's nanoseconds Let me get one time stand more off -6 Which is one microsecond. So next we want to consider the explosion process right the physical explosion process of the uranium. And compare it with the time needed to fishing the uranium. So to do that we gotta need to find the speed of sound uranium. They will tell us how fast he uh physical pressure wave will transmit across the uranium. So the speed of sound. The material is given as the square of the bulk modelers divided by the density You know Casey but with this is given a 152% of nine turns per meter square. And the density has given us 18.7 I've said bo tree kilograms per meter cubed the C. D. C. For natural uranium get one in three times stand of three m plus second. No to find out what is the time needed. Right to for a wave a compression wave or physical way for like a sound wave to travel across the radios of our sphere of uranium. Well first you find what's the radius of the sphere. So the mess it's given us rule times volume. Right? So we have the mess. We have the density, we can find the radius so radios we just have to rearrange the equation a little bit gadgetry. And for for buying role put off one third Subsequuted fellas seen right. M is 5.5 kg role is the same density used over here. Should get 4.13 Understanding to AR -2 m. So the time required to travel for the sound waves to travel from within the sphere and center of the sphere to the surface of the sphere. The time and quiet. We take the radius divided by the speed of sound And we get 14 1.46. 6 Time stands Power -5 seconds And this is actually 14.6 micro seconds. So this the time required for the explosion of a force to travel from the center of the sphere to the surface of the sphere. And this actually tells us that it is possible for the entire bomb to fishing because 14 microseconds is much greater then one microsecond. So one microsecond is all the time that's needed in order to fish in the entire Product by the entire 5.5 kg of uranium. And so we expect the and I want to be able to efficient probably before the bomb actually explodes right and the objects actually starts to you know move apart. And so we can calculate what is the energy that will be released from this. Now we are given that the amount of energy released proficient energy proficient Is 200 m. e. We know the number of new clients there was calculated before. Maybe a number of new clients Is 1.4. 1 time stand about 25. So the energy released you take 250 Multiplied by 1.41 To understand power 25 and we're going to convert that into jews in order to compare TNT because our TNT, we also have it in terms of jews. So the commitment to Jews from MTV multiple about 1.69-10 power is 13 jews per M. V. P. So this gives us 4.51 time stand power 14 juice. We are given that the energy per TNT per ton of TNT is four point two do you go jaws? She is 1012 9. So you can calculate what is the number of tons of TNT for more energy cripple it tons of TNT. So you take 4.51 time step up 14 to fight by, I went to step off nine. Get one point to eat some tests time stand The power of five tons of TNT. So this is the equivalent amount of t n t four, our five kg of of uranium 2 35. So this is very, it's a very large number.

Hello. So uh this is the solution to uh the question. So this is for the uh a park with N. G. D. B. In 441 respectively respectively sorry. And then for the B. We also have that uh the does the N. G. D. Four B. 652 in the C. Part? We got uh they're supposed to be It's supposed to be a. three. So you got a straight three. Okay. So that is not uh for the scholarship question, hopefully you understand it. Thank you very much.

Welcome to this lesson In this lesson. We'll plot the the position at that time. Then after that we'll look at the relationship between the Y. And the T. Okay so let's blow that. Mhm. Yeah they told me. Oh yeah. Okay so this is the flocks with why as the position as the Y. Others and at that time on the X. As is the time to be on the X. Others here. Yeah. We ask you that the Y. The UAE is related to the T. S. A. T. And wear A. And then uh Constance. Okay. So would have to plot their lock of both sides in order to see if it's a reasonable assumption it was a reasonable assumption we should get the street line. Okay. Okay. Okay. So with that said I would. Mhm. Flood the lock of tea and a log of why? So we have locked the T. S. A. Two A. Look into the trolley for a base tan so we can't drag through. All right then. Also the love for a while that is B. Two. We are looking for fiscal year baseline. We can drag through to complete the arrest for us. So we have the log of T. The log of Y. Let's delete this one so that we plot for the locks and see if there's a straight line. Yeah. Yeah. Mhm. Okay. So it's indeed a straight line. Mhm. So the next question access to find values of A. And T. So we took the lock. Okay. Okay. The law of wilder's go to log A. T. To the far end. So here look why would be called to when we expand that to have look of a plus long of T. To the park. And so look why would be called to lock a blast and Love T. Okay. So here since you are plotting low T. A log y. Look it becomes the why interest becomes the intercept. Okay. So this is the Y intercept and and becomes the slope. So the y intercept. Looking at it as between 0.8 and 0.6. So we have look a that is equal to 0.8. Black. 0.6 Oliver too. So this is equal to 0.7. Okay then we have and that is equal to the slope of the of the graph. So the slope we cannot stand it. Us this so that this becomes the change in why and this becomes the change in T. Okay so they changed in why is from here? That is 1.3 minus zero. Okay. And this is from this side to that side. So 0.3 minus negative zero point 35. So this becomes 1.3. All over 0.65 and this is equal to two. Okay so we have then clear this side. So we have log equals two there on seven. So many that A is equal to 10 to the past 3.7 and A. Is equal to so in the cap 10 to the power. 0.7 and that is 5.1 Okay. Into the small places. Then we have an US two. Okay. So empirically we can right going home the why the Y. S. Five. So why is he called B. T. The bar And so 5.1 T squared. Ok, so that is why we can try to see if we got any closure to the values that we Yeah. Mhm. So this is the empirical from. Mhm. Yeah. Yeah. Okay. So let's have 0.5 51 joe won t to the bath too and see if we get any closer to why. Yeah. So here is 5.1 I'm the T. S. A. Two where we are looking at the power of it T. Something things he gave. And let's drop through. Okay. So let's take their variants. See how close we've been able to predict or how appropriate assumption was. So here we will take mm 2 to 82 minus Y. To be well drug through. Okay. So see that it is very very close with the highest being. Is there a 0.1 variance? Okay. All right. So this suggests that yeah, it is very reasonable to assume that why does it go to E. T. To the park and you can show your time this is the end of the lesson. Mhm.


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