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WIS (0) =('0 So) (0)f = x0 utSuOpnsqns 341 MOY MOYS *SnonuJuoj J1 FBP (Jrssaj3u J4 1e4} SuunssV anj Jjod 0 Jo 41buz7 'I6 saiduexg pue Ko3yl"aunay But...

Question

WIS (0) =('0 So) (0)f = x0 utSuOpnsqns 341 MOY MOYS *SnonuJuoj J1 FBP (Jrssaj3u J4 1e4} SuunssV anj Jjod 0 Jo 41buz7 'I6 saiduexg pue Ko3yl"aunay Butkuedu1o3oe 341 UI uor3al ] popeJUI[ 34 J0 14311 341 01 Gz SoJ JS3 € = / Jul[ 341 JAOcSOJ 341 Jo doo[ JJUUI 341 episino PLE door0 WS D=4 (2)Soj p = . (4) p =1 (B) :(0 < p) S?[JJIJ 3uIMO[ Aequalejunjl? 124 341 ?1e[nopej 01 (1) uouenbg U! e[nu 30 Bel aouauadxa Ised YILM Ju9JS-SLOJ SljnSaU SJN5 Tealqo Jellluey Ia uO 1! KJ1 01 B?pI

WIS (0) =( '0 So) (0)f = x 0 ut SuOpnsqns 341 MOY MOYS *SnonuJuoj J1 FBP (Jrssaj3u J4 1e4} SuunssV anj Jjod 0 Jo 41buz7 'I6 saiduexg pue Ko3yl "aunay Butkuedu1o3oe 341 UI uor3al ] pope JUI[ 34 J0 14311 341 01 Gz SoJ JS3 € = / Jul[ 341 JAOc SOJ 341 Jo doo[ JJUUI 341 episino PLE door 0 WS D=4 (2) Soj p = . (4) p =1 (B) :(0 < p) S?[JJIJ 3uIMO[ Aequalejunjl? 124 341 ?1e[nopej 01 (1) uouenbg U! e[nu 30 Bel aouauadxa Ised YILM Ju9JS-SLOJ SljnSaU SJN5 Tealqo Jellluey Ia uO 1! KJ1 01 B?pI pOO3 1 S! J! 'einw ~paoey u3ym '[ensn SV 53pj4pj Jo Saju3jaunjiij 'Op 2A 2 20 =0 07 SOJ +A =1 JAnJ ?4L '68 742 =0 =0 0z UIS =4JAinJ J4L "8€ IA +/2 =0 =0 ~(E/0) JuInJ JUL 'Le -ul E503 -6 "(0 $03 Dz = 4uju33s ooqead J41 9f =0 {6 503 V/9 JuauBos J!OquJed J4L St 4-0 E 0 UIS / JAInJ ?L TTi "(2/o) Po'u) J T 50J Soj € + [ = LOJELTI J J0 cooi = 4JpJ11J 341 apISInO PUE # SQo - 22S) ?plino PLE ~ Jn PIOIpJEJ J41 [p = 1 JpisIi pur + * 4 PpJ13 J41 a V PiIp V pue Soj Piduee' Pqe SOJ V M - FuE | P 7 s = Jo 3o {ymojb Jeiv



Answers

I just posted a question 12 minutes ago and I am not sure if it will be understood because of the way it was typed. So, this is a correction to the question so that it is understandable I need to find the volume and density of the hollow brass cylinder with a mass of 106.23g +- 05g. And then after the errors for each calculation as described in the attached sheet.

Data


.Measurements with caliper. Margin of error +- 0.0025cm.

Data: Trial 1 (Length:
7.77cm, outside diameter: 2.12cm, inside diameter: 1.59cm, thickness:0.53cm). Trial 2 (Length:
7.80cm, outside diameter: 2.18cm, inside diameter: 1.56cm, thickness:0.62cm). Trial 3: (Length:
7.78cm, outside diameter: 2.12cm, inside diameter: 1.58cm, thickness:0.54cm). Mean: (Length:
7.78cm, outside diameter: 2.14cm, inside diameter: 1.58cm, thickness:0.56cm). Standard deviation: (Length:
0.007cm, outside diameter: 0.014cm, inside diameter: 0.007cm, thickness:0.021cm). Thank You.




Having done all Unicorn Dance Trip room is cool to room. Not one plus ex upon end. So if we plot the rules grab that access called to deal with having that is to do not. Then it starts increasingly nearly like this in second part. If we consider the smaller limit DX. Mhm. At a distance X. The mask of it will be all right ruined. Two dips well not want less X. Upon end in two D. X. To find its total Mazz. Uh huh. We have to integrate it for the limit 02 L. Mhm. Or not one plus X. Upon end. Mhm. The X. Mm. Mhm. So you will get a cronut X plus X squared by to it. For the limit zero to win. So you will get root note and killers and the square going to work minus zero. So on solving it you will get three will not end fight too. They're so thanks for watching it.

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.

So we're giving information for two different sets of data. And so let's call one set of data set one and information for the other set of data will be called set to. So range is defined as the maximum value in the data set of mine is the minimum value in the data set for So for the first day to set the maximum value Is 6.10 minutes. And for the other data for this dataset. Also The minimum time is 0.38 minutes for the other day to set the maximum value Is equivalent to 10.49 minutes. We'll have a minimum value Is given by 3.82 minutes. So you can find that the range first at one is Is equivalent to 5.72 minutes while the range were sent to is equivalent to 6.67 minutes. And really the only conclusion you can really draw from the range of the data is that there's really a greater for the second set of data there is essentially a greater difference between the maximum and minimum values. Branch isn't often the most particularly useful set of data since basically it's only also it's if you're only given the range and you're not given the maximum and minimum values range isn't particularly useful. Us US parameter basically to interpret. So mean is defined as the sum of everything in the data set divided by the number of elements in the data set. So for set one, if you plug everything into the one very little stats function, you can find that the mean is approximately, so you find that the mean for set one is equivalent to 3.88 minutes. While the mean first step to similarly using one variable Stassi in or plugging in Basically directly into the formula is equivalent to 7.02 minutes. So you can see here that there is essentially a significant difference, significant difference between the waiting times of basically these two sets of data, to be conclusively sure though that there is a significant difference between the two sets of data. Technically you would have to run a two sample T test, but just by eyeballing it, there seems to be a significant difference in the waiting time based on the mean of the of the two different sets of data. However, it's also important to remember that essentially mean can be easily skewed to the right or skewed to the left by different sets of outliers. So we have to be careful, especially with mean and often the median is a better way of approximating. So in this case were also asked for the sample standard deviation and in this case we're using specifically the sample standard deviation rather than the population standard deviation, Since we're only given basically a subset of the basically the total number of customers that enter basically this location. So the sample standard deviation for the first set of data is equivalent to Equals 1.55 minutes. While the sample standard deviation for the second step of data is equivalent to 2.24 minutes. So the sample standard deviation gives us a very good idea of basically the variability of data and since basically set to has a larger sample standard deviation, essentially this implies that there's typically greater variation in data for basically the second step second, basically the second set of data. And we're also asked to find the coefficient of basically the coefficient of variability mhm which is typically defined as this population standard deviation over the population means. But in this case, since we're treating a sample, we have the sample values. So the coefficient will just call in this case we can call it new or something like that is equivalent to the sample standard deviation divided by the mean of the sample. Okay, so for the first step of data were given that essentially The sample standard deviation is 1.55 minutes Divided by the mean, which is three eight minutes. So that gives us a final value for the coefficient of variation Of about 0.4. Well, for the other South data are mean is given us seven points 02 Minutes. While our sample standard deviation is 2.24 minutes. So this is basically has um you of about 2.24 by the seven point there are two, about 0.32. So the importance of essentially the coefficient of variation is basically adjusts the mean, basically the sample standard deviations for the different means. So although it appeared initially that essentially we had greater greater variation and sent to based on just eyeballing the sample standard deviation in actuality, we have a lower coefficient of variation for sent to. So the lower coefficient of variation means for that specific mean, for set two, basically you have lesser variation around the mean, while for basically it's actually greater for someone. So also it's important to be careful not to be misled by essentially the sample standard deviation, basically just by the value of the standard deviation itself unless basically the only case where basically you can directly interpret the sample standard deviation is if you have two different samples with the same mean. So in that case, basically the coefficient forum that isn't really, you don't have to use the coefficient formula since you can directly interpret it from the sample standard deviation right? And the last piece of information were being asked is basically the scariness of the two different sets of data. So for these sets types of data, probably the best way the best way to check for squareness is to use basically dot plots since basically Top Plus can give a good indication, like you have a good indication of the shape of the data. So essentially we can draw dot plots for a set one and set to So the dot plot for set one, the range of values is typically from From the lowest is about zero to the highest, which is about six. So so 123, four and 5. And we're just going to approximate. So 4.21 5.55, 5.13 4.77, 2.34, 3.54, 3.20 4.5 6.1, And 3.79. So, you remember in this case that the average is essentially about 3.88. Yeah, So we can just nearly draw a line here for 3.8. So we can see that approximately the number of values on both sides is relatively equal. So basically this shape of the distribution is nearly symmetric in this case. So there's not really a significance here, nous. And so we're going to do the exact same process for the other set of data in this case for the other set of data being everywhere ranging from Essentially 3.82 To about 11. Maximum of 10.5 or 11. So we can still label or box plot dot plot here. Yeah, Just approximation will be fine in this case. So we have 10.5. About 6.68 5.64 4.08, 6.17 9.91 5.47 9.66 5.90 8.02, 3.82 And 8.35. So we're essentially given that the average in this case is about 7.02. So once again, we can draw a line for 7.02 here approx. So we're we can see that this basically is left. Uh This is less symmetrically distributed, it seems since, but it's still pretty close to the metric, although this has more of a skewed to the right distribution. But it's basically cement it's nearly symmetric basically to about skewed to the right, since the mean seems a little bit higher than it should be compared to the other values. It seems like the mean is being pulled towards a higher value by basically the higher values of the 8910 11 Region.

From this expression for the sinus tidal wave, we can immediately immediately see that pemex or the amplitude Is 1.2 seven. So since this is an A. C. Unit, we can read off from here that delta P max is equal to one 0.27 paschal's. And that's because delta P is equal to delta P max. Sine K. X minus omega T. So here we have delta P necks. Yeah, checks minus omega T. Great. So now that we have Mhm. Down to the max the absence of the pressure. Can we find the angular frequency omega? Well, sorry not the angular frequency but can we find actual frequency? Well, we know omega Is equal to two pi. Yes, so we have omega from expression for our way. So if you re arrange to get the F. Is equal to Omega over two pi. Omega is 300 and 45 divided by two pi, Gives us the frequency of 100 and 70 huts. So all these values are already in S. I. Units. Okay, so that we have the frequency, can we find the wavelength? Yes, we can we know that the wave number K. Is simply two pi over lambda. The wavelength lambda. So hence lambda is equal to two pipe over K. And that is to buy and K. In this case is just high and this gives us a wavelength lambda of two meters. Lastly can we find the speed in which this wave is travelling? Well we know we is equal to the wave length times the frequency of the wave a lot of times F, and this is simply two m, two m, and the frequency of 100 and 17 cuts. This gives us the speed of the wave To be 340 me just a second.


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