5

PAGE 3 (25 pts): You have a string with a length of 2.9m and a mass of 12.5 grams_ You attach it to a vertical wall. After travelling horizontally for 2.1 meters, t...

Question

PAGE 3 (25 pts): You have a string with a length of 2.9m and a mass of 12.5 grams_ You attach it to a vertical wall. After travelling horizontally for 2.1 meters, the string passes over a pulley: A mass M hangs down from the pulley, and the distance between the pulley and the mass is 0.8 meters When you pluck the string, it oscillates with a frequency of 224 Hz. It oscillates in the following pattern: There is a node at the wall, 2 nodes at intermediate positions along the string; and a node at

PAGE 3 (25 pts): You have a string with a length of 2.9m and a mass of 12.5 grams_ You attach it to a vertical wall. After travelling horizontally for 2.1 meters, the string passes over a pulley: A mass M hangs down from the pulley, and the distance between the pulley and the mass is 0.8 meters When you pluck the string, it oscillates with a frequency of 224 Hz. It oscillates in the following pattern: There is a node at the wall, 2 nodes at intermediate positions along the string; and a node at the pulley: There are a total of 3 antinodes along the horizontal length of the string: The vertical part of the string is not vibrating: What is the amount of mass M which is hung from the pulley? Give your final answer in grams. 2. Tos



Answers

Three Pulleys, One Mass In Fig. $17-34 a$ string 1 has a linear density of $3.00 \mathrm{~g} / \mathrm{m}$, and string 2 has a linear density of $5.00 \mathrm{~g} / \mathrm{m} .$ They are under tension owing to the hanging block of mass $M=500$ g. Calculate the wave speed on (a) string 1 and (b) string 2. (Hint: When a string loops halfway around a pulley, it pulls on the pulley with a net force that is twice the tension in the string.) Next the block is divided into two blocks (with $M_{1}+M_{2}=M$ ) and the apparatus is rearranged as shown in Fig. $17-34 b$. Find (c) $M_{1}$ and (d) $M_{2}$ such that the wave speeds in the two strings are equal.

Okay, so in this question, we have kind of a weird set up. We have a string that is attached to a surface at the bottom, and the top end is kind of attached to this metal cart that is able to roll around. And so it's the top is able to move freely. Um, so it's a little bit of a weird set up, but that's honestly the most difficult part of the problem. The rest of this is pretty simple stuff that you've dealt with throughout this chapter. Um, so the mass of this spring or string sorry, um, is equal to 5.50 g, which of course, is 0.550 kg. Um, the length of the string is 86 0.0 centimeters, which is 0.860 m. And we're also told that the we can assume that there is a constant tension in this string. Um, of 1.3 years, mutants, um and in part a were s to find the, uh, speed of transverse waves in this string as it moves. And so this is pretty, pretty straightforward part. Um, we just need to remember that, um the speed of, uh, wave in a string this under tension gonna be equal to the square root of the tension for us two divided by mu, which is Mu, of course, is just equal to theme ass of the string, divided by its length so that our form that just becomes, ah, tee times l over and And if you look at the variables that were given straight away to begin the problem, they are m l and T. So we have everything we need to do to solve this part of the problem. So we just go ahead and plug in everything. So our calculation is the square root of 1.30 times 0.860 divided by 0.55 zero. And if you could calculate that out, you should get Ah, the speed of the waves is 14.3 m per second. So that's your answer her A. And then in part b um were asked to find the noted tow anti no distance, um, for the three simplest modes of vibration. So that means for the first three modes of vibration, we want to find the distance between a node and it's a Jason anti. So to do that, it would be helpful to draw the the first three months of vibration. So we're told in the problem that the, uh and uh, that's attached to the platform of the bottom, um is going to have a note and then because the top end that's attached to the car can vibrate, Really, that's going to have an anti note, um, when it has a standing wave. So if we draw that the way that we draw like sound this thing, we live, we die like sound waves and other, uh, staying waves that we've done with this book the first node is gonna have, or the first mode of vibration is gonna have a note at one end and a note at the other end of the string, which gives us a diagram like this. And so the distance from Ah, a no to any Jason anti note is, of course, just the whole length of the string. Um, so for your first one, I'm going to number these Just 123 The first one, um, were the note Anti no distance is just, um, And then for the second motor vibration, we need to add a set. We have to add another node in between there. Um, so are, uh, drawing is going to look like this. And that's not quite to scale, obviously. But we do know that from a no to an 80 note and from an anti note back to a node, all those distances are going to be the same. So we have a note going to an anti note. That's one that's one. And then from 19 0, back to note and then from a note to an 18 So we actually have three lengths issued. Drew this accurately, these would actually be all the same size. Um, but that means that the note to anti no distance, um, from here to here is going to be one third of the length of the two. So l over three, and then we just do this again for the third motor vibration. And so we just add another note in there before we get to the end of the tube. So look this and I'll draw one more time. So we have a no to an anti note here and, you know, to a node, No, to an anti note anti no to a node and then notes the anti No again. So we have five lengths now. Five distances of a note in 18 0, that means that the distance the Noto anti no distance is 1/5 of the tube, so l over five. So for the first, the three simplest, most vibrations the distant node to anti no distances are l l over three and l over five, which then we just need to calculate out. So l, of course is still 86 10 centimeters. Um, and then l over three is equal to 28 0.7 centimeters, and l over five is equal to you. 17 0.2 centimeters. And I feel like all of that stuff explanatory. You just take 86. Divide by three in divide by five. So these are our answers for part B. I don't feel like it needs much more explanation than is there. All right. And then in part C, we need to find the frequencies for each of these three modes of vibration. And this part is also fairly straightforward. We're just doing you know the same calculation three times. Um so of course F is going to be equal to V divided by Lambda. And Lambda is going to be, in this case four times each of the links that we had in part B. Um, because the distance from a note to its adjacent anti note is always 1/4 off the full wavelength. So we just take these three distances 86 centimeters, 28.7 centimeters and 17.2 centimeters, and we multiply them by four to get the way of links for each of the three modes of vibration. So, um, if you want to, you can also calculate it directly from the length of the tube or not that you don't like this string. Um, And if you do that, you will just say that you will say that, you know, for the first motor vibration, Um, the wave like this going to be four times l ah, for the second one is going to be four thirds times l. And for the third one, it will be or fifths times l. And that's what I did when I was calculating this out so that is how our write it. So for F one, uh, the frequency of the natural frequency, the first motor vibration. You'll have 14.3 for your speed and then in the denominator you have for l is equal to the wavelength. So you have four times 0.860 liters. If you calculate that you should get 4.14 hurts where the first motor vibration. Um, And throughout this problem, it's important that you carry all the decimals until you get a final answer. Because as we're building on our previous work so much, if you make a rounding error, they will compound as you go through your work, and it will make your answers in part C a little bit off. Um, if you you know, round if you don't carry out the best in Ulster. So, um, that is the frequency of the first ah, motor vibration. And for F two, we have 14.3 Still, for the speed of the wave. Um, and then this time, uh, E Linda will be four thirds l, which means that if that's in the denominator, then there's gonna be, uh, you may take the reciprocal multiplying By taking typical, you can say that it's times three in the numerator and then four zero point 860 in the denominator. So we enough with just an extra three in the numerator and then your frequency for the second motor vibration is going to be at 12 point or hurts. And for the third one last but not least, F three will be equal to 14.3. And now the Lambda is, um, 4/5 times l. So we end up with a five and numerator because Lambda isn't a denominator. So and then you end up with four times 0.860 in the denominator one more time. And so frequently that out, then your frequency for F three should be 20.7, and then we're all done. There you go.

Hi, everybody. So this one's gonna be a long one. So for this one, A is finest, be the transfer breathes on the top of the straying. So we have a and it's is gonna Eagle Teoh Velocity equals the tension over you, okay. And the linear density which is the new It's the equal to mass over length. And for that we have 5.5 times 10 to native, three kilograms overhead and that's gonna be divided by 0.86 meters. And that is given to us that we plug it into our calculator and we get six 0.3953 times time to negative three kilograms for a meter. God. So now I'm just gonna go back up here and we have attention of 1.3 Newtons and wants them divided by the 6.3953 times 10 to negative three kilograms per niedere and want to square root this entire thing. And from this we end up getting 14.2 57 meters per second, Ok? And now for B and B is asking us the different states. And so what it wants is the string's vibration possibly there sat saying waves each with a know that fixed bottom And in answer the grand find the no anti distances for each of the three simple estates of high So we have the first state and the first day is gonna look somewhat like this guy. And this trip since their length And we got our why you have X. And then here is our lambda before that. And from this we have the length is as Manda before is you can clearly see And I can put the l like this helps. Okay? And it's because of this, your why is equal to Orlando means your length, which is 86 okay, And 86 it was 86 centimeters. So that's what they gave us into the problems of a sits centimeters. And because of this, your lambda is gonna equal to four times 86 centimeters. And that is equal to, um, 344 centimeters. Or for this one, he could convert it to 3.44 meters if that helps. And I mean, I think that will nicer so that we can read. It's a three points for four years And now for the second state. So put second State. Okay, so for the second state, we have this bad boy up and we have this guy going this way. And then we have this guy going this way, and we have this as entire length. And this guy here is Lambda over two, and this guy is landed. Four. So the open is Linda for and the hoses claimed up to. And because of this, we have relaying, which is gonna be 23 Lamba over before God. And you can clearly do that with just the math here. And, um so Lambda is gonna equal to four times a lane three, and that is that the four third's climbs 86 centimeters. Okay. And this is gonna Eagle Teoh 114.667 which is equal to wind 0.14 on easy seven meters over. And so for one No, here we have at why equals zero. The wise one equals two Lambda over to which is gonna equal to 1.4 Agassi Six who's 11.14667 by the bites you or you can use the round up one. It doesn't really matter. And it's gonna be 57.33 meters. Okay? And as it's no. So then we have anti No, we have that. So we're into it will be the open one. So we have. Why equals land over four, And that's gonna be 1.14667 divided by four. And we end up with 0.28667 meters. Okay. Or you can round this to be, um, 28.667 center years time. It doesn't really insidious. Um, how you like it. And wary, professor. Once it and you have. Why? Because l which is 86 centimeters. Okay. And now for the third states, get the third state. We have, um, this bad boy here, and we're gonna go 123 And I hope I did that. I think I did that wrong once, When Stew's love it slow. 00 no, not need to do that. Okay, So what? Ideas will slow? There we go. Some should have three of So, as you can guess, this is the close ones of them being land over to and the open one is Wendell before? No, but And this whole thing is just the length. And you are, but that is OK. Cell, okay. And so we have the total. All the total is going gold. Five land of four of the total length. And of course, slammed. Uh, is the legal Teoh four bits times laying, which is, um, 56 86 centimeters, which gives us 68 0.8 centimeters, which gives us 0.688 meters. And so we need to dio um, the one a node here. Um, why's that equal to zero and four seconds? No, we have Why equals lambda over to And we're some plug. And the centimeters of 68.8 centimetres divided by two taken do the meter one if you want. And only 34.4 centimeters. I'm just choosing which looks nice. Um, and then the third node does he have wine? Teoh One in three here. So then the third node we have Why equals just lambda? And that is gonna be 68.8 centimeters. And we can just It's for me. Yeah. And now for the anti note, I'm anti okay. And we get anti one. Okay? And that's what you y equals? Um, Landau before And that IHS it's 8.8 centimeters the vital by for and and that we get 17. Appoint Teoh son in years. Go by and then we get the seconds and it would get Why equals three Lambda over four. And for this wine, we get three times 17.2, which is 50 wine of 500.6 centimeters. And then our third wine is gonna be why equals the length which iss? 86 cents meters. Okay. In an are, um, last question is asking a spine of flippancy. Oh, each state. So, um, we have the frequency of each state's of a frequency of the first day alums put first. All right, here is gonna be velocity over lambda. And that is gonna be 14.257 meters per second, divided by the lander that we found says 343.44 meters and as gonna be 4.144 hurts. And now for the 2nd 1 um, we have frequency equals two again. 14 point Teoh 57 meters for seconds. Divided by 1.14667 Eaters and as Navy 12 points for 334 boats. Oh, by And now for our 3rd 1 we have the frequency is given to us that 14 0.257 divided by 0.68 eight meters and got meters per suck it here. And this has been equal. Teoh 20 0.7 Thio Thio three hurts. All right, Thank you, guys.

My friends as soon figure but end up a light sleep with fourth cross strait 100 neutral permitted Attest to the verticals off wall light spring It's tried to gather and off the article spring on the spring changes from her gentle toe vertical as it passes over a solid police. So radios off Police give it four centimeter The police is three to turn on the fixed a smooth exit The vertical section of the strict supporting off rate Oh, much Oh, 200 grand, That is point to Katie on file 04 m. The sprint does not slip as in contact. Uh, but find frequency off for sedition if we have to find host Marshall Negligible second MOSOP clip. Go 50. Get up. Seen Masa Police 7. 50. Get up. See? It were too broad. A free body diagram. See it the weight and developed here. Patient deep on attention to this. And well, foot speak. All right. We can write trees. Got two cakes. They access the extension for block you and G minus students is Puerto m e. Yeah, for Willie Neck talk. Don't. I am to help him treat us in tow. Art minus steel toward oh, so it can be realized Are in tow. Tedious. Minus t. Hi. Hi. Into hey upon art. Yeah, I believe me. Um Ali Square by toe. Uh huh. From 12 and three way can write em Pless Mbai to tow a bless p x respecto empty things could be recognized. Empress on by toe can be recognized g two x over d Tito plus k existed toe energy So solution off equation for between the form it ex Bilby a sign off Omega t Yes, MGB kid. Oh, so frequency will weep ball upon to fight through top k upon and bless, um, by two. Mhm. Yeah, a part. Yeah, for my soft treat, v zero have filled even upon to fight. He is? Yeah, hundreds upon months. Um uh huh. Just a moment, please. Yeah, So it is to me. Oh, mhm. This is the answer off a part. But for beep Marcia, please give it to the program. Yeah, 100 upon point to pless happened toe by toe fight. So it is to be 2.79 Hearts see part for most off little B 70 program brought upon to fight hundreds upon five to plus 5, +75 So, frequency, in this case, you will get to appoint one hearts, That's all. Thanks for watching it. Oh!

All right, so in this question, we are given a string, and we're told some information about it, and we're supposed to figure out what its fundamental frequency is. Um, more specifically, we're told that it has, you know, weight hanging at the end of it, and that is providing tender force on it. And we're supposed to use that information to help figure out what the fundamental frequency is. So I'm gonna just redraw kind of the situation that's in the book a little bit. So here we have, uh, kind of a countertop, and then the string is attached to a wall here, Um, and then at the edge of this counter, there's a pulley. The string is draped over the pulley and then goes straight down. And at the end of the string is attached a mass, um, of mass big M Andi. We're told that the length ah of the string from the wall to the poli is a distance D, um, and then were given a bunch of values. So for little MD mess the string, we're told that it is 8.0 grams, um, which I'm going to convert into killer graphs 0.8 00 kilograms. Um, then we're told that the total length of the string l is five meters, 5.0 Um, distance D is equal to 4.0 meters and the mass of the weights Big M is 4.0 kilograms. And from this, we're supposed to figure out what the fundamental frequency of string is if it's plucks between the pulling the wall during that over the distance. D. So to do that, let's start with the equation. Um, if is equal to V divided by Lambda. And this comes, of course, from the equation that V is equal to 1/2 times. Linda, um just rearranged so that we can sell for, uh so for F So we don't quite know what V or Lambda are right now, but we can figure it out by looking a little bit more closely at the problem. So because we're dealing with a plucked string, let's go ahead. And because there's some special conditions for, um, special properties of the wave Speed V um, when we're dealing with, like a standing wave on a plucks string, um, specifically V is equal to the tension force in the string divided by its mass per unit Length new. Um, so we have that weaken. We have information to figure out both of these values. Um, because the tension force in this string is going to be, uh, equal to the tension force acting on the weight here. So because the weight isn't moving, it's not like it's not like on a spring or anything. The way is just standing still. So the tension forces in the spring in this string, um, the tension portion, this string holding the weight up is equal to, you know, detention exerted on the whole string. Um, but because the weight isn't moving, we know that the tension force T has to be equal to the weight of the mass. So t is equal to big M times G. Um, you know, there's no net force acting on, um, the mass, so yeah, no force is equal to zero. Therefore, there has to be some the tension for us, and it has to be perfectly balanced. The force of gravity on it. Um, and then from you, um, we have the full length of the string and we have the full weight of the string. So to find the linear weight, the linear mass density of that we just take the weight of it and divided by how long it is. So m u is equal to little m divided by So we have, you know, everything we need to know to figure out V. But what about Lambda? So, Lamda, since we're trying to find the fundamental frequency of it Um, and since we're dealing with, you know, plucks string, um, we know that the wavelength of the standing with will be equal to two. Uh, normally would be to out that's the general form. But because we're plucking the part between the wall and the pulley, it's actually gonna be equal to D because there's a node. There will be, you know, the one quotable end of the string is at the wall on the other end of the string that will freely vibrators at the police. So the length that we have that convey vibrate is actually only deep instead of l. So we so for this formula will actually use D instead of l. So lambda will be equal to two D divided by N where n is the number off the harmonic. We're dealing with the fundamental harmonics and is equal to one. So lemma is then equal to two D. So now we can go ahead who school down a little too fast. So now we can go ahead and replace um, all of these values in here for our formula for F. So F is then equal to the square roots, um, of tea. But we know that is equal to big M times G, and then that's divided by mu. But we know that Mu is equal to little M, divided by l, and then that is divided by Lambda. But I'm going to say that's multiplied by one overland. Oh, Linda, just make things a little bit clearer. So it's multiplied by one divided by two deep, and that's a family of four frequency. And we have all these values so we can go ahead and just plug them into the equation. So F is equal to the square roots of I M, which is 4.0 kilograms times G, which is 9.80 that is divided by the mass of the string. 0.800 and divided by the length the total length of the string, which is five meters. We'll go ahead and extend these lines a little bit on. There's a dust one point there and then that is multiplied by 1/2 times the distance of the string that can vibrate, which is four meters, 4.0 And if you plug all this into your calculator, you will get that f. The fundamental frequency of this string is equal to 19.6 hertz and there you go.


Similar Solved Questions

5 answers
Conddru Inpunrh hnl anduerdu Ilu EA LLelnubaul _uathe cenchr Apeeen Tundeutdtaudud 100 cecundtetudrd Ilin ork notn nlthnuchie [haeae EAlarAaE TALEli MATn7 mmdadltt mum Ceeodeel aea tein a Cenbi pontonAun c ElnlAnta~onut4hlteFrnuGhhyulla nectrhlangta deihifintattlrachMesundte dutu cetnleezFettee dIRounalo Lr_ Denclior_KetttamenKniKetmam FAna-nelnJunaS LlaenDloporionthah
Conddru Inpunrh hnl anduerdu Ilu EA LLelnubaul _uathe cenchr Apeeen Tundeutdtaudud 100 cecundtetudrd Ilin ork notn nlthnuchie [haeae EAlarAaE TALEli MATn7 mmdadltt mum Ceeodeel aea tein a Cenbi ponton Aun c Elnl Anta ~onut4hlte Frnu Ghhyulla nectrhlangta deihifintatt lrach Mesundte dutu cetnleez Fet...
5 answers
Fzrd piano sl Jes 2.9 ocan 178 incline ardr Kcpl trcm &220 craing bya man#tJ pushna b23i On pwalk ncane (580 Mgure(Fiouc lanane IricuanPin BDcarinacondon onthe OlnnaIeTInExpross Your answcr Io tro siqnitictni (IquresInciudctnc tonronrinic unlcValueUnitsFigureStomiRcaualnetPar CDuleinneRon conuMnnodetaahErpacs: volAetoiwo -inoilcant #4ltetInr VotMturnarinitnlValtc"Rui
Fzrd piano sl Jes 2.9 ocan 178 incline ardr Kcpl trcm &220 craing bya man#tJ pushna b23i On pwalk ncane (580 Mgure(Fiouc lanane Iricuan Pin B Dcarina condon onthe Olnna IeTIn Expross Your answcr Io tro siqnitictni (Iqures Inciudctnc tonronrinic unlc Value Units Figure Stomi Rcaualnet Par C Dulei...
5 answers
Prove by Strong Induction: If 01 = 1 and (2 5 and (n (n-1+2an-1 for all integers n 2 2, then On (-1)" +2"_
Prove by Strong Induction: If 01 = 1 and (2 5 and (n (n-1+2an-1 for all integers n 2 2, then On (-1)" +2"_...
5 answers
20.4 Lesson 20 Problem Set Calculate Ra; right endpoint approximation, for f (x)for the interval [0,2If 01 3,02 -1,03 = 2, and a1 5, calculate the Sums .5 5 (a;+ (-11";) 2 (careful: the denominator is @k+l- aL+l 31 Evalaute XXOZ JimFind formula for Rv the right-endpoint approximation for f(2) =22 +1 on the val [0, 1] Then compute the area under the graph by evaluating the limit of Rv as N5) (Bonus question): Using the formula given in the text forevaluate
20.4 Lesson 20 Problem Set Calculate Ra; right endpoint approximation, for f (x) for the interval [0,2 If 01 3,02 -1,03 = 2, and a1 5, calculate the Sums . 5 5 (a;+ (-11";) 2 (careful: the denominator is @k+l- aL+l 31 Evalaute XXOZ Jim Find formula for Rv the right-endpoint approximation for f(...
5 answers
Wuhat6 1 Concentretion H3o+ sclv3 4 3 + 5 J8 '8 X 0*1 2 Qase W 3
wuhat 6 1 Concentretion H3o+ sclv 3 4 3 + 5 J8 '8 X 0*1 2 Qase W 3...
5 answers
Solve the following initial value problem for & first-order linear differ ential equation with discontinuous coefficients: dy -y = f(2), dx y(0) = 0, whereez f6) = { = 0 < t < 1; 1, I > 1zer 0 <I <1; Answer: y(c) = J e" +et-1 _ 1, I >1
Solve the following initial value problem for & first-order linear differ ential equation with discontinuous coefficients: dy -y = f(2), dx y(0) = 0, where ez f6) = { = 0 < t < 1; 1, I > 1 zer 0 <I <1; Answer: y(c) = J e" +et-1 _ 1, I >1...
5 answers
Question 151 ptsConsider the PDEy UrT T 1 Uyy Tu L Uy = 22 +yPIn what region of the xy-plane is this PDE elliptic?~8 y > 0~ y < 0~ y > 0~x y < 0
Question 15 1 pts Consider the PDE y UrT T 1 Uyy Tu L Uy = 22 +yP In what region of the xy-plane is this PDE elliptic? ~8 y > 0 ~ y < 0 ~ y > 0 ~x y < 0...
5 answers
22 (e [6-21}&xxA -?-x)Jdx
22 (e [6-21}&x xA -?-x)J dx...
5 answers
ChlerendlBealMeaecnnnLe 10" NI; slon ncrstt ,-4. Unt_ tlac Io'Nc Jung Nreivt , LLAantm
Chlerendl Beal Me aecnnn Le 10" NI; slon ncrstt ,-4. Unt_ tla c Io'Nc Jung Nreivt , LL Aantm...
5 answers
QuESTiOn 26 2 PointsCalculate the concentration of hydronium ions in pure water at 0.00 C, given that the ion product ofwater at 0.00 %C is L14 X 10-15Your answer should include three significant figures Write your answer in scientific notation. Use the multiplication symbol rather than the letter 2 in your answer:Provide your answer below:FEEdbACK
QuESTiOn 26 2 Points Calculate the concentration of hydronium ions in pure water at 0.00 C, given that the ion product ofwater at 0.00 %C is L14 X 10-15 Your answer should include three significant figures Write your answer in scientific notation. Use the multiplication symbol rather than the letter...
5 answers
Solve the homogeneous differential equations.$$ rac{d y}{d x}= rac{x^{3}+3 x y^{2}}{3 x^{2} y+y^{3}}$$
Solve the homogeneous differential equations. $$\frac{d y}{d x}=\frac{x^{3}+3 x y^{2}}{3 x^{2} y+y^{3}}$$...
5 answers
2) A ball rolls off of a table with velocitv of 4 m/s. It lands 5 ft from the base of the table:What are the initial and velocities of the ball when it leaves the table?How long does it take the ball to reach the ground? How tall is the table? and direction of the velocity of thc ball right before it hits the ground? What is the magnitude
2) A ball rolls off of a table with velocitv of 4 m/s. It lands 5 ft from the base of the table: What are the initial and velocities of the ball when it leaves the table? How long does it take the ball to reach the ground? How tall is the table? and direction of the velocity of thc ball right befor...
5 answers
Find Question the 95.5 76.5 67,5 58.5 49.5 Class Mean to another 94.5 05,5 76.5 67.5 58,5 Boundaries for the 1 [ 5141 JaPS frequency 1 Frequency distribution (Roundup OALL decimal digits)10
Find Question the 95.5 76.5 67,5 58.5 49.5 Class Mean to another 94.5 05,5 76.5 67.5 58,5 Boundaries for the 1 [ 5141 JaPS frequency 1 Frequency distribution (Roundup OALL decimal digits) 1 0...
5 answers
QUESTIONTop Materlal; Glass Top Indox Rolractlon (n1) =Botlom Matorial; Wator gorlon Index of Relraclion (nz)= Incident Anglo Roliccted AnglaRetracted Ang 0Now play wlth the slmulatlon Ilnd the critical angle for tolal intornal ailection (which Inc Incident angle where you lose Ihe retraction)
QUESTION Top Materlal; Glass Top Indox Rolractlon (n1) = Botlom Matorial; Wator gorlon Index of Relraclion (nz)= Incident Anglo Roliccted Angla Retracted Ang 0 Now play wlth the slmulatlon Ilnd the critical angle for tolal intornal ailection (which Inc Incident angle where you lose Ihe retraction)...
5 answers
A metal has 5.89 x 1022 conduction electron per cubic meter then its Fermi velocity is (h-bar)=L.OSxiO-34 J.S, me =9.11x I0- kg)Select one: 8.84x106b. 1.39x1067.25x108d; 3 x1020
A metal has 5.89 x 1022 conduction electron per cubic meter then its Fermi velocity is (h-bar)=L.OSxiO-34 J.S, me =9.11x I0- kg) Select one: 8.84x106 b. 1.39x106 7.25x108 d; 3 x1020...

-- 0.022189--