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Consider the functions $E_{1}=3 \cos \omega t$ and $E_{2}=4 \sin \omega t .$ First prove that $E_{2}=4 \cos (\omega t-\pi / 2) .$ Then, using phasors and referring to Fig. $P .7 .10,$ show that $E_{3}=E_{1}+E_{2}=5 \cos (\omega t-\varphi) ;$ determine $\varphi$ Discuss the values of $E_{3}$ wherever either $E_{1}=0$ or $E_{2}=0 .$ Does $E_{3}$ lead or lag $E_{1} ?$ Explain.

Okay, So I'm just going to give you a quick synopsis, because I know you. If you're in calculus, you know how to grab these points. Um, it looks like the highest value for the wise. Go to 42 10 2030 40. I don't wanna spend all my time doing the easy part against my point. And the curve goes 2025 30. Uh huh. 25. 30 and they're half in commences. So I'm going pretty slowly. And then 35 30 30 that they found 30 30. There should be more smooth on DE then 35 40 40. So 35 up to 40 and 40. Something like that. Uh, it doesn't have to be perfect. But then in part B, you find in the mid 0.3 months sums and the first part is saying and equals two. Well, if you're only creating two rectangles, eso let me split up. I think it's pretty obvious that the first one would be 0 to 2 in the second would be 2 to 4. When you're doing midpoint, you want the height of the X value that's in between. So you're looking at X equals one you want to find that values height. I hope it is clear that the width of each rectangle was too. And then this is the, um I'm just gonna x equals one. I forget which variable they use. Oh, they use t t equals one. And the height when t equals one is 30 on, then add to it two. Because that's with, um, this rectangle when t equals three. So go back to your table. Just double Check that here. You know, the right one 35. But if I go up right here yeah, it looks like I drew a little bit iron. I'm multiply that out so it gets 60 plus 70. It is 130 for that first answer. So how does this change when I put in four rectangles? I'll do this right here and equals four. It's still mid point, but now we need the heights of halfway between zero and one. And now the wits air also each one one times plus one times plus one times. And this is why they gave you all those decimal numbers. So right here will be when t equals 0.5 t equals 1.5. And I think I've exhausted 2.5 and 3.5, but I'm just gonna go and start plugging away. This one was 25 35 30 and 40. Now it's nice about this is multiplied by one is really easy. So 40 50 60 70 80 9100 and 5, 10 2030. Hey, look, that's gonna be the same answer which doesn't always happen, but it did this time.

And this problem are given a table of values that represents the speed of some object. And you're asked to graph it and then we have to take the midpoint Riemann sums of both of them using two sub intervals and then four sub intervals. So as you can see, I have drafted same graph on both sides. This is showing two sub intervals and this is showing four sub intervals, both using the midpoint rule. Let's get to solving. So as a reminder, the area under the curve is equal to the sum of delta X times fx for your amount of sub intervals. So if we are doing sub intervals of two, then we know that our delta X, It's going to be equal to 4 0 from this ban over to delta X is equal to that means we can just split our spans in half and find the mid points for our function. That means we're looking for the midpoint of 0-2 In the mid point of 0- four. But that's easy because here at this midpoint is going to be one. This midpoint is simply going to be three, but we're not interested in the X values for this case where it shouldn't F of X is so it turns out We will be using f of one And F of three. F of one is equal to 30 And f of three is 35. So using that formula in blue people do two times 30 plus 35. So that will give us our area using two sub intervals in the mid point rule To be 130. And that's the first part of the problem. Done For the second part of the problem, we have to use four sub intervals and equals four in the same way to find our delta X. We will do delta X is equal to zero for 0 from our bounds Over and equals four To find the Delta X is equal to one great. So that means that we can split up our intervals as follows, 01, 12, 2 to 3 And 3- four. And because we're using midpoint role and that means we need the midpoint of these bounds. So for this one it is 0.5. This one is 1.5, It's when it's 2.5, This one is 3.5 just like before. We're actually interested in the f of all of these. 1234. So we will take our numbers from the chart F of 0.5 is 25. Half of 1.5 35. F of 2.5 30. Finally, after 3.5 it's 40 plug those into our calculator. We will find That this area is actually the same as the last one. and 30. And the units for this will be meters because the velocity was meters per second. So this is a distance 130 m and that's it

Okay. I have blotted these points based on the graph based on the table given. So now for part A, we will just draw a quick, smooth line past these points. Awesome. Now we can move on to part B where and equals two if equals two. We want to petitions so we can divide them up like this on day. Have these mid points here. Okay, Now we can write the some for our mid points. So we have f off one plus f off three. We multiply that by Delta X, which is this length. So that's two. I'm not plugging in our values. F of one equals 30 while f or three equals 35. Okay, so 30 plus 35 s 65. Once you try that by two, you get 1 30 Now for an equals four meet. You raise this real quick. We'll divide our domain into for petitions so it would look like this. And it has thes mid points. So I was some equals f of 0.5 plus f of 1.5 plus f off 2.5 plus f off the 0.5 oops multiplied by daughter X which in this case is just one. So we don't have to write that. So here we have. So here f off 0.5 is 25. Next we have f of 1.5, so that's 35. Next, we have 30 and finally we have 40. So adding those numbers up, you should get 1 30 as well.

Hello, everyone. This is a salute from two problem. The revival Chapter 24 Off college business. You were given the way forms for the Autry feel in the magnetic field and a Mila in an electron. That wave and the general form looks like this. We're told that the e feel this oscillating in the Y direction, whereas the magnetic field is oscillating in the beater in dissent direction also looking at the way form, where s to find as many of the properties of the wave as we can? So let's see what's happening here. So we see that the general form has E. Max s the amplitude off the electric field and be max as the amplitude off the magnetic field. And these sit outside of the actual wave modulation off the field, which is this periodic function or any other period of functions. Could be exponential is complex. Explain shoals were a signed waiver. Koh Samui. There could be some face here, but whatever it is, the amplitude of the wave sits out front. That's what's modulating the envelope of the waves, so to say and so looking at the values that were given, we see that E Max is equal to, um, 2.4 2.4 times 10 to the five. Excuse me, it's a bit far. So 2.4 times 10 to the five units per cool and were also given Be Max and me. Max is equal to eat when zero times 10 to the minus four, I believe 10 to the minus. For it's tentative minus four Tessler. So these are the amplitude of the electric and magnetic fields, respectively. Now, looking at the way form, we can also see that, um, in the ones your factories to pie. The argument in terms off the argument of the oscillating part of the function, um, in terms of time is divided by the period where, as the modulation in terms of space is divided, by the way, So this directly tells us that the period is equal to whatever is sitting under Thea the variable t, and this is 1.5 times 10 to the minus 15 seconds. So this is the period of the oscillations. Then we can bureaucrat from this with the frequency is because the frequency is simply one over a period. So this is 1/1 0.5, uh, times 10 to the minus 15 seconds. So she worked this out. Then the frequency is 6.7 times density. My 10 to 14 hurts, so that's the frequency of a wave. Now, as I said under the space modulation of the argument off the oscillating function or the purity function, says the Grapeland. So you can also just directly read off when the wavelength is. And this is 4.5, 4.5 times 10 to the minus seven meters. So that's the wavelength of the way Now, Additionally, you can tell that the waves travelling in a positive direction because there is a relative minus sign between the time and the space modulation of the periodic function in both cases. So you know that these waves actually correspond to each other. You can work out, you know, if I have an actual like we've who's electric. Bill is also heading in the right direction, and there's ah, there's a magnetic field is also lacing. Isolating is that direction, I noted. Thes tube is actually correspond to the same thing. It's the same. We've, um it's the same military medical leave, but the different fields inside it. Now, additionally, you can work out what the with the speed of the wave is because you can multiply lambda with F. This is true for any wave. So if you do this, then you get 4.5 times stands in a seven. That that my seven, um times 6.7 times 10th e 14. And this is roughly 3.0 times 10 38 meters per second. So you know that it's traveling at the speed of flight. Now, one additional thing that you could look up is to see what part of dr on the Spectrum this frequency corresponds. Do. But we're not gonna do that here. So as far as we're concerned, this is all of the things that you can read off. You can read of the maximum electric field, the maximum magnetic field, the period, the frequency, the wave length and the speed


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