Okay. Can you hear me? I'll got to right. So we return to the. These multiple choice questions for the moment and then we'll see where I will go from there. The last one I think we did was this one here. Just a reminder, not not really a reminder, although this does appear in igcse or has so in the past, that the formuly for Young double slits, diffraction through just one slit and diffraction through a diffraction grating, which is a whole series of slilits, all have their own formula. And you can't use the wrong formula in the wrong place if you get what I'm saying. So you're gonna to make sure you know what formula it like. And the one here, this comes with a number of different variables or letters, that is a fringe spacing. Equals. Lambda d over little d where. Little d is the spacing. Between the senof, the two Snitz and capital d is the distance to the screen and the fringe spacing then is the spacing between the fringes, the patterns of lies and dark excuse me, that we get on the distance screen. I forget what the answer was for that one. That's it out. So still talking about waves, stattionary swave is created in air and. At this point, let's just look at how stationary waves are produced because it's something that a lot of students find quite difficult. In order to understand this, you've got to appreciate that if you have a so we have a piece of stretched material, okay? So there's an elastic rope or something connected to a wall or to some point, which is is fixed, and that's important. Then if you. Send, if you move the end of this up and down, say, up and and then down, you produce A, A pulse, which moves along the string. Now if this string was not fixed at this point, that pulse would just keep going until it lost energy. Refraction, what have you. But at this point, we have reflection. And that's a key word to use in describe anything like this, not echoing, not bouncing off. You've got to have the word reflection. So as I said to you when we were talking the other day about how Polaroid sunglasses work and how light changes its plane of polarization when it's reflected in a similar sort of way, the guys get rid of this doa minute and the, what are you doing, God adme? Sorry henwhen, it's reflected it's reflected back. Like that. Now if you get the distances and the timmings, in other words, the frequenis just right, you can have the situation where a peak moving in that direction meets a aque moving in that direction, in which case we get nothing at all. If you have a peak moving in this direction and another peak moving in the opposite direction, they add, and of course, we get a large peak. So not very good diagram. Yeah, large peak. So if you put all that that all those little sort of ideas together, you end up then with the vibrations on a stretch string as. Either looking like that and half a cycle later it looks like that, or we could have a vibration that looks like that and so on. If this only works, say, if we've got a fixed point at that point there. So these are called stationary waves, just not really accurate, because the waves are made up from a wave traveling that way at the same time as a wave being reflected. And the appearance, just the appearance is that of being stationary. And if we just remind ourselves the vocabulary in the situation where you have points of zero disturbance, we call those nodes n for short. Maximum disturbance are annodes. So we got node. An annode describing the positions of zero and maxima along the line. So let's see what this question is about. Sprin's got a force constant now in most syllases wecall that a spring constant. We could tell from this, however, that they mean it a spring constant, because the units are newtons per meter. The idea being that if you wanted to produce an extension of the spring which is a meter long, you would need to have 4900 newons. Of course, the spring may only be this long, that's all proportional, but we still deal in newtons per meter, not small numbers in newtants and centimeters, or whatever forces apply ed to the spring, causing it to compress by 0.5m. And this is a reminder that spris compressing obey the same rules as springs which are extending. That is not always obvious, because most of the springs that you deal with in a school laboratory are springs where the turns are. The spring are already so tightly connected to each other that they couldn't be made smaller. Used to have one of these in my desk drawer, but I've lost it. They call expendable springs. Otherwise, words we use them once and throw away, they can't be made any shorter. But if you look at the springs of a car suspension system. Then you can see that you could make it smaller as well as make it bigger. So. The change in elastic potential energy you've stored in the spring. Now if you plot a graph, if I've covered this before, I apologize, but it's probably worth saying again, if you plot a graph of the extension of a spring and if you look in various books, you'll see there's no consistency in how the axes are labeled. Let's say this is force and this is extension, and I'm going to use extra extension then as long as we don't pull it too far, as you know, we get a straight line and you, Saido behookks law. And if we think about the energy stored in that, then if we consider the area under the line, the physical significance in that area is actually energy because it's for work done. It's force times distance. So that must present represent energy. And if we have a value of x with extension and a value of f, there is a triangle. So that energy is a half fx or half fe or half f delta l. You've got to get used to different people, different teachers, different books, uses and different symbols, but they mean the same thing. Now, hooks law tells us that the relationship between the force and extension, there's f equals kx, where k is the spring constant. So what we find then, if we substitute then for f in this formula here, is that we also have elastic potential. Energy is a half k squared. And that's. So I'm going to have to take that. I'll add the time at the end. Hello hello. Fine. Thank you. You're not selling me anything, are you? Now bye bye. Bye. I thought that was something important. As soon he said he was some sky networks, I said, you're not selling me anything, are you? And he immediately put the phone down. So he obviously was. And Yeah, so about that. Henry, right. So if we look at what we've got here, it's all in nice units, half a meter. Okay. So the change in elastic potential energy looks to me like it's going to be a half times 4900 times 0.5 squared. So that's two, four, 50. I can't do that in my head. 49 zero times point five squared which is 610 joules. It's actually 612.5, but. There is the answer, okay with that nice simple one, nothing too complicated. Now there are some words that you may have come across or may not have come across if we are talking about the mechanical properties of materials, what they do when you try to change their shape, in other words, and there are. Two. Main. And largely fairly opposite meaning type words. One is ductile. And the other is brittle. Now these words do not relate, so the strength of the material. They relate to how the material behaves when the stresses are large. If you had to define a ductile. It used to be said that it's the ability. So draw, in other words, pull a fine wire from the material. And that's an old fashioned but very graphical explanation. So if I took a piece of metal. And I had a machine which was powerful enough to pull it. So stretch it longer and longer and longer and longer and longer and longer. As it gets longer and longer, it must get thinner and thinner because the volume is constant. That's how copper wire is made. You take a piece of copper and you stretch it. If you could stretch it easily without it breaking wesay, it's ductile. So it has a large region of its stretching behavior before it breaks. So a good example of ductile is copper. It's a material that you can stretch and stretch and stretch and stretch. And very often we use copper in the lab, as when we are doing experiments to determine the Young modulus of various materials. Copper is a common one. Brittle means it is strong, hugely means it stronger. To a point. It deforms very little. It doesn't increase the strength, the force of the stress, the strain does not increase very much. And all of a sudden it breaks. Now, I don't know whether you've come across whether you have a type of biscuit in Mota called a rich tea and biscuit. Probably have, but most English students would know what a rich tea biscuit was. It's a biscuit that is quite stiff, but if you try to bend it, it breaks its snaps. And that's what Gretel means. If although that it might be quite strong, when it breaks, it breaks. There is no prolonged extra change in its dimensions before it finally breaks like would be of a copper. So another example of brittle is iron. Iron is brittle if you take an old piece of iron. And a typical example would be the sort of older design of covers that you see on roads and pavements would cover a drain or cover pipes or something like that. And we call the manholes in English usually, because their design ought often for men to go down inside. If they're made from iron, you could pick one up and drop it, and it will probably break like a piece of glass would break. So glass is another example of material which is brittle. So if it's ductile, we're looking for a large amount of strain. If it is brittle before it breaks, if it's brittle, we're looking for a small amount of strain. So the answer to question 16, because, a, if you look at that and they've got the axes around the other way to how I had it, that doesn't matter. You could see that as the stress increases quite a bit, the amount of strain and the words amount by which it changes its shape is relatively small. But there's no sort of this is the one to compare it with. This could be copper. There's no period and no part of the stretching that where the strain becomes bigger and bigger and bigger without it breaking. Okay, does that make sense? Yeah, good duck or ducactile rtle to very good words. Now this is one of those things I wouldn't necessarily expect you to know, but let's have a look at it anyway, to which the following is a correct statement about the emf of a cell. So let's just visit that again. There's a cell. If we could measure. The potential difference across the cell without taking any current from the cell, then we would call that the emf. In practice, we can't because the cell has an internal resistance, which and our simple model of the cell we have as a proper sort of fixed little resistor there. But it's not a resistwithin the cell. It's the resistance, which is a property of the materials that make that cell on and the connections to the cell. The only bits that we can connect to are outside of this box. Those are the only connections we can make. So as soon as we start taking current from that cell, we get potential difference dropped across the internal resistance and those often referred to as lost volts. Because we've lost them, we don't know where they've gone. And the potential difference therefore decreases as you take more and more current. And we have this formula that v so terminal potential difference is e, the emf minus the current times, the internal resistance. And that's a very, very useful formula. Which the following is correct about the emf. It is equal to the energy transfer from chemical energy per volt. No, that's a rather subtle one. The emf. Is all to do with the conversion of chemical energy to electrical energy. And there's doesn't quite as not full enough explanation. Energy transfer to thermal energy and the load resistance, no. Pd measured across internal resistance. No, because that's ir. So that just leaves d equal to the pd measured across the terminals, the cell when there is no current. So in fact, the the simplest possible answer there is, in fact, the correct one. There are various definitions that you need to learn at a level, whichever board you use. They vary little bit, it seems to me, from exam board to exam board. But emf will be one of them. Potential difference will be one of them. And you really need to make sure you're okay with that. And I'll let you have a look at this, but I'll rethrough it. The diagram shows a semmicirlar glass block with a refractive index of 1.5 glass blocks surrounded by air. A ray of light follows the path shown from x to p. Which path will the ray follow after it arrives at p? Now that. It's a badly drawn diagram. Wait, no, no, it's probably. I would need a calculator for this. Funnily enough, I'm just looking at the March scheme here. They've got it wrong. Let's see why this happens. Sometimes the formula that you need is to know what the critical angle is and the we know that the refractive index 1.5 in this case is one over sign of the critical angle. So if we. See what the critical angle is then c is equal to sign minus one of one over 1.5. And I think this is going to come out at 42 degrees. So sminus one. 42 degrees so. C is 42 degrees, so 50, the Anlo instance here is greater than the critical angle. So the light is totally, in turn reflected. But if you look at this diagram, the arts, yes, they say the answer is a, but it's a bad diagram. This angle there should also be 50 degrees. Doesn't look to me like it is anyway. Maybe it's what happens when you make something look a lot bigger. I'm sure it's not the same angle each side, but that's the only only possible answer. So well done. A so Yeah work ethic and you're given the a refragial index, you can work out what the critical angle should be. Now I'm not sure what you've done with uncertainties and percentage uncertainties at igcse. Are you aware of the difference in how you can use them or not? Oh, I don't know. No. Okay, not worry. It's one of the things that even sort of very bright students often get wrong in the calculation. Surprising number. Now, there are two types of uncertainty. And uncertainty is a measure of the confidence that we have in a numerical answer being correct. So if we have an absolute uncertainty, and over the years we've called these errors, but the term these days is uncertainty, because the word error implies that somebody's done something it wrong, which is not what it should be, because uncertainties are a fact of life as far as science is concerned. Absolute uncertainties, let's say that you've got a. You're measuring MaaS. And you can only measure to the nearest. One gram, say. So if you measure 230 grams in reality, if it says 230 grams in reality, we know that's 230 plus or minus one. Graso, that's an absolute uncertainty. The term absolute is implying that the uncertainty has the same unit as the measurement. On the other hand. A. Percentage uncertainty. Actually, as it happens, usually ends up being more useful. To take my numbers here, the equivalent percentage uncertainty would be one, which is this one here divided by 230 times 100. To make it into a percent. So. So that's 0.43 is said. That's the minus. Now so in expressing our 230 gram MaaS, we now say it's 230 grams plus or -0.43%. And we wouldn't usually we go to a lot of significant figures with percentages because they are after that were the uncertainties because we're not certain of their value in the first place. So it would be ridiculous to have uncertainties to 506, seven significant figures. And often, often we only have one significant figure for uncertainties. In my case here, we've got to now there's an advantage to doing it this way because when you do a calculation and. You have numbers in that calculation which represent different physical quantities. Then you can't combine uncertainties, absolute uncertainties of, let's say, MaaS, with uncertainties of absolute uncertainties of acceleration or uncertainties of force or whatever. So if I take an example here and then we'll do the question, let's say that we are we've got MaaS up here. So let's say that we're calculating an acceleration based upon force divided by MaaS. And let's say that they we know that the force is equal to one Newton plus or -0.2 newtant. Now before we can do anything with that, we need to convert that an to a percentage uncertainty. But then you'll see what I do with these percentage uncertainties. And now it's so easy when we look at to find out what the uncertainty is in the acceleration because that's what we're looking for. How accurately do we know the value of the acceleration based upon the force and the MaaS given their individual uncertainties? So this is then also 0.2. Over one size 100. Which is plus or -20%. So we could say that f is equal to one Newton. Plus a -20%. Now if we do the calculation, a. A is going to be one Newton. The MaaS is going to be, let's say, 0.23 because we also have kilograms. But this is plus or -0.43%. And this is plus or -20%. Now what what happens then is you do your calculation of one divided by point 23, which is 4.34, which I should have known. 4.3 say. Plus or minus the sum of these two uncertainties I've chosen to bad numbers in not quite realistic because plus the -20.4. So. When you have things being multiplied or divided, you add the percentage uncertainties. You can't add the absolute uncertainty, as you cannot, in a physics equation, add a MaaS or divider a kilogram. Biyer, sorry, Newton, by the absolute uncertainty in Newton to, by the way, kilograms, you could just can't do it. You've got to look at them, the percentage uncertainty. So I'm not expressing myself very well. So whenever you multiply a divide, you add the percentage of uncertainty because then there's a little bit further. Before we look at the question, let's go back to our MaaS. Of 0.23 kilograms plus or -0.43%. If we have an equation where. We have m squared. Then m squared is 0.23 squared which is 0.053. Thatbe plus or -0.86%. Because you've multiplied two masses. So you add the percentage uncertainty if you wanted root m. So a square root of 0.23 is 0.48. That would then be 0.48 kilograms plus or -0.21% or 22% round it properly. Okay. So you're happy with that. So let's look at this question. So the first thing we need to know is we're finding out a value of energy. From the pd, the current in the time. So the first thing to appreciate, of course, is what is the formula. And the formula is that the amount of energy W, I would always refer to use e, but doesn't matter as equal to the current times, the time times, the potential difference that of course, comes from power equals current times voltage. Energy W equals power over time. So that's where that comes from. So you could see that when we look at these three quantities that having found the percentage uncertainty of each one of those three, we're simply going to add them together. So the percentage. Uncertainty of the potential difference is zero ught by two over twelve times 100. This is one over 80 times 100. This is 0.01 over 60 times 100. So if we're going to take those, we're going to add them together. First one's 1.67. Let we seem straight away, that means it can't be either of those two, one over 80 is zero, 125, isn't it not a bay. 125 yes. Point zero one. One over 60 basically divided by 60. There's 0.0. 17 plus 1.25 plus 1.67 equals 2.93. So we can see it's answer c. So these questions are always easy, but you'll be surprised while a ma level of students make them when they get them because half them don't understand the difference between an absolute uncertainty and a percentage uncertainty, sometimes called a relative uncertainty. That percentage uncertainty does not need to be represented in percentage by point 43, could be point zero zero 43 as a number, as a ratio. Okay. So you've learned a bit there. I hope just a little bit more about that before we go on to the next question. If you don't mind, it looks it might look to you that absolute uncertainties are a fairly useless for anything a nasdaq converted to percentage. But there is one situation whether or not no, in most physics equations you might multiplying or dividing. In some physiquestions you are also adding. Remember, you can only add or subtract. Physical quantities, which are the same type of quantity. So for example, v equals U plus at. If v is a meters per second, then both of these must be in meters per second, otherwise we couldn't add them. It might sound obvious, but we're adding us to tracting and only add and subtract the same thing. I talked to the students about apples and oranges. You got say, what's three oranges plus two apples equal to? You can't ask that question. You could say you got five pieces of fruit, but that's not quite the same thing. So where do absolute uncertainties come into play then? Is exactly in a situation where you're adding and subtracting and you can't use percentage uncertainties in that situation. Now one classic example of this, of the use of absolute uncertainties is in the measurement of the whole series. Let me see if I can get this right. How do I copy Strike Command c? Ground, Hey. Come to b. B right, okay, let's move those so they are beside each other. Here are four coins. You've been given the task to measure the diameter of a coin and how you can do it most accurately. Well, you can take your coin, you can take your ruler, or better still, something called a vernier caliper, which is tool that you'll meet them at a level to a way of measuring distances with. No, are you okay? Good. Much greater precision. So let's say that you're using a vernier caliper, vernier calipers. You ke still there have lost vision. Blush your video. Henry, can you hear me? Oh Yeah, the Internet was I was wondering whether there might be some problems with the Internet day. The sun has he met Ted? A lot of radiation today. And only about a an hour or so ago, there was a very large, two hours ago, a very, very large solar flare. And what they called a corronal MaaS ejection, a cme now and cmehit earth, they can knock out satellites temporarily. Usually when big cmehit earth, they can knock out power systems. They can cause all sorts of damage. I don't think this one is big enough, but I did, I did wonder for them. Mother, anyway, let's say that your vernier caliper can measure, and this is, think the writer answer, third permit, right? Number plus or -0.01 mm. But I can remember the Vernet calical rthere's, another instrument called a Oh God. It's not called Oh God. Oh, never mind. Itcome to me in a minute that I could be getting confused with that. But let's say that we could measure to plus -0.1 mm. Let's say then that we measure the the coin and just one coin, and it comes out of 20 mm. But we know that's going to be plus or -0.01 mm. So that gives us a percentage error of 0.01. So it's got a big problem between zero over one. Point zero 5%. So that is plus or -0.05%. However, I've drawn four coins there because if instead of measuring one coin, we measure the diameter, rule four, and say they're all perfect. And our vernier caliper micrometer was the tool I was looking for. The other thing that measured anymore precisely or for, let's say I'll measure 80 mm, but we've still got the same absolute value of the uncertainty. So as a percentage, then we are now four times better off. So this now I measurement now if we divide that by four to get just one coin, we don't divide this by four but the percentage then uncertainty then becomes plus or -0.0125. So in other words, you got a much more accurate value if you measure across many things then divide by the number of things you've got because you're uncertainty is a fixed uncertainty and that's what it is for. An absolute uncertainty is a fixed value. With good. Object is completely immersed. In water upthrust acts xon the object. I've talked about this before. The origin of that thremember is the fact that because pressure increases with depth in a fluid, that if you immerse an object into the fluid, the pressure on the bottom of it, it's going to be bigger than the pressure on the top. The difference in the pressures times the areas creates the unthrust. It happens to be that if you put an object into water, that the volume, the MaaS of the volume or the weight, sorry, the weight of the volume of the water displaced, pushed out of the way is equal to the up thrust. So we're looking for something here, which is weight. Which answer would you go for? You've got to think about what physical quantity you get when you do these calculations. So let me help you with that. People might multiply the density by g. You don't really get anything useful at all. The same, of course, here. The fall over the object by g means nothing but the MaaS of the water times. G is the weight. And it's the weight of the water displaced, so it equals the upthrust. Okay. Yeah. Okay, right. Good. Now I haven't looked through these longer questions because just like the paper we looked at week before last. Starts off with multiple choice questions and then leads on to structured questions. So just how suitable this will be. I don't know. Let's not go with that one. Okay, let's go with this one. Grab technicians moving boxes. Technician pulls a box using a rope with a force of 32 nutants. The force acts at not 50 degrees to the horizontal fox, moves a horizontal distance of three and a half meters along the floor at time of 6s. So there's a lot going on here. The fundamental. Term here is good work done. It's equal to four size displacement. Now there's another way of writing that we could say there's the force multiplied by the distance moved. In the direction of the force sorry, in the direction, of course. Equally, we could say that it's a distance moved. Times the component. Of the force in that direction. All three of those are essentially saying the same thing. So the first thing we need to do to this, it's to work out because the box is going to move along the floor horizontally, we need to work out the horizontal component of that force. So. This is the hypouse 32 we're looking then. For the value there of the force. Which is 32 times cosine of 50 good Yeah, 32 cofifty. Which is 20.6. Remember not to limit yourself to two few significant figures in the middle of a calculation. What a lot of exam candidates do. It's so sad because they lose Marks as immediately caught, some would say straight away, that's 21. If they do that, then they're not going to get the right answer. Keep it at least a 36 figs, if not more. Okay, so force times distance. So our honanddeths is 35m. So the work done to begin with then is 20.6 times, Oh, 3.5. The cleverest thing to do, actually, even if you write down an intermediate answer like this, is to keep the whole answer in your calculator, you to ten significant figures or whatever it is, which is what I've done as that. And that actually works out to be 28. 28 watt what have I done is work. So that's 28, Jules. That's not right. I to Press the wrong button. How can that be right? 72. Point one and then what happened then? Sorry about that. Well, we need the power. P is W over t so can take a 72.1. Dear Oh dear dear. 72.5. And divide it by six. So twelve bots. So that's a nice simple starter question there, which I hope you would have been able to do. So we've got working because four stands, distance, power because worked and divided by time. And it all falls out very nicely. It's a very bad idea, not bending your knejust looking to see what's going on here. Okay, once to keep moving backwards and forwards between the two, technician lift a box on the floor without bending their knees. The diagram shows the force W due to the weight of the box. Okay, H you can see is a pivot point. And. We've got to calcullator. As we're told, this is five kilograms. Bring the data up here. That should be 5.0 kilograms. Actually, I don't know, give it to 1s figure. That really odd. But d is 0.6 and remains constant. Still, one sieffect can get the moment's backpoint H when theta is 90 degrees. State the unit. So ages. This is very straightforward because we're now in a situation where we've got weight aing downwards vertically. We've got a distance d the average is 0.6m. This is a pivot point. This is five. Some students will forget that's a MaaS noof weight. So we're going to multiply that by 9.8. And because they're saying, let's do this for theta equals 90, of course, we don't have to do any further calculation. So that's just five times 9.8 times point six, which is 29.4 litmeters. So that was quite straightforward. The next part of it is a. So this is where you don't forget new ton meters as well. But I gram, show how the technician can pick up the box while bending her knees. This keeps the spine more vertical. Explain why bending the knees is less likely to cause damage to the spine. So this is trickier. Excuse me, because we're got to write something. We need to think about those diagrams and the point of view. Of the moment produced by the box. And as we can see, the moment produced by the box at 90 degrees is 29.4m and that's going to be its maximum value. Excuse you a. Yeah. Right? So. I'm going to just copy what it says in the mark scheming. You can see how their minds are working. I hope. Yeah. So. Doesn't allow me to place that on top of the. So you can see that we've got three Marks here. So these are the three points they're looking for, perpendicular distance between the line of action of the weight. That is you know this line here, the five times 9.8 line is reduced. Because theta is not 90 degrees anymore. So that's your first mark. So the second mark comes from saying if that distance is reduced, the moment is reduced. And the third mark, quite logically, says that if the moment is reduced, then the force must be reduced. So you can see how these questions work. The Marks go one thing after another, exploring what you know and what you can write down in a readable, sensible sort of fashion. So the perpedicular distance is smaller, which makes the moment smaller, which makes the force on the spine smaller. When you see in marsk's words in brackets, it means you haven't actually got to write that, but you mustn't write anything that contradicts it. So if you said the force on the leg is reduced, you wouldn't get the mark because although you said the force is reduced, you talked about the leg. If you say the force on the spine is reduced, fine. Or just the force is reduced, fine. Yeah the way these things work. So it's knowing these little things that make the world a difference in getting a top Marks at the end of the day, not just getting something right properly, right. Okay. We've only got a few seconds left, so look at the next one anyway. Okay, this is. One that is worth looking at. For next lesson, let's just think about it to begin with. A thermester is a device where the resistance of the device changes with temperature. You're expected to know that as resistance goes down sorry, as the temperature increases, the resistance goes down. A lot of students think it's bitformisted. It's changing the temperature. Well, if the Mr ster can get hot, and you've got to make sure that that's accounted for. But really, we're talking about the thermister as being a censor. But when it's hot, because the environment around it is hot, the resistance of it is smaller. And that means you're got to be very careful with using thermisters, because the more current they candle, the hotter they get, because that's the way current works. After they get, the resistance gets smaller, the resistance gets smaller, the current gets bigger. So you end up with a thermister actually exploding if you're not careful. Okay, we'll talk about this in more detail. I shall be at my daughters next week, so you'll see a different room, but I'll see you next wednesokay. Bye, bye. Yeah, thank you. Bye. The.