 There we go. So the odds are your first question is they roughly follow the order. So you're going to have either projectile, forces, ramps, or at-wood machines with ramps, or some kind of an energy momentum. So here we go. A 1.5 kilogram ball moving east at 72 meters per second. By the way, hopefully you just kind of glanced at this and you've done enough of these that you're going, I'm pretty sure there's a collision with kind of momentum. Collided with a stationary, I would probably underline the word stationary. That's very, very important. Wooden sphere, the ball rebounded at 43 meters per second in the direction 55 degrees north of west. What were the speed and the direction of the wooden sphere after the collision? Of course, I'm going to make sure my calculator since I wrote math 12 earlier today is in degrees. We're not going to pull a Dan here. He did do that on my first mock, about halfway through. We suddenly heard him moan. Is there a collision? So what kind of a question is this? What topic is this? It's momentum. And I like to always start out with a similar approach. So I'm going to start out by writing the sum of all the initial momentum equals the sum of all the final momentum. The amount of momentum before equals the amount of momentum after. Now this is going to be vectors and triangles and stuff. Before the collision, it looks like I have that. There's my very rough sketch. How big? Momentum is what times what? It's mass times velocity. Mass times velocity. This is stationary. So it looks like my initial momentum is 1.5 times 72. Looks like my initial momentum is 108. Units are mass times velocity, kilogram meters per second. I think I told you they're talking about naming that an Einstein, but they haven't yet, so we'll go with their units. Afterwards, looking at my picture, it looks like this is going to equal that plus that. I guess. First of all, I know this angle here is 55 degrees. I know this magnitude. It's the same mass of 1.5 jasmine, but the velocity this time is what? Say 43. Got you zoning out with me. Same mass. Velocity is 43. So the final momentum of the little ball at an angle of 55 degrees is 1.5 times 43, 64.5. And here's the momentum of object 2 final that I don't know. 1, 2, 3, here are my vectors. I'm going to add these two together. How do I add two vectors together, draw them? And I know what the answer is going to be. A horizontal line, see it? So as a vector diagram, it's going to look like this. 64.5 plus my mystery momentum. And I know to stop there. I know Vitaly don't stop here. I know Vitaly don't go further. By the way, if you guys are writing this down, you can. I can print this up for you afterwards if you really want to or email it to you because if you're looking down, you might miss stuff. Or if you learn better by writing stuff, go ahead. But do whatever works for you. I have to stop right here, though, because I know that my initial momentum was a horizontal line of 108. Now, I can already tell my scale is garbage because this is 64. This is 108, but they'll both look about the same size. Oh, well, I won't freak out about scale. Justin, how big is this angle right here? Sorry, you guys don't have this in front of you. How big is it? 55. So really the question is, how big is that angle? Because that's what I want in my triangle, 180 minus 55, which I'm pretty sure is 125. Check me. Yes. And this is where we said the cosine law, Sean. Now, you could also do this with components. If you want to, you can, Leslie, break this 64.5 at an angle of 55 degrees into up and left. You can break this into right only and figure out how big this would need to be down and right to cancel out the up and left. I like the cosine law. Now, the cosine law is this on your formula sheet. And it's the momentum of object 2 squared equals 64.5 squared plus 108 squared minus 2 times 64.5 times 108 cosine of 155. We use the cosine law. Sorry, that's terribly messy. We use the cosine law when it's two sides in the angle between them and we want the one across from the angle. Pardon me, 125. What did I say, 155? Because I got the 55 on the brain, 125 degrees. And those of you that were here this morning, make sure you correct my mistake, because my brain is not at its best this morning today, for some reason. 64.5 squared plus 108, whoop, not log, there you go, squared minus 2 times 64.5 times 108 times the cosine of 125. I did go degrees. Yes, yes, I did. Enter. Now, that's momentum final 2 squared square root. I would write that answer down on the provincial, but I'm trying to cut corners here. And then there, final momentum of object 2 is 154.32. That's not what the question asked, Vitaly. The question asked, what were the final speed? How can I find the speed if I know an object's momentum? Momentum is what times what? Mass times velocity. I'm going to divide by the mass of object 2, which is 8.3. So v of object 2 is going to be this number divided by 8.3. I would write that down in my work that I was showing the marker, but I'm cutting corners. 18.6 meters per second. But they also wanted the direction. What's the direction going to be? Well, I would like to use this angle right here. This angle right here would be what of what? South of west, east of south, north of east, west of? South of east, yes? So I'm going to go south of east. Now, how can I find this angle? The problem is it's not in my triangle. But I did want to use it, because it is next to a horizontal line. I can use that angle right there, because it's not next to any vertical or horizontal line. It's just hanging there midair. Oh, remember the Z rule? I really can find that angle in the triangle. How can I find that angle in the triangle? Sign law, the sign of my mystery angle divided by what's across from it, 64.5, equals the sign of the angle that I know, 125, divided by what's across from it, which was I figured out 154.32. When I cross multiply that times that divided by that, I'm going to get 64.5 times the sign of 125 divided by 154.32. And I get what is the angle 0.34? No, no, no, no. That's the sign of theta, Vitaly. How do I find theta if I know the sign of theta? How do I find an angle if I know the sign? Oh, you've got to know this, folks. How do I find this basic trig? No reference angle, no cast rule, it's physics 11. Yeah, second function, sign, right? You have to. OK, I'm going to be asking you on a test. Oh, I hope it is. I hope it is. Please let it be. 20 degrees. You cheer for Montreal. You should be used to people staring at you and thinking you're a little, well, anyways, OK. There's a momentum question. That's a fairly tough one, by the way. Usually you don't get this ugly a triangle. Usually it was one goes this way, one goes that way, and the triangle was fairly obvious. But OK, I dealt with it. And apparently I had two pages to deal with it. Yay. What topic is this? Can you glance at the diagram and tell me what we're going to be doing here? Darn right, it's going to be torque, OK? Let's read the actual question. A 1.5 meter long beam supports 120 kilogram load. The beam is suspended by a wire connected as shown. The wire has a tension of 3,700 newtons. What's the mass of the beam? So usually we were wanting to find the tension. Here, they're telling me the tension. Find the mass of the beam, OK? First thing I'm going to do here absolutely is I'm going to label my forces on the beam. What are the forces acting on this beam? Get the obvious ones. OK, I'm going to have mg of the beam. But that's going to go at the center of mass. So I'm going to go m beam times g. What else? M1g, I'll call it M1. What else? That can't be all, because two forces down means this has to be accelerating down. OK, I have tension at this wonky angle here. And that can't be all either, because which way is tension pulling? It is pulling up, but which other way is it pulling? To the left, I'm pretty sure the wall is pushing to the right. I'll call that fx. And I'm pretty sure the wall is also doing some of the lifting work. I'll call that fy. But why can I ignore those? Why is torque so nice? Why can I ignore those when we're doing torque? Because how far are they from the pivot? Zero. That's the beauty of torque is you can get rid of multiple unknowns. Otherwise, you can't solve this. This is torque. How do I start every torque question? The sum of all the torques clockwise in this direction equals the sum of all the torques counterclockwise in this direction. But I can't solve this yet. I'm jumping the gun. None of these are acceptable, because what of the forces have to be in relation to the beam for you to use torque? Had to be perpendicular. I'm going to add mass of the beam g perpendicular. I'm going to add m1g perpendicular. And I'm going to add tension perpendicular, where this angle here is 90 degrees. This angle here is 90 degrees. This angle here is 90 degrees. Now, how big is this angle right here? How big is this angle right here? See the Zed? How big is this angle right here? There, I've got my 70 in the bottom. This is going to be 70. This is going to be 70. Let's try that again up top. How big is this angle right here? 18. Can you see the Zed? How big is this angle right here? 18. You could also have gone, hey, these two add to 90. You could have put a 72 right there. You'll just be using a different trig function. I always try and use the angle that they give me, only because I know on the answer keys that I give you from an eventual exam, they'll use the angle that they were given. You don't have to if it's tougher. I just do. And so really, my equation is going to look like this. Clockwise in this direction, the mass of the beam G perpendicular times its distance from the pivot. Oh, the beam is 1.5 meters long. So where will the mass of the beam go? Center of mass, what's half of 1.5 meters? I think 0.75 meters. Is that right? I hope I think. Plus, this is also a clockwise pivot. Mass 1G perpendicular times its distance from the pivot equals tension perpendicular times its distance from the pivot. What are we trying to find? That bad boy. Let's do some trig. These two triangles are basically the same shape. It's going to be the same trig function that goes here and here. Vitaly, it's either going to be sine or it's going to be same for both. This is a different triangle. I'll analyze it separately. But Vitaly, I would pick up one of these triangles. I'll pick up this one because it's not as cluttered. And I would probably just draw it over here and say, okay, I have M1G. I have 70 degrees. I have M1G perpendicular. Which trig function relates those guys right there? Troy. Sine. Sine of 70 equals M1G perpendicular over M1G. There, I cross-multiplied. I have an expression for perpendicular. You know what trig function it's going to be, Justin? Sine. Okay? So this is going to be mass of the beam G sine of 70. And then don't forget to drop down the point 75. Plus mass one G sine of 70. And don't forget to drop down the point 15 equals. So far so good, Leslie. I'm doing lots of trig real fast. But yes, I need to find an expression for this one here. This is going to be opposite hypotenuse. Is this sine as well? Sine of 18 equals perpendicular over tension. Yeah, this is sine as well. It's kind of nice. It's not often it's the same trig function for all of them. This is going to be tension. Sine of 18 times 1.5. Is that okay so far? I'm going fast, I know. What am I trying to find? Mass of the beam. I'm going to minus this and then divide by. Do the G's cancel? No, because there is not a G over here. Sorry. The mass of the beam is going to be the tension times the sine of 18 times 1.5 minus M1G times the sine of 70 times 1.5. So I've minus this term over Jasmine and then I would divide by these three. Yes, I hope G sine 70.75. And I would type that very, very carefully into my calculator. They did tell me the tension, 3700. They did tell me mass one, 120 and I could crunch this. I'm not going to bother going to the answer. I'm going to assume if you can get this far, you can carefully, carefully type on your calculator. Yes, yes, yes, yes. So there's a torque. Me too. Circuit. Okay. What is the terminal voltage of the battery? Now the terminal voltage is your EMF minus whatever voltage you lose going through here. It's the chairlift minus little bump getting off the chairlift. So really it's going to be 12 minus I times R because that's what voltage is. Although this question is saying find the terminal voltage, really I'm probably going to be finding other stuff. I always look, first thing I always check for did they tell me total current? They did? Yeah. That's very nice of them. Very thoughtful of them. How's that help? Well, I'd like to find the voltage drop through here. Okay. Now there's a sneaky way of doing it and the sneaky way of doing it is to realize this resistor is twice as big as this resistor which means it's going to get half the current of this one. So if you divide this by three, you'll have two here, one here. Now that's the sneaky shortcut way. Otherwise I would say this, Vitaly, these two resistors, are they in parallel or are they in series? They're in parallel. That means I can replace this with one single resistor if I want to. One over R parallel is one over 20 plus one over 40. One over 20 plus one over 40 is that. Take the reciprocal. That's the same as a 13.3333 and I'll carry a few extra sig figs. That's the same as a 13.333 ohm resistor. Mathematically, here's what that means. These two here could be replaced with a single resistor of 13.33 ohm. With those not being there. With these temporarily not being there. How many amps will be going through this resistor, this single mathematical resistor? All of them. How many amps is all of them? So if this mathematically took 0.77, how many volts high is this hill right here? From here to here, how many volts is that? You get 10.267, 10.267. Now that's not the actual circuit Leslie, this is the actual circuit. But you know what? If from here to here is 10.67 volts, you know how many volts from here to here? 10.67 volts. You know how many volts from here to here? 10.67 volts. How high was our chair lift? 12. Can I go on a ski run like this and get back to zero? Then if I start 12 high and I lose 10.67, how many volts must I lose going through here? 1.73, now I actually don't even need to figure that out by the way, I'm gonna label it because it's gonna come in handy later, 1.73 volts. But can you see my terminal voltage is 10.67. It's chair lift minus if you're only going through one ski hill and that's the bump, then the one ski hill must be your terminal voltage because there's the bump. My terminal voltage is 10.7 volts. How much do you want to bet? Part B is find R, let's see. Oh, no. Instead it says explain what happens to the terminal voltage of this battery when switch S is opened. So let me erase this and I'm gonna erase all this because I can't, it's convenient. If I open this switch, what's your total resistance? You know what? We know the voltage here and we just calculate, we know the current, it's 0.77. Just for the heck of it, let's find R, the internal resistance of the battery because this is a physical construct. It doesn't change at all. What is the internal resistance of this battery? R equals V over I. It's going to be 1.73 divided by 0.77. The internal resistance of this battery is 2.25 ohms. With this switch closed, what's my total resistance? This plus that, 15 point something. Is that okay? With this switch open, that means this is no longer part of the circuit. You know what my total resistance is? So closed, R total equals, what did we say? 13.3 plus 2.25, 15.6. Open, R total equals, so this is no longer part of the circuit. My total resistance is 40 plus 2.25. Opening the switch, when you have resistors in parallel, removing a resistor in parallel actually increases your resistance, not decreases, even though you're losing a resistor. You're closing a ski run, which means more skiers got to go down the same hill. Which means they're going to get crowded. What did I say? 42.25. Now, I total equals V total over R total. My total current is that, where V total is my EMF of 12. R total is changing. Will I have more current with switch open or with switch closed? Jacob, I total decreases. So what's gonna happen to the voltage that I lose going through here if this current gets smaller? Less, so what's smaller bump off of my chair lift? That means my terminal voltage is larger. You don't lose as much going through the battery because not as much current is going through the battery. You could crunch the numbers and do it that way. I did it algebraically, although I used some numbers. Moving on like crazy. I'm looking at this. I'm thinking solenoids, right hand rule transformers and all that stuff. Let's see. Two solenoids, place side by side as shown, are functioned together as an ideal transformer. So I'm thinking of my transformer equation as an ideal transformer. As an ideal transformer. Thank you. Okay, we gotta keep it going. It's luck or whatever and I don't wanna stop now, right? So, sorry for those of you listening at home. Someone said more than meets the eye. It's a tradition in my class for those of you who are watching this. Yes, I'm not weird. It is an AC power supply, which means the magnetic field's going to be going left, right, left, right, which is gonna be giving you a huge changing flux, which is going to generate a current. Now, it wants the power dissipated in this resistor. I think what I'm gonna need to do is find the current in this resistor. Let's see. They gave me the number of coils. So I'm gonna use the number of coils primary divided by the number of coils secondary. And they told me the initial current. So someone who has a formula sheet, how are coils related to current? It's NP over MS equals, yeah, it's backwards, IS over I primary, right? Number of primary coils, first one, 230. That's a nice picture, actually. I should steal, that's a nice graphic to steal because I could chop off the resistor as well and that way I could use that on path. I'll have to remember that. Over 46 equals secondary current, I don't know, over initial current of 0.35. Can you guys right now cross multiply? That times that divided by that. What's my secondary current? I'm trying to keep this all on one page so I don't wanna show one more line of work. So what would you get? 1.75. So how many amps are going through this resistor then? 1.75. Now, I know the resistor and I know the current. The only formula they give me on my formula sheet is that however, can you look at that one and also use V equals I times R and give me one that only has I's and R's in it? Yeah, plug this in for V and you'll get I squared R. The power is gonna be 1.75 squared times 15. That's a pretty easy, how many marks is that? Five marks, okay, I'll take that. What's the answer? 45.9? Yes, yes, happy joy. Here's our graphing type question. So during an electrostatics experiment to investigate electric potential energy, okay? Right away I would be saying we had this P E equals K Q 1 Q 2 over R and we also had P E equals Q times voltage, thinking those. A massive, a positive point charge Q 1 is moved gradually closer to a 10 micro coulomb charge that is fixed to the tabletop. I'm kinda looking at this, I think, I think, I think. The electric potential is determined at several distances and they graph it. Now this has created a linear graph. How the heck could you get a linear graph from here? It says in the box on the graph below, write the function, including units, of the separation distance R that must be used on the horizontal axis to produce a linear relation from this data. Because Vitaly, simple question, where is the R right now on the top or on the bottom? Hmm, that's no good. How the, oh, I'm gonna erase this one here, you don't wanna use that one. What the heck are they doing here? I'm gonna wimp out, curious to see how they explain this. I was gonna use calculus, but I can't do that. Give me a second, overventure exams. No, I think what they're doing is they're graphing the reciprocal of R instead of graphing R along the baseline. I think they're graphing one over R along the baseline, which gives you a straight line, but I just wanna see how they want me to phrase this. Yeah, okay. So here's what they're saying. If you wanna straighten this out, you don't wanna graph R along your x-axis. You want to graph one over R, because if you graph one over R, the reciprocal, that will take the reciprocal in your data and that will give you a lovely straight line. What they're really saying is how is energy directly related to the radius? And if you look at my original equation, sorry, my voice is finally dying. Potential energy is related to something over R, so one over R and including the units meters to the negative one or one over meters. Pardon me? Did you? Okay, what does it want? Oh, explain how you could use the slope of this graph to determine the unknown charge q. So slope is rise over run. Now on my y-axis, it's potential energy. On my x-axis, it's not R, what do I really have on my x-axis? One over R, so if I go divided by one over R, how do I divide by a fraction? Flip it and multiply. Let's see, potential energy times R is, now if I move the R over to this side from my equation, what's left? K, mystery charge and the charge that I know. So how could I use the slope of this? I think if I then divide by the charge that I know and divide by K, the mystery charge that I don't know is gonna be the potential energy times the radius divided by K, nine times 10 to the ninth, and they told me that the second charge was q2. These are tough. They have made the graphing questions, I think, tougher on the newer exams. Let's see, am I right? Slope divided by K, q, there you go. Last question is going to be a using principles of physics right to explain. Let's see, okay, you're pulling a crate across a smooth constant, a concrete floor with constant horizontal force, and I notice the crate speed is increasing. What that means is my force doesn't equal friction, it's bigger than friction. Says, using principles of physics, explain why your power output is also increasing. Okay, power, not in terms of a circuit, but the way we define power initially, way back in working energy, Sally, what is it? Work the way we defined it originally. God, good, good, good, because now I got a force appearing. What's this say? Can you see speed in my equation anywhere? It's hidden, but it's there. Ah, this is meters per second. Is it not? My force is constant, but it's told me that my speed is increasing. So even though this is staying the same, what's happening to this number right here, Justin? Getting bigger, getting bigger. So what's happening to my power? So I would now say as D over T increases, P increases. By the way, in our notation, I've used up arrow for increases, and down arrow for decreases, don't on the provincial or right off the word, okay? You should find, by the way, also, you have three full hours. Those of you that wrote the math one today, was time an issue at all, okay? Physics, time will not be an issue at all. And there's no non-calc section, so as soon as you come to a question, you don't get right away, circle it. Don't even let it throw you off. Come back to it later. Do a dry run through should take you about 90 minutes, and then spend the next 90 minutes wrestling with all the questions that you're wondering about. Again, don't leave the exam early. Stay the whole time. You finish in two hours. Let your mind wander. You'll be amazed at how often an answer pops into your head. I think you were saying that for one of the probability, for one of these ones, I think, no, the trig identity, you were saying initially you're like, what, what, what, and then it pops into your head, right, yeah. The brain's worked that way. You and yours, maybe. Okay, so there's one written section. Kind of jog in your memory. Let's go back a year, 2009. Well, these are a lot of things, but. Oh, you mean positives, sorry. Let's see, you made it. A little late, but better late than never. Of course, better never late. All right, so I said to you question one. It's gonna be from one of the first three units. Forces, kinematics, projectiles, or momentum work energy. So this one here, forces. We've done a bunch of these. Is this at an angle with a ramp? Oh, even nicer. Is there friction? Okay, piece of cake. I generally start with the hanging mask. So here's M2. What are the forces acting on this mask? Get the obvious ones. All we're doing is working through some provincials that I haven't given you yet. So kind of follow along. What about right there? Tension. What are the forces acting on mass one? Get the obvious ones. M1g. Gonna have a normal force. And these are equal because we're on level ground. If we're on a ramp, they're not equal. Or if they're tugging up at an angle like a lawn mower, we're not equal. What else? Tension and friction. Who's winning? You know what? It can't possibly be moving to the left because friction doesn't generate motion. If it's moving at all, it's moving down into the right. Oh, and if I get a negative answer for my acceleration, I'll know that friction is so big, it's keeping this up, but I don't think it is because this mass is bigger than this mass. It's moving to the right. Winner minus loser plus winner minus what else is along the rope here? Friction. And I've done both masses, Troy. So it's gonna be M1 plus M2 times A. Both masses times A. At this time of year, it's always good when tension vanishes. Okay, enough of that. Friction is what times what? I don't know the normal force. Oh, but look, look, look, look, look. I know another force the same size as the normal force. So it's gonna be mu. I'll be careful of Jasmine, M1g, because I got more than one mass. I'll be really careful with my notation. And they want me to find the acceleration. I'm gonna get the A by itself right now and I'm gonna divide by M1 plus M2. Do you know M2? Check. Do you know G? Yaw. Don't put negative 9.8. We decided in this unit, we would use the winner minus loser. We would let winner be positive to avoid having to decide which way is up, which way is down, especially if you had stuff moving up and down. It was less confusing. Mu.4. What are the units from you, by the way? Aren't none. It's a unit list measurement. M1, 3g9.8 divided by both masses. Could you find that? Then we're not gonna bother. I'm going to assume if you've got that far, you could carefully type that in. Now, another follow up question might be, find the tension. Once you find the acceleration, you can either go M2g minus tension equals M2a. And that's probably what I choose because it has less forces. And then I would say get the tension by itself, plus that over, minus that over. You know M2, you know g. You just figure it out, A, and it's plug and jump. Oh, here's your principles of physics question. Student B tells student A that mass M2 has to be greater than mass M1 if this system is to accelerate. Explain why this need not be the case. You know what a simple solution would be? Take this, and instead of having mass 2B5, try plugging it into two and see if you still get a positive acceleration. The issue is, as long as friction is small and what determines friction is mu, as long as you have a small mu, it's gonna move to the right and accelerate. In fact, if there was no friction, if this was ice, even if this was 0.1 kilograms, it's gonna accelerate because there's nothing holding it back. Right? Does that make sense? Yes? Good? Lying? Okay, I don't mind. I can do this thing, no problem. Right click. Copy. That's so cool. Paste. I would write that down, and then I would just say, let M equal one kilogram, sorry, let M2 equal one kilogram, let M, jeez, do I keep your numbers right? Let M1 be three kilograms, and I think if you crunch the numbers, you'll find you still get a positive acceleration. What's slowing it down? This, as long as MG is bigger than this, and which means really, Leslie, what's making this small is your coefficient of friction. This is 0.1 or 0.2? Even if this is really big, it's still gonna be a small force. In fact, if this is close to zero, like ice, you're gonna have no friction, which means this can be almost microscopic. It's still gonna move. Right? Yes, yes? Good, good, good, good. Hey, really quickly, I'm glad to answer this question. What topic is this? Pretty sure it's torque. You know how I can tell so easily? It's a beam. Now I'll actually read the question, but it is nice to be able to, so I know where I am to be able to say that when you glance at the question. So two vertical wires with tensions indicated support a uniform 14 kilogram, three meter long beam containing a mystery object of mass M as shown. What distance L from the right-hand wire is the object located? Mm. Okay. This is a bit of a trickier one because I don't know the mass. Well, you know what the first thing I'm gonna do here is? Label my forces. So I have mass of the beam times G, capital M, G. Oh, hey, wait a minute, wait a minute, wait a minute. Are these two forces straight up? So the first thing I can say is this, 625 up plus 348 up has to equal mystery mass times G down plus mass of the beam times G down. I can solve for mystery mass M actually quite easily because all the forces up have to equal all the forces down. And in particular, this is nice metallic because everything's vertical right angles. No trig. I like that. Mystery mass M is going to be, oh, turn it on, Mr. Duke, 625 plus 348 minus 14 times 9.8 equals divided by 9.8. I've gone really fast but you're okay in getting the big M by itself, yes? So I get a mystery mass is 85.3 kilos. That helps me now. Now I can solve this with torques. Where am I going to put my pivot? Leslie, I could put it anywhere I wanted to. I could put it here, here, here or here. Although since they gave me this L here, I'll put it right there and that way I'm solving for L directly. I would write the sum of all the torques clockwise in this direction equals the sum of all the torques counterclockwise in this direction. This is a bit nicer, Justin, than the previous tort question that we did about a half hour ago because right angle, right angle, right angle. Everything's perpendicular, no trig. So which force is causing this to spin clockwise? Can you give me its name? Pointing doesn't help me. It's going to be a letter L for left and an R for right. Tension left. Yes, I know it's small. You guys sat in the back row. I can't. So the only clockwise torque is 625 times its distance from the pivot, three. And that equals, there are two counterclockwise torques, this one and this one. 85.3 times 9.8 times the mystery distance from the pivot plus 14 times 9.8 times, how far is the mass of the beam from the pivot? 1.5 center of mass. Could you solve that for L, Jasmine? Yes, how? Minus this, right? And then divide by that and we're done. You can do that, then I'm not going to bother. And somehow those ended up down there. Okay, where the heck are they from? No idea. Well, I'll do this. The software is pretty good. I mean, it hardly ever glitches, I have to admit. This program here called one that whoever wrote it knows how to write software. A small sphere having a charge, ooh, I bet you were in electrostatics. Is suspended from a thread hanging between two charged plates as shown. What's the mass of the small sphere? You know what, I've seen this diagram with a weight hanging off at an angle so often, I'm pretty sure we're going to be labeling the forces and getting some kind of a vector triangle. So you ready? What are the forces acting on this sphere, Justin, get the obvious one? Darn right. What else? Tension. What's pushing it to the right? Something is. A force. Which force? Magnetic, I see no magnets here anywhere. I think we'll call this electric. It's got a charge. And in fact, you know what they've done? They put it between parallel plates that have an electric field. But oh, I think I can do this. Okay, I can do this. What's the net force on this sphere? Is it accelerating? So you know what my net force is then? When I draw these three together, I'm going to add them tip to tail, tip to tail, tip to tail. I should end up with a triangle that comes back where it started from. How do I draw these? I always draw the easiest one first. What's the easiest one? MG. I always draw the toughest one next. What's the toughest one? Tension. How do I know to stop right there and not go any further? I know my third force is exactly horizontal. Goes like that. Oh, and I'm pretty sure this is 18 degrees. Can you all see the Zed? Yes, I'm going to get rid of that line because that really clutters things up. No, yes. Yes. What do they want me to find? Can you see? I know it's at the bottom of the screen. Mass. I don't know tension, but I'm willing to bet that I can probably find electric force. Which trig function relates these two together? Yeah, here's what you're going to solve. The tangent of 18 equals opposite over adjacent. The mystery mass is the electric force divided by g tan 18. I move the m up to here, Justin, and I move the trig down to the... I'm going to stop moving diagonal. Hey, if I've taught you nothing else, by the way, that's a great shortcut to use for the rest of your mathematical life. You know what I'm really going to spend most of my time doing? This. So on our formula sheet for electrostatics, do I have two charges in this diagram? So I'm not using the first force equation that has q1, q2. What's my other force equation? Eq? And that's when we got one charge? Okay, excellent. The electric force is equal to, I usually wrote it as qe. How's that helped me? Do I know the electric field here? Nope. Hey, find something for electric field on your formula sheet, preferably that has maybe, maybe voltage, and maybe, maybe, maybe some kind of a distance in it. That would be lovely. Sorry. Oh, electric field is voltage. So if I hear you correctly, this is q voltage over distance. Okay. What was q? 2.3 micro coulombs, what was micro? Question Battali or Leslie? No, we're good, we're good, we're good. Yes, Leslie, we're good, Leslie? Awesome. 2.3 micro coulombs, so 2.3 times 10 to the negative six, voltage 62. I have to be a bit careful. What distance did they give me? Can you see it at the very, very back? Centimeters, I'm not gonna fall for that, right? Once I have electric force, and I probably carry it to like five sig figs because it's not my final answer, put it there, divide by g, 10, 18, and you've got the mass. Am I expecting a big mass for a charged particle? No. If you get like five kilograms, you're wrong. Okay. So there's a nice combination of forces and electrostatics. With a little bit of equilibrium mixed in, okay? That's what they started doing since they've shortened the written. More and more of the written questions are multi-unit questions, which is why I kind of preferred the old one just as its structure. The nerd within me likes these. These are more and more of the scholarship type questions. Sorry. The ones that most of you didn't do that I did put on all your reviews for a reason. Usually my nerds get into those, but this year, okay. What unit did we just do? Electrostatics. I'm gonna guess either magnetism, circuitry. Let's see. Magnetic field. So right away, I'm thinking right hand rule. Okay, I'm looking at that section of my formula sheet. Okay. A block of metal moves north at 180 meters per second. So right away in my magnetic forces, I'm looking at the ones that have a lower case v for speed in them. Perpendicular to a magnetic field. If an electron in the block experiences, so here's our electron right here, experiences the force vertically upwards. What are the magnetic field strength and direction? Direction we're gonna be using our physics yoga. Let's see if we can get the strength. So what's the equation that has force, speed, and magnetic field in it? Oh, how did you know not to use bill? Cause they're talking about an electron, which is a charge as well. Okay, so f equals q v b. v is gonna be the force divided by q. Oh, I don't want v, I want b, Mr. Duk. B is gonna be the force divided by q v. Do I know the force? Yeah, 7.2 times 10 to the negative 18 divided by q, it didn't, oh, electron. 1.6 times 10 to the negative 19, right? That one I do finally have memorized, we've used it so often. v, they said was 180. Well, geez, that's not, that's like two lines of work, that's not bad. I can crunch that and get the magnitude. By the way, what's a big magnetic field? You remember? Like 100. If you get an answer bigger than 100, you're like either at the large hydron collider or you've made a mistake, okay? Most electric fields are in the 10 to the negative two, 10 to the negative one ish, sorry, magnetic field. That's electric field, magnetic fields. Now, they also want the direction. Which way is an electron feeling a force? Which way is the electron feeling a force? Our rule doesn't work for electrons. Which way would a proton feel a force then? Downwards. So although I drew an electron, I'm gonna make him a little proton. He would feel a force downwards. So I would point my palm, I'll face in the same direction as you downwards because palm is my force. Which way is this proton traveling? Which way is the block traveling? Upwards. So I'm going to go palm down. Gotta be careful, Mr. Dewey. Vertically upwards out of the page. Hang on, do they mean vertically upwards towards me? Yes. Sorry. I don't like the vertically upwards. I think it should just be out of the page. So okay, force is out of the page towards us. Now that's the electron. Which way would a proton feel a force? Into the page, I'm gonna have to do this. So into the page, the proton is traveling upwards as is the electron. Which way are your fingers pointing? Which way is the magnetic field? East. Okay. Leslie? Yes? East. I don't like the wording of that. At least you knew something was wrong when you tried to point your palm downwards and bend your thumb, unless you're double jointed or something. So you can tell, right? Justin's looking at this, oh, I can, watch this, but I can bend it. So what did I just say, magnetic field? Next one, I bet you it's gonna be the graph because the magnetic field was last unit. Oh, oh, there's a part B. What's the potential difference developed across this block? What's another word for potential difference? Voltage, can you find me an equation that has little V velocity and EMF symbol for voltage? BLV, let's see. E equals BLV, I'll go up here. Do I know the velocity? Check. Did I just figure out the magnetic field if I'd crunch the numbers? Check, which length, they gave you two, well actually they gave you three, which length is moving perpendicular to the magnetic field? Now the magnetic field we said was to the east, that's the length that's moving perpendicular. Sorry, Mr. Dewick, careful here because it also has to be perpendicular to the velocity, Mr. Dewick. I think this one, isn't it? Don't believe me? That's the only way that this could be coming out of the page towards us, right? It would be moving from the bottom to the top. You're not gonna get much of a voltage. Oh, 15 centimeters, okay, it's a big block. So there's your part B. Graphing, during an electrostatics experiment, a small positive, another electrostatics, okay, a small positively charged plastic sphere is accelerated from rest through several voltages, V. So right now, and they're talking about the final velocity, how is the final velocity, the voltage related? I remember this one, potential energy was QV, where we had all of your kinetic becoming all of your potential. From rest, so what's your initial kinetic? And your final kinetic is whatever you've used up going through the charge here, okay? So the final velocity of the sphere is possible to create a linear graph and then obtain the slope. How is voltage and V related? They're not directly related, what do you see here? You know what, if you wanted to, you would graph that. If you graphed V, you'd get a lovely parabola because it's a V squared graph. I'm just gonna double check on that one as well, I'm that uncertain of this. They've changed these ones, I'm not gonna include some of these in my reviews. Right, the function include units. So there you go, they want you to get the voltage by itself, voltage equals M V squared over two Q, instead of a one half I'll just put a two in the bottom. There's the slope of this graph. Y equals M X, okay? So I guess the total function would be this. The function of little V including units would be M, well it'll be this. What do they want me to write, Mr. Duke? M over two Q V squared. Units are meters squared over seconds squared. Last one, I'm not that thrilled with these new graphing ones. I'll clearly explain how you can use this to find the charge. Oh, forgot that. Get the Q by itself, go back here. Get the Q by itself. Last one, oh no I want this one. Student A states that there is no force due to gravity acting on the International Space Station, the ISS. He states this has to be so because he's seen television coverage showing astronauts and equipment floating around in the spacecraft. Student B, my students, explain that this is a common misconception for orbiting spacecraft such as the ISS. Using principles of physics helps student B convince student A that there must be a force due to gravity acting on the ISS. Is there a force due to gravity acting on the ISS? Then why do they feel weightless? They're not weightless, we call it apparent weightlessness. What's really going on there? You're free fall. We just give you enough sideways velocity so that you're free falling and moving sideways at exactly the curvature of the Earth. So here is their fancy schmancy answer. For a spacecraft to orbit the Earth, it must be an accelerated motion because it's moving in a circle. If it's moving in a circle, there must be an acceleration on it. What force is moving it in a circle? Gravity. This force is a gravitational force produced by the Earth's gravitational field. You could say it's in free fall, I mean all sorts of explanations they would take. Pardon me. Technically, the force out at the edge of the universe would also be zero, which is why you have no potential energy because there's no force to tug you down. You could argue that and they would take that and then they would probably go, hey, check out this nerd here, she's going all calc on us. Truthfully, as we were marking, we probably, like I marked one of these one, I think I marked physics twice and I marked one of these questions one year and you would see some kids pull out limits and calc. Oh sure, oh wow, that's really, that's even a better explanation. You could assume that anybody who's marking it has done calculus in the past. Pretty much comes with the physics territory. So we're an hour and 10 minutes in, I'm gonna try getting through one more. Fair enough, January 2008. I may skip questions if it's the same topic. Oh, projectile, good. Cannonball was launched at 85 meters per second, 50 degrees above the horizontal toward a hill as shown. What horizontal distance D does the cannonball travel before it impacts the hillside? Components, components, components, components, components. First thing I'm gonna do is take this 85 and this 50 degrees and then break it up into VX and VY. And VY, oh, I'm gonna break this up into horizontal, and vertical. Now VY is gonna be an initial. VX is not gonna be an initial. Why does VX not have an initial but VY does? Because of gravity, thank you for not explaining it by just using a word, sorry. Horizontally, if we ignore air resistance, no acceleration. Vertically, there is an acceleration. In fact, I would also write AY equals negative 9.8 and I would write AX equals zero. VY initial is going to be sine, trust me, do the trig and it works. So I'm gonna do this really, really quickly. 85 sine 50, 65.11, there's your initial vertical velocity and my initial horizontal velocity is gonna be 85 cos 50 which is 54.64. That's pretty much a given for almost any projectile. You start out doing that and then you pause. If you're lucky, they gave you time of flight. Did they tell us the time of flight? I say no, no. Then what you will do is you'll say somewhere they gave you a displacement in the diagram in the paragraph somewhere. They did give me a displacement in the diagram. Can you see it? How big? 72, vertical or horizontal. Above or below from where we started? Above, so positive 72. What if it was below? Negative, that's if you're on a cliff. DUI equals 72. Now I can find time from here. You have that option. Now to solve this, you're going to need to use the quadratic equation and I'm noticing that some of you are a little shaky on the quadratic equation. So you do have a second option. Your second option would be to find VY final. Could you find VY final from those four things there using VF squared equals VI squared plus 2AD? You would have to decide whether VY was positive or negative because when you square root you will always get a positive. Looking at my diagram I think I would assume it's negative but once you find VY then you can use find T using VF, VI, and A. But I'm going to do the quadratic formula because I'm a nerd. D equals VI T plus a half AT squared. 72 equals 65.11 T minus 4.9 T squared. Justin, what kind of an equation is this? You were elsewhere in nice recovery. How do I know? You weren't elsewhere, you're actually writing things down. I stand corrected. Usually when I saw that look in the drill that you were elsewhere, you're here good, good. How do I solve the quadratic? First thing, before I do anything else. Make it equal to zero. Zero equals negative 4.9 T squared plus 65.11 T minus 72. Find the roots. Now I encourage you to write down the quadratic formula which is on your formula sheet but then sheet, use your quadratic solver which I've installed on almost all your calculators I hope. Where is your quadratic solver? I don't have it on my virtual one but if you press the apps button you should have a polysmult right there. Blue apps button right here. Do you have a polysmult? Good, run it. And then it should say polynomial equation solver option number one. And then it should say degree question mark. The degree is two because it's quadratic. And then it's doing A, B, and C but I think on yours it's not doing A, B, and C. What's it doing instead? Yours is A, B, and C? Yeah, yours has A, two, A, one, and A, zero. So you guys have polysmult number two. You have polysmult number one. They rewrote the program. Anyways, type them in and I think solve lines up with graph, does it not? Yes, good, fine, what are the roots? And if you don't know how to do that, I will show you but later. What are the roots? Yeah, you know why? Because it's 72 meters high right here and right here. Which root do I want? Which time do I want? The bigger time, 12.07 seconds. Time of flight is 12.07 seconds. Okay, if you're not able to do the point, I'll show you later but don't let it freak you out. I don't want you to zone out trying to just focus on that one thing and miss what we're doing. So we solved it, we got the time of flight. Less than we took the bigger one, we got two positive answers. We said, oh yeah, I guess it's 72 meters here and it's 72 meters high here. I want the bigger time because that's when it's over here clearly. And they want me to find horizontal distance D. They want me to find DX. What's an equation that has V, A, and T in it? This bad boy, but what's A? So DX is just gonna be VX times T. 54.64 times 12.07. There's your classic projectile range type question. Is that okay with telling? Can't type it in wrong. Again, I'll do it later then. Just leave it, but I'll do it later. Remind me at the end and I'll scroll back. Next one. So we've done momentum. We've done torque again. Are you guys okay on torque? You don't want me to do one more? Can you see the eight? Yeah, you know what? I'll get the equation then I'll quit. You do it every other time too. Okay, fine. This one only has one angle. Okay, what do they want us to find? Tension. Which tension? Oh, cable. They want us to find this. Tension T. What are the forces acting on this beam? Justin, get the obvious ones. Darn right. So I'm gonna have mass of the beam times G. Now I'm gonna need to find the perpendicular component. What was that? Good, Tshush. Okay. Ready? You wanted the angles. There's the only one they gave me. So I start here. What letter? Zed? How big? Okay, remember that? Yes, yes, yes, yes, yes. What letter? How big? 50. Two Zeds. So I would go, that's 50. And it's gonna be the same kind of an idea here. You have MG perpendicular. That's 50. Here, tension perpendicular, 90 degree angle. I realize this goes through the, I don't care. By the way, we're missing a couple of forces. There's a couple more forces acting right here as well. But they're right by the pivot, so who cares? All right. What's the only angle that they gave me? 50? Okay. How big? Why? Because there's a triangle here. How big? How big? If I wanna use the 50 or I could use the 40 right there. Only advice I can give you is if you need to, get lots of scrap paper that you can tear out, make a little sketch and extend vertical and horizontal lines. Oh, and it often helps to extend your beam. You can often spot an extra angle or two. Like, oh, that's 50. This here is 130. Oh, this here, you can find other angles too. Yeah, that great 11 geometry, woo-hoo! Yeah, maybe. Usually though, it's not this ugly because this force would be up and at yucky angles. It'll almost always be a horizontal beam if they're gonna do that. And almost always be vertical weights and only be one tension at an angle. So there's torque, real quick. Here's our using principles of physics. Explain how this tension will change if the beam is lowered to a more horizontal orientation as the angle increases. The cable remains horizontal and connected to the same point on the beam. I always go to extremes. What if the beam was right there? What would the tension on the cable be? Zero. So you know what, as this goes down, what do you think happens to the tension? It gets bigger. And honestly, that would be my explanation. A vertical beam would have a tension of zero as theta increases, therefore tension increases. The other thing you could do is you would have a lovely equation for tension right here. You could simply rewrite it here, look at where the angle is, and just do a couple of examples with a bigger angle and show that tension is bigger if you wanna crunch it numerically. Hey, what topic must this be? Glance at it, and you know right away what? Circular motion. I see a circle. Okay. A blue ball is swung on, I don't know why we need to know that's blue, on a black and white diagram, on a black and white test, really? Well, honestly though, they probably printed several hundred of these exams. That's one extra word, that's extra laser ink, that's money, and I'm willing to bet the government probably pays by the word. Really? Anyways, a sphere is swung in a horizontal circle, and it completes a single rotation in 1.2 seconds. You know what? They just told me the period, and they gave me a strong hint as to which equation I'm probably going to use here, along with the period in it. The 0.44 meter long cord makes an angle of 35 degrees with the vertical. What's the centripetal acceleration of the ball? Okay. I know to use that one because of that. Do I know four? Yeah, do I know pi? Yeah, do I know r? No. Well, that's r, yes. Can I figure out r using grade nine trig? Yep, opposite, hypotenuse. Ooh, it's going to be sine. In fact, I'll be honest, for five marks, this seems really cheesy, really wimpy, unless the next two are nasty, in which case then I would say great, and I'm kind of actually now. And while the first one, did we use the quadratic formula on the first question? That's considered fairly difficult, so okay, that was a bit tricky. Although this torque one wasn't too hideous, although okay, fair enough, then maybe they're okay with it. Could you guys all find r, okay? And then it's just, oh, by the way, why didn't I put an m here? Did they ask for this centripetal force? No, they just asked for this centripetal acceleration. A proton enters a 6.5 Tesla magnetic field. Oh, I'm in magnetism. Apparently we skipped circuitry. Or circuitry is going to be using principles of physics right to explain. That's my bad. Horegraph. The velocity is perpendicular, causes it, travels in a central arc. Ooh, what's this asking us to find? What is the, I gave this to you on your test. This is one of your questions. Are we moving in a circle? Then you know what? We have circular force. What force is pushing us in a circle? Troy, which force is it? Not electric force. Which force is it? Which force is it? The one that you said last time before you said electric force. Ah! But we use the symbol b for magnetic because m is mass. Okay. We're gonna use mv squared over r. This time it's a force, so I better have the m in front of the a, yes. Equals, what's the force on a charge in a magnetic field? Like, help me out, find it. Which one is it? Bill? No, oh QVB, okay. Conveniently Vitaly, one of my b's cancels. What did this question ask us to find? Momentum, problem is, I don't know the mass or the velocity. Oh, wait a minute. Just get the mv by itself. There's the momentum. And you know what it equals? Q, b, r. Do I know Q? What's the charge on a proton? 1.6 times 10 to the element, the elementary fundamental charge. Do I know b? Do I know r? Boy, I almost said do I know r, but no, that's math, 12. Wrong subject. Although, Justin, you're the only one that's not in math 12, so the rest of you go with Joe. There you go. For apparently five marks, one, two, three lines, and now crunch the numbers. Now, what you do have to clue in is to recognize I can't find the momentum, but I can find what it's equal to. Nene, you know, this though, this is really wishy-washy, because four lines, two lines, there's gotta be a nasty coming here. Let's see. Electrostatics, again? Are you serious? Okay. Electrostatic experiment to investigate Coulomb's law. Right away I would say that's KQ1Q2 over r squared. I think the function of r is going to be KQ1Q2 over r squared. Anyways, this is almost the same question as we did before, but they had voltage. It's the same charge. It almost looks like somebody just changed this and then hit paste into another exam, because I think it was also 10 microcoulomb last time, too, if I recall. Okay. Function of r, including units, what do they want us to write here to do? Yeah, one over r squared. Pretty simple. Okay. Explain how you could use this to determine the unknown charge. The slope is rise over run, and then once you've done rise over run, get the unknown charge by itself. Oh, and we'll just do this here because we're running out of steam. Here was the using principles of physics. It said, two bricks above they have equal mass and they are initially positioned at the same height, which means they have the same potential energy. At rest, the surfaces of both inclined planes are frictionless. Okay, don't need to worry about friction. Using principles of physics, explain why both masses have the same speed when they reach the floor. You know what? I would just do it as straight conservation of energy. In fact, let's do it on the actual one that I can write on. I would say, you know, kinetic energy initial plus potential energy initial equals kinetic energy final plus potential energy final. And over here, same thing. Which of these are zero? They're hitting the ground. Mgh initial equals a half MV final squared. Mgh initial equals one half MV final squared. Now, I'll be honest, if it's frictionless, they don't need to have the same mass. It's nice if they said that they had the same mass, but what do you notice about the masses in this equation? What do you notice about the masses here? Cancel. Anyways, then I would probably, well, let's go to what they want you to write, but I would finish by saying, hey, V squared, there you go. Both bricks have equal mass so same potential total, all PE. Neither brick produces any heat energy through friction. So I guess because they specifically said no friction, you might add in your answer, no friction, no friction or something like that. And then you're good. The only thing I'm running through my mind, I haven't done any orbiting stuff. You guys want me to? I've still got a half hour circuitry. Let's see. Let's run through one more really quickly. So here's one curling stone, collision momentum. We did a momentum already. Torque diving board. Here, this one's a bit nice because there's two unknown forces. No problem. Put the pivot at one of these. Well, if they want you to find the force B, put the pivot at force A and then you can find force B using torques. You'll have to think about which way your forces are as it turns out force B is pushing up. You know how I know? Because there's a force down here and a force down here, which is causing it to spin this way. Which way, both these forces down would mean this would have to be pushing up and this is actually pushing down. You look at the end of a diving board, this rivets on the end of a diving board, holding the diving board down, but who cares? That's where you put your pivot. Electrostatics says a potential difference of 360 volts is required to hold this stationary, find the magnitude of the charge on the sphere. What are the forces acting on here? Get the obvious ones. Yeah, I think it's gonna be mg equals fe. Are there two charges here? So I'm not gonna use kq1q2 over r squared. I'm gonna use qe. Do I know the electric field? No. But I do know that it's voltage divided by distance. Sorry what? I never heard a word you said. Cathode ray tubes? I've seen them occasionally as you can principle physics right to explain question. Okay. Usually all they're gonna ask is tell us what the solenoids are doing. Having said that the technology is not as common, so it's like it used to be great. It was like, hi, we're doing exactly how a television works and all of you have a TV that works just like this. Stupid flat screens. Sally, what letter? Magnetic field. Oh, I guess this bar is being pushed out at an angle. And there is a current in the rod. What are the forces acting on this rod? Get the obvious ones. So I'm gonna do an end view. Gravity down. Tension, which I don't care about. And magnetic force. Which magnetic force? This is the same question Troy as the one we did a couple of questions ago. We had the particle, except this magnetic force is gonna be through a rod with a current. They've told me the length of the rod right there. They've given me the mass of the rod right there. What do they want me to find? They want me to find the magnitude and the direction of the current. So I could find the current by doing my triangle again. Which direction? Which way is this rod being pushed to the right? So point your palm to the right. Which way is the magnetic field up the page to the arrows? So palm to the right, magnetic field up the page. My thumb is pointing. I have the current running from X to Y, yes? So there's the direction. Magnitude and direction. Next we're on the graph. Oh, you get to graph this one. Graphing final, oh, they want you to graph V squared versus distance. For me, maybe. Slope of the line of the line of best fit. I guess, yeah, bring a ruler. I'll have a few up there, I guess. Okay, let's look at this bad boy as well, just cause. I wanna see if they want you to actually graph it. Oh, we did this one already, Mr. Duke. So yeah, now when you graph it, the points won't line up quite bang on. What you wanna do is split the difference with your line, some above, some below, and then find the slope of your line. In other words, you could not use that point right there for slope, cause it's a little bit above the line. Specifically, you couldn't use that point right there. If you use that as part of your rise over run, it would take a half mark off. Rise over run, extend the line to the Y axis and use the Y intercept to determine the initial speed of the car. So with your ruler, right about there, 20 points something. 24, they're saying, and they mark these with a fair ability way, I mark this one, you're two. What's using principles of physics? A motor and a light bulb are connected in a series with a battery. Specifically, this should say an electric motor and a light bulb are connected in series with a battery. That's where they should use the extra word, not for blue, but for electric. Using principles of physics, explain why the light bulb gets brighter when the motor is prevented from spinning. What happens when we prevent a motor from spinning? No feedback, which means your overall net voltage goes up. Your overall current goes up and that means the current through the bulb goes up. So what's gonna happen to the bulb? Brighter, you know what's gonna happen to the motor? Get hotter, overheat. As soon as you let it spin, the net current goes down because the net voltage goes down because you're generating a back voltage in the opposite direction. Okay, any specifics now? 20 minutes, what are you wondering about? Still didn't find a good orbital question, but oh well. There's a bunch of exams that I gave you. You making fun of me? Is he making fun of me? Because I'll give him a detention, such a detention. I think you'll have more time. I think it flows better. The students who wrote it already found it. It was just nicer to do everything and be able to go back and forward. They found the written easier. I'm gonna get three if I can. I had a bunch right Monday, and I think every one of them approved it right. No, never mind. Well, Kowski did. He sat there and I didn't realize he hadn't read my email because I sent out the emails on Saturday and he was at 99% overall. He didn't know that, okay? So he sat there for two and a half hours and I finally mentioned it. You know, unless you get perfect, you can't improve your grade. What? You're at 99. Unless you get perfect on this mock, you can't improve your grade. You could just see him. I've been sitting here for two and a half, and you didn't. I'm sorry, I thought you read your email. Yeah, he was pretty perturbed. So he's the only one that did worse then because I think as soon as he found that out, he was so mad he didn't go back and you know how he likes to redo everything, but he didn't do that. Any specifics? Then folks, get a good night's sleep, okay? Between the English and the physics, although I'll be here, but I would really encourage you, go for a walk, clear your brain, do something different to flush your brain out, so to speak, okay? Believe it or not, when I was in university, what I would do between final exams with a friend of mine, a plate chess, because I had to concentrate on that for half an hour, which meant whatever the previous exam was gone, and then I could study for the next one. And yes, I was that much of a nerd too, but still. And I'm terrible in case you're wondering. Right click.