Three. And we did this loop and this loop to get our two expressions. So what I got, and I'll just reduce it a bit, was I one equals I two plus I three. Then I got for the first loop, I got the. Six belds is equal to three I one plus two I two. Okay. And then for the second loop, the red loop, I got the. Six plus four. So because I have two going in the same direction, so it's the short along the short along. So they're going in the same direction so we can add the emf's in the second loop, okay, is equal to. Three I one. Three times a. One. So this let us work out that I, one is equal to ten divided by three, which is 3.33 amps, which we got. And then. I substituted the value of I three n to this equation to get I two. So six is equal to three times 3.33, etcetera, which is ten. Plus two I two. So we we were on track except that we needed to add these. So six equals ten plus two I two. So six minus ten equals two I two. So. Four is equal to two I two. It should be minus four. Or minus two is equal to I two. And it's okay to have a negative value because it means the current is going in the opposite direction. So if we substitute minus two plus 3.33, we can get I one. So I one is negative. Negative two plus 3.3 equals 5.33 amps. So I one is 5.3 amps. Okay, so the only thing you you went wrong with was to add the two potential differences, okay. Do you want to try this one for tomorrow? Number six, try it again. Now see, it's a long ashalong, a long ashore. So this means the two long ends will be facing each other. So we need to subtract the emf's so it be six minus four at some stage. So if you do two closed loops, if you do this loop has Lupe and. This as lop b maybe, and then you get two expressions, but this time, because these long ends are facing each other, you will subtract the emf. Okay, so everything would have worked out except for the emf's there. Okay, so today we were going to go on to materials. And in materials we have viscosity, density, Stookes law, and then we have the spring hooks law and property of materials and Young modulus. So I thought wefocus on viscosity, density and Stookes law today. Does that sound okay? So. So with viscosity, viscosity is how what is how reluctant a fluid is to flow. So basically viscosity honey is quite viscous, water isn't. So we know density is MaaS over volume. The fluid upthrust is equal to the weight of the fluid displaced by an object. And. So when we were doing this equation up, thrust is the weight of the fluid displaced, and we can derive an equation for upthrust. So up thrust is equal to. If we have a. If we have a sphere falling through a viscous liquid, the up trust will be equal to the density of the fluid times, the volume times gravitational pull. So the upthrust, the force that acts up on the fluid is equal to the density of the fluid times, the volume times g, and the volume is equal to, if we're talking about a metal sphere, four over three ir cubed. And density about ruv. So what's that single row R H O so row. I mean ris the I mean what what equal to rv J so the uptrust force. When the sphere is falling, the utruforce, if you like, it's like resistance and electricity. So the weight is acting downwards, that we have a drag force, which is due to the density of the material. So the an object that is up thrust. Upthis the resistance of motion of the object through the medium. So if we have density as MaaS over a volume, but density of a fluid is equal to rho gh. So essentially your object will experience a resistive force. So oppressed and wait. So fiscous drag acts on objects moving through a fluid. So when an object moves through a fluid, we get a frictional force. And the viscous drag is the same as friction between fluid elements that move past each other. When you calculate force due to the viscous drag on a spherical object moving through a fluid, we get this relationship where f is the viscous drag force, eta is viscosity of the liquid, R is the radius of the object, and v is the speed. So. Up. When things float up, thrust is greater than weight, but if things don't float, then theymove through the fluid with a particular velocity. So basically, how they move through the fluid depends on whether they can experience streamlined laminar flow or turbulent flow. So laminar flow is when the fluid makes uniform streamline layers, the velocity of the fluid at a point is constant. There's no crossing of the lines. There's no change of direction of the fluid lines. They are parallel. So with Stokes law, we assume that an object is moving with laminar flow. Turbulent flow is the opposite. And viscosity depends on the temperature of a fluid. In most fluids, viscosity decreases as temperature increases. So when your fluid gets warmer, the viscosity decreases. This is a bit like your thermistor, the hosher it gets, the lower the resistance. So with stoke law, it's possible to calculate the resistive force on an object if it's moving through a fluid using this relationship, and this equation will be given to you in the back of your booklet. So R is the radius of the sphere, eta is the viscosity six pi, and then v is the velocity. And we assume that itbe a terminal velocity. So when your fluid starts to fall, it will speed up till it reaches a constant velocity. So. Drag force viscocity radius velcity laminar flow. So. We have buoyancy, we have drag and we have weight. So. Yeah. So if you measuring the term, the viscosity of a liquid, you would have to measure the radius of your sphere or the diameter, you would have to measure the density of the metal, the density of the fluid and calculate the terminal velocity of the object. And then you plot R squared against b, and the gradient is your viscosity. So. In this equation set of questions, we have the density in a fluid. The change in pressure is density gh, up thrust is weight is density times volume times g, and density is MaaS over volume. So basically, the up thrust depends on the size of the object, the density of the fluid. Freer is density times g, times H. There isn't much pressure in this curriculum from what I remember. So let's try a simple question first. Let's try question one. A block of metal has a MaaS of that and a volume of that calculated density. So wesimply use this equation. Yeah density equal to MaaS divided by the volume. So that is three divided by 0.02. So you get 15 hundred kilograms per meter. Cute. Actually that may be 1500. Yes. 15 hundred is the same good. A tank is filled with water with a density of a thousand kilograms per meter cubed. That's a useful constant to a depth of 5m. Calculate the pressure at the bottom of the tank. So pressure, so weuse this equation, we're working out the pressure at the bottom of the tank, because pressure tends to increase with depth. Density depth, okay density. Yeah, I see. Death a death also H 9.81 and the pressure density pressure okay, 1000 times, five times 9.81. 4000 no. Five times ten to four. Yeah 1234 so 4.9 so you rounded it up, that's fine. A liquid has a density of twelve hundred kilograms per meter cubed, and the volume of this calculate the MaaS of the liquid. MaaS of the liquid MaaS. Density volume, the MaaS. Yeah 1200 times 2.5 times ten to the minus four. Not point three. Kilograms, Yeah, kilograms. Pressure at a certain depth in a lake is measured to be 820 kilopascalves. So pressure calculate the depth of the water in the lake. The density is a thousand. Kilograms per Mesia cube. So we know the density, we know g, we know the pressure. We're finding the depth. 8.2 times ten, two os of five. Pascal divided by 1000 and then divided by 9.81. It is 3.6 good meters. Yeah. A rock has a MaaS of 5.6 kilograms and a density of 2800 kilograms per meter cube to calculate the rock's volume. 5.6 divided by. 5.6Yeah 5.6 divided by 2.8 times ten to the five. No no no just the thousand 82800. Two times ten to the minus two. Yeah, minus three. Two times ten to the minus three. Good meters, good meters. Q Okay. A sphere has a MaaS of this and a diameter of this find the sphere's density number seven. So we know the MaaS, we know the diameter of the sphere, so we have to find the volume. Yeah volume diameter is 40, so the radius is is 20, so 20 but that is centimeter. So 20 cm is not point 2m and mid and meters q times pi not equal to cube times pi times four over three is a volume. And then this number, 1000 divided by this number. 29.8 times ten to the three. Yeah good. A submarine is submerged at a depth of 100 mein seawater, which has a density of 102, five kilograms per meter cube, so seawater is slightly more dense than ordinary water. It rises from this depth to 50m. Calculate the difference in pressure between these two depths. After acting after. Abacting equal to row good. The density shouldn't change 100 times. One, zero, two, five. Okay it rises from this Depto 50m so that is 50 times 1001O2 five and times 9.81. Five zero 2762.5Yeah. Hello, la pascalals good? Okay, let's look at question eleven. So up thrust is row vg. So density times, volume times g. So a rectangular block has these dimensions and it floats in seawater with half of its heights submerged. So only half of it is displacing the water. Calculate the upthrust on the block. So. G we know density, we know. V, we have to calculate. This 12m times one meters samnot for 5m and half of his height subupon and there's density. Upstretwo. Times one, times 0.5, equal to one. And then so that is one, two, five times 0.81, sorry, 9.81, that is the answer. One Oh o five, 5.25, half of its height is submerged. Do you think that's important? Maybe times two. So. Half of its height had submerged. So half of its volume is submerged. So we have to instead of using one, we use 0.5. Everything else is fine because if our block is only half submerged, it means. An object will displace this volume of water displaced will be equal to the weight of the block. So that if we collected that volume of water, that would be the same weight as the weight of the block. So twelve, a solid object of volume point zero 4m cubed. So we know volume experiences an upthrust of. So we know you, when fully submerged in a liquid, find the density of the liquid. Okay, up rest 300 newtons, divided by 0.4m cube, divided by 0.81 km kilo per kilograms per meter square and equal to. 764.53. Okay, let's try number 14. A Boge floats on seawater. The total weight of the boat is that number of nutons. Calculate the volume of the boat that is submerged. If the density of seawater is given. So we're looking for the volume. We're looking for the volume of the boat that's submerged, the volume and volume density to the weight, the volume of the boat total weight and the density. So seven, four, zero, zero divided by zero, one, 9.81, equal to seven, seven, eight, seven, seven, nine, nine, 1.85, but that's not the answer because that is just the kilograms. And we use one, two, five divided by this number and 0.13, that is the answer. Meters cube seven. 7.8m. What's wrong? Maybe that is not correct. 1.25 times ten to let me see. Yeah, correct. So maybe Yeah trust is equal to weight, which we know. So weight is weight divided by density, divided by g will give you the volume of the boat that's submerged. A sphere with a diameter of this is fully submerged in oil with a density of 800 kilograms per meter cubed. Calculate the upthrust acting on the sphere. So we know the density, we know v we have to calculate and we know g. So what is the volume of our sphere? There zero 0.14q times pi times. Power three. 0.11m q and then times 100 times. 9.9 point 81 equal to 90.2. Newtons? 90.2 perfect good. Let's do number 17. A plastic cylinder with a MaaS of ten kilograms is partially submerged in oil with a density of 800 kilograms per meter cubed. If the cylinder's total volume is 0.025m cubed, calculate the volume of the cylinder submerged under the oil. So we're given the MaaS, not the weight. Weight equals up thrust. We know density, we know g, we know v so 800 times point zero two, five times 9.81. Hello. To 196. But if that is only 50% density, so the volume should times okay. So you get the idea. So then we have the. Stookes law equation, so basically Stokes law. Involves us using this equation. Jackson, so the force, the resistive force, if you like, depends on the viscosity eater, the radius, the terminal velocity and six pi. So viscosity. High viscosity leads to greater resistive forces, greater drag forces. So you can imagine probably salty water has a slightly greater viscosity than normal water. The velocity of the sphere influences the drag force, higher speeds, higher drag force. So if you think of a submarine traveling through seawater, the faster it goes, the greater the drag forces itexperience. And stoke law only applies if we have laminar flow. If our flow is laminar, it's not chaotic, turbulent. So Stokes law is obeyed when we have uniform lines that don't cross over, they're in layers, laminar flow. So. For this flow, we, there's the drag force, the resistive forces, there is Stokes law. And this, the velocity depends on the difference in densities. And we can use this equation to find the terminal velocity of an object. So. Looking at these two equations, if we have a small sphere with a radius of 1 cm moving through a fluid with a viscosity of 0.89Pascal seconds at a velocity of 0.2m per second, what is the drag force acting on the sphere? Viscous drag force. So six pi doesn't change eer. Is the viscosity. We know the radius, we convert to meters, we know the velocity. So we know the radius, we know the viscosity that is eta and and out of a velocity terminal velocity v are eta pi six. Okay, so the joke force. So firstly 1 cm one a noone is not de one. Times not point 89, times not 0.2, times six, times pi, 0.034 newtance and when you're given numbers to work with with two decimal places, that's how many decimal places you should give your answer to in physics generally. Take that as a rule. Yeah no one, no three. So no point zero. Yeah take take give it just three. Maybe that is two significant figures. Yes. Two significant figures. Okay, a droplus with a radius of 5 mm is moving through an oil of this viscosity. At this speed, calculate the drag force exerted on the droplet. So millimeters. So we have the radius, we have the viscosity, we have the velocity. What is the drag force? So five x minus three times velovelocity and R five milliter. Okay, 5 mm and that's okay. Chofor on okay. Five times ten to the minus, three times 1.2 times zero, one, one, five and that is answer. Nine times ten to the minus four. Did you month 0.017? 17 point five times minus three, 1.5Oh Yeah Yeah I forget two times five times, six times pi times six Yeah okay. Okay, here's question three, a sphere with this radius. We know the force this time 0.06. We know viscosity. Find the velocity of the particles. So this time, we're finding v. What will the velocity of the particle be? So zero? Point zero six divided by six times pi times point zero three. Times point eight. Despite? With drug force 0.06 divided by 0.03, equal to two, and then divided by 0.8, and as the velocity is 2.5m per second, we multiplied by six pi, or divided by six pi, or divided by six divided, divided by six pi. No point 13, no point 13Yeah don't forget those six pi. I don't like a Pino. I know it's an annoying number, but it's very meet at number when you think of it, all the ratios and relationships. Okay, let's try number five. A small sphere is moving through a fluid. With this speed, it experiences a drag force. So we know the force. If the viscosity is 1.5 pascalseconds, find the radius of the object. It is it is okay. It is. With gusity drug force zero zero five divided by 0.25, divided by 1.5, divided by pi, divided by six, and remember that good. 7.07 times ten to the -3m. Number five. 7.1 mm. Okay. So if we look at this type of processing. I'll just find enough. So with this relationship, this terminal velocity can be rearranged from these two equations. The weight is equal to the uthrust plus the viscous drag. So there's our viscous drag equation. The upthrust is the v rogh H. So rogv, that's our volume times, our density times g and that's. The volume times the density of the this is the fluid, this is the solid. So that's v times, rho times g. So that's up thrust. Is. The drag force plus the viscosity Yeah drank force. The way is trust plus up, trust plus viscous drag. Gives us weight, okay, so rearranging all that, we can solve for a terminal velocity for our object if it's falling through a viscous liquid. So two R squared g times the density of the solid minus the density of the fluid, over nine times the viscosity eater. Okay, so. Let's find. No. A metal bead with a density. So that's density of the solid and the diameter falls through a liquid of terminal velocity v the density of the liquid is that so the density of the fluid is that what is the viscosity of the liquid? So using this equation. Viscosity is eta. So that's what we're trying to find. 2G. Are you going home or are your parents here for this holiday? Yeah, my parents live here. My parents. Okay, that's nice. R squared. Over nine feet. So question ten. So that's what we're trying to find out. We'll do this one and we'll finish up for the day. So. So we know terminal velocity 0.02 from the question. It's equal to 2G two times 9.81. Times density of the solid. 78 block not. Minus density of the liquid. -11 zero zero. Times R squared a diameter so point, so times one by ten to the minus three. 1 mm. Squared. All over nine heer. So. So if we work all this out. We can find eta. Two times 9.81 times 78 zero zero ught -11 zero zero. Times one x minus three squared. Divided by. Nine times. And if I swap eta and v going not to. Point 73 zero three, so we should get an answer. Or eta is 0.7. Three. Oh, three. And the units for viscosity are Pascal seconds. Number ten. Point 73Pascal seconds, okay, so it's just substituting the correct equations. Okay, so tomorrow we will look at springs and Young modulus. Okay, books oklaw, okay, Jackson, good work today. Bye bye. We're doing it.