 five exam or 2000 I think 2006 I gave you guys the 2004 2005 exams because that year still the written sections were longer there was 60 marks worth of written and 60 marks worth of multiple choice those of you that are writing the provincial now you'll find I think it's uh 70 40 or something like that or 70 50 though they made the multiple choice longer because they don't want to pay teachers to mark exams anymore I suspect in about five years the exam well the grade 10 exams already completely multiple choice so here is the first question says a hammer hopefully you can read if not move forward a hammer slides down a roof sloped at 35 degrees reaching a speed of 4.6 meters per second before falling off how much time does it take to fall the 15 meters to the ground this is a projectile-ish question now usually Sam when we did projectiles they were launched from cliffs at upwards angles looks like this is launched from a cliff at a downwards angle okay I won't let it freak me out what we still said you want to do is you absolutely want to go components I'm just finished teaching projectiles with my physics 11s this year and so what I've taught them as soon as they deal with a projectile question they draw a big line down the middle of the page and they go horizontal and they go vertical and I've yelled at them right away they go horizontal acceleration equals vertical acceleration equals and they would fill this in and I'm going to ask you to fill this in let's jog our memory horizontally for a projectile if we ignore air resistance what's the acceleration horizontally for a projectile you remember zero vertically for a projectile if we ignore air resistance what's the acceleration vertically for projectile I got to be really fussy because now we're even vectors a negative 9.8 okay then hopefully Christina you would also say to yourself self if we're going horizontal and vertical why don't I take that velocity of 4.6 meters per second and why don't I break it up into horizontal and vertical I'll call this vx I'll call this vy initial why did I put an initial next to the vy but not next to the vx why did I put an initial here if vx is since there's no acceleration v final is v initial is v you know it's just the horizontal velocity the whole time the y1 will be changing in fact I can tell you it's going to happen the y1 is going to be negative and it's going to be getting bigger negative as it falls oh the 35 well the 35 is right here so I'm pretty sure the 35 ends up right there with our good old alternate interior z angles rule we're pressed for time so I'll take a few shortcuts I think vx is cosine let's see adjacent hypotenuse opposite vx is going to be 4.6 cosine of 35 and vy is going to be 4.6 sine of 35 let's crunch that and see what we get so I'll add that to my little list here vx equals and vy initial equals what do we get for vx I know why things are running slow this antivirus thing is trying to scan and it slows the machine down like crazy abort what's vx my calculator is not booting up it's going to take ages so someone needs to type this in and the faster you type this in the more review we get so chop chop let's participate what's vx 3.8 give me a few extra digits because I don't want to this is not my final answer I'm going to be using it 3.768 what's vy initial 2.634 okay 2.638 like that okay always carry a few extra digits if it's not your final answer don't be rounding off early um what do they want me to find in this question how much what time to find time they have to give me some type of a displacement did they give me a vertical displacement or a horizontal displacement in this question vertical so I'm going to solve the rest of this vertically now also they can give you a horizontal displacement we call that the range or sometimes they can sneak it in if you launch from the ground and end up on the ground what's your vertical displacement zero because you're changing height is zero don't let them scare you with that one they'll do that one once in a while too but I know that my vertical displacement is what is my vertical displacement we have not 15 and also does anybody see my mistake right here it's not yet no it has to be it's down I have to go sorry you're saying yeah my mistake was that it's not negative yes I thought you're saying it wasn't supposed to be negative and I want to find time do I have an equation that has those in it oh I didn't write negative 15 good gosh thank you is that what you're saying hey I was marking till 10 30 last night sorry hey what equation I think this one the common mistake kids make is they put vertical into horizontal or horizontal into vertical so keep them separate I think we're going to probably have to pull up the quadratic equation let's see I think we're also going to need more room Mr. Duke negative 15 equals negative 2.638 t minus 4.9 t squared where the negative 4.9 come from okay half a and yeah you have to use the quadratic formula oh how do I know this is a quadratic equation Megan so make it equal to zero up plus this over zero equals negative 4.9 t squared minus 2.638 t plus 15 or you could have plus both of these to the other side and had less negatives you'll get the same answer no matter what now probably many of you have built in quadratic solvers and things and my understanding is if you went straight to the if you wrote down the quadratic formula right now which we will it's on your formula sheet by the way if you haven't noticed or you can use the german pq one as long as you can as long as you can solve an equation like that I'm good I don't care how you do it okay okay so we would go like this this is the quadratic equation if you haven't seen this before where that number there is a that number there is b that number there is c and then you just plug them into there it looks like this t equals I'll use t equals instead of x negative b negative negative 2.638 is going to be positive 2.638 plus or minus the square root of negative 2.638 squared there's b squared minus 4 times a negative 4.9 times c positive 15 all over 2a which is negative 9.8 I can do 2a in my head if you can solve a quadratic using the the p and q one then I'm fine I don't care how you do it just as long as you can now I've given most of you that built in quadratic solver and you're welcome to use it now because they can't tell I'll do this by hand just in case some of you've forgotten how to use the quadratic solver or have deleted it or don't have a graphing calculator I would do the inside of the square root first bracket negative 2.638 close bracket squared minus 4 times negative 4.9 times 15 the inside of the square root is 300.959 that's probably good enough to round off so it's going to be 2.638 plus or minus the square root of 300 point what did I say 959 all over negative 9.8 and now let's actually get an answer bracket 2.638 plus the square root of 300.959 close bracket for the square root close bracket for the top because there's two terms on the top spencer so I got to put the whole thing in brackets divided by negative 9.8 one of my roots is negative 2.03 negative 2.0 of course we can't have a negative time so I'm going to ignore that one then I'll redo the same equation because it's plus or minus 1.50 and there's my final answer if they wanted to make this question a little different usually what they would do here Sam is give you an upwards velocity and then ask for the range so here they just tweak them for five marks probably a lot of kids would have missed putting a negative right there okay other questions that we did there's one where they if you look at the review the last question on their review this one shows that fairly often where they fire it into a wall you know how from the ground you know how high up it hits the wall and they want you to figure out the range or they tell you the range how far to the wall how high up does it hit the wall they have to somewhere though give you some kind of a displacement either vertical or horizontal you can use that to find time of flight and once you know time of flight then just be very very careful breaking up horizontal and vertical and solve for whatever you needed the key to remember though Savannah is horizontal ax is zero vertical a y negative 9.8 oh yeah this is we usually what they give okay something like that I think if you guys are doing okay I'm going to be bad and skip that one if you want to try it at home when I post this online you can try it yourself so they've given you 20 meters per second they gave you time in this case if they give you the time the question really falls apart because you don't need the quadratic equation anymore just break everything into horizontal and vertical again Spencer be very careful horizontal acceleration zero that means dx is just vxt not plus a half at squared because a is zero vertical acceleration negative 9.8 torques I think is what I need to review for a bunch of you so here's what it says a four meter long steel beam is supported three meters from a hinge by a cable attached as shown find the mass of the steel beam there's a fairly tough one I there was an easier one I picked this one instead what we do here well we would label our diagram with all the forces it's a free body diagram but instead of redrawing it I'll just draw it right on here what are the forces acting on this bar get the obvious ones where okay so this is four meters long I'm going to put straight down mg where this distance here is two center of mass what other forces are acting on this bar well tension I also suspect sam that there is a vertical force right here because I don't think tension is cancelling out all of gravity and I notice tension is pulling to the right a little bit is the bar moving to the right then there has to be a force to the left but because I'm using torques and I'm going to put my pivot right there how far are both of these forces from the pivot zero so what will their torque be zero I can ignore them but I did that with you just in case as a part a for one mark they said draw a free body diagram yeah uh not with one this tough if they gave you a nice horizontal bar sure and what you would do is you would find all your vertical forces on using torques and you'd say what's missing this one has to counteract that vertical force and then you would find all you would find well in this case you would find the right hand component of this guy because that's the only sideways force that's how big that is and if you added those two together tip to tail you'd get the resultant okay that's a fairly tough one yeah it showed up once in a while don't panic if it does um I don't think I'm giving you one like that on your written that it shows up on the real provincial yeah okay on your written your written I'm going to be doing and I think I've said this already one kinematics or forces those are combined as one unit so you're either going to have a winner minus loser question or a projectile kind of a question one energy or momentum or a combination of both like one of the ballistic pendulum collision thing okay and then it'll be one equilibrium it'll either be the uh traffic street sign the one that looks like this with the weight hanging or torque hopefully you picked up the hint there uh then that's question four question five will be a circular motion probably it'll be a some kind of a gravitation one how much work to get an orbit or orbital radius or something like that okay if it's orbital radius and speed fc equals fg if it's work think work equals change in potential plus change in kinetic and then read the question very very carefully are they just lifting it up is it stopped then change in kinetic is zero oh are they putting an orbit then there's also a change in kinetic off to figure out a velocity that's unit five unit six there will be one electrostatics question okay unit seven there'll be a circuit for you guys to solve unit eight there'll be a magnetic forces then there's going to be a question nine question nine is going to be interpreting a graph I'm going to give you some kind of a graph I may ask you to graph the data it'll be a little chart I may ask you to draw a line of best fit I may not what I'm going to specifically ask you to do is either do something with the slope or the area and interpret it remember to do that what you do is for slope divide the units you can usually figure out what the heck it is for area multiply the units you could usually figure out what the heck it is and then the last question will be a using principles of physics right to explain so there's I think nine or ten questions okay that's what the old provincial exams were I like those because that way you kind of knew oh I'm on question five it's circular motion the question it goes in the same order that I taught you you kind of know where you are they've shortened the written as I already said but that's okay let's continue here so we would oh you know what though here's the problem both of these are useless to me because we said that torque had to have perpendicular components of force so I'm going to go like this there's mg perpendicular there's parallel here's tension perpendicular there's parallel Christina what angle is right here what angle is right here 40 that one I can do pretty easy I think the z rule I've got an angle in that triangle this one's going to be a lot tougher way tougher let's see what angle is right here what angle is right there yeah we called that she said savannah said 65 we called that you may recall from math 11 corresponding now can you see a z as it turns out this angle here is 65 degrees in this bottom force hypotenuse opposite now I'm just going to do this just to clear up my diagram I'm going to put that 150 newtons right here because that's where it is right unfortunately it just overlaps with the way we drew it and I'm pretty sure this is going to be hypotenuse oh also opposite both of these happen to be signed just kind of nice how could they make this tougher I'll be honest this is probably because we got two different angles considered tough if they gave you a horizontal bar they'd add an extra mass on the end but usually the mass hands can hang straight down and there's no components a little nicer so we're going to say this the sum of all the torques clockwise in that direction equals the sum of all the torques counterclockwise in that direction and congratulations you just got one mark right they will give you I've yelled at you guys about never leaving a question blank even if you're able to write down what you think is the correct formula usually you get a mark for that because you've recognized what it is they value that clockwise what would cause this if that's my hinge what would cause this to spin clockwise now remember we're looking at this and this because those are my only perpendicular components which of those would cause this to spin in this direction mg perpendicular so it's going to be mg perpendicular times its distance from the pivot two and I think that's it equals tension perpendicular how far from the pivot does say three the scale looks a bit weird but times three over often the margin I would write this as a trig function let's see opposite over hypotenuse sine of 65 equals the perpendicular component over mg so Megan can you get the perpendicular component by itself what's this equal to from this so I'm going to replace this with mg sine 65 common mistake Megan for some reason kids forget to drop the two down I don't know why but often it vanishes equals and often the margin somewhere let's do the trig for this guy so we have tension per sign sine of 40 equals perpendicular over tension Megan could you rewrite this and get tension perpendicular by itself okay t sine 40 times three what are they wanting us to find in this question what is the what if they ask for the weight what are they asking me to find mg I did that to you guys I think on your unit test on this unit it doesn't show up very often but that's just like cheap marks don't throw that away read the question carefully so it looks like here the mass is going to be the tension times the sine of 40 times three divided by the sine of 65 times two tension was 150 times g thank you sine tension was 150 sine 43 divided by sine 65 times two times 9.8 don't need to do negative 9.8 we took care of them in our in our torques we took care of the negatives make sure I'm in degrees I am 150 sine 40 times three divided by this it has to be in brackets sine 65 close bracket times two times 9.8 close bracket did I miss anything 150 sine 40 by the way why don't I have to put this in brackets your calculator assumes things are on the top if you don't tell it specifically right I'm getting 16.3 kilograms jessie center of mass is weird but yeah here's another one that shows up I'm just going to get this one started and then we'll talk about this can I get it on one page the ladder question fact I think in your notes I called this the famous ladder problem they want us to find the minimum force of friction between the ladder and the floor that's required to keep the ladder from sliding okay let's list all of the forces on this ladder get the obvious ones okay we have the mass of the ladder um let's see the ladder is five meters so um the mass of the ladder would be right about here I'll call it mass of the ladder times g and then we also have the mass of the person I won't draw it quite that big mass of the person times g we would also have a normal force right there because it's touching a wall it's leaning against the wall so the wall will push back there will also be a normal force right here so you know what I'm going to do I'll call this normal force number one because it's the one from the floor that we're used to I'll call this instead of normal force I'm going to make the n into a w I'll call it force of the wall hmm this can't possibly be it force of the wall Christina what direction is this force are there any other forces to the right then this ladder would have to be accelerating ah wait a minute I think friction is that way not only that I think normal force number one cancels out the mass of the person times g and the mass of the ladder times g and the force of the wall equals friction to do this question we would go components components use torques to find this once you know that you know force of the wall and that's friction in other words even though this question is saying find the minimum force of friction you're not you're going to find the force of the wall which is friction okay problem is this question would take me about 10 minutes I haven't got that much time we did a couple like that in your homework if you look the lesson was called the famous ladder problem I think it was lesson four of the equilibrium something like that momentum wrong button a curling stone is sliding along the ice when it hits a stationary 15 kilogram bucket of sand after the collision the curling stone's velocity is three meters per second east I better draw compass and the bucket has the velocity of 22 meters per second 40 degrees south of east what direction was the curling stone moving before the collision so this is a collision collision momentum let me say this again collision momentum I'm going to say this one more time in collision momentum hopefully you remember that in a couple of days okay and here's what we said the sum of all of the initial momentum the sum of the momentum before the collision was the same as the sum of the momentum after the collision unfortunately this was not like energy where you went also kinetic energy initial plus potential initial equals kinetic energy final plus potential energy final those were scalars those you could solve but solve by doing straight math and cross multiplying and you were done these momentum is a vector we're going to carefully draw pictures after the collision what do I have moving curling stone which way so I'm going to draw that like this there's the curling stone how much momentum does it have momentum was what times what okay so you by the way you might want your formula sheet in front of you just a thought is that good learning did I really need to tell you guys that I guess I did oh I weep it's going to be 10 times three there's my mass times velocity and I have something else moving bucket what's the velocity of the bucket direction south of east okay I'm going to go like this it's 40 degrees south of east so 40 degrees south of east and it has a momentum of 15 times 2.2 and this question wants to know the direction that the curling stone was moving before the collision that's going to be this here I don't know it wait a minute wait a minute how do I add two vectors together if I draw these tip to tail I should be able to find the direction of the resultant by doing some trig okay so we're going to have this 30 plus can someone go 15 times 2.2 for me 33 I think but don't quote me on that is it almost an isosceles triangle now this angle here was 40 degrees how big is that angle there okay 140 degrees and looks like my resultant is this can I erase that so it's not quite so cluttered now this question doesn't want me to find the resultant it wants me to find the direction it wants me to find theta can you already tell me if that's which way my curling stone must have been moving initially what of what north of south south of west south of what okay this is going to be south of east now I just need to find the degrees let's see what I'm always looking for if I want to find an angle if possible it's nice if I can find use the cosine law or the sin lot find a pair sin law is when you have a pair I don't know those two I don't know those two I don't know those two that's not going to work very well I'm going to make this 140 a little smaller so that takes up a little less room 140 degrees I think what I'm going to have to do first is use the cosine law to find that and then once I know that I can go sine of this divided by x equals sine theta divided by 33 and go shift sine and find theta so the cosine law which is on your formula sheet remember the cosine law can you read it to me jesse x squared equals 30 squared plus 33 squared minus 2 times 30 times 33 cosine of 140 this doesn't find you x what does this find you so please don't forget to square root 30 squared plus 33 squared the nice thing is you can type this in in one line if you have a good calculator times or minus 2 times 30 times 33 cosine of 140 square root I get 59.21 for x and now I have a pair I know both of those so I can say sine of 140 over the side across from it equals the sine of our mystery angle over the side across from it that's the sine law also on your formula sheet by the way like how would I get the sine theta by itself here what times what divided by what do the whole thing so I'll give you in 33 give you another sign 140 divided by what do that on your calculator oh this doesn't give you theta what does this give you sine theta how would I find theta at the very very end inverse sine and I get oh I bet you the answer supposed to be I bet you the answer they started out with to make this question up was exactly 21 degrees so for momentum for collisions if they're straight and linear just make sure you let to the right be positive into the left be negative when are they straight and linear uh the ballistic pendulum where you're firing a bullet into a block that was a collision and then a change in height oh change in height energy or we did the ones where the roller coaster rolled down the hill crashed into another roller coaster and then how high would they go that was a nice linear collision as well with the roller coaster stuck together circular motion a blue ball I don't know why they told me that it was blue a ball is swung in a horizontal circle and completes a single rotation in 1.2 seconds 0.44 meter long chord makes an angle of 35 degrees with the vertical during the ball's motion what's the centripetal acceleration okay I'm thinking whenever they give me one of these ones at an angle free body diagram probably a triangle well let's see what are the forces acting on this sphere get the obvious ones gravity down what else tension what else nothing else wait a minute it wants circular acceleration remember circular acceleration circular force mike was the net force it never actually appeared on a free body diagram it's what you got when you added the forces together how will I add these forces together well how will I add vectors together draw them tip to tail now to do this sam I always draw the easiest one first what's the easiest force to draw here I would go like this I always draw the toughest force next what's the toughest force tension now I know not to go to there I know not to stop there how do I know I want to stop exactly right here what path is this tracing out a circle where must the net force be toward the has to be dead horizontal what we're saying then sam is this plus this this plus this is what gives you your circular force your centripetal force oh we need an angle looks like right here is 35 degrees right so I'm pretty sure right here is 35 degrees because I'm tight they want me to find centripetal acceleration well on my formula sheet I have this ac equals v squared over r or it equals 4 pi squared r over t squared they told me how long it takes to go around once that's the period I think I'm going to end up using this one here so let's see 4 pi squared r oh I don't know the radius oh I know the period though what's the period 1.2 squared so you know what I don't think I needed to necessarily draw this right now I think I just need to figure out the radius r now I drew this with you guys anyways because often they're asking you for a mass or attention this will be your approach next time I'll read the question a little more carefully how can I find what r is well if I know that angle opposite adjacent to a hypotenuse in this triangle here how about here what's this hypotenuse sine 35 equals the radius divided by 0.44 Lee could you get the r by itself that's what I'll put right here so actually this question wasn't as bad as I thought this was basically write down the equation and do some grade 9 trig 4 times pi squared times 0.44 sine 35 divided by 1.2 squared I get radius of 6.92 meters acceleration mr duke what did I say radius acceleration of 6.92 meters per second squared I was going this doesn't look right because if that's that long the radius can't be 6.92 that's why I was pausing there and going what I've done something silly okay so other twists on this question they might ask you to find the centripetal force or the mass because now that you know the acceleration if they told you the mass you could find the centripetal force mass times acceleration or they could ask you to find the tension because if you knew either one of those you could find the tension using trig from this diagram electrostatics wrong button again come on mr duke electrostatics alpha particles with a mass of blah blah blah and a charge of blah blah blah are fired towards each other from a great distance they're both positively charged will they attract or will they repel each other repel if they each have a speed of that much to start with what will be their minimum separation distance I guess what we're saying is this they each have kinetic energy the closer they move together the more that kinetic energy gets turned into electric potential energy they're going to slow down they're going to slow down when they come to a stop for a split second their minimum separation distance that's when all of their kinetic energy has become electric potential energy then they'll start to repel each other does that make sense all of our kinetic is going to become potential does that make sense all of our initial kinetic is going to become electric potential kinetic energy what's that half mv squared now we have two particles moving thankfully they're the same mass and the same speed so I'm just going to go like this two times a half mv squared and conveniently Christina the half's cancel take that now electrostatics this was the one that had like eight formulas that all looked identical so you might want to very carefully look at your formula sheet and ask yourself which of those is the potential energy between two point charges I'll give you a hint if it's point charges it always has a k in it are both charges the same do you mind can I save myself some time and just do this because q1 times q2 is just q2 if they're the same less typing what do you want me to find you know the mass they told you you know speed they told you you know k that what is that it's it's a constant it's on your sheet nine times 10 to the nine okay did they tell you the charge yep could you find our in fact I think this has suddenly become straight cross multiplying I think the radius is going to be k q squared all over mv squared I'll assume you can type that what are the questions will they ask with electrostatics another classic one is uh two charges finding the electric field between them or at sort of electric field at a certain location finding the voltage between them that was what I asked you on your test okay the key to remember the tricky part with electrostatics unit was making sure you were finding what they wanted you to find a number of you found the voltage when asked for electric field or found energy when asked for electric field and I know Spencer it's unfortunate because energy begins with an electric field begins with an e and they use the letter e for both of them e p for potential energy that's why I always use capital p capital e for potential energy I know but you just have to keep them straight oh remember the first row in the formula sheet those were the vectors we don't put the positives and negatives in we said for forces and fields we decide the direction like charges repel unlike charges attract hey what was the direction for electric field which way would a positive charge want to go if it could if it was sitting right there how big a positive charge such a small one that it didn't have its own electric field because otherwise that would change the question okay uh there was a part b to this question I think the part b was uh using principles of physics right to explain question I think it said something like uh if the speed of each was doubled what would happen to the separation distance using principles of physics right to explain if that was doubled you'd have a four coming out of there because it would be two squared the separation distance would be one quarter as big magnetic forces I don't have a circuit here sorry a 0.75 meter metal rod is suspended as shown a current of 13 amps then flows as indicated is the tension in the springs increased or decreased so it looks like if I'm reading this properly we are running a current this way this way this way which way is the current which way is the current traveling in this bar to the right which way is the magnetic field so point your thumbs to the right fingers out of the page which way is the force on this bar down so will the tension in those springs increase or decrease by how much does the tension change I think it's going to change by the magnetic force and I think the magnetic force is going to be did they tell me the magnetic field did they tell me the current did they tell me the length well here's the question I need to ask I guess does the current go just from here to here or do I include the whole length for me so the portion of this rod that's actually getting the current is just this stretch right here because there's no reason for electrons or positive charges the current to flow in either of those directions so point four five and the answer is uh come on point two two times point one three times point one three mr. do it times 13 times point four five one point two nine newtons another magnetic force's question so an 18 centimeter long metal rod of mass 35 grams is suspended from the ceiling with light wire a uniform magnetic field is directed vertically upward when there is a current in the rod it swings outward as shown so if it swings outward like this which direction is the force acting on this current carrying rod to the right point your fingers sorry point your palm to the right because that's which way the forces which way is the magnetic field up so can you figure out is the current going from x to y or is the current going from y to x which way is your thumb pointing right so here's taking the right hand rule in reverse they're telling you the forces that way magnetic field is up the page so my palm is pointing to the right my thumb is pointing this way I think from x to y must be the direction of the current so the direction x to y then the question is uh how big anytime they gave you something hanging at an angle do a free body diagram so here's my mass what are the forces acting on it get the obvious ones gravity tension and I guess the magnetic force is exactly to the right let's draw this as a equilibrium triangle draw the easiest one first sam gravity draw the toughest one next tension I know though that magnetic force is exactly to the right otherwise this wouldn't be hanging there so I know to stop right there and I think I can find magnetic force this is 15 degrees right here they did give me the mass 35 grams so 0.035 I know G I think I can go opposite and adjacent which trig function links magnetic force and gravity tangent in fact we would say this tangent of 15 equals opposite over adjacent the magnetic force is going to be mg times the tan of 15 ah but wait a minute they want me to find the direction of the current the magnitude of the current so I'm going to take this magnetic force which is equal to mg tan of 15 I'm going to recognize this is also b i l right get the i by itself the current is going to be mg tan 15 divided by b l did they give me the mass 35 grams so 0.035 g is 9.8 tan 15 divided by did they give me the magnetic field 0.22 did they give me the length of this 18 centimeters so 0.18 2.3 amps from x to y I said to you that on your written on Monday there's going to be some kind of a graph question so here's an example of one an experiment was performed on the surface of an asteroid a mass was dropped from various heights and the time taken to fall was recorded a says plot a straight line graph of d versus t squared they gave me d and t so I guess I'll have to square these really quickly on my calculator takes a couple of seconds 1.31 squared 1.72 thank you 1.72 oh lee what's zero squared 1.56 squared 2.43 1.77 squared 3.13 2.05 squared 4.20 2.15 squared 4.6 okay let's see if I can graph this goes through oh I need to pick a good scale let's see my distance goes from zero to 1.3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 squares 1.3 meters I bet you each square 0.1 0.2 0.4 0.6 0.8 1 1.2 1.4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 22 23 24 25 26 I don't know each square be 0.2 maybe 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 2 0.2 0.4 0.6 0.8 3 0.2 0.4 0.6 0.8 4 graph goes through zero zero point five comma point sorry point one seven two comma point five so there's one point six one point seven goes through point five about there 2.43 goes through point seven 2.43 goes through point seven about there 3.1 goes through point nine 3.1 goes through point nine about there getting roughly a straight line 4.2 goes through 1.2 4.2 goes through 1.2 about there and 4.6 goes through 1.3 4.2 4.4 4.6 1.3 about okay freehand it mr. duick use a ruler but I don't I can't put one on the screen there's my line of best fit b find the slope so slope is rise over run it if you're graphing a line Megan you have to use points from your line I would not use this point right here because it's not on my line I'll use give me a nice point how about right there that looks like the rise is 1 over the run is 3.4 units meters second squared hey what do you think the slope of this graph is if the units are meters over second squared so when it asks me to find the acceleration due to gravity I'm pretty sure that's just the slope 1 divided by 3.4 I get point 29 0.29 meters per second squared what's the acceleration due to gravity on this very very small asteroid 0.29 meters per second squared that's what I mean by giving you some kind of a graph to interpret okay and then there's going to be some kind of a using principles of physics right to explain so here's a circuitry question says when checked with a volt meter an old six volt lantern battery shows the expected reading of six volts however the battery fails to light a low resistance light bulb identify the property of the battery that must have changed as it aged and explain why this changed to the property results in the bulb no longer lighting I'm going to draw a little circuit there's the battery there's the bulb now it's saying that it's finding six volts across this battery so it's saying that before we hook it up the volt meter measures six volts but as soon as we hook it up nothing well there's one more part to this battery remember we said at the very end that we said batteries have an internal resistance normally before you hook the battery up since no current is flowing you don't lose anything through here but if this is measuring six volts from the battery and we're getting zero volts here where must I be losing all my voltage right there you know what must have changed the internal resistance I'm guessing some of the chemicals inside the battery have corroded or rusted and so now the current can't travel through the battery unimpeded anymore explain why this changed I would draw this circuit and I would say using kerkov's laws six volts from battery zero volts in bulb therefore six volts in internal resistance voltage is what times what v equals what times what v equals i little r that's what we call the internal resistance I don't think your current has changed what's happened is this has increased that's why old batteries stop working even if you never use them that's what I have for a quick review of the course I hope that helped