Physics Math Review | Geometry, Algebra, Trigonometry Problems & Solutions

Transcript

hi hello how the heck are you I'm Dr Karen Deegan and if you are watching this video I'm assuming you're here for the solutions to the practice problems I posed earlier this week this video's questions are for the physics one math review practice problems that I posted and I already know why you're here so let's get into the solution so for the first question I did I didn't give the answer to this in the actual questions when I posed it because this one was multiple choice so I asked which of the following arrows is perpendicular to the purple line and I did say that there could be two answer choices to this so there are or I guess I said there could be more than one there are two answer choices to this so the correct answers are going to be B and D so this is going to come in I think it's a little bit easier to identify which of these lines is perpendicular once you're once they're drawn already for you but something we want to keep in mind as we start getting into physics is you're going to be drawing diagrams especially ones that kind of look like this one that's on an incline and you'll have to start with say you know you have something that's on an inclined plane so maybe you have a car on a ramp and you need to identify a force that's going to be perpendicular to this and we want to keep in mind that what we're drawing that that perpendicular it's a little bit off but should be pointing in this direction whereas the way that I have it pointed in the option C is that it's pointed just straight up along like if you had a y-axis if I was positive and negative it would be pointing along the positive y but the positive y would not be perpendicular to the ramp that I have drawn here so something more so that I feel like is going to come up when you're actually drawing these pictures yourself but it is going to be important to be able to know and distinguish your perpendicular from your parallel and be able to use and like understand those descriptive words if they're just used um in problems because professors and teachers are going to assume that you know what perpendicular and parallel mean and they'll probably just go ahead and use them without giving any sort of conditions or explaining a lot of things further okay so for this one we want to isolate the variable K which is down here in the denominator so we want to rework this to get that K by itself which is going to involve some ugly math so let's get to it so the first thing I would say maybe not the easiest thing to do but it's the first thing that I see to do and that's to divide both sides by 2 pi so we'll get those two pies canceling out in front of that square root so we'll be left with capital T over 2 pi equals the square root of M over k and then I would go ahead and square both sides to get rid of that square root that's on the right and when we do this you can leave it just as that parentheses squared or you can or you can distribute it so if we distribute it we'll get t squared the 2 is going to become squared to become 4 in that denominator times pi squared and that'll be all equal to M over k and then when we get here I'm going to take the slope also I'm going to multiply both sides by K to get it out of the denominator and then yeah I should have enough room that I'll do this in two parts so we'll get K times that t squared over 4 pi squared all equals m and then to get that K by itself we have to multiply by the reciprocal of everything in those parentheses so it'll be four pi squared over t squared 4 pi squared over t squared so that basically you can see that that 4 pi squared will cancel that one this t squared will cancel that one so we'll be left with and then we can simplify k equals I'm going to put the 4 pi squared out front times m over that t squared your final answer there making sure that you're Distributing that squared to everything inside those parentheses making sure you're remembering how to even get rid of that square root with the squared in the first place those are all things that you need to be able to keep in mind to solve this this is one of the more complex equations you can have okay and then last one we need one to find the length of each unknown side of the triangle below where I gave you the hypotenuse it's a right triangle I gave you the hypotenuse and the angle so they're kind of two perspectives you can take on your right triangle trigonometry or physics you can kind you can deal with or approach it from the unit circle perspective which the unit circle perspective is always if this is your unit circle going around your angle is always going to be defined with respect to the positive x-axis so 0 is always along the positive x-axis and then your X component um call that radius r this would be your X component this would be your y component so in this case you're always going to have your X component be the radius times cosine Theta and you're going to have the Y component always be the radius times sine Theta but you'll see in physics a lot of the time the problems that you can get you'll have this angle to find not always with respect to this reference line that I'm going to highlight so in that case I tend to advocate for using sohcahtoa method so the definition that sine is equal to opposite over hypotenuse I'm summarizing here cosine is the adjacent side over hypotenuse and then and they should be absolute value bars um and then the tangent is the absolute value of opposite over adjacent oops so you have that otherwise the shorthand which I'll pretty much write on any solution that I need to use sokoto for which is a lot in physics unfortunately I would just use that abbreviation abbreviation so since we have the Ang the hypotenuse and the angle and we want the adjacent side which is X and the opposite side which is why I'm going to use cosine and sine so I'm first going to set up that sign of 41 is going to be the opposite over hypotenuse where the opposite is y and the hypotenuse is that nine so we can multiply both sides by nine and then I'll just simplify 9 times sine of 41. is going to be equal to that vertical component and then make sure your calculator is in degrees and not radians because that angle is in degrees and I'm going to round to one Sig fig and I get 5.9 and then I'll set up something similar for the X component so for the X that's the adjacent side I'm going to use cosine 41 equals adjacent which is X over the hypotenuse which is 9. oops and then all again multiply each side by 9 so we'll get 9 times cosine 41. equals X and nine times cosine 41 degrees is equal to 6.8 answers and that sums up our physics one math review for some of it and tune in next week for the physics 2 math review some of that might also be useful for physics one I'm sure some of this stuff is also useful for physics too if you need a refresher so much for tuning in and watching the solutions for this week if you're interested in working with me or getting more resources from me feel free to click the link in the bio so that you can learn more including how to work with me as your tutor and until next time happy physicsing