Intro, Units & Conversions

Concept: Intro to Physics

11m
Video Transcript

Hey guys, let's get started. So, in this very first video I wanted to a brief overview of some proof physics topics you're going to need to know so let's jump right into it. So, physics is the study of natural phenomena and I'm not going to bore you with a long definition it's basically lots of measurements and lots of equations. I like to think of physics as math with rules. it's basically, a math class with a story line. So, for example if you drop something it falls, that's a rule, that's a physical rule, thatÕs physical phenomena and there's an equation that describes that. So, that's why I think of it as math with rules. In physics and in any other science, we measure physical quantities. PhysicalÉ quantities and measurements must have units for them to make sense, ok? Now for equations to work, these units must be consistent with one another. And I like to think of it as they have to speak the same language, so that they can communicate, right? So, for example here is an equation F = ma. This is probably the most important equation in all of physics 1 and it means that force equals mass times acceleration. WeÕll see it later, but for now IÕll tell that force is measured in newtons, mass is measured in kilograms and acceleration is measured in meters per second squared. But this equation only works if you have these three measurements, these three quantities in those units, right? If you plug in a mass in grams and acceleration meters per second square, you don't get a Newton, you get something else, this equation breaks, right?

So, units have to work together, they have to be consistent; and a group of units that work together form what's called a system of units, right? So, there's lots of systems of units; and in physics we use the international system, now it's abbreviated SI instead of IS (International System). ItÕs kind of backwards because it's in French, right? And I'm not going to try to pronounce that, you can try it. So, many other conventions and standards are using physics, we'll talk about those as we go, let's look, let's look into the SI unit for now. There's three basic quantities you want to measure in physics and they are length, mass and time. The unit we use for length is meter, which is abbreviated as an (m), for mass it's kilogram (kg) and for time is second (s). Notice that mass is kilograms like grams if you know gram is the base unit in chemistry you'd use grams but in physics we use kilograms. So, we want our units to be in the international system SI units. We want to use these units, not other units but if you are given non-SI measurements then you have to convert them into SI measurements before you can use them in equations, right? So, I have an entire video just to talk about that and you should watch that after this one. So, equations must also be dimensionally consistent. kind of mentioned this briefly. It means that the units on both sides of the equation have to be the same. most professors will worry about this, they won't do problems with this, but if yours does IÕll post a video in that as well so there's sort of an optional video on unit analysis.

But a quick example velocity is displacement over time, velocity is measured in meters per second and displacement is measured in meters and time is measured in seconds. So, on the left I have meters per second, on the right I have meters per second, so this equation is dimensionally consistent; it has unit consistency, the units on both sides are the same. OkayÉ And then the other thing we're going to look into is precision sig figs and scientific notation. If you've taken chemistry which most of you have by now, you've been lots of this and physics were much less picky about that stuff but you still need to know some level of it. Measurements must be precise, obviously, umÉ precision has to do with how many significant figures. I'm just going to abbreviate significant figures; and this is also sometimes called significant digits that a measurement has so the more precise the more significant digits. When you're adding, multiplying, dividing, subtracting measurements, you must use significant figure rules; now, again some professors are going to be more picky that others, most typically don't care about this stuff in physics. If yours does I'm going to have a video, an optional extra video on significant figures and their rules so we can get a little bit more specific on that, for now we're just doing a brief overview. UmÉ what else? If we're going to, those two ways we can order several ways we can compress numbers, two important ones are to use the power powers of ten or to round them.

In physics were usually going to round things to two decimal places. That's just another convention in physics, right? And the reason to compress numbers, you don't want to write a long ass number, right? So, for example if I have something like this, that gets really annoying to write, so I can write it as 10 times 10 to the power of 9, because there are nine zeros here, okay? And we can also use what's called scientific notation to represent numbers and it's going to take the following form: a dot B C times 10 to the D that's the format for scientific notation. So, the idea is that before the dot you want only one number, one digit. After that, you want as many as two digits, two decimals and then this is yours, D over here is your exponent. Okay? If you have a number that has, so notice how there's three digits here total, if you have a number that has four or five, six digits then you're going to have to trunk it down by compressing by rounding, okay? And remember that when you round if the digit is 5 or greater you round up, if it's four or less you round down. Obviously. So, IÕm going to show you quick some examples, you should have done this, you should have known this, from before; but just to kind of refresh your memory; um... So, I'm just going to kind of quickly go through this. so, for each one of these how many sig figs do it have? And we're going to rewrite them in scientific, scientific notation with two decimals. So, let me do that really quick. This one, you might remember trailing zeros don't count for

significant figures, so this number only has one significant figure and if I want to write this in scientific notation I want it to look like that, but I only have one significant figures so I can only have one decimal; so, it's going to be 1 times 10 to some exponent and in this case it is, if I have an imaginary dot there, which I can't add, because that gives it more scientific notation that's why I said imaginary, I would move one, two, three, four, so one times ten to the fourth, and this answer has also one significant figure. If you have, if you start with one you can't have more than one otherwise it would gain precision in the process of rewriting and you can't do that. Here, because I have a, because I have a zero in the decimal this counts as sig figs so all of these numbers count, except a leading 0, right? The leading zeros never count. So, I have one, two, three, four, five, six, so six significant figures, but when we write your

scientific notation you only want three digits; so, I'm going to do one point zero, zero, and the rest gets rounded, which is really easy because it's just a bunch of zeros, times 10 to the, let's see, one, two, three, four, times ten to the fourth. This has three significant figures. Notice how now this number is less precise, there's less significant figures because we're rounding to conform to scientific notation. So, these examples are kind of annoying because it's kind of the same number over and over, but I want to show you the slight differences in these. So, here are trailing zeros and leading zeros don't count in the integers, so I have 1, 2, 3, 4 significant figures and if I want to rewrite this, I only want to use three so here are the three that I'm going to use, I look at this fourth number, the number that kind of gets left out, the first number that gets left out, and if it's bigger than 5, 5 or greater rather, then it would round. In this case it doesn't cause rounding so I just have one, point zero, zero, times 10 to the fourth, this ended up looking just like the number before. Let's do three more: here I have one, two, three significant figures, because these numbers are decimal, decimals, so, theÉ if I were to write this, I would move the period over here and I only have two numbers, two decimals so it would be 1.0 times 10 to the, 1, 2.

To the negative 2, because I'm going towards the decimals. So, this answer has two significant figures. Notice how in the process of doing this it lost one significant figure. and those are sort of, those are kind of annoying but these are sort of more realistic numbers you would see in physics without a bunch of zeros to make it tricky, so this number has one, two, three, four significant figures. These two don't count, so four significant figures; but if you want to rewrite them we only, if you want to write this number we only care about the three decimals. So, three-point 82, notice that the fourth one, the guy that gets left out, does not cause a rounding, it stays as it is, right? Times 10 to the, get the imaginary number move it over 1, 2 ,3, 4, 5, 10 to the 5th, the answer has three significant figures. I want you to try doing F so you could either pause the video or see if you can do it really quick, and I'm going to just keep going here this number has one, two, three, four, five, six significant figures; again, zeros after decimal do count as far as significant figures are concerned and, but when I write in scientific notation, I only want the first three numbers. So, I look at the first guy that gets excluded and I say: does this guy cause 5 to rounds up? And yes it does; so, it's actually two point three, six, times ten to the, one, two, three, four, times ten to the fourth, and this number has three significant figures, notice how it also lost precision. So, that should be enough for most people but again, if your professor is more picky about this stuff, I'll post a video on that.

So, that's it for now.

Concept: Converting Units

13m
Video Transcript

Hey, guys. Now, we're going to talk about converting units. Remember that units have to be in the same system so they can talk to each other. So, if you have a bunch of units and some of them are in a different system, then you're going to have to convert those, so that they're all in the same system. Let's get started.

This usually is going to mean converting units into SI units. Remember SI units are meters, kilograms and seconds. So, I'm going to do three examples and then I want you guys to have an opportunity to try three practice problems as well. So, let's jump into it. So, first question says convert 44 pounds into kilograms; and I've built here a little bit of a pattern to help you out with this. So, this first part here is what I call the starting point or your given and IÕm going to put 34 pounds and I want to go from pounds into kilograms and this says at the end here, this is our target and my unit is going to be in kilograms, because IÕm going from pounds to kilograms, and what I want to know is what number goes in front of that, and to do this we're going to use conversion ratios. However many we need to get this done, all right. So, I'm going from pounds to kilograms so I have to use this conversion here. Now the idea here is that these numbers are the same. So, if I were to write them like this: one kilogram divided by 2.2 pounds, even though the numbers are different, one and 2.2, the quantity is actually the same. So, these two equals to just one. So, IÕm going to write them as kilograms/pounds or pounds over kilogram. And the decision of what goes on top or what goes in the bottom has to do with how you want to cancel things. So, I want to get rid of pounds, so IÕm going to put pounds in the bottom, therefor kilograms goes on top. And one kilogram equals 2.2 pounds, notice how pounds cancel, I have kilograms left, that is what I want, I want kilograms left, so I have nothing else to do here other than just multiply. I have this here, is sort of an extra One, we're not going to need, and now I can just multiply these numbers. I get 44 x 1 you don't have to multiply by one but IÕll just put it there divided by 2.2, and if you plug this into your calculator you're going to get a 20. So, the answer is 20 kilograms. So, that's how we're going to do that, you just follow the pattern. now in example 2, I have 100 meters squared into feet squared so the novelty here, the new thing is, that I have a square. You have to be careful how to do those as well. So, starting units 100 meters and I'm going into squared feet and what I want to know is what number ends up here. And IÕm going to use as many conversion ratios as needed. So, one feet equals 0.305 meters, so meters is on top here, so I want meters to go in the bottom there for feet is on top. Now this is actually meters squared and if you look at this, I'm only canceling 1m from the bottom with 1m from the top, so I actually have to do this twice, so that the second M cancels with the second m here. So, that's what's different about converting squared units. So, one feet equals 0.305 meters. You have to do this basically twice. On your calculator, you could do 100, if you want to set this up as a fraction would be like this, and what I recommend you do is always do this stuff very slowly is multiply the bottom first and then do division. I have this number already worked out here and this is 10.75 comes out to 10.75. So, the answer is 10.75 square feet. All right, let's do one more and now we're going to convert velocity, which you're going to do a lot of this in physics. So, we're going to convert 30 meters per second into miles per hour. So, the first thing here is I'm going to write instead of m/s like this, IÕm going to write m/s like this. It makes a little bit easier. So, what I want you to realize is that going from m/s to miles per hour is basically, two things, two conversions at the same time. I'm going from meters to miles and seconds to hours. So, we'll just do one at a time. Let's start with meters, it doesn't matter, meters are on top so I want meters to be in the bottom. And I'm going from meters to miles, so I have miles up here, and I do have a conversion for these two: one mile is 1600 meters. So, that cancels meters, and notice that IÕm left with miles, which is what, I want miles on top and hours at the bottom. So, this part I got it out of the way. This part is already done. And then the next thing I'm going to do is convert the bottom. So, seconds in two hours, seconds is in the bottom, so I'm going to put it on top, same process, and I want hours in the bottom, and one hour equals 3600 seconds, notice that the unit's check out, I have hours in the bottom, hours in the bottom, so, miles per hour, we're good to go, now I just have to multiply all this. So, in the calculator you would do 13 times 3,600 / 1600 now I have this number here, it is 67.5. 67.5 miles per hour. ThatÕs it. Now, I would like for you guys to try these practice problems. the way this is going to work is you can pause the video, give it a shot and then see if you got it right. I'm just going to keep going and hopefully you try them, hopefully you got the right answer. So, 10 feet into kilograms, I am going to write 10 feet and IÕm going to write a conversion ratio here, I have feet up top so IÕm going to put feet in the bottom and I want to go into kilograms. I want to go from feet into kilograms, but notice that the conventional given is from feet to meters, so IÕm going to have to go feet to meters then to kilograms. ThatÕs a two-step process. so, one feet is 0.305 meters, notice that now I have meters left, but I don't want meters, I want kilograms. So, I have to get rid of meters convert into kilograms, so meters would be in the bottom and, not kilograms kilometers, sorry, kilometers go on top. So, meters cancel and I have one kilometer is a thousand meters, and that's it. I have kilometers left over and all I have to do is multiply all this, I have 10 times 0.305 divided by 1000. And the answer here is option A 0.00305 kilometers. 10 feet is that many kilometers. hopefully got that right and if not, then hopefully at least you identified where your mistake was. So, again, you should pause now and try to do this problem as well and I'm going to jump right into it. So, we're going from 2 milliliters, 2000 milliliters into cubic meters. Again, I want to kind of do a little bit of planning here, I'm going from milliliters to cubic meters but the conversion that I'm given is that one liter, is from meters to centimeters cubed; so, these two guys are related, so I'm going to have to go from milliliters to liters, and these two guys are related as well, so IÕm going to have to go from this to this. So, it's sort of a three-step process. So, let's start at 2,000 milliliters, I'm going to get rid of milliliter so it goes on the bottom, now when I go following the pattern here, I want to go into liter. And one liter is a thousand milliliters, so, these cancels. I'm now right here: liters. And I want to go into centimeters cubed. So, one liter is 1000 cubic centimeters, according to the conversion that was given to us, and now liters cancel and IÕm left with cubic centimeters. So, now IÕm over here. So, I just have one more step to go, and IÕm going to say that IÕm going from centimeter to meter. The problem is IÕm actually not going from, I have centimeter cubed here to get rid of and this is just centimeter. So, the way to do this is to do with three times. Now I could do this, and kind of put the numbers over and over, or you can just cube the whole thing. This makes this a cubic centimeter, which will get cancelled, and then what you have left over is cubic meters, which is what we want it. Now one meter is 100 centimeters. So, if you multiply all of this, it would be on the calculator, we can kind of simplify some stuff, this 1000 cancels with this 1000, so, I have 2000 divided by 100 cubed and I have this here, the answer 0.002 cubic meters. So, it's A again. So that is your answer. This question is certainly a little longer.

Again, you should pause the video, give this one a shot, this one's very important for physics; specially for early chapters, you're going to have to know how to convert. So, go ahead and pause it, hopefully you did it, I'm going to keep going. I'm going from km/h to meters per second. Remember what this means, this means you are going to first convert kilometers to meters and the hours to seconds or the other way around, it doesnÕt matter. IÕm going to write 80 km/h, remember to sort of stack the up like this, makes it easier to see, and lets just pick any, it doesnÕt matter. I am going to go with km first, km is on top, I want km to be on the bottom so they cancel. 1 km is 1000 meters. Now I have meters which is what I want in the end. Now letÕs get rid of hours. Hours are in the bottom, so itÕs going to go in the top over here, and in the bottom I want seconds. So, IÕm going to put seconds over here. So, I got my units, I got m/s As the leftovers there, and 1 hour is 3600 seconds. I can multiply everything. In the calculator, it will look like this: 80 x 1000/3600. I have this number here is: 22.22. so, its 22.22 or, most of the time, you will see 22.2 which are 3 significant figures and this is the correct answer.

Concept: Dimensional Analysis

10m
Video Transcript

Hey, guys. WeÕre now going to talk about unit analysis, which is the idea that equations must have unit consistency. So, unit analysis which is also known as dimension analysis. Now, dimension analysis in chemistry means an entirely different thing. It's the way by which you convert units, but in physics, dimension analysis often refers to unit analysis which again is the idea that the equations must be dimensionally consistent. The units on both sides of the equation have to be the same, they have to check out. So, here's a quick example.

Here's an equation: speed is distance over time. so, if you move things around, if I move this time to the left, I get that distance is speed over time. So, because distance is measured in meters and time is measured in seconds, I have that speed, which is distance over time, has to be measured in distances, in meters, and time is in seconds; so, speed is measured in meters per second. That equation is dimensionally consistent and so most this variation here. Distance equals speed over time. Let's see how this is dimensionally consistent. Speed is meters per second and time is measured in seconds. What happens here is that you're multiplying them so the seconds cancel and I'm left with meters. Distance is speed over time, distance is measured in meters, so both the left and the right side of the equation have the same units, so this equation is dimensionally consistent. That's the idea, it checks out. So, let's do an example. It says here suppose Wikipedia says that the distance Y measured in meters that an object free falls from rest in T seconds is given by this equation. This is actually correct. That's not just Wikipedia making it up. Where g is the acceleration due to gravity, we'll talk about that later, but for now, it suffices to know that the unit is meters per second squared. And the question is: is this equation dimensionally consistent? Let me just show how this works. You're going to write the equation and you're going to replace all the variables with their units so it says here that Y is measured in meters, I can just cancel out half, because half is not a measurement it's just a multiplier, a factor. G says here is measured in meters per second squared, so I'm going to do this: meters over seconds squared; and time is measured in seconds. And then I have to see if this stuff checks out. So, second squared here multiply by second squared here cancels and I have meters equals meters. So, this equation is dimensionally consistent. That's all it is. So, let me do another example that's similar to this. Another kind of question would ask you not if it's dimensionally consistent but it'll give you the quotient and say: what are the units for this one particular variable? So, for example, HookÕs law. Which we'll see later, says that the restoring force F and a spring is related to the displacement from equilibrium X, blah, blah, blah. Basically, this is the equation. F equals negative KX it says if F, is measured in Newtons and X is measured in meters, what units must the force constant K have? What you do here is solve for K and by the way, just like how I ignore the half here, it's just a multiplier, I can ignore the negative sign as well. So, if I solve for K, move them out of the way, it looks like this. And that must be, for this equation to be consistent, that it has to be the unit for K. that is the correct answer. So, k is measured in Newton's per meter; and I got that by moving things around in that equation.

Now, I would like you guys to try practice 3 and practice 4, they're very similar to examples one and two, so what you would do is pause the video, give it a shot and then come back, and hopefully you got the right answer. So, I'm going to just keep going. Practice three says distance X is measured in meters, time is measured in seconds, velocity is measured in meters per second and it's just going to tell you the units for all of these different things, so you can plug them in and check if they're dimensionally consistent. Let's jump right into it. velocity is measured in meters per second but that velocity is squared so it's meters per second, the whole thing is squared, equals to, remember that 2 I can just drop, you only look at the units; acceleration is meters per seconds squared; and x is distance which is meters. I have to see how these two are the same. well on the Left I have meter squared, second squared, and on the right, I have, there are two meters here, so I have meter squared over second squared. they are the same, so yes, these are dimensionally consistent. If you haven't tried B, you should give it a shot now and try to make sure that you're getting this right. Let's do it. Meters per second squared, I'm going to go ahead and put a square already here and here, because I know that they're both getting squared, the 2 goes away; acceleration is meters per second squared; and then time is seconds, so one of these two seconds down here cancels with this, so I have meter squared second squared equals meters, only one meter, and only one second. This is not the same, so this is not dimensionally consistent, because you have different units on the two sides. Even though they're the same, meter/second, meter/ second; ours are squared and these arenÕt. So, that's different enough. And let's look at this other one here. f, it says your f is measured in Newtons but a Newton is a kilogram times meters per second squared, that's all that's on the left, and on the right I have m, m is mass which is in kilograms, then I have velocity 2 minus velocity 1. Let me show you how to handle that. IÕll show you by doing the example. 3 meters per second minus 2 meters per second is 1 m/s. 3 apples minus 2 apples is one apple. The units for this whole thing is just a unit for velocity, which is meters per second. And then I have, this whole thing has been divided by T, so this whole thing is being divided by seconds. Another way that you can do this that might look a little bit easier is, whenever you are dividing by something; A divided by B it's the same thing as A times 1 over B. Notice how the BÕs are just on the bottom. So, if I want, I can kind of do this as: times 1 over s. The reason I'm doing this is to make it easy to see that these two S's actually go together. So, this is kilograms meters seconds squared and this is kilograms meters and also seconds squared. So, they're actually the same and this is dimensionally consistent. Hopefully got that right. You should try practice for if you haven't yet, I'm going to jump right into it and it's very similar to number 2. so, let's go. NewtonÕs law of gravitation states that the attraction between two objects is given by this equation and it tells you what the units for all these different things are and it's asking: what is the unit that G must have? This big G over here. Again, I write the equation and I replace everything with their units. F is kilograms meters per second squared. G is what we're looking; for this is kilograms times kilograms divided by, r is, it says r is in meters, so, this is meters squared. Now, what I have to do is solve for G move all units around so let's do that really quick. Notice that one kilogram cancels here, this mass has to go up, so I have mass times mass squared divided by second squared; and this kilogram up here has to go to the other side. it's at top so it's going to go to the bottom. Then I have this. I just have to clean this up a little bit and I get cubic meters/ kilograms seconds squared. Obviously, the order of these two guys in the bottom doesn't really matter, so cubic meters kilogram second square its option B. So, again hopefully you got that, and that's it for now.

Intro, Units & Conversions Additional Practice Problems

For the curve in the following figure, find x as a function of t.

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If 2 cos2 θ - 4 cos θ + 2sin2 θ = 0, find θ in degress (0 < θ < 90°).

 

 

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If 2y = 10 +15x is the equation of the straight line. Find the intercept on y-axis.

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Find the numebr of radians in 120° 

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If f(x) = x2 and g(x) = sin (x), what is g(f(x))? 

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x = -3y - 1 and y = 3y - 9 are equations of straight lines. Find the coordinates of the point where these lines intersect.

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In the figures angle, ABO and ODC are right angels. What is the relation between angle OAB and OCD?

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The lengths of the sides of a triangle are 6, 8 and 10 meters. What is the lenght of the longest side of a similar triangle whose shortest side is 15 meters?

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The magnitude of the gravitational force between two planets is given by F = G(m 1m2 / r2), where m1 and m2 are the masses of the two planets, r is the distance between them, and the force F has units of a mass times an acceleration. What are the SI units of G?

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Suppose the volume V of some object happens to depend on time t according to the equation V(t) = At3 + B/t2, where A and B are some constants. Let L and T denote dimensions of length and time, respectively. What is the dimension of the constant A?

1. L/T

2. L2/T

3. L3 • T3

4. L/T3

5. L3/T3

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When an object falls through air, there is a drag force (with dimension M· L/T 2) that depends on the product of the surface area of the object and the square of its velocity; i.e., Fair = C Av2, where C is a constant. What is the dimension for constant C?

1. [C] = M/T

2. [C] = M/L3

3. [C] = M/L2

4. [C] = T•L/M

5. [C] = T2•L2/M

6. [C] = T•L2/M

7. [C] = T2•L/M

8. [C] = T/M

9. [C] = M/T•L2

10. [C] = M/T2•L2

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Albert defines his own unit of length, the Albert, to be the distance Albert can throw a small rock. One Albert is 52 meters. If an object is travelling at 4.16 Alberts per hour, what is its speed in meter per second? How many square Alberts is one acre? (1 acre = 43,560 ft2 = 4050 m2)

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237 m (meter) is the same as

A) 237000 km

B) 2370 km

C) 23.7 km

D) 237 km

E) 0.237 km

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There are 1609 meters in one mile and 3600 seconds in one hour. If you drive your car at 75 miles per hour, what is your speed in kilometers per hour?

A) 75 km/h

B) 34 km/h

C) 1684 km/h

D) 121 km/h

E) 120675 km/h

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If you are driving at 50 mph, what is your speed in m/s?

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