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01:24

Keshav S.

(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 $g$ of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?

0:00

Aditya P.

(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

04:39

Muhammed S.

00:48

Averell H.

(I) What force is needed to accelerate a sled (mass = 55 kg) at 1.4 m/s$^2$ on horizontal frictionless ice?

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welcome to our first section and ray optics in this section. We're going to look at reflection and refraction of rays off of interfaces. So in this case will be considering an interface for example, that might be and one and then two. But for now, we're just going to think about the light that is reflected. First of all, given AnAnd Gle fada serve I, which is the angle of incidence and an angle theta are, which is the angle of reflection notice here that we're measuring these angles from the normal to the surface, not from the surface itself. But given these two angles, the law of reflection states that data I is equal to theta are. So we know that the angle of incidence will be equal to the angle of reflection. This is probably something that you've observed yourself before. If not, you can go kick a ball at a wall. As long as it doesn't have too much spin on it, then the incident angle will match the reflected angle of that ball. Um, incidentally, if we were thinking about this in the three dimensional mode, then we would have to say that the to raise are in the same plane as the normal to the surface. Um, here it's fairly obvious because we're in a two dimensional plane. It's the only option. But if we were in three dimensions, it could vary. There is. Well, it just means that data I the normal and theater are will all be in the same plane. Um, looking at this in, ah, larger picture here and starting to talk about some of these very important words vocabulary and variables that we need to look at here. Um, we can draw this picture on the right where we have an object at Position P, which is a distance s from a reflecting surface. It comes in at data is reflected at data are to some detector up here. Now the detector appear. What it sees is it's going to see as if the light has traveled in a straight line all the way back to hear a position p prime. So this is the position of the image we call it. The distance from the object to the mirror is called s. And the distance from the image to the mirror is s prime for a plainer mirror. that is to say, a mirror that is two dimensional. It's not curved it all we will find that s is equal to s prime. It's worth noting here that some books will have a s equals negative s prime because they have a different convention for the signs in this. Um, I'm going to go with this particular convention here. As long as you're consistent with one convention, everything will work out fine. So what we're saying, then, is that the object distance is equal to the image distance. It looks to you as if this object P is a distance s prime, which is equal to s inside the mirror. You go stare in a mirror for a while and look at different objects that are being reflected to convince yourself of this. Uh, now a couple of of vocabulary words we need to mention here. Uh, there are two types of images that we can observe in optics. There is a virtual image and a really image. A virtual image means that lights does not pass through it, so light isn't actually passing through the image, whereas a riel image light will pass through it now, in the examples for this section all we're going to see our virtual images. You can see here that light doesn't ever actually pass through p prime. Therefore, it is a virtual image. Um, but we will run into some examples in later sections off really images moving on, then to the idea of refraction. So why reflection is fairly easy. That data incident equals state reflected refraction is the idea that some of this light doesn't get reflected. It gets transmitted into the new material. Now if the new material has a different index of refraction again, if you don't remember what index of refraction is, you can see some previous videos. We talked about it in one. We mentioned electromagnetism. And also it's mentioned back in physics 101 But the basic definition of index of refraction is this. That is the ratio of the speed of light to the speed that light travels through the material. So if we have an index of a fraction like this, where N two is greater than and one, we see that the fate a two is actually going to be smaller than they do one. On the other hand, if anyone is greater than end to. We see that data to ends up being greater than data one. So it's either bending towards the normal or away from the normal, depending on how n one and N two compared to each other. Now this relationship is something known as Snell's law, and it's been confirmed many, many times throughout history. Though Snell is the name assigned to it, and this is how it works. We have n one times the sign of theta one. That's data incident here is equal to end to time. Sign of data to where theta two is the refracted angle, Um, a little more complicated than the law of reflection. But this refraction law here is still not too challenging to deal with. There is one interesting thing that comes about because of snails law and some idea known as total internal reflection. Notice here that if they to to were to equal 90 degrees, we'll sign of 90 degrees. That's one which means that we would have n one time science data. One is equal to end two times. One. If we were to calculate the angle at which that occurs, they the one is equal to the inverse sign of end to over and one. We call this angle the critical angle. When light is incident at an angle equal to the critical angle, it means that instead of refracting into the material, all of that light instead never actually enters material and to. Instead, all light gets reflected back into N one. It is impossible that that angle for the light to enter into material, and two, it seems like an odd bit here. But it is definitely a real effect and is used to great effect in things like fiber optics, where, if you have the light, come in at the correct angle or greater than the correct angle. So you have stayed a critical here, or if you went farther in that direction. Then instead of refracting into the rubber sheeting around the glass wire, it is reflected back into the fiber optic cable. If you're unfamiliar with fiber optics, there again some great videos out there that you can go on watch about how they work. But the principal concept of them is that you want total internal reflection, so you have no loss of light as light traverse is down the cable now. Lastly, we're going to take a look at how images work with refraction. It's a little hard to see this, so we have to exaggerate it. You can see I've tried to draw here with an image of a fish underwater, and you're above it trying to see where it is. If you've ever looked at an object underwater and then tried to grab it, you probably had some difficulty. The reason is that because when you have a refracted object here in the water, the light rays get bent. As it comes out into air, you can see they get bent outwards. We've kind of exaggerated it here so we can see all the details. And what ends up happening is we see an image here at P prime instead of where the object is that actually here a piece. So it be like seeing instead of seeing the fish down here, we'd see a fish, right? Here's That's fish prime. Um, and this is because of the bending of light waves as they come out. Now again, it's hard to see if you pull back too far. But if we zoom in, we can kind of make sense of what's going on again. We have an object in an object. Distance s an image and an image distance as prime. And then we have our angle of incidents, Theda one and then are refracted. Angles faded, too. So looking at all of this and drawing Cem pulling some triangles out, we can see that because both the refracted raise and the actual raised from the object are passing through the same point. We can draw these two triangles that have the same height along the Y axis that we're calling H. So since H is equal to H Prima's, I've drawn it here. That means s times, Tangent of data One is equal to s prime times tangent of data to well, for this convention that we're using right now because s prime is on the same side as s we're going to require that s prime is less than zero. Thus, we're actually going to write that s prime is equal to a negative s times tangent of data, one divided by tangent of data to now. Because we have blown it up so much if we were to consider, raise that are closer to this access here. This is called the optical axis. If we were to consider raised closer to the axis into the point where they're almost Perak sealed, that is they are parallel with the axis. Then that means that tangent If data one and data to well, if they do one and data to will be extremely small, so they will actually approach. Sign if data one and sign if data to which we can rewrite using Snell's law giving us a new expression for the image distance. In terms of the object distance that is as prime, the image distance is equal to negative s the object distance multiplied by end two over N one. We're not gonna worry too much here about this convention that when s prime is on the same side as s, we need s prime to the to be negative, but we will start to consider it more and more as we talk about lenses in future videos.

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