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(II) A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 $g$'s. Calculate the force on a 65-kg person accelerating at this rate.What distance is traveled if brought to rest at this rate from 95 km/h?
I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?
(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?
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welcome to our next unit in physics 102 and this unit will be looking at the topic of ray optics with ray optics rather than representing things with wave fronts. We will just directly draw them as vectors as in light is emitted here, and it goes in thes two directions. Uh, though generally speaking, they will go in many different directions. If it's a spherical source, it would in fact, go in every direction, and we would have to draw it something like this. Okay, eso. The advantage of Ray diagrams is that they're easy to draw when it comes to reflecting off of surfaces. We've already done that in the previous unit, and they're easy to draw when it comes to diffraction through lenses, Um, which will get to at the end of this unit because of the simplicity they are favored in terms of teaching basic optics and also where the favored model for how light worked for hundreds of years. Um, however, we know now that it's it's not the most accurate when it comes to certain situations like the interference situations we just looked at in the previous unit. However, for now we will begin to treat light as a ray. Notice that this does not mean that, for example, if there is a source of light and it shines down towards something that you will see the ray and long its entire path. Just because you can't see the ray, though, doesn't mean it's not there. If you've ever stood in a room with a direct the light source like a lamp and then had dust appear in the room, chances are you were able to see the light as it reflected off the dust as it traverse the room. And in that sense, you're kind of able to see the light ray, though generally speaking, the way you're going to detect a light ray is about where it lands, where it ends and what effect it has on the object that it's shining on or the image that it's creating. There's a lot of very specific vocabulary that goes with this unit, and so I very much encourage you to pay attention closely to the words that are used. I'll try to highlight them for you, Um, because they are. They do have precise meanings and will be associated in many cases with meaningful variables that will show up in our equations. So make sure you understand all the terms that are used to describe the distances between objects, um, how things are shaped and all the angles that will come across.
Condensed Matter Physics