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Rutgers, The State University of New Jersey
University of Winnipeg
(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?
(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?
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welcome to our third example video. Considering centripetal forces in this video, we're going to start to think about gravity as the centripetal force all by itself. So think about the earth and we have satellites orbiting all around the Earth. Thes satellites have some of them have very circular orbits at least. And we might ask ourselves, How fast are these satellites moving? Well, um, satellites are a certain distance r from the center of the earth. And knowing that then and that are centripetal force is just equal to gravity, which we'll just call em G right now, we'll find out later that little G here is not 9.8 meters per second squared, especially when you're out this far away from the Earth. It has changed, and we'll figure out how that changes later on when we talk about Newton's Law of gravity. But for now, we're just going to call this m G. So we have mass times gravity of the satellite, and that has to be equal to M v squared over our, which means theme EMS would cancel and fee would be equal to the square root of our times G by the same token, If we knew how fast the satellite was traveling, we could calculate what little G is at the point of the satellite by taking its velocity squared, divided by our, um either one works. This is then the orbital motion velocity or the orbital motion speed for a satellite circling a planet. Or even it works for a planet circling a star as long as the planet has more or less a circular motion that it follows a circular orbit which most planets have very circular orbits. Okay, so we have V equals square root of RG. We could ask some other questions here, for example, we could say, um what if we want to calculate how long it's going to take for a satellite? Thio go around the earth? We could say, Well, I know that we have circumference divided by velocity here and I know that velocity is equal to two is equal to the square root of RG. Therefore, we could write this as the period going to be equal to the square. Davar over G noticed that the units will work out because we have meters over meters per second squared. So then this would be the period that is the amount of time it takes for the satellite to go around in orbit around the earth. Um, there is a specific kind of orbit called a geo synchronous orbit, which says, I want the satellite to stay at a particular point above the Earth, which means it would need to have the same period. So we would say, Okay, well, has a period of 24 hours. That's gotta be equal to two pi screwed of our over g here. Notice then what we could do is we could find out what the R would have to be foraging grow synchronous satellite. That means that how far it is from the center of the Earth would dictate whether it has a period of 24 hours or a different period. Now, as I said, G actually does vary with are a little bit here, but we won't be able to pull that apart until we start looking at Newton's Law of Gravity
Equilibrium and Elasticity