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Simon Fraser University
(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?
(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?
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welcome to our first example video on drag forces Here we will consider terminal velocity in a situation where we use the drag force que times v two model. What's going on? So this would be a situation where we drop a rock into a lake and it falls very, very slowly. Okay, so again, when we draw the free body diagram of the Rock and the lake, we have M g down and we'll say we have f drag up. Now again, we're still missing a force here because there would be a buoyant force on the Rock. But that's a topic we'll get to later, so I'm not going to include it here. But we will later, and that will give us a better model for what's going on. So, uh, given this problem, we have F drag minus M G is equal to negative mass times acceleration. If we say it's reached a terminal point of A's equal to 0 m per second squared, then we find F drag is equal to M G, or that V is equal to M G over K. All right, and then all we would need to know is what's going on with this rock. We know that G is 9.8. Mass would probably be on the song on the order of something like, um, we'll say 0.2 g and then K, as we saw before, has to have units of kilograms per second. I am. And you can go on, look up for a sphere what the value K would be, um, so that that's something you can look up online or in your book. If your book uses this model, then it would provide it to you. Uh, the other option is we could say, given some terminal velocity, what's K? So we could say K is equal to M G over V. And we know that V is equal to some particular speed in meters per second. Um, which again, if you check that, you'll come up with units of kilograms per second, which is exactly what we would hope for eso. That's kind of the utility of this very simple model. When we're looking just at terminal velocity
Equilibrium and Elasticity