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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 constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow.What is the skier's change in velocity?
(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 fourth example. Looking at the Doppler effect in this example Will look at what an echo works like. So imagine that you have a train that's coming into the station again. You are walking away from it, but there is a wall in front of you, and the sound from the train is bouncing off the wall and back towards you. So you are hearing the frequency of the train when it blows its whistle, it comes at you this way. But then it also comes at you from this way. So you're gonna hear two different frequencies and we'll be able to calculate what's called the beat frequency where beat frequency is one frequency minus the other. We'll talk about this more in the next unit on interference. Okay, so we'll call this frequency one in this frequency to, and we need to figure out how to solve for it. Okay, So Frequency one coming in is going to be equal to the frequency from the train. Will call that f s still so fs from the train. The train is coming in with some speed V train and you're walking this way with some speed we'll just call it the not to be consistent. So we have speed of sound, plus or minus your speed, divided by speed of sound, plus or minus speed of the train. Okay, well, we see here then that you are walking away from it. It is moving towards you. So we're gonna end up with the same signs as we used in the last video. So F one equals F s times V s, minus V not over V s minus B train. Okay. And then we also have to think about frequency to So where's frequency to coming from? Well, frequency to comes from the wall, which is stationary, and that frequency also comes from the train. So we're going to say that at the wallet experiences a frequency f not. Okay. So f not at the wall is going to be equal to frequency of the source times V plus or minus veena veena zero, because it's the wall divided by the plus or minus be trained. Well, in this particular case, we see that velocity of the train is towards this, which means is trying to increase the frequency, so it wants a small denominator. We have f not equals F s times B s over V s minus V train. Okay, And then we consider F two here, then will be equal to this frequency, which is now the source frequency for F two multiplied by the plus or minus. The not divided by we have the speed of sound plus or minus the speed of the wall. Well, the walls still not moving. So it's there. You are moving towards it. So that has the effect of increasing the frequency as it reaches you because it's in the denominator. Then you'll say F two is equal to f, not times V s plus V nut over v s. Okay, notice here that because of the algebra when I plug in F not well f S V s is will cancel and I'll be left with Frequency two is equal to F s multiplied by V s plus B not divided by PS minus B train How? Okay, so these are the two frequencies. And as I said before, you can actually calculate the difference between these two frequencies and you'll hear a little bit of a flutter at that beat frequency that affects your hearing
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The First Law of Thermodynamics
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The Second Law of Thermodynamics