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04:27

Kai C.

(I) A 110-kg tackler moving at 2.5 ms meets head-on (and holds on to) an 82-kg halfback moving at 5.0 m/s. What will be their mutual speed immediately after the collision?

03:04

(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?

03:38

Keshav S.

(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?

02:55

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welcome to the next section in our unit on alternating current circuits. In this section, we're going to discuss how power is delivered to the A C circuits. Uh, well, we're going to start off by thinking of power in the same terms that we thought about it before, as the current in the circuit multiplied by the applied E M F. The only thing is, if we think about on RLC circuit with an alternating time e m f, then we are going to have different action than what we had before thinking about the capacitor and the induct. Er, though their power is going to look something like this, where it's going to oscillate back and forth because they're constantly storing energy and their electric and magnetic fields and then releasing it back into the circuit. And so the average of the power delivered and received from these two circuit elements is going to be zero. So we don't really have to worry about the power delivered to them, because on average it's going to be zero. Meanwhile, for the resist er, it's going to look more like this where it's oscillating around some constant value, and that constant value will be the average amount of power that is delivered to the resist er So that's what we're after here In our discussion about power now, thinking about the resistor itself, we know that PR is equal to the current across it multiplied by the voltage across it or the other two forms that we found before. Well, if remember, the function for current as a function of time and that resistance is constant, we can plug it in and we find the equation I squared are times cosine squared of omega t multiplied by our If we use a trick in a metric identity to expand out cosine squared of omega T we come up with this expression here in which the second term notice we have one half times i r squared times are times coastline of two omega T coastline of two. Omega T is also going to average out to zero over time, which means the second term contributes an average of zero power. And this first term is going to be our average. So right here, that's our average one half i r. Squared times are Okay, so this is the amount of power that is on average, delivered to the resistor. Uh, now, oftentimes you'll see books write it out in a different way. They'll combine the I r and the with the route to hear and square both and call that term the arms current the root mean square current. The reason this is popular is because it turns out you can also write out a root mean square voltage which is equal to the maximum voltage divided by route to just like how we had the maximum current divided by route to was the root, mean square current. And you can also have a root mean squared applied e M f, which can be written as Thea ply the maximum plied e m f divided by square root of two. So this is very convenient because when you go to write down, then the power delivered to the resistor in this circuit, you have i rms times v r m s is equal to I r m s squared Times are which is equal to v r m squared, divided by our So we have lots of different ways to calculate this. And we also remember that if we were in a situation where the resistor was our Onley element in the circuit, Then VR would always be equal to e not. And we would be able to write the power from the delivered from by the source z equal to i r M s times er mess. And in fact, this will be true even often average in an R L C circuit because again, the only contributing power sink in the RLC circuit is the resist er because the induct er and the capacitor are going to conserve energy and feed it back into the circuit. Now the other way to think about power delivered to a circuit is by thinking about something known as the power factor. So the power factor comes from considering our phaser diagram for an R L C circuit, which, if you recall, let us does know that the current and the E M f the applied e M f. We're not in the same at the same angle here. They differed by a phase constant fi where the tangent if I was equal to the the re inductive reactions minus the capacitive reactions divided by the resistance in that circuit. So looking at this then we can try and make an analysis by going back to power from the circuit is equal to i times applied MF. We can plug in our funk, our current and rmf as functions of time. So we have cosine of omega T minus five for our current and co sign of omega T for r E m f. Now, when we combine these again and I use another trig and metric identity were able to obtain this term here where you'll find that we have one factor multiplied by cosine squared omega T and then another factor multiplied by sine of omega t times cosine omega t Well, if you plot sine omega t cosine omega T, you'll find that it oscillates around zero. But cosine squared of omega t will not oscillate around zero. It'll look something like this instead. So we know then that we have a term on the left here. Our first term is going to on average be one half times I epsilon not times co signify where all we've obtained is this one half we can separate out the one half and we see here we have our root mean square current multiplied by our root mean, squared applied E m f times this cosign Fine. Now cosign fives, your power factor noticed. It's the only thing that makes this equation different from what we had before, but thinking about five then if we were in, say, a circuit that had on Leah resist er in it, we know that we have XL minus X c, which would be zero. And we would find that fi then would be zero Or rather, it would be five would be zero, and we'd have CoSine of zero is one. So your then your power source suddenly becomes I arm s times e r m s exactly as we had before. On the other hand, you could also find this if you just run an RLC circuit at its resonance. Omega, remember that Omega are came under the condition that excel was equal to X c. And when we run at that residence, angular frequency. Then again, we'll find that we have a 50 So we have co sign of zero is one. And again, we are applying that same power to our so our circuit of I r M s times E R M s. So we're gonna do a couple of calculations and a couple of problems with power here, and then we'll be moving on to our next topic.

Electromagnetic Waves

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