ðŸ’¬ ðŸ‘‹ Weâ€™re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Like

Report

No Related Subtopics

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

McMaster University

04:39

Muhammed S.

(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?

01:24

Kai C.

(I) What is the magnitude of the momentum of a 28-g sparrow flying with a speed of 8.4 m/s?

01:40

Keshav S.

(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?

03:02

Averell H.

(II) Superman must stop a 120-km/h train in 150 m to keep it from hitting a stalled car on the tracks. If the train's mass is $3.6 \times 10^5$ kg how much force must he exert? Compare to the weight of the train (give as %). How much force does the train exert on Superman?

Create your own quiz or take a quiz that has been automatically generated based on what you have been learning. Expose yourself to new questions and test your abilities with different levels of difficulty.

Create your own quiz

welcome to the next section in our unit on alternating current In this section, we're going to consider two circuits that combine the circuit elements that we've looked at so far with our alternating MF source. In this first part will look at an RC circuit where we have an alternating current source that's attached to a Siri's resistor and capacitor. Now, in this particular case, using Kirk Offs Lyles, we find that the current through each circuit element would need to be the same. And then the some of the voltage across the major across the resistor and the capacitor would have to be equal to the MF provided by the source at any time now. It can be helpful here and to look at our phaser to gain some insight into what this all will mean. Remember that for a current in this direction, given them by Omega T, then we would have toe have a voltage in the are here at this point. So at the same angle, it's parallel tie, and here is the maximum voltage. It's the size we are remember that the magnitude of these angles stays the same as they rotate around. Meanwhile, VC would have to be perpendicular to this according to our previous analysis, where we found that V. C. And I see we're off by an angle of 90 degrees. Given this and the fact that they must add up to equal our EMF, it means that when we add VR plus V c, they must add up to equal our total MFR maximum e m f. This is interesting. It's a right triangle with three sides v R and V C and high pot News e not writing that out using the Pythagorean theorem, then were able to find an expression for the maximum current in this circuit, the maximum voltage across the resistor and the most maximum voltage across the capacitor. Now, looking at all these, you can see that we have this similar denominator R squared plus XY squared here and we can ask ourselves, well, what happens in the situation where we have the same potential across each of these. Remember these air maximum values, not values as a function of time. Well, in the situation where they're the same, we know that VR would have to be equal to V. C. Or, in other words, IR would have to be equal to I times the capacitive reactions. The eyes canceled, and we find that our must be equal to one over Omega time. See, so we see that there's actually a specific frequency which, if this circuit is driven at that frequency, we will get the same potential across each of these circuit elements. So, uh, this isn't a very interesting circuit. It finds applications in what are known as frequency filters. We have high filled high pass filters and low pass filters thes air filters, which allow you to hook this circuit up to any sort of any sort of signal analysis and find out whether the signal is above or below a certain frequency. For example, you can filter out all the high frequencies, or you could filter out all the low frequencies below a certain point. Um, there's some great application videos for that, but it's a little beyond the scope of this course to really talk about that. The next circuit that we want to talk about is are slightly more complex circuit, where we've added in an induct er again we can look at Kharkov laws finding that the current must be the same through each circuit element at any time because they are in Siris. And then also we would add up all of their potentials that we measure across them in order to find out what the actual applied potential is going to be. Okay, so this is at any particular time. So when we draw that again in our phaser, the way it'll look is we have a particular I at some time that again, is that some Omega T and the current through the resist er will still be parallel to that again, this is our total I and the capacitor will be perpendicular to that. But then, remember, the induct er is also supposed to be perpendicular toe I in the phaser, which means it will be pointed this way and bond again. All of these must add up. Two are applied E m f Now notice here that v, C and V L R in opposite directions. So the way we could write this out then would look like a triangle, but it would have a shape like this VR b l minus B c. And here would be Are we not okay, so given all of these we know that then e not squared would have to be equal to B r squared plus B l minus B c quantity squared. If we convert all of these Volta statements for voltage into our statements for current, we can have he not squared is equal to I squared r squared Plus we'll have i times x l minus I times xy again remembering that excel is the three inductive reactant and XY is the capacitive reactant. Okay, um, we can pull eyes out of each of these and again find current for this system. The maximum current for this system will be equal to n not divided by the square root of r squared plus excel squared minus X c squared. Now the denominator here actually has a special name. Very often, it will be called the impedance of the circuit impedance being equal to variable Z, which is the denominator r squared of r squared plus x l squared minus x c squared. Notice that the impedance can be made to look like simply are if we have excel equal to X c, which again is going to be a case of finding the right frequency. So in order for excelled equal XY, we need omega l to be equal to one over Omega time. See? So that would be Omega is equal to the square root of one over l time. See? So if we're able to apply this frequency with our source, then what we will have is the maximum possible current we will see because this term will completely disappear, making the denominator very small. And in fact, I will simply look like it's on Lee passing through the resist earth. This looks like alms law all over again. So, uh, this idea that we're able to find a resonance frequency where we gain a maximum current is really interesting. It's something that we've seen before. So this is an important frequency for you to remember the resonance frequency oven rlc circuit as thes air titled. Now, if we want to dig into this a little more, we can analyze what happens if we have frequencies that are equal to less than or greater than well, if it's equal to, then our plots of voltage and current are going to look the same. They're matched with each other. If we look on the other hand for at voltage and current when omega is less than omega Not you can see that here that we have current is leading. So I leads E. Meanwhile, if we have omega greater than our residents frequency or you cannot than we have I lagging behind. Okay, so here these air some of the basic ideas that support RLC circuits there is one more that we should consider going back to our phaser. If we look at our phaser here and we say I know that my e is going to be pointed because of my addition here, my e not is gonna be pointed somewhere up here. And I were to ask I know I have some angle between the current and my n not. What would that angle B? How could I write that out? Well, just looking again at the triangle that this has to make, we have absolutely not. And then we have v l minus V c. And then we have our v r here. So if we want to know this angle, then what we can say is that tangent of fi is opposite over adjacent. So that's V l minus V C over Jason Is VR then plugging in to get our currents out? We can actually find that tangent if I is equal to x l minus X c all over are so this angle phi here is also an indicator just like the impedance. If x l is equal to x C, we would find that we have fi of in five is equal to inverse tangent of zero which would mean that if I would be zero and we'd have a matched e not. And I in other words, this angle here fi zero means Enoch points in the same direction as I. On the other hand, if we had Justin induct er or just a capacitor, the same thing would happen And we would find in a pointing in the direction of V, C or V l here on our phaser eso these air all the important points that should be gone over about RLC circuits in your course. Hopefully this will help you to determine what's going on there

Electromagnetic Waves

03:28

01:50

01:51

03:03

07:02

02:27

02:33

02:56

02:08