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Rutgers, The State University of New Jersey
University of Washington
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?
(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?
(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?
Suman Saurav T.
(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 Looking at R, C and RLC circuits in this video, we're going to attach an A C source to a simple RC circuit. And remember when we did this before we found that we had the maximum voltage around the resistor was this function and the maximum voltage around the capacitor was dysfunction. Well, given that X C is equal to one over omega time, see Or in other words, in terms of the frequency 2, 1/2 pi f time. See, what if we have a very large frequency? So for a large F, that means we have a small XY. Well, a small XY would mean that this term is approximately Epsilon not in this term is very, very small. So we're for a large f. We have a V R. That's approximately equal to eat. Not what does that mean? Well, it means that almost all the produced E m. F is across the resistor. It's almost like the capacitor isn't there. On the other hand, if we have a small F which would lead to a large XY, then this term is the large one we have. VC is approximately equal to he Not Okay, so this is very interesting. We're able to say that most of the voltage across CO is across the pastor when we have a small frequency and most of the voltages across the resistor. When we have a large frequency based on this information, let's go ahead and try and design a low pass filter. So for a low pass filter, it means that for high frequencies, we want zero output current we want for high s V out equals zero volts. Okay, In order to draw that, what I will need is I'll need this circuit. So I have some oscillating source and resistor and a capacitor, and I'm gonna have my output voltage be across the resistor. Okay, So here is my the out of a problem. I apologize. Not across the resistor across the capacitor. If the output is across the capacitor, then what that means is when I have a very large frequency. Then there's very little voltage across the capacitor, which means in my output, voltage would approach zero as f becomes large. So this will allow me to filter out any high frequency signals that air coming in through my source, which would be an antenna in this case. On the other hand, if I had, if I want a high pass filter so for a high pass filter, I would want all my low frequencies to be filtered out. In which case, as you might guess, I'm just going to flip it around and I'll have my resist er here and my output. And I didn't even really need to change the position of the resistant capacitor just because of space constraints. I want it this way. But my vaulted out voltage out here will be large when I have a large frequency, and it will approach zero when I have a small frequency. So these air, my low pass filter and my high pass filter