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University of Winnipeg
(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?
(I) What is the magnitude of the momentum of a 28-g sparrow flying with a speed of 8.4 m/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?
(I) What is the weight of a 68-kg astronaut ($a$) on Earth, ($b$) on the Moon ($g =1.7 m/s^2$) ($c$) on Mars ($g = 3.7 \,m/s^2$) ($d$) in outer space traveling with constant velocity?
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welcome to our third example video. Looking at simple circuits with alternating current, this video will consider a simple inductive circuit with an induct er that has the size of 100 million. Henry's well say here that we have a frequency equal to 500 hertz and we have a maximum applied voltage equal to six volts. Given all this, we want to find what is our inductive, reactant sar inductive reactant, remember, is going to be equal to omega times l Omega Times l. And once we have that calculated Remember, Omega here is two pi times f times out. So plugging this all in again we can do the same trick is last time where we say I knowing this. Then I confined My maximum current is going to be equal to V l over Excel, which is just six volts over two pi f times l Having plugged that in, we can then write our equation for current as being equal to I l times. Cosine I l times cosine of omega T minus pi over two. And our voltage, as we've known from the beginning, was VL co sign of omega T. So again, we found the pertinent equations here. Other questions that could be asked in your book are things like at what time do we reach a maximum current? If at time T equals zero, we don't have a maximum current. Or at what time do we reach a maximum voltage? Or we could switch it around and give us the maximum current, asking us then to find the maximum voltage. Either way, using the reactions could be a very powerful tool to quickly find what you're missing and then plug it into the equations that you know.