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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Sheffield

University of Winnipeg

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?

00:39

Averell 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$^2$ ?

03:27

Kai C.

(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?

04:27

(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?

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welcome to our fourth example video. Looking at simple circuits with an alternating current. We're going to in this example look at two plots of voltage and current and try to determine the properties of our circuit just by looking at the plots. Um, first thing we want to determine is what type of circuit we have here. Notice that we see a crest for voltage before we see a crest for current. That means that current is lagging behind the voltage. That means that we have a simple inductive circuit. Okay, eso questions we might ask. Could be things like What are what's Omega? What's L what's excel and so on? Um well, I see that I have avielle and an I l V l is equal to one volt, and I l is equal to 0.5 amps. Um, for Omega, remember how omega's related to period? The period for both of these is 0.4 seconds. Since T is equal to 0.4 seconds. That means Omega is going to be equal to two pi over 0.4 seconds for our Elvin. What can we do here? Well, we know that Excel, for example is equal to Omega Times out. And we know that I l is equal to V L over Excel, which means we have some relationships here that we can work it. If we plug this in, we have I'll equals plover omega times l so I can solve for l then as equal to one over Omega Times VL over i l And so now I have my l and I can also go back and solve for X L So l then is equal to one over Omega Times VL over I l and Exhale. When I plugged that in, we'll end up looking just like bl over. I also I could have solved that from the beginning as well. So, um, lots of different ways to analyze this and to go about solving things because we have so many different equations. Important things to remember are that current lags behind voltage in an inductive circuit. If it had been a capacitive circuit, we would have seen the current peaking before the voltage. It would have lead the voltage and then also remember all the different relationships we have in particular the inductive reactant being a very useful one

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