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

Simon Fraser University

University of Sheffield

02:55

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?

0:00

Aditya P.

(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?

03:04

Kai C.

Muhammed S.

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 our third example video looking at power dissipated in a sea circuits in this video will return to the circuit We considered in our first example and ask ourselves what is the power delivered to the resist er And then we'll ask ourselves What if we were to change the frequency? What would what frequency would we have? A maximum power delivered to the resistance? Um, as you might have already surmised, What we're going to need to do here is take into account our power factor which remember, is going to be equal to I r m s times e r m s times the cosine of fi where cosine phi remember is our power factor. In order to calculate our power factor, we need buys equal to inverse tangent of x l minus x c divided by our. So I'll need to calculate Excel and XY remember that Excel is equal to Omega times. L xy is equal to one over Omega time, See, and omega is equal to two pi f So plugging in all that we have five is equal toothy inverse tangent We should of two pi f times l minus 1/2 pi f times see, all divided by our we have our our our C and R l. Okay, so now that we have five, it's still not quite enough. We need r r r m s and R E R M s Well, e r m s is just going to be equal to nine volts divided by route to I. RMS, on the other hand, is a little more challenging. We have I rms is equal to i r divided by route to which is going to be equal to one over route to multiplied by e not over the square root of r squared plus x l minus X c Great. So luckily, we already have excel in X'd I can plug those in find I r m S e r m s plug in fi and I'll be able to find my power dissipated now if I wanted a maximum power noticed that for a maximum power I need cosine of fi to be equal to one For this to be true, I need fi equal to zero. And that means that I need excel to equal xsi the condition that we've seen before. So I have omega l equals one over Omega Psi. Or, in other words, omega is equal to the square root of one over L C. An omega is equal to two pi f Therefore, F is equal to 1/2 pi times the square root of one over l C. And that would be our resonant frequency for this for this RLC circuit.

Electromagnetic Waves

02:08