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This demonstration is a fundamentally critical example of how understanding one’s audience absolutely makes or breaks the effectiveness of “teaching” Sadly, it is often the case that failures to educate come down to: 1) lacking the ability to perceive things the way others might, and; 2) appreciating or fundamentally understanding why/how/what factors should be addressed I’ve come to appreciate immensely, the difference a good or bad teacher can make in “unlocking” understanding Bravo 👏

Im shocked not more videos explain the current overrides with this simplicity. Thankyou

clear and short , i have a better idea about this topic than watching a 30 min lecture THANKS

Omg I finally got it, thank you so much! ❤

Amazing.

no indepent X axis data sets. How utterly incompetent is the google sheets team. Unbelievable. Thanks for this workaround

Wow great 👍. Thank you ❤

In the last question(bullet question), if the rod is unhinged, when conserving momentum, in initial momentum are we supposed to take the angular momentum of the bullet wrt to the new com after collision or the old centre of mass before collison?

Thank you for posting this. Very concise and helpful

Thank youuu❤

thank you

I am from India

Thank very much mam I understood very easily and very well

Good work. But, mu is forgotten to be mentioned in the last acceleration.

Thank you sir❤. U earn a new subscriber 😊

In which way will current flow if we have two discharging battries?

thank you so much it helps

Do you think I could take AP physics C mechanic and E&M without taking AP physics 1

This video could've save me from my first quiz disaster lmao

Thanks

actually we said x equal to five so, dont we have to add five to the 70.4, so the answer became for c is 35.4

For problem 3, L Hopitals Rule can be applied. Apply the derivative to the numerator and denominator. Plug in 2 for x. The answer is 12.

Thank you this was very helpful!

u man, my hero❤

You explain it so clearly. Thank you. 🙏🏽

THANKS!!!

Thanks so much!

I didn't understand why torque = radius * tension for atwood machine problems, but now thanks to u i understood it

We never got taught Fc = mac?

you are AMAZING!!!

tysm!! this saved me for my physics lab

what the.. I cant solve the last problem I have tried different method, it justcant be solved

Is it P=V^2/R as stated earlier, or P=V (V/R), as stated at the end? So, if it's 12V & 3 Ohms, that would be 12x12 / 3, rather than 12 x (12/3)...?

Since series circuits have the same current everywhere, why cant we directly use P=IV for the last part with 2A and 12V??

how do you choose which kinematic equations to use?

It will be clearer if you do a KVL in the circuit.

thank you i love u

Thank you very much

omg google this is so stupid.... I might just completely stop using google sheets, not having the ability to just add Xn and Yn to plot independent data on the same plot is frustrering. And fair this workaround makes it possible, but imagien have a lot more data than in this example. omg. Google should force there employs to use this until they fix it. 🤦🏼‍♂

Why is anything relatively simple end up being a time consuming pain in the c*****g arse with this crappy, dogsh*t software

OH GOD THANKS UUUUUU

thanks for your help, last question on assignment was evaluating limits by hand. normally we just use maple. I got the question right

I went through so many vids and only got to understand other concepts here. 🙏🏽

In 3 ohm resistance u have current also because in series current doesn't change so you can also use p = i²R for finding power in 3 ohm resistance 😊

This explanation of why the static friction force does no work is popular on the internet, but I'm just not buying it. The point of contact is the base of the wheel, but the bonds in the wheel transmit the force to the rest of the wheel. Translationally, we apply Newton's 2nd law to the center of mass. The static friction force opposes the motion of the center of mass and therefore does negative translational work on the wheel. The static friction force also produces a torque that does positive rotational work. Presumably these two cancel out (didn't work it out yet) so the net work is 0. In your example, the ball needs some force to the right.

Thank you. Exactly what my students needed to learn and an easy way to understand and do at the same time.

Very helpful this was 🤞🏽

I finally understand this thoroughly thanks to you!

dudeeee, thank you :)))

You sound like Badger from Breaking Bad, great vid btw :-)