15.2 Coulomb's Law | General Physics

[@birdis9476]

At around 6 minutes you said dividing by .01 is the same as multiplying by 100, and gave the answer as 8.99e^7. If we multiply 8.99e^9 by 100, isn't it 8.99e^11? I appreciate you time and efforts thank you for all your videos, you are a great teacher!

[@NightStar942]

Hi Chad, I’m a student at ASU in Phy121, how relevant are these lessons to the curriculum. Your chem lessons were very helpful and I was hoping to use these as well for my class. Thank you.

[@ChadsPrep]

They will help you 80-90% of the way. But being algebra-based, there will be a few areas where they will fall short lacking the calculus. But overall, you should still find them very helfpul!

[@AlanDaniel-ls8dv]

Also isnt the first example 8.99x10^11

[@ChadsPrep]

(8.99x10^9)(2x10^-6)(1x10^-6)/(1x10^-6)^2 = 1.798x10^10.... this one?

[@AlanDaniel-ls8dv]

@@ChadsPrep Yea, i was trippin my b

[@ramneekk6036]

This is what I got too

[@fermiumfrancium-pi5kr]

gigachad

[@ChadsPrep]

Happy Studying!

[@abdulazeezshuaibu8570]

in second example , by 11:40, how did u get that ans when i pressed it in my calculator i got 17980 and if we are to change it to standard form is 1.798 x 10^4

[@aaronmann1578]

He said to use absolute value for the negative charge, be sure to also square the radius. It will become 1E-12m^2 instead of 1E-6m. This is crucial.

[@LifewithAmairanny]

I'm so confused how you got 8.99x10^7 instead of 8.99x10^11 for the first problem?

[@ICantSeeYourRepliesDickhead]

Fr, please Chad let us know how you got this! I am praying I did it right, without calculator for MCAT LOL please tell me I did it right!!

[@ramneekk6036]

same!

[@AlanDaniel-ls8dv]

Hey chad, my school is on trimesters so Im about 5 lessons ahead right now. Is it possile to see the videos that arent released yet somehow. You got me through orgo 1 & 2. Thank you

[@ChadsPrep]

Shoot me an email: [e-mail]

[@alibilginler]

Can it be done if I do it like this: Tcos30-mg=0 for x-component Tsin30-kq*2/r*2=0 for y-component?

[@ChadsPrep]

Unfortunately, it won't work out. The angle provided in the problem was the angle with respect to the vertical (the y-axis). To use the convention that we use cosine to determine the x-component and sine the y-component, you have to have the angle with respect to the horizontal, the x-axis. Now you could have still used the 30 degree angle, but the x-component would have been the opposite side causing you to use the sine function and the opposite for the cosine function which would have yielded; Tsin30-mg=0 for x-component Tcos30-kq*2/r*2=0 for y-component Compare this to how I solved it in the video and you might realize it is an equivalent set of equations as sin30=cos60 and cos30=sin60 (the sine and cosine of complimentary angles are always equal). Hope this helps!

[@alibilginler]

I didn't know about this relationship between complementary angles, thank you.@@ChadsPrep

[@unitedwisdom36]

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