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I'm a physics graduate student.


Aug
23
answered Is there an intuitive explanation of the work formula?
Aug
22
revised How is pressure an intensive coordinate?
added 5 characters in body
Aug
22
answered How is pressure an intensive coordinate?
Aug
20
revised How can Kinetic energy formula and the Work formula be derived without assuming the other to be true?
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Aug
20
comment How can Kinetic energy formula and the Work formula be derived without assuming the other to be true?
I suppose what you want to know is something along the lines of, "What motivated people to come up with that definition of work in the first place?" I don't know, but that's a different question than the one I was answering, which was an accusation that the logic is circular.
Aug
20
comment How can Kinetic energy formula and the Work formula be derived without assuming the other to be true?
I fixed a mistake in the derivation. I'm not sure what you're asking for with "how do we know the work formula"? I am taking that as the definition of work.
Aug
20
revised How can Kinetic energy formula and the Work formula be derived without assuming the other to be true?
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Aug
20
revised How can Kinetic energy formula and the Work formula be derived without assuming the other to be true?
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Aug
20
answered How can Kinetic energy formula and the Work formula be derived without assuming the other to be true?
Aug
2
answered Why don't all free particles lose their kinetic energy?
Jul
31
awarded  Popular Question
Jul
27
awarded  Notable Question
Jul
24
awarded  Good Question
Jul
22
awarded  Nice Question
Jul
7
awarded  Favorite Question
Jun
22
comment Intuitive explanation for why centripetal acceleration is $\frac{v^2}{r}$
No, that is completely wrong. Nothing in my answer says anything remotely like that. It is only because both position and velocity had constant magnitude and rotate at the same frequency that we can take their derivative in the same way. It is obviously not true for arbitrary functions.
Jun
22
comment Intuitive explanation for why centripetal acceleration is $\frac{v^2}{r}$
The answer does assume understanding some basic properties of derivatives, such as linearity.
Jun
22
comment Intuitive explanation for why centripetal acceleration is $\frac{v^2}{r}$
Yeah, I'm really not getting your point. Maybe you can think more about it, phrase it more clearly, and post it as a separate question.
Jun
22
comment Intuitive explanation for why centripetal acceleration is $\frac{v^2}{r}$
Sorry, but I can't understand your point at all. There's nothing about polynomials here. It's an argument based on geometry.
Jun
21
answered Intuitive explanation for why centripetal acceleration is $\frac{v^2}{r}$