- #1
Obliv
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I know that energy is recognized through motion. Even in the mass-energy equivalence a velocity is present even though it is a rest-energy (Not really sure if this would count as a potential energy since there is no 'field' of acceleration that the mass is in)
So does kinetic and potential energy make up all other forms of energy by definition?
also as a side tangent: it is described here https://en.wikipedia.org/wiki/Time_in_physics that time is used to derive kinetic energy. I tried defining [tex] v = dx [/tex] without respect to time
[tex] \frac{dv}{dv}{v} = dx [/tex]
[tex] \frac{dv}{dx}v = dv [/tex]
[tex] m\int{a}dx = m\int{v\frac{dv}{dx}dx} = m\int{v}{dv} = m\frac{v^2}{2} + C [/tex]
actually it might be easier to just say [tex] a = dv [/tex] therefore
[tex] m\int{a}{dx} = m\int{dv}{v} = m\frac{v^2}{2} + C [/tex]
How come I can define energy in terms of position but it says time is used to derive energy?
So does kinetic and potential energy make up all other forms of energy by definition?
also as a side tangent: it is described here https://en.wikipedia.org/wiki/Time_in_physics that time is used to derive kinetic energy. I tried defining [tex] v = dx [/tex] without respect to time
[tex] \frac{dv}{dv}{v} = dx [/tex]
[tex] \frac{dv}{dx}v = dv [/tex]
[tex] m\int{a}dx = m\int{v\frac{dv}{dx}dx} = m\int{v}{dv} = m\frac{v^2}{2} + C [/tex]
actually it might be easier to just say [tex] a = dv [/tex] therefore
[tex] m\int{a}{dx} = m\int{dv}{v} = m\frac{v^2}{2} + C [/tex]
How come I can define energy in terms of position but it says time is used to derive energy?
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