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henpen
Group Title
Power and work
\[ W= \int P dt = \int Fv dt= m \int \frac{dv}{dt} vdt=m \int vdv= \frac{1}{2}m(v_1^2v_0^2) \]
Is this true?:
Did I oversimplify somewhere?
Is this unphysical?
 one year ago
 one year ago
henpen Group Title
Power and work \[ W= \int P dt = \int Fv dt= m \int \frac{dv}{dt} vdt=m \int vdv= \frac{1}{2}m(v_1^2v_0^2) \] Is this true?: Did I oversimplify somewhere? Is this unphysical?
 one year ago
 one year ago

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henpen Group TitleBest ResponseYou've already chosen the best response.0
\[ W= \int\limits P dt = \int\limits Fv dt= m \int\limits \frac{dv}{dt} vdt=m \int\limits vdv= \frac{1}{2}m(v_1^2v_0^2) \]
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
it's fine.
 one year ago

03225186213 Group TitleBest ResponseYou've already chosen the best response.0
correct
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.0
Similarly, \[W=\int\limits_{}^{}Fdx=\int\limits_{}^{}madx=m \int\limits_{}^{} \frac{ dv }{ dt } dx=m \int\limits_{}^{} \frac{ dv }{ dx }\frac{ dx }{ dt } dx = m \int\limits_{}^{} v dv=\frac{ 1 }{ 2 }mv^2\]
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
@Kainui That's what I did.
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.0
No, you did something slightly different. You used power, I did not.
 one year ago
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