Recent content by JohnnyGui

Okay, so the energy input that you subtract from Q is different from the one to overcome the Coulomb barrier. So in the case of fusing Fe-56 to get As-75, what is the type of energy input that costs you resulting in a negative Q? Surely it's not the input to overcome the Coulomb barrier since...

@QuarkyMeson Sorry but I'm struggling with your two statements which I find to be contradicting. Here it states that the net energy is the output minus the input. This should mean that energy input costs you and therefore you subtract it from the output to get the net energy out. Here it...

Okay, one last question if you don't mind. When fusing Fe-56 to get As-75, obviously there's a net energy loss as shown. Based on your previous statement: What type of energy input outweighs the output in this case such that energy is lost from fusing Fe-56 to get As-75?

So the ##E_{input}## gets released along with ##Q##? And that's what the graph also includes? How can ##E_{input}## be added to the released energy ##Q## while it's energy that costs you, not energy that you receive in addition to ##Q##?

@QuarkyMeson Thank you, I still have some confusion though based on the following: Okay. If the graph only shows the binding energy that you need to break atoms apart, I'd understand why it doesn't include the input energy to overcome the Coulomb barrier since this input energy is needed for...

My question is about the following graph: I keep on reading that fusing atoms up until Fe-56 doesn’t cost energy and only releases binding energy. However, I understood that fusing atoms also require energy to overcome the positive charges of the protons. Where does that energy go after...

Thank you. How about the upper limit of the equation in pbuk's post of this thread if we substitute ##\delta x## by ##\Delta x## giving: $$\lim_{\Delta x \to 0} \int_a^{a+\Delta x} f(x) dx \approx \lim_{\Delta x \to 0} f(a) \Delta x$$ I'd expect you won't agree with this upper limit, even if...

Yes but not insisted though. I only said it in 1 post because the source ultimately used ##\Delta t \rightarrow 0## for his conclusions, which I thought is the same as ##dt##. After your correction about this I kept using ##\Delta t## instead. @PeroK @fresh_42 Thank you for the further...

Please look at the source carefully because you said I didn't understand the earlier posts and stated the source completely wrong. I did understand it which is why the source's ##\Delta t## actually confused me.

That is literally a ##x+ \Delta t##, not ##x + \Delta x##.

@fresh_42 Thanks, but I have no idea where you're seeing ##\Delta x## as an upper limit in the source. He is clearly stating ##\Delta t## in the upper limit and says: $$G(x+\Delta t)-G(x)=\int_x^{x+\Delta t}f(t)dt$$ He is constantly using ##\Delta t## in his expressions afterwards. He might as...

@PeroK @Frabjous Thank you, this cleared it up for me. I have one last question because this source seems to say otherwise, or so I think. It says that if $$G(x)=\int_a^xf(t)dt$$ then one can deduce $$G'(x)=f(x)$$ Which I can grasp. However, the poster also says that from... (note the ##dt##...

Thank you, I'm trying but I fail to see how this answers whether my conclusion in post #14 is correct or not about getting the same ##f(a)## regardless of using ##dx## or ##da##.

Thanks. I have one other question that came up when approaching this another way. ##f(x)dx## is also equal to ##dF(x)##. Now, if ##x## takes on a value ##a##, then ##dF(a)/dx=f(a)##. But we just saw that ##dF(a)/da=f(a)## as well. Because ##da## is not equal to ##dx##, such a similarity is only...

So different variables that don't have a relation between them, since ##x## ultimately becomes ##a## after the integration? Also, what is your stance on the upper limit being ##a+dx##?