Recent content by Valour549

Yes I was aware of the fact you mention regarding the that we cannot integrate wrt to x and also have x in the limits of integration. So what about the rule Stewart wrote (the one boxed in red), does that only apply when the limits of integration are actual numbers? Or anytime when the limits...

Let me see if I got this right: if any of the limits of integration contains a variable, then it cannot be separated from the integrand which integrates wrt that variable?

I am trying to do the double integral. And I remembered there's this formula that says if the integrand can be split into products of F(x) and G(y) then we can do each one separately, then take the product of each result. Taken from Stewart's Calculus 9E. So I tried to do the integral two...

Except Rx turns out to be negative and -Rx turns out to be positive right? How do you represent the terms ρQ*V1 and ρQ*V2 in the FBD?

OK so... since ρQ*V1+P1*A1 turns out to be greater than ρQ*V2 (P2=0), the force acting on the CV (or nozzle) by the pipe must be to the left (negative X direction). And therefore the force on the pipe by the nozzle is to the right (positive X direction)?

So combining what the two of you helpers have said: The pipe applies a force to the nozzle through the flange, and then the nozzle then applies it to the fluid flowing through it? And this is necessary for the fluid to have a higher velocity exiting the nozzle (as compared to the slower...

I know P2=0. When you say P1 acts on the left side of the CV, that could be one of two possibilities below and I really don't know which. I know that you guys are trying to help users derive the answer without telling them, but the issue here is I already have the answer and the equation from...

The mass at the nozzle exit because it has a higher velocity? When you say the force the nozzle exerts on the fluid, is that at point 1 or point 2? The question says the axial force of the nozzle assembly on the pipe flange, isn't the pipe flange that rectangular piece between the pipe and the...

Let me start off by saying that I have found (or is given) all of these: ρ, Q, V1, V2, P1, P2, A1, A2 (V being the velocity here). So no problem with Bernoulli or the Continuity equation calculations. I am just struggling with drawing the FBD in order to evaluate the axial force, Fx I know we...

Homework Statement 2. The attempt at a solution Am I doing something wrong? I know it's just voltage division with AC phasors and we have to equate the real parts and imaginary parts to find the unknown, but I feel like the question shouldn't be this hard.

xy'' + 2xy' - y = 0 Honestly no clue where to start, Wolfram Alpha gives a rather complex answer lol (http://www.wolframalpha.com/input/?i=xy%27%27%2B2xy%27-y%3D0)

While the proof on Wikipedia and the proof I am working on both ultimately aim to prove the statement of FTC2, namely ##\int_{a}^{b}f(x) dx = F(b)-F(a)##, I think we go about it in different ways. The problem is I see no reason why my way shouldn't work. It proves the above statement from left...

That is exactly the problem! I don't want to start a proof with the derivative of ##F## from the the first theorem (where ##F## is defined by an integral of ##f##), integrating it, then acting surprised that the result returned in the form of ##F##. What I want is to start with: ##G##, the...

Yeah, in this respect I'm now convinced that ##\int_a^x f(t) dt## is a definite integral. The definite integral ##\int_a^x f(t) dt## = ##F(x)##, a single function- a particularly useful antiderivative of ##f(x)##. The indefinite integral##\int f(x) dx## = ##F(x)+C##, a class of functions-...

What would you call the LHS of the first line then? According to the scanned text (in the spoiler), it calls the LHS function "an indefinite integral of the integrand [on the RHS]". Do you agree with this? Because succinctly put, that quote basically says "The definite integral ##\int_a^x f(t)...