Desperately Seeking Modelling Help: Expert Assistance for Homework Questions

[magicuniverse]

Homework Statement

I need some help with all of the questions in the attatchment but would love it if you could provide some help with the first question please.

Homework Equations

In the file.

The Attempt at a Solution

Dont be silly, I don't have a clue. If I could do it I wouldn't be posting here!

 

[malawi_glenn]

I don't like your tone: "Dont be silly"

 

[magicuniverse]

Well I was talking to the computer and trying to be lighthearted. Sorry if I really offended you.

 

[HallsofIvy]

I would also point out that there is an enormous gap between "I don't have a clue" and "If I could do it". What you have posted appears to be a test (perhaps a practice test) for a class in mathematical modelling. Surely, you have had some instruction in this?

Problem 1 appears to be a matter of replacing some terms in the formula for Q by their expression as a function of T, the temperature. That is, the formula for Q involves viscosity \eta; \eta itself is a function of \overline{\nu} which, in turn "is proportional to the square root of temperature". Q also is proportional to the radius, R, to the fourth power and (1/R) (dr/dT) is a constant.

For problem 2, you are given how the Velocity depends upon position, x, V(x).
Use F= ma. a= dV/dt= (dV/dx)(dx/dt)= V dV/dx.

 

[Shooting Star]

For problem 2, you are given how the Velocity depends upon position, x, V(x).
Use F= ma. a= dV/dt= (dV/dx)(dx/dt)= V dV/dx.

V(x) represents the potential here, not velocity.

I don't have a clue.

We don't like that here. If you show a bit of effort, a lot of help will be readily forthcoming. I'll give you some hints anyway.

F = m*(-dV/dx). At equilibrium, F=0, so you can find the value of x.

For plotting the graph, take the derivative and see how it changes signs. Consider how V(x) behaves for fractional values and for x>1. Remember, it’s an even function.

For prob 1, you must have understood by now that what you have to find is dQ/Q in terms of dT.