the first step makes sense: ln[(x^2+1)(x-1)]=ln(x^2+1)+ln(x-1)
but then you continued: ln(x^2+1)=ln(x^2)+ln(1)
You can't do that, but you can do something else to the x^2+1...
Do you mean you need to reproduce the theoretical calculations present which involve a step function? I'm not exactly clear on what you're trying to do.
This is a really cool question. I was stumped for a few minutes before I started thinking about what additional effects are present when the capillary tube is vertical that were not there before. Do you remember in which direction pressure acts?
one of the key differences in your choices are the numbers 12 and 24. What do they represent, and which one is relevant in the case of a full rotation of the hour hand?
Yes, that is what I meant, except that I used the second derivative of temperature. I know there's a major difference, is your first derivative just a typo or is there something I missed?
By assuming that T(x) << T(outer) I can get a temperature of approximately 175K. Thanks for all of your...
Perhaps I'm not approaching the problem the same way you are:
\alpha = \kappa / c_p \rho
C = \epsilon \sigma A * T(outer)^4 / (m*c_p)
--> this is the delta T as contributed from external radiation, adjusted for the mass (which will have the inner and outer diameters in the...
you're right, there should be a constant in front of the second derivative of temperature wrt x:
\alpha d^2 T(x)/dx^2 = C - \epsilon \sigma A*T(x)^4
However, I'm changing the problem into a 1D heat conduction problem because I want to solve the simpler version first. In...
Homework Statement
Let's say you have a 3m long copper pipe, 3mm in thickness with a diameter of 170mm. You fix one end at 1K and insulate it to prevent conduction or convection between the air and the pipe itself. There is still radiation. Assume that the inside of the pipe has no effect...
I don't understand where you're getting your angles from, unless one of the masses is also on a ramp? Can you describe the geometry of the setup a little better?
three hints:
-in which axis is "range" measured?
-what forces do you have acting in this axis?
-based on these forces, formulate your acceleration and displacement equations.
delta PE is just your change in gravitational potential energy, delta KE is your change in kinetic energy. Does this formula seem different than the one you used?
Have you included gravity in your FBD? Once you have all of your forces for the FBD, look at which way the box could possibly move. You have the values of all of the forces including coefficients of friction, so how do these come into play?
If you put up your images on a hosting website (like flickr or photobucket, etc) it will be much easier since your attachments still haven't been approved and we won't be able to see them until they are.
Or you could just describe it?
double check the way you're assigning your forces, is the Fny component which is cancelling vertical gravity the hypotenuse of the triangle? I find it helpful in these cases to rotate the coordinate system so that the x axis is actually the sloped road. That way your Fn has only 1 component, and...
Think about (and set up) your free body diagram. You know that tension is an internal force (i.e. it doesn't happen by itself, something causes tension). And you know your system is isolated. Just imagine intuitively, what's stopping the 5kg block from sliding down the ramp if the ramp is very...
for c I was thinking of even more simply: d = v*t, which is just a simplified version of the one you used.
d) you still need to take y velocity into account here...
when i say equated the two, captFormal assumed that the system is isolated and that the total energy is conserved from point to point. I.e. he said PE2 = KE1. This is not true since the drag force does work on the system.
you equated the 2. This is not correct because drag exists. your difference in energies will be equal to the energy lost through mechanical heat (aka drag in this case). You know the value of this because it's a force acting on the system (i.e. changing the total energy).
In an isolated system...
i haven't done the calculation so I don't know the actual number. but your acceleration for the wheaties box will be just the force on the wheaties box - friction on wheaties box, not the entire 29N - wheaties friction. Since force wheaties = - force wheaties on cheerios, you can set up your...
you also have the force of the cheerios box on the wheaties box. i.e Fcw = -Fwc. Your actual effect on the Wheaties box will not be the entire force, but will be something smaller. so what will it be?
Your step to assume that the acceleration is equal is correct.
Are you summing your forces properly? What are all of the forces that lead to a 3.0 m/s^2 acceleration? You're right about the normal force, but you seem to be disregarding that whole 320 N that's also being applied. Unless I'm missing something in your attempted solution.
post picture as a link, we still can't see the attachment. What I do in these cases is redraw the circuit to actually show group the series and parallel resistors, that way it's easier to see where you can simplify.
use the right hand rule to determine current flow direction. Which way is the induced force going to be pointing, and in which direction does the current flow for this to be true?