Recent content by conquertheworld5

Thanks!

I recently inherited some legacy code written in fortran77 which I'm going to be modifying quite a bit, and at this stage I find that I'm continually getting stuck on small fortran quirks... Just wondering if I could get some advice on this particular problem. I'm outputting the values of...

Homework Statement You've got N marbles, and N bins. one by one, a marble is randomly placed in a bin. What is the probability that there will be no marbles in a given bin. Homework Equations P=N!/(n1!n2!)*p^n1*q^n2 : binomial probability The Attempt at a Solution since we're...

Homework Statement A sphere full of air at room temp and 1 atm is placed in a chamber filled with He gas at room temp and 1 atm. The sphere is permeable to He only. What will the equilibrium pressure in the sphere be? Homework Equations Variables: Po=initial pressure V=vol of...

Hey,beginner programmer with a simple question... is there any way I can use the where function to search for a string in elements of an array? I have a structure with two arrays, the first for the date (a string in the form of dd-mm-yyyy i.e. 11-07-2007) and the second has data. I want all the...

this is rather embarrassing, but it's late, i have a headache and can't figure this out. so i have this complicated function i need to sum (not sum as in add up - sum as in sum a series). my formula that needs to be summed: (2/(n*pi))*sin(n*pi*z/L)*e^(-(n^2)*pi*pi*k*t/(L^2)) pi, z, L, k...

I think it's just your conversion for the k value; you have to multiply 1.2N/cm*100cm/1m = 120N/m my bad... didnt see that someone already replied.

I believe the factor for your moment of inertia is off. I=(1/2)mL^2 is the moment of inertia of a disc. I think you should be using the moment of inertia for a rod, which turns out is just a different factor... it may be 1/3 instead of 1/2. You should look this up though, I'm not entirely positive.

I do know a little bit about geology... and I too have no idea what the question is asking.

well... looks like I'm handing in my lab report as is whether it's right or wrong... just printed it. Still would like to get an answer though!

hmm... I'm not sure what you mean by this. You ARE leaving the acceleration in m/s^2 because by definition, acceleration is in m/s^2. Doing anything to that acceleration would change the formula. You don't want the final answer in m/s^2 because you have a distance, x, which should be in meters...

okay, here's the units: acceleration = m/s^2 t^2 = s^2 so a x t^2 = m/s^2 x s^2 = m so it's just xf = .5(-5.2)(4^2), this will give you your answer in meters. a is already in m/s^2, if you square it again, your units would be m^2/s^4

heh, take a look at the acceleration...after everythign cancels, you have xf=.5(a^2)(t^2) I have xf=.5(a)(t^2)

xf=xi +vixt +.5a(t^2) compare that to your equation "xf=0+0+.5(-5.2)^2(4)^2" you have something I don't...

If you're not given an acceleration, I think you can assume that there is none and the particle is moving at a constant velocity in the y direction.