Recent content by Taiki_Kazuma

Homework Statement Hollow Circular tube of Length (L) 600 mm is compressed by forces P (axially). Outside diameter (d2) is 75 mm. Inside diameter (d1) is 63 mm. Modulus of Elasticity (E) is 73 GPa Poisson's ratio (v) is 0.33. axial strain (ε) is 781 x 10-6 Find shortening of tube (δ). (This...

Sorry. I guess I didn't show my last step. the series further simplifies to: -1 + 1 + ... Which is a basic alternating series of (-1)^(n) This series is a known to diverge. No further justifications needed

You are correct that the alternating series of the form (-1)^n diverges. So, this series for sine is just a different form of an alternating series. Now, the nth-term test is a waste of time. This would be a last resort test, because there are so many other (more simplicities) ways to...

Let me see if I can help... If the problem is asking to solve for f(x)... Then, I'd first ask myself, what function is equal to its derivative. I only know of one function: e^x Therefore, the answer must be of be form: e^(kx), where k is some constant. Does that help?

(My apologies for not using the pretty font) Now, rho = D dh = delta-h dF = delta-F I think I see the logic behind it: P = F/A F = P A P = (D g dh) A = (w dh) dF = [D g dh] * [w dh] dF = D g w (dh)^2 I think we can simply rewrite the end as: (dh)^2 = h * dh Therefore...

From what I understand, you don't need to know the sizes of the other walls. Pressure is delta-Force over delta-Area. And for flat surfaces, Pressure ends up simply being P = F/A (no integration needed). Besides, the problem states that it only wants to know the pressure difference of...

This question has been asked several times, but I don't want the answer (I can see the stupid answer...) I need help with my solution... Homework Statement A large aquarium of height 5.00m is filled with fresh water to a depth of 2.00m. One wall of the aquarium consists of thick...

Question: Why does my calculator state the limit of the following function is 1, but my calculations state it does not exist? There's a gap in the graph between -1 and 0. I noticed that there are a few points in that section that do generate real values which appear to move towards 1, but...

What's the process that needs to take place to be able to graph the following equation (not-using a graphing calculator): x2 + 12xy + 36y2 + 2x - 3y - 9 = 0 I know to use the B2 - 4AC formula to identify the equation as being parabolic, though I don't understand the formula... Looking at...

I want to know how one purchases pure elements. Like Caesium or Neon or Chlorine. Do you need license to purchase some elements? Or do you need to be affiliated to some sort of organization? Where is a site that has prices for the elements? Just curious how one gets a hold of these...

GREAT! Everyone here has been a big help with the terms that I'm looking for and the equations =) Ok I think I have it now =) http://www.physics.ubc.ca/outreach/phys420/p420_04/kenneth/theory.htm Based off of that site I think I have my answer =) ^____^...

Ok, so the shape of the liquid matters... Then let's say it's a cube of liquid. Not a gallon. 1 m^3 will be the size of liquid. Now how do I figure the time frame? I want this to freeze in 5 seconds...so how cold would it need to be outside for a cube of water to freeze starting from 25...

---- http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html This states that rougly the specific heat (I think that's what C is) for the range we are working with is roughly 4.182 kJ/(kg*K) 4.182 KJ/(kg*K) So let's resolve for Q1 again... 3.79 L = 1 Gallon m =...

http://www.Newton.dep.anl.gov/askasci/chem00/chem00005.htm Based off of this Density of pure water at room temperature (saying room temperature equals 25 C) is 997.07 g/L 3.79 L = 1 Gallon m = D/V = (997.07 g/L) * (3.79 L) * (1 kg / 1000 g) = 3.78 kg c = 4.182 KJ / Kg DelT = T2 -...

Ok...Let me try solving for kW so we have a number to work with. Then we'll go from there. It'll probably take me a few minutes. I will have a reply before 15 mintues are up from this time. Thank you so far for your help.