Recent content by AnotherParadox

I like this reasoning Here's some work I thought up with considering this more {T}'+n^2T=0 \frac{{dT}}{dt}+n^2T=0 \frac{{dT}}{dt}=-n^2T \frac{\frac{{dT}}{dt}}{T}=-n^2 \int \left ( \frac{dT}{T} \right )=\int-n^2dt from there it's ln -> exp Let me know if this checks out I agree. I wasn't...

Also the term for 1/(2k+1)^2, right?

Thanks for the response Here's my thought process It can't take 0 since sin(0*pi/2) and sin(0*x) is 0 and that leads to the trivial solution. It can take 1 since sin(1*pi/2) does not equal 0 it equals 1 It can't take 2 since sin(2*pi/2) = 0 and the trivial solution It can take 3 since...

Solve the boundary value problem Given u_{t}=u_{xx} u(0, t) = u(\pi ,t)=0 u(x, 0) = f(x) f(x)=\left\{\begin{matrix} x; 0 < x < \frac{\pi}{2}\\ \pi-x; \frac{\pi}{2} < x < \pi \end{matrix}\right. L is π - 0=π λ = α2 since 0 and -α lead to trivial solutions Let u = XT X{T}'={X}''T...

Given ln(ab) = b⋅ln(a) Then ln(1x) = x⋅ln(1) Also ln(2x) = x⋅ln(2) Say ln(2x) = ln(1x) Then Also x⋅ln(2) = x⋅ln(1) But, dividing both sides by x ln(2) ≠ ln(1) Similarly, x⋅ln(2) = x⋅ln(1) Dividing both sides by x and ln(2) 1 ≠ 0 But we know x = 0 as per the original statement...

Oh I see, the energy from the collision goes towards deforming, embedding, and the heat associated with that. Any insight on how to solve this problem with momentum conservation and not energy conservation (if possible?) or how the solution manual author came up with their energy balance equation?

This took a lot of time and effort and I understand if you wish to skip past everything and just read my questions about it in the The too long didn't read summary (TL;DR) at the bottom. Homework Statement The 10-g bullet having a velocity of v = 750 m/s is fired into the edge of the 6-kg...

Ah yea I used the wrong equation in my original post. However even with the other one I couldn't come up with any results. Here's the other one τx'y' = - ((σx-σy)/2)*sin(2θ) + τxy*cos(2θ)

If a plane is at some angle and the stresses are transformed such that one is normal and one is parallel with the plane then wouldn't the vector that is parallel be the shear component and thus not be 0 ? I'm definitively missing something here

If I understand correctly the definition of a principal direction of stress is the stress vector at which the transformed normal stress vector is maximized. According to the book If I substitute the relations for ThetaP1 or ThetaP2, which are 90deg apart the shear stress will equal 0 and thus...

Homework Statement "In a component under multi-axial state of stress, the ratio of shear stress to normal stress along principle places is _____. A) 0.0 B) 0.5 C) 1.0 D) 1.5 E)2.0" Homework Equations σx' = (σx+σy)/2 + ((σx-σy)/2)*cos(2θ) + τxy*sin(2θ) σy' = (σx+σy)/2 - ((σx-σy)/2)*cos(2θ)...

On first thought, this would only be true if it were the only bearing on the rod. If you put two bearing mountings that are self aligning on the same rod (such as on the example problem) then self aligning wouldn't be able to happen since the two mountings are fixed and pointing at each other by...

Ok, so this is an undergraduate level question probably but consider this R. C. Hibbeler's Textbook on Mechanics of Materials says this about thrust bearings and smooth journal Bearings : In other words, it says nothing about reaction moments. Now this chart has other 2D connection types...

Photographs are taken by cameras. "Definition: Camera - a device for recording visual images." "images, created by using information from instruments that gather spectral data that has been processed by computers [utilizing] various software" All digital cameras work in this manner in some way...

Oh man was I wrong. (and so were the online solution help sites) I can determine s2 from the specific volume by looking at the table and finding where the specific volume vg matches v1. Pressure is increasing several times and temperature by around 15% (abs scale)