Recent content by anlon

Blood behaves as a fairly viscous fluid, so it is reasonable to think that it would lose energy due to frictional effects as it flows. Additionally, you're right in that fluid cannot flow against an adverse pressure gradient - not for long, anyway. Consider this: if blood were to flow based...

What kind of (specific) physics/mathematical background do you have? It depends on where you're starting from. For intermediate physics, I learned using "Classical Dynamics of Particles and Systems" by Thornton and Marion. For electricity and magnetism, "Introduction to Electrodynamics," by...

Ah, so it would be ##(a_+)^\dagger = \frac{1}{\sqrt{2 \hbar m \omega}}(ip + m \omega x)## Thanks, I'll need it.

Just doing some studying before my final exam later today. I think I've got this question right but wanted to make sure since the problem is from the international edition of my textbook, so I can't find the solutions for that edition online. Homework Statement The Hermitian conjugate (or...

##\LaTeX## is a formatting system for writing equations and papers that look nice. For example, instead of writing: Ym=2ym.cos(π/10.0) therefore Ym/ym=2 cos(π/10.0)=1.902 You can write $$Y_m = 2y_m \cos{\left(\frac{\pi}{10}\right)} \Rightarrow \frac{Y_m}{y_m} = 2...

Boneh3ad is correct. The explanation I gave is actually only applicable for inviscid, incompressible flows, a fact that I overlooked when replying. When viscous effects are accounted for, surface roughness does indeed have an effect.

Yes, this is correct. You may want to use LaTeX in the future to make your work more legible, but good work.

As it turns out, solving it this way ends up giving the equation $$\frac{\hbar^2 k}{m\alpha} = 1-e^{-2ka}$$ which only has a solution when ##\alpha > \frac{\hbar^2}{2am}##. Thanks for the help! Also, an error I made in the first post: ##u = rR##, not ##u = \frac{R}{r}##. The work is all...

I would advise you not to choose your major based on what is intuitive to you or not. Choose it based on what you would like to do for a career. It's much easier to justify the long sleepless nights you'll go through studying engineering when you're working towards a goal than when you're taking...

Thanks, I'll try solving it that way tomorrow and report back my results.

The general solution is ##Ae^{-kr}+Be^{kr}##. So now I'm confused. I've always taken statements of solutions to these differential equations at face value, never bothering to work them out myself, but it seems like the solution is exponential for positive ##k## and trigonometric for negative...

The differential equation is the same, but there are two different kinds of solutions: ##A\sin{(kx)} + B\cos{(kx)}## for solutions where you want to have a "node" (wave function equal to zero) at a specific point, i.e. a place where potential is infinite, and ##Ae^{-kx} + Be^{kx}## for solutions...

The velocity multiplied by 1 second is your distance. The math is up to you.

$$L = \rho_\infty V_\infty \Gamma$$ Where ##\rho_\infty## and ##V_\infty## are the freestream density and velocity, respectively, and ##\Gamma## is the vortex strength over the cylinder: $$\Gamma = -\oint_C \textbf{V} \cdot \textbf{ds}$$See Anderson, Fundamentals of Aerodynamics, Sixth Edition...

Are you accelerating the bullet with force or power? They're two different things, and only one of them causes acceleration. Doing work on the bullet over time equivalent to 100000 horsepower will put energy into the bullet; if all of this energy becomes kinetic energy, you can integrate the...