Recent content by John004

Homework Statement A vector field $\vec F$ is defined in cylindrical polar coordinates $\rho , \theta , z$ by $\vec F = F_0(\frac{xcos (\lambda z)}{a}\hat i \ + \frac{ycos(\lambda z)}{a}\hat j \ + sin(\lambda z)\hat k) \ \equiv \frac{F_0 \rho}{a}cos(\lambda z)\hat \rho \ + F_0sin(\lambda...

Is there some resource you can point me to so I can learn how to type out symbols and equations neatly like you just did? I can't really picture it in the way you're asking me too. What if I substitute z-hat = r-hat cosθ - θ-hat sinθ?

Homework Statement The angular velocity vector of a rigid object rotating about the z-axis is given by ω = ω z-hat. At any point in the rotating object, the linear velocity vector is given by v = ω X r, where r is the position vector to that point. a.) Assuming that ω is constant, evaluate v...

well if I plugged that in for r, wouldn't I just get E = - (∂[p*rer/4πε0r3]/∂r)er = (p*er)/2πε0r3er ? I haven't done vector calculus in a long time, idk if I am forgetting something obvious or what

Homework Statement The problem statement is in the attachment Homework Equations E[/B] = -∇φ ∇ = (∂φ/∂r)er The Attempt at a Solution I am confused about how to do the derivative apparently because the way I do it gives E = - (∂[p*r/4πε0r3]/∂r)er = 3*(p*r)/4πε0r4er

Ah ok, good. I thought I was going crazy for a second there.

Homework Statement So in the attachment you'll see a picture taken from a linear algebra book where a linear system of equations is presented in the equivalent augmented matrix form. I'm confused about the representation of the first equation in the augmented matrix. What happened to the...

m1<u1x, 0> = m1<v1x, v1y> + m2<v2x, v2y> <m1u1x, 0> = <m1v1x, m1v1y> + <m2v2x, m2v2y> <m1u1x, 0> = < m1v1x + m2v2x, m1v1y + m2v2y> therefore, m1v1x + m2v2x = m1v0 m1v1y = -m2v2y 3v1cos(φ) + ½v2 = 3v0 v2 = -(6/√3)v1sin(φ) still got the same thing for this one

Ok so this is what I've done so far m1<u1x,u1y> = m1<v1x, v1y> + m2<v2x, v2y> where u1 is the velocity of mass 1 before the collision, v1 is the velocity of mass 1 afterwards, and similarly for mass 2. I end up with (√3)/6 v2 + v1cos(φ) = v0 where v0 = u1x these are the magnitudes of the...

Oh right! its there now.

Homework Statement A super cue ball is made of the same material with the target ball (radius r) but slightly larger: rc = (3)^(1/3)*r The cue ball collides with the target ball on a frictionless table, as shown below, with initial speed of v0. The collision is not head-on, as shown below. a)...

Not sure if you worked out the whole problem, but if you did, could you take a look at what I got and let me know if it's ok?

It is a constant with respect to m isn't it?

I just noticed that you can do the following also Fext dt = u dm + m dv, since dm/dt = -α -mg dt = -uαdt + m dv -g dt + (u/m) α dt = dv (-g + (u/m)α)dt = dv. Then using the same relation from before (-g + (u/m)α) (-dm/α) = dv (g/α)(m-m0) + u ln(m0/m) = v, then using m - m0 = -αt -gt + u ln...

Ah ah ok, i see. anyways, i was still using the equation without knowing why it was right. So it is valid because "u" is measured relative to the rocket?