Recent content by whatdoido

ODE can be made to look like this: ##k\, z'(t) = -F_0e^{-z(t)/h}-m\,z''(t)## I lack experience in studying asymptotic behaviour, so this is what I gathered. The time approaches infinity; ##t \to \infty##, so velocity approaches some constant ##v_1## and ##z \to \infty##. The deceleration...

The potential energy field ##U## is as far as I understand $$U = hF_0 \int_0^{z} -\frac{1}{h}e^{-z/h} \,dz = hF_0 (e^{-z/h} - 1)$$ ##((e^{-z/h} - 1)## not ##(1 - e^{-z/h})##, made a mistake earlier##)## To be honest, I am just confused right now. U = 0 when z = 0, but I thought finding escape...

The original text is not in English and my translation was not carefully done. The second part asks: "How does the situation change if the object is also affected by drag force that is proportional to velocity?" Previously I said it was "a velocity dependent drag force" which, yeah I see it...

Me being busy with other work, this problem got neglected and it is still unsolved. Thus returning to get this done, there is a part that bring difficulties. The second part is treated as a energy = work problem, where the change in kinetic energy is ##\Delta E_z = -\frac {1}{2}mv_0^2## Then...

Yes I hopefully understand this better now. It is more fitting to word it as "when object does not return to xy-plane it has to approach infinity" and refrain to mention potential energy. It is very logical, if it would not approach infinity then the object would be returning back. The clause...

Okay, so the answer is infinity meaning that the object can only escape if it reaches for infinity. A limit can be taken \lim_{z_0 \rightarrow \infty} \sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}} = \sqrt {\frac{2hF_0}{m}} I want to understand this clearly: the object still has some potential energy...

I'm understanding that this should be evaluated with potential and kinetic energies At the distance ##r## the object has potential energy ##E_{p1}## E_{p1}= hF_0 (1-e^{-r/h}) This is the limit after the object will not return to the xy-plane. The object returns when its kinetic energy...

Would something like ##z_0 sin(\theta)## suffice? Where ##\theta## is the angle between object's initial trajectory and ##xy##-plane That is true, I just fancied putting it there

Homework Statement [/B] An object of m-mass is to be thrown from xy-plane with an initial velocity ##\mathbf v_0 = v_0\mathbf e_z \, (v_0 > 0)## to a force field ##\mathbf F = -F_0 e^{-z/h}\mathbf e_z\,## , where ##F_0, h > 0## are constants. By what condition does the object return to...

Okay simply with Pythagorean style: ##B_{total}=\sqrt{B_B^2+B_A^2}=\sqrt{(\frac{μ_0 I_1}{√2πx})^2+(\frac{μ_0 I_2}{√2πx})^2}=1.886...*10^{-4} T \approx 0.19mT## Thanks!

Homework Statement In the picture at points A and B are two thin parallel wires, where traveling currents are 15 A and 32 A to opposite directions. The distance between wires is 5.3 cm. Point's P distances from A and B are the same. Calculate the magnetic flux density at point P. Homework...

Now that I thought about it, simply adding the change does not make so much sense. But then I got an idea to take partial derivates since it is about change. Adding those partial derivates together should give the overall change in voltage. ##U=E(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})##...

Right now I'm trying to figure out why this would not be possible: ##E(\frac{R_3+\Delta R_3}{R_2+\Delta R_2+R_3+\Delta R_3}-\frac{R_4+\Delta R_4}{R_1+\Delta R_1+R_4+\Delta R_4})=\Delta U## I can simplify it a bit, but is this the right way to go

Yes that is true, ##\Delta U## is zero before the change of resistances.

Homework Statement Prove the following equation: ## \Delta U=\frac {R_1R_4}{(R_1+R_4)^2}(\frac {\Delta R_1}{R_1}-\frac {\Delta R_2}{R_2}+\frac{\Delta R_3}{R_3}-\frac{\Delta R_4}{R_4})E## This is used in Wheatstone bridge Homework Equations [/B] U=RI The Attempt at a Solution This has...