Recent content by Rugile

Thank you for the help!

Sooo, the pressure of the gas is caused by the atmospheric pressure and mercury, thus ##p_0 = p_{merc} + p_a## instead? I'm confused

So do you mean that ##p_a = p_{merc} + p_0##? But what about the fact that the bottom end is sealed?

Homework Statement We have a tube with the top end open and the bottom end closed. There is some oxygen gas in the tube and on top of the oxygen there is 10 cm high column of mercury. The initial temperature is 20oC. Then the tube is flipped over and heated to 40oC. The column of mercury shifts...

Okay, so here is the final solution: the change in momentum (in the interval 0+ to ∞) will be from mv to (m+M)V, thus $$V = \frac{mv}{m+M}$$. Change in kinetic energy thus will be $$\Delta E = \frac{(m+M)V^2}{2} - \frac{mv^2}{2} = \frac{m^2 v^2}{2(M+m)}- \frac{mv^2}{2} = \frac{-Mmv^2}{2}$$...

But how can we apply the balance if we know nothing about the impulse? Initial momentum will be (m+M)v I suppose, so is the final momentum?

Sorry for my late reply, yes, I'm still here :) And thank you for all your replies. Minus work done on the gas is the same as plus work done by gas, right? Zero? Gas in one side expands as much as the other one contracts (don't know the right word for opposite of expand ? So $$ \Delta U =...

:) okay, so if there is no heat exchange, it means that the internal energy change of the gas will be equal to the work done by gas. But it is still quite complicated, since both volume and pressure may change (referring to work done by gas) and we do not know anything about those..

I'm not sure what is IG volume - I'm guessing, insulated gas volume? Adiabatic processes you mean?

Homework Statement A thin-walled cylinder of mass m is filled with monatomic ideal gas and is lying on a horizontal frictionless surface. The initial internal energy of gas is U = cT. There is a thin insulating piston in the middle of the cylinder of mass M. The piston and the cylinder are...

Homework Statement A voltage source is connected to a series LC circuit. The frequency of the source is resonant. The voltage amplitude of capacitor is 1V. Find the average power in the circuit. Homework EquationsThe Attempt at a Solution I realize that if there is no active resistance the...

Accidentally posted the post above before finishing it, and for some reason can't edit it, so I'll finish here: |R- j\frac{1}{\omega C}| = | R - j(\frac{-1}{\omega C} + \frac{\omega L}{1 - \omega^2 L C} | => \sqrt{R^2 + (\frac{1}{\omega C})^2 } = \sqrt{ R^2 + ( \frac{\omega L}{1 - \omega^2 LC}...

Got it... |R- j\frac{1}{\omega C}| = | R - j(\frac{-1}{\omega C} + \frac{\omega L}{1 - \omega^2 L C} | => \sqrt{R^2 + (\frac{1}{\omega C})^2 } = \sqrt{ R^2 + ( \frac{\omega L}{1 - \omega^2 LC} - \frac{1}{\omega C} )^2 } ## ##

The problem is not from a textbook - I don't have the answer (I didn't make it up myself, either ) - and it's under a more broad topic - electromagnetism. I just though that resonance is probably the key to the problem. And yes, the problem statement is correct in the first post.

Okay, so Zlc = -2Zrc ( if we say that Zlc is negative and the modulus of difference of Zrc and Zlc has to be equal to Zrc, Zlc has to be twice as big). Then after rearranging we get: L=\frac{2(i-R\omega C)}{3i \omega^2 C}. What do we do with the i? Should it be rearranged to L=a + ib, and L...