Recent content by Rain10399

Yes, now I managed to solve the problem! Thank you so much to everyone who responded ^^. I love this forum.

Ohhh. In the book it said balloons so I thought they actually were balloons. Thank you! So because there's no heat exchanged and no work done the internal energy remains the same? I think I understand this part now:D Then U1=UA+UB and (nA+nB)CvT1=nACvTA+nBCvTB, like Chestermiller said Thank...

nA=PAV/RTA nB=PBV/RTB nC=PCV/RTC first law energy balance: U1=-W1 But I wasn't sure what formula to use for W here, so I searched on the internet for one. nACvT1 + nBCvT1=-K[(2V)1-γ-(V)1-γ]/1-γ I will assume that the balloons are filled with helium, which is a monatomic gas, so: (γ = 1.66)...

It means that the energy is transferred to the surroundings only as work, and the system doesn't transfer Q with its surroundings. U = nCvT and U = -W (because Q=0). U1=(nA+nB)CvT= nACvT + nBCvT= UA + UB

Homework Statement An adiabatic isolated system is formed of three identical balloons (of unknown volume). The balloons are joined by tubes of negligible volume. Each tube has a faucet/tap that is initially closed. The balloons have different quantities of the same ideal gas. After opening...

Oh, yes, you're right. Thank you for helping me with this problem!

So can h be the difference between A and D (or E. D and E are supposed to be at the same height, but I'm not sure if that can be seen on the drawing)? Then h = sin α * AE and AE = cos θ * l. So mg * sin α * cos θ * l = mv2/2, and from there I can express v in terms of θ. However, I had to make...

Oh! Then what is the relevant height? At the start there's no kinetic energy, only potential energy, so I still need to express the h in mgh.

The relevant height is from the base of the plane to the position of A or to m's initial position (which have the same height)

I'm trying to find the height at which is m in terms of angle θ, so that I can replace h with the angle in formula mgh = mv2/2. I don't know how else to solve the problem.

So I found h = sin α * AC, and now I'm trying to find AC in terms of angle θ.

I laid down the coordinate system on the surface of the plan at the origin A, but I didn't manage to express the first position of m in terms of the angle θ. I drew a straight line between the first position of m and B, but the triangle with the angle θ in it is not right, so I'm not sure how to...

Homework Statement Find v(θ), the dependence of the speed of the object on the angle θ. The string is ideal and there's no friction. Homework Equations - The Attempt at a Solution I'm quite sure I should use the conservation of energy mgh = mv2/2, but I don't know how to express h with the...