Recent content by hurreechunder

dG (note: should have used the deltas in the original post) = 0 in the limiting case. I'm trying to find the corresponding value of ln Q = pCO2/pCO dGo = dH-TdS; where for every reactant and product, H = Ho + Cp(T-298K). It's a bit complicated because the Cp itself changes with temperature...

Thanks - I took a second look at the calcs, and it appears that I was dividing the Cp in KJ/Kg by the molar mass instead of multiplying it. With that adjustment, the temperature adjustment of the enthalpy is no longer negligible. I am still getting a number different than 2.3, but at least Ln(Q)...

Some numbers below: these are the standard molar enthalpies and entropies at 298K that I got from a website. In the reaction FeO + CO = Fe + CO2, the deltas are: delta H = (0+ -393) - (-266.5+ -110.5) = -16.3 KJ/mol. delta S = (27.2+213.8) - (197.9+54) = -10.9 J/mol/K Now at 1200 K, Go = -16.3 +...

The point is that with the CoG raised to the same height, the Fobsbury flop can clear a higher bar, which is what wins in high jump. This video from the 1980 Moscow Olympics shows both jumps. The straddle looks a lot harder because the CG needs to go above the bar and hence higher:

Specifically for the case of toppling over someone squatting low vs high: toppling happens when the centre of gravity goes outside the base. Why? Well, as long as it is inside the base, gravity tries to turn it the other way and back to its sitting position. When the CG goes outside the base...

You’re right, it’s possible for the intersection point to be on the falling side as well. I didn’t see that. The “envelope of safety” formula should provide the answer in that case.

Sorry, I meant from 90 deg to 45 deg, not from 0 deg to 45 deg. Hope that clarifies

It's actually trickier if the distance to the base is <183 m. In that case, the max height does not correspond to the 45 deg projectile path and a different approach has to be used.

The max range for 45 deg launch is 366 metres. The halfway point is thus 183 m. Therefore, the projectile will hit the hill on its way down rather than up. Up to 45 deg, all projectile paths intersect the 45 deg trajectory behind its point of max height (or its halfway point along the x-axis)...

Perhaps it would help for the OP to specify what is a 'given' in the problem. If theta1, theta2, and theta3 are givens, there is no need to use torque balance as you have two equations and two unknowns. If one or all of them is not a given, then there is a need to know the location of B. In that...

So you have two unknowns and two equations along the rule and perpendicular to the rule for the forces. The torque balance will be a redundancy

Thanks, Borek. Makes sense.

I’d see it thus - with salt, at the same temperature there are more configurations of water + salt molecules (more disordered, more entropy) where the salt “gets in the way” and prevents the formation of ice. So the water needs to cool down even further to have the same probability of h2o...

Thanks for the link. My question is precisely around that: in LiC6, the Li exists as an Li ion. It's not as if it's forming a covalent bond with the carbon. The electron in the outer shell of the Li atom is delocalised in the graphite, so it's actually Li+ and C6- forming an ionic bond. So on...

The centre of mass is at L/2 and (yo/2) respectively