Recent content by votingmachine

If you plot the numbers that can be squared to equal a result, you would have at X=1, two Y values, 1 and -1. At X=4 you would have two Y values, 2 and -2. At X=9 you would have two Y values, 3 and -3. At none of those is the positive number equal to the negative number. The "square root" of 4...

Fair enough. I saw the question "can Australia remove fire from their energy mix, if they've eliminated nuclear from the non-fire options?" and did not think it was being answered. I compared their aggregate energy demand to a single energy resource and that single resource appears far in...

I assume the per capita energy use includes industrial and commercial use, which dwarf home use. I googled: australia per capita energy consumption The "answer" in my search results is putatively from the World Bank. (5,483.82 kg of oil equivalent (2015)). Commercial power use HAS to be...

There are alternatives to batteries. I recall a project from the 70's where the TVA was planning to pump water to a reservoir on a mountain top, with energy recovery via hydroelectric generation. The efficiency is quite good, but the energy density is not great. Back of the envelope, say a...

Google says that Australians use about the energy equivalent of about 5500 kg of oil per year. It also tells me that oil is about 12 kWhr per kg. So an Australian needs roughly 66,000 kWhr of energy per year. If you were strictly relying on solar and storage, and make the assumption of 3000...

For the equation x^2-1=0, there are two answers, 1 and -1. You can solve the equation by quadratic formula: x= (0 +/- sqrt(0-4(1)(-1)) / 2(1) =+/-sqrt(4)/2 x= +/-1 or solve by simpler factorization (x+1)(x-1)=0 which is true for: (-1+1)(-1-1) and for (1+1)(1-1) But definitionally, the...

I skimmed after page 1, which was where I thought my comment was a useful suggestion. I see now your comment #45, which is VERY close to how I chose to approach it in a spreadsheet. When I wrote the equation all in terms of height, it looked troublesome to solve. Just tossing the formulas into a...

My method comes up with 11 km. I used a spreadsheet to calculate the densities at every km using your formula. Then to calculate the moles based on: density * 850 m^3 / 29 kg/kmol That is the outside moles in 850 m^3 for that density (at that height) 200 kg / 29 kg/kmol = 6896 mol. So the...

What are the comment numbers that you are referring to? I could look at those. An idle thought that just occurred to me was that you make the statement that the molar mass of air doesn't matter. But consider if the atmosphere was hydrogen. At STP, 850 m^3 of hydrogen would only weigh about 76...

I should have read the 5 pages, and I still have not. But look at your original post and carefully try to determine where the 200 kg load is in your final reduction. The force = 200 kg x g You have it correct that the buoyant force reduces to the difference in densities. But the diffence in...

I have not looked at every reply. I am proposing that in the ideal gas law, you can treat V as a constant ... the 850 m^3. An you know the pressure inside equals the pressure outside. So the ideal gas law with P and V constant reduces to constant x n1 x T1 = constant x n2 x T2 The delta-n is...

At equilibrium the weight of the gas inside the 850 m^3 will be 250 kg less than the weight of 850 m^3 of the outside gas. As you point out there is virtually no change in temperature or density of the outside air with the the altitude change you solved for. The mass of the internal 850 m^3...

I would question whether you are putting "information" into the box. Say it is a single radioactive atom with a long half-life. If you stimulated it's decay, you have information about a previously unknown decay-or-not. The information is not stored in the box. Anyone coming along and seeing...

Sorry to slip across the line. The "trick" is that trial-and-error is a valid and easy approach. It is how I always approach it. It has been too long since I learned to remember how it was taught to me. It is somewhat like successive approximations. You want to start with an informed starting...

I tend to think it is a process rather like finding a common denominator. Usually, you can look at the few fractions and see it. The most rigorous way is to write every equation: (a)NaNO3 + (b)NaOH +(c) Zn -> (d)NH3 + (e)Na2ZnO2 + (f)H2O Then do a balance for each element, ie: Sodium: a + b...