exchemist

OK, so you maintain they did know π, contrary to what most Egyptologists say. That's fine, I just wasn't clear what you were saying earlier.

Looks OK to me. Just remember the -1 superscript means "per" whatever the unit is. You may find if you rewrite it as a "/" it is easier to see what cancels. And try not to just follow the formula you have been given mechanically, but think what you are trying to do, i.e. the number of moles in a certain volume of solution and then then the number of grams that corresponds to.

I agree it seems a strange* question in some ways but I found this which seems to help: https://www.thoughtco.com/definition-of-conversion-factor-604954 So it says a conversion factor is between different units for the same quantity. In the case of the mole, it is a unit for numbers of particles, expressing them in more convenient form for use in chemistry, so I think I would agree Avogadro's number can be considered a conversion factor. But grams to moles?........ *actually damned silly , in my view. What's the point of tripping students up with pedantry like this, that has no impact on their understanding of physical science? Just puts students off, in my view. But don't tell anyone I said that.

Again you refuse to address the simple case of you jumping in the air. Until you do so, there is no point whatsoever in moving on to more complicated systems that are more challenging to analyse. I suspect @swansont is on the money when he refers to the Gish gallop. You did the same thing on the previous thread. Whenever we got to a position in which you had no option but to understand what we were saying, you switched the subject by introducing a new aspect, to prevent us bottoming it out properly. I don't know whether you do this deliberately, as a form of trolling to wind us all up here for your amusement, or whether it is a psychologically motivated thing you do without being aware of it, because you can't bear to face the fact that your ideas are wrong. You've been at this for around a decade now so I would not be surprised if it is the latter. If you want, you can go your grave believing your have proved Newton wrong, but guess what will be your epitaph. So can we please sort out what happens when you jump in the air, then we can apply that to the vibrating ring and then we can see what is making it move on the table? In that order. Otherwise, I'll assume you are not interested in analysing your experiments properly.

mol/litre x g/mol DOES cancel. And gives you an answer in g/litre. Which actually is OK since you've been asked to work this out for a litre, i.e. on a per litre basis. But in fact, to be pedantic, I asked you to consider how many moles there would be in one litre, the answer to which is "0.025 moles", not "0.025mol/l". So you can plug that into your equation, can't you? You've nearly got this, so don't fall at the last fence. Have confidence!

No dice. I am still trying to get you to be clear about what happens when you jump in the air. So far, you are not managing even that.

OK, so does that mean you recognise your body, when you jump into the air, is not an isolated system? And that, when a system is not isolated, internal forces can cause it to move?

Because, unlike you, we like to get one thing at a time straight, instead of continually lobbing in yet more complicating factors in order to obscure the analysis of the problem.

How can the centre of gravity of your body move, just by the action of internal forces? Haven't you spent the last fortnight telling us Newton's laws say that can't happen?

But it's not an isolated system, that's the point.

When you jump in the air, you use your muscles. There are connected only to your bones, i.e. they act purely internally. How do you account for your ability to jump in the air?

This is now so garbled as to be impossible to tease apart and correct. Almost every word is wrong or meaningless in the context in which it appears. Are you now really now claiming that Newton's laws say the ring can't vibrate, in spite of all the previous explanations?

You are wrong in asserting that " internal forces may not be transferred outside the ring", if the ring is in contact with something external. If you jump up and down on the spot, your muscles exert purely internal forces on the bones of your legs, causing them to extend. If you were not in contact with the ground, nothing would happen. But you are in contact with the ground, so as your legs try to extend they press on the ground and the reactive force from the ground pushes you up. We have already explained why the AC current makes the ring vibrate. What is a vibration? It is a repetitive up and down motion. So exactly analogous to your legs when you jump on the spot. The ring is in contact with the table so every downward motion of the vibration makes the ring jump up in reaction. The motion then changes to an upward one, but the ring is how higher off the table than it was before, so it is free of the table for an instant. There is thus a point in each vibration cycle at which the ring has jumped off the table and is free to move in response to any tiny force. We can get onto that in a minute, but first, do you now understand how a vibrating ring in contact with a table experiences varying forces from it? If you still can't understand this point, we can't move on to the rest of it.

So, let me understand this. The Egyptians, so you say, did not know π, but the pyramids builders who, you say, were Egyptians, did. How does that work? Do you mean that while most of Egyptian society didn't, this hypothetical priestly caste of yours did? Also, you'll have to explain to me why the cutting of the pyramid core into steps precludes them making it 280 cubits high by design.

OK. So, on the right hand side, you know the molar mass. The other thing you need to know is the number of moles of substance, right? How many moles would there be in one litre of the solution, at that concentration?

To put it another way, we know @John2020's (a) is not correct.............because of Newton's laws. @John2020 does not seem to realise they are so well established that the first thing we do in analysing a mechanical system is to apply them and see what they tell us. So the logic is that the ring starts to move, ergo there is an external force responsible. It is then just a matter of identifying how that force arises.

I'm not quite sure what this is about but regarding the effect on g of the Earth's rotation, this once came up elsewhere and I worked out that even at the equator, the reduction in apparent g due to the Earth's rotation was only about 0.3%. It went as follows: - The radius of the Earth is ~4000miles, so its circumference, say at the equator is 2 x π x 4000 which comes to ~25,000miles. - The tangential speed at the equator is therefore 25000miles/day, which is close to 1000mph. - The centripetal acceleration needed to keep an object moving with tangential velocity v, in a circle of radius r, is v²/r. So, for an object at the equator, the apparent centrifugal acceleration it experiences, counter to the acceleration of gravity, will be 1,000,000/4000 = 250 miles/hr². What we need to know is how this acceleration compares with g, the acceleration due to gravity, which is about 10m/sec². To do that, we need to get this result into the same units as g is quoted in:- 1 mile is ~1600m. And 1hr is 3600 seconds. So 250 miles/hr² becomes 250 x 1600/(3600)² = 25 x 16/(360 x 36) = 25 x 4/(360 x 9) = 100/3240 = ~ 0.03 m/sec². So the centrifugal force at the equator, as a proportion of the force of gravity, is of the order of 0.03/10 =0.003, or 0.3%. At higher latitudes it will be less, presumably by a factor of the cosine of the latitude, falling to zero at the poles.

You can never get more energy out of a device than it consumes. Most attempts at doing so involve misunderstandings about...... things like gears on a car. In a gearbox, you can choose what combination of torque and rotational speed you want. If you double the speed you halve the torque. But power is torque x speed, so the power output stays unchanged. The same is true of any system that uses a lower speed element to drive a higher speed one. Nothing will multiply the energy for you, I'm afraid. In your proposed device, you seem to have an electrical power input from these magnets. The output you get will be no greater than the electrical power input, regardless of how you arrange fans of different sizes etc. Energy is conserved, i.e. it cannot be created or destroyed. This seems to be a fundamental principle of the universe. (Back at the beginning of the last century a very clever woman called Emmy Noether proved this must be, so if the laws of physics don't change with time. Or so I understand - I'm not a mathematician.)

I'm not sure I follow. Are you saying the Egyptians didn't know π but the builders of the pyramid (I presume you mean the designers) did? Are you then claiming the builders or designers were not Egyptian, but brought in from somewhere else?

Yes, exactly.

The surface area of a sphere is 4πr². If the radius of Kepler 22b is 2.4 x that of Earth, that would be ~9600 miles. So the surface area will be 4 x π x 9.6² x 10⁶ sq.miles ~1.2 x 10⁹ sq. miles if my arithmetic is right. Or in "ordinary" notation that is 1,200,000,000 sq. miles. So you can divide that between land area and ocean area as you wish. If you want ocean mass, however, as opposed to area, you have to calculate the volume of water, which means you need to say how deep you think the ocean is on average. You can then use the formula for the volume of a sphere 4/3 πr³. Since the water is a spherical shell around the planet it will be the volume at full radius, (call it R) minus the volume at the radius at the ocean floor (call it r) : i.e. 4/3 π(R³ - r³). You can then apply the density figure you want to use to get the mass. That would be for a planet whose surface is all ocean , so then you would pro-rate for the proportion that you want to be ocean.

That's interesting. However that would have been written about a thousand years after the pyramid was built, to judge by the most recent estimates of its age.

As it happens, only today I was reading a very simple explanation for why the base perimeter of the Great Pyramid divided by its height is very close to 2π. It seems the Egyptians may have used a trundle wheel, one cubit in diameter, to measure out the base at 280 revolutions from one corner to the diagonally opposite corner, and then built it 280 cubits high. So they automatically got a ratio involving π, without anyone at the time needing to know anything about it. I thought this was rather neat.

Yes the banned list is very entertaining, isn't it? Some of the entries seem to have been crafted with some care for humorous effect.

This seems to be incoherent rubbish. I'm off to bed.