[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 190

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Weight and mass are related by a constant ratio in resting frames of reference
In this case we need to know the volume of the weight, which is the main difficulty of the task.
We measure the diameter, we can also measure the circumference.
Finding the length of the spring requires only one weight.
In physics problems it is possible and necessary to make approximate calculations.
In many cases calculations within an order of magnitude can be considered exact
ну тогда какая плотность гири, ответте уже?
7.4 g/cm3Yes
Ais >> В физических задачах можно и нужно производить приближенные вычисления
Right. The formulation of any problem, even an Olympiad problem for an 8th grader, implies reasonable reticences. If the problem statement explicitly says "given a piece of paper in a square", but does not explicitly say that the size of the square is not necessarily 5 mm, then it is 5 mm, as it is used to it. Just measured it - very accurately (although I have encountered a cage not 5mm but, say, 4.8).
Hence the caged sheet can be used as a length gauge. What is the point of a clock, then, if we can already measure lengths with paper - I don't understand.
That's the point!
The fact that the caged sheets were given out could have distorted the problem, either intentionally or accidentally.
There's no need for paper or squares to solve this problem at all.
Richie, did you study physics at school - or did you take a walk? How is it that a pendulum doesn't care about weight? Do you remember the formula for the period of vibration of a pendulum?
It's elementary. There is the "mass of the oscillating weight", there is the "mass of the resting weight", we calculate the volume of the weight. How? I won't tell you.
You're on your own, physicists.)
Элементарно. Есть "масса колеблющейся гири", есть "масса покоящейся гири", есть часы, вычисляем объём гири. Как? Не скажу.
Мучайтесь сами, физики :)
Uh-oh, he invented a wire without a magnetic field. And he probably made fun of the students with this wire ;)))The cages were never 5mm because they were printed on Soviet equipment where the plates were made by eye and there was a lot of sheet slippage. And the printers were getting drunk at work. They're more accurate now than in '83.
In '83 there were different cages too, there were cages for the visually impaired, but I don't remember the size, I remember they were bigger.
Клетки всегда не 5 мм, потому что печатались на совеццком оборудовании где и пластины на глазок делали и проскальзывание листа было большим. И печатники квасили на работе. Сейчас уже поточнее печатают чем в 83-м.
gip, we are not talking about reality, but about how to understand the conditions of an olympiad problem, in which idealizations and reasonable defaults cannot be avoided anyway. The condition of the problem posed by Ais is understood exactly as the size of the cell is exactly 5 mm.