[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 524
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A couple of problems on the basics of STO:
It is known that the volume of water in the ocean is 1.37*10^9 km3. Determine by how much the mass of water in the ocean will increase if the temperature of water increases by Δt=1 °С. Take density ρ of water in the ocean to be 1.03*10^3 kg/m3.
Answer: dm = 6,57 * 10^7 kg.
At what speed of the spacecraft will the mass of food increase by a factor of 2? Will the widow's time of using the food stockpile increase?
Not exactly so: if the base area of the parallelepipeds is the same, then the Archimedes force will be the same, because the Archimedes force is the net result of the surface forces acting on the body from the medium, which has 1. mass, or more exactly the density gradient - (at different levels the density is different) 2. fluidity. If the body stands on a support, the Archimedes force is not directed upwards, since the medium does not act on the bottom - that is only presses down. If the base areas are the same, then the vertical component is also the same and the horizontal component is compensated. For arbitrarily shaped bodies - this will not always be true. Yes, that said, we are assuming that the density of air at the upper boundaries is the same. And, by the way, if the bodies are suspended, that's when the weight will be different and precisely because of Archimedes' force.
Archimedes' law is formulated as follows[1]: a body submerged in a liquid (or gas) is subject to a pushing force equal to the weight of the liquid (or gas) displaced by the body (called Archimedes' force)
F A = ρgV,where ρ is the density of the liquid (gas), g is the acceleration of gravity and V is the volume of the submerged body (or part of the volume of the body below the surface).
So it does not matter what is the surface area of the body. It is the volume that counts.
There are two parallelepipeds. One is made of steel, the other is made of foam. The mass of each one is 1 kg. The base area is the same. Which parallelepiped will exert more pressure on the support?
Since the problem does not say anything about inhomogeneities in the materials of the parallelepipeds, we have to assume that the material is homogeneous in both cases. For equal mass, the foam parallelepiped will have a larger volume, and thus will be more strongly forced upwards by Archimedes' force.
And since the base areas are equal, the foam plastic parallelepiped will exert less pressure on the support, which is calculated as follows:
P=(m*g- ρg V)/S; P is pressure on the support.
From the formula, you can see that the pressure for the foam one is less because it has a larger volume.
In our case the volume of both bodies is completely immersed in liquid (or gas).
But if the test is carried out in vacuum, the pressures on the support in both cases will be equal.
sanyooooook:
Update the page - forgot to insert the link.
it's too late.
And massometers are used to weigh
Actually, it's a dynamometer.)
ZS: I just said that out of the blue.)
Strictly speaking, the highlighted is wrong. If both bodies stand on their supports, then
the correct answer is that the weight of both bodies will be the same. The presence of a side wind is not considered.
Hint: The Archimedean force is a surface force.
VladislavVG, in the formula for Archimedean force, the immersed body has only volume and nothing else. I absolutely don't care how it appears there. But there are no surface properties in the final formula. There is only an indication that the body must be completely immersed in the medium.
I once saw the derivation of an expression for Archimedes' force. Well, yes, it takes the integral of the surface. So what? In mathematics, they already know how to move from the surface integral to the volume one and vice versa (you must know the Ostrogradsky-Gauss theorem by heart, even at 3 a.m.). And the equilibrium is not surface forces, but simply pressures from the environment on elementary areas of the body surface. As far as I know, surface tension forces are usually considered to be surface forces. But we are not talking about them here, it's obvious.
The weights of the bodies will be different because their rest masses are equal, but the Archimedean forces acting on them are not. Point.
Not quite so: if the base area of parallelepipeds is the same, then the Archimedean force will be the same, because the Archimedean force is the result of the surface forces acting on the body from the medium, which has 1. mass, or rather the density gradient - (at different levels density is different) 2. fluidity. If the body stands on a support, then the Archimedean force is not directed upwards, since the medium does not act on the bottom - that is only presses down. If the base areas are the same, then the vertical component is also the same and the horizontal component is compensated. For arbitrarily shaped bodies - this will not always be true. Yes, that said, we are assuming that the air density at the upper boundaries is the same. And, by the way, if the bodies are suspended, that's when the weight will be different and precisely because of Archimedes' force.
Yeah, let's complicate the problem to make it quite strict and for a schoolboy it becomes "unsolvable": air density depends on height above sea level exponentially and at several meters height (for Styrofoam) it has different temperature and, therefore, our bodies are surrounded by gas with variable properties. Let's take into account convection, which will definitely be present there. And, yet, let's screw in the problem statement a little side wind, which you so thoughtlessly neglected (now this is a real hydrodynamics, together with the Navier-Stokes equation). And, finally, let's not forget that, strictly speaking, the gravitational field of the Earth is inhomogeneous. That's it, now we can begin to solve the problem.
In 50 years, when you, probably, will solve this complicated problem (numerically), taking into account all variables, the same Archimedean's law will still be present there (probably, in a form of some horrible integral on the body surface), but with the correction not exceeding 1% due to all these complications. And this correction won't affect the final answer anyway!
What do fluidity and density gradient have to do with it? Why involve all your knowledge of hydrodynamics to explain elementary things known in school? It's a school hydrostatics problem!
P.S. From the article Archimedes' Law:
So, for example, Archimedes' law cannot be applied to a cube that lies on the bottom of a tank, hermetically touching the bottom.
Well, well. Who there is in this problem hermetically attaching metal and foam to the surface?
2 Mathemat & Joo the wording of Archimedean's law you mentioned is for a body floating (or rather submerged to its upper limit) in a liquid/gas medium. Then everything is correct. But if the body is completely in the liquid/gas medium, so that the medium is also above the surface of the body - then the equidistance of all forces will not be numerically equal to the weight of the liquid/gas it displaces (I am not talking about the surface tension forces that will act in the liquid, but about those forces that act from the liquid/gas medium on the body itself - they are also called surface forces). The equilibrium will be equal to the difference in pressure of the medium under and over the body multiplied by the base area - this is for the parallelepiped shape. Look up what zero buoyancy is and how this effect manifests itself. I don't know if it's in the secondary school curriculum: I graduated in physics and mathematics and we went through it, as far as I remember, back in 9th grade - if I'm not mistaken, see Landau's textbook.
By the way, if we assume that in both cases, the scales stand on one level and are accurate enough to weigh the volume of air equal to the difference in volumes of bodies, then, yes: the iron one will be heavier as its upper limit is lower and the pressure above it will be higher. Numerically, the difference in weight of the parallelepipeds will be equal to the weight of the indicated volume of air.
Astronaut Anton Shkaplerov and astronaut Daniel Burbank of the ISS-30 expedition show the 'Janibekov effect'
http://www.federalspace.ru/main.php?id=189
Who can explain this effect in terms of physics?
As far as I remember, this effect is connected with relativistic dynamics, and is simply explained, though such "tricks" look fantastic.
Though, who knows what "spirit of enlightenment" prepares for us, but scientists are not in a hurry to explain impossibility of entering orbit of small bodies by spaceship...