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In keeping with Archimedes principle: The Buoyant Force = The upthrust in a fluid.Thus treating air as the fluid which it is, $\rho$ $V g$ = ($M + m$) $g$ Where $m$ is the mass of the cargo and $M$ is the mass of the air balloon itself. $m$ = $V$ $m^3$ ($1.2 - 0.95$) $kg$ /$m^3$ This is because the net force is equal to the difference of the mass of ...


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Basically, it has to do with the density of the material as a function of temperature. The density of iron increases as it cools, that is, solid iron is more 'packed tight' than when it is melted. This is understandable, since the kinetic energy of the iron atoms decreases as the temperature drops (ie: the average velocity of the atoms decreases), allowing ...


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Solid water (ice) is one of the few known solids whose density is lower than that of its liquid form. Yes water is special! (but very much so in its chemical properties too) Due to the crystal structure of the solid phase of water, the molecules arrange themselves in a rigid, ordered fashion and end up being, on average, farther apart from each other (than ...


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Ok, trying my luck with a physics answer. Let's first look at the boundary conditions given in the movie, since we're particularly talking about that here: The water planet is said to have 120% of earth's gravitational acceleration on the surface. So we have \begin{equation} g_W = 1.2 g_E \end{equation} This is a given and not to be violated. And in fact it ...


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Yes, it would be easier. . . . But only if the planet was similar in size to Earth. The escape velocity depends on the mass of the body. For a sphere, it's $$v=\sqrt{\frac{2GM}{r}}=\sqrt{\frac{2G}{r}} \sqrt{M}$$ Earth has a mean density of roughly 5.514 grams per cubic centimeter; liquid water has a density of roughly 1 gram per cubic centimeter, or ...


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A degree of closeness is called density, and number of field lines passing thorough any closed surface is one method to find density...


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From the Archimedes principal we know that the object will displace the water of same mass as it. So the object will displace 500kg water and 500kg water = 0.5m3 water. We also know that the lost weight of an object = weight of water displaced by that object. The object does not lose any weight. It is pushing down with its weight. The waters is ...


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In this case you show that the net force acting on the object is zero in this situation, that is, half of the object is within the water and another half on the water. This position of the object is thus equilibrium position. If it would come totally on the water surface or within the water then the equilibrium will be lost, i.e. it comes to in-equilibrium ...


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$ \newcommand{\r}{\mathbf{r}} \newcommand{\F}{\mathbf{F}} \newcommand{\g}{\mathbf{g}} \newcommand{\t}{\boldsymbol\tau} $Let me start by defining $\g(\r)$ to be a position dependent force per unit mass. Then the force per unit volume $d\F$ is given by $d\F(\r) = \g(\r) \rho (\r) dV$. The torque per unit volume $d\t$ is given by $d\t = \r \times d\F$. The ...



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