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2

One should always specify whether one is talking about rest mass per unit rest frame volume, $\rho_0 = m_0/V_0$, rest mass per unit observer-frame volume, $D = m_0/(V_0/\gamma) = \gamma\rho_0$, or relativistic mass per unit observer-frame volume, $(\gamma m_0)/(V_0/\gamma) = \gamma^2\rho_0$.1 (I can't imagine the fourth case, relativistic mass per unit rest ...


0

The best case scenarion would be zero density. Even so, for the boat to float, it has to have a surface large anough to displace 80kg of water, otherwise the buoyancy will be never enough. So, as far as the density of the boat is smaller than that of the water, the variab;e has to be the size of the boat (or the shape), such that it can displace a volume of ...


-1

should be able to realize that it is necessary to keep the center of gravity as close to the midline of the ship as possible. Once the center of gravity is beyond the deck of the ship, it will tip over (just as the towers tipped over once their centers of gravity got beyond their bases).


2

So I am guessing you have a a 2 dimensional density? say, kg/m^2? In this case, you will need to get the area of your box -> say it is 2m^2. The mass of the box is 2 * 0.5 = 1 kg. But the mass of your person is 80kg. So your new mass is 81kg. This means the new density will be 81/2 = 40.5kg/m^2. The item will only float if the item has a lower density ...


0

Escape velocity is only applicable for non-powered projectiles. A powered vehicle can leave another body at any speed, it only needs to provide a greater force than gravity. To leave a body near to a black hole, you need only sum all the available forces, and make sure the spacecraft can provide a greater force, in which case it can leave the system at any ...


0

If the black hole creates those huge tidal waves, then the black hole's gravity must have an impact on the planet and therefor it would be easier to escape the planet, if you were on the side of the planet that faces the black hole, wich the tidal wave indicates. If the people were on the other side of the planet the effect would be reverse, minus the fact ...


1

If we take neutron star material at say a density of $\sim 10^{17}$ kg/m$^{3}$ the neutrons have an internal kinetic energy density of $3 \times 10^{32}$ J/m$^{3}$. So even in a teaspoonful (say 5ml), there is $1.5\times10^{27}$ J of kinetic energy (more than the Sun emits in a second, or a billion or so atom bombs) and this will be released instantaneously. ...


-4

ice is a crystal pure made up of water after frozen on a certain temperature,as it is lighter than water hence its density is less than water............this can be taken as in simple words


2

Assuming you are talking about exoplanets, I'll offer this. To obtain a density you need a mass and radius. Masses come via two methods - either measuring the radial velocity variations of the star it orbits (the bigger the RV variations, the bigger the planet mass), or so-called transit timing variations. This latter works in multiple "transiting planet" ...


5

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 ...


6

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 ...


8

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 $130\%$ of earth's gravitational acceleration on the surface. So we have \begin{equation} g_W = 1.3 g_E \end{equation} This is a given and not to be violated. And in fact ...


1

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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