New answers tagged

1

Prahar is correct, but here are two more things to note. If spacetime is an $n$-dimensional Lorentzian manifold $(M,g)$, let $\{E_1,\dotsc, E_n\}$ be an orthonormal frame, i.e. $E_i\in\Gamma(TM), T_pM=\mathrm{span}\,\{E_i\lvert_p\}$, and $g(E_i,E_j)=\eta_{ij}$, where $\eta=\mathrm{diag}\,(-1,1,\dotsc, 1)$ is the Minkowski matrix. Then, if $M$ is orientable ...


0

It's all density. Gravity should start to immediately pull it down. The density of the cloud nor atmosphere can hold it up. More than up draft would be needed


0

Have a look at the answers to Pressure and altitude as they explain how the pressure:altitude equation is derived. There is nothing wrong with our working, but you have assumed that the temperature is constant and in reality the temperature falls with altitude (in the troposphere at least). That means the pressure falls more rapidly with height than your ...


-1

The coordinate invariant volume element/measure on a manifold with metric $g$ is $$ d^d x \sqrt{|g|} $$ By coordinate invariance, I mean that if I choose to work in a different coordinate system $x'$, then both the metric determinant changes as does the measure $d^dx$. But they change in a way so as to cancel each other out. In other words $$ d^d x' ...


0

Density here is a bit misleading. For example density of galaxy cluster is low because there is so much space between it. It does not mean all the material inside has low density.. We cannot observe beyond Event horizon, so we use diameter of event horizon to measure volume. Actual matter will be in a smaller volume. To understand why large objects need ...


0

This density is not possible that's why earth cannot become Black Hole. Larger masses need less density to convert into black hole. For example super massive black holes in the centre of glaxy need density of water to become black holes !!! link However stellar black hole have much less mass, thus very high density. Even removing all empty space in inside ...


0

Simplify the equation to make it more convenient to you by using the definition of a mole. A mole , generally denoted by 'n' is the mass of the substance taken divided my its molecular weight. On solving the only unknown in this equation , you get the mass of air contained in the volume you obtained. Now , you can just plug in this value into the definition ...


0

The ideal gas law says you something about $p$, $V$ and $T$ in terms of the number of particles. That's nice, because it holds approximately for all gases. For a given $p$ and $T$ you know the number per volume - it is independent of the type of gas! Then why is the density of gases different? - because the mass per particle is different. So now you have to ...


0

BP = 101.325 kPa STP = 101.325 kPa, 0°C NTP = 101.325 kPa, 20°C Molar mass of oxygen = 31.9988 g [1] Molar volume of gaseous oxygen at STP = 22.41 L [5] Molar volume of liquid oxygen at BP = 28.04 cm³ - Molar volume of metallic oxygen = 23.5 cm³ [4] Density of gaseous oxygen at STP = 1,429 g/m³ [2] Density of ...


0

You can probably take advantage of measuring pressure at two different heights. Air pressure is caused by its weight, similar to liquid pressure. Therefore, it can be shown that pressure difference of two points can be related to air density: $$\Delta p=\rho g \Delta h$$ This equation gives $\rho$ (average mass density) as a function of $g$ (gravity ...


0

Okay - probably you know, that weather physics is serious business and having even lot of measurements in many points you still can't calculate long enough evolution of a system to get results for e.g. weather in the next month. But you are asking for something else- just equation of gas density in open system depending on local temperature and pressure. ...


0

K is the bulk module, P is the pressure, and V is the volume. If K is constant, then we have: ρ is the density.


-4

I was considering this question as well. Neither matter/energy can be created or destroyed, only converted from one form to the other. Consumption of star stuff by a singularity would convert that star stuff into star energy, but with no other way of expressing that energy there seems no choice but for nature to convert with 100% efficiency that matter into ...



Top 50 recent answers are included