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Pressure at the Earth's surface is p_(a)=10^(5)Pa and the density of air at Earth's surface is rho_(0)=1.4kg//m^(3). At height h from the surface of Earth the density of air is reduced to (rho_(0))/(2) the value of h is (Assume that the temperature is constant through out the earth's atmosphere and let In (2)=0.7) |
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Answer» 10,500 m `p_(h)=p_(0)e^(-mgh//kT)` where, k= BOLTZMANN's constant g= gravitational acceleration and `p_(0)`= prsssure at the earth surface |
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