Asked by Y.Abdulsalam
Show that the atm.pressure of 1.013x10^5Pa will support (a)10.3m of water (b)760mmHg of Hg.If the density of water is 10^3kg/m3 and density of mercury is 13.6x10^3kg/m3.
Answers
Answered by
Damon
water mass = 1000 kg/m^3 * volume
volume of 10.3 high by 1 meter square base = 10.3 m^3
so mass of water sitting on that square =
10.3*10^3 kg
weight = m g = 9.81*10.3*10^3
= 101 *10^3 = 1.01*10^5 N/m^2 or Pascals
do the HG the same way.
volume of 10.3 high by 1 meter square base = 10.3 m^3
so mass of water sitting on that square =
10.3*10^3 kg
weight = m g = 9.81*10.3*10^3
= 101 *10^3 = 1.01*10^5 N/m^2 or Pascals
do the HG the same way.
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