Asked by Chris
One atmospheric pressure is that which will support a column of mercury 760.0 mm high. Show that 1 atm equals 1.013 x 10^5 N/m^2. Take the density of Hg to be 13.60 g/cm^3 and the gravitational constant to be 9.80 m/s^2. (You may find unit analysis helpful!)
I have no idea where to begin. Any help would be appreciated.
Start with figuring the weight of a column of mercury ( cross sectional area A), then the pressure (weight/A). Weight= mass*g=density*volumecolumn*g=denstiy*A*h*g
I have no idea where to begin. Any help would be appreciated.
Start with figuring the weight of a column of mercury ( cross sectional area A), then the pressure (weight/A). Weight= mass*g=density*volumecolumn*g=denstiy*A*h*g
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