Respuesta :
Answer:
The atmospheric pressure is [tex]9.845 \times 10^4\ Pa[/tex]
Explanation:
There are two ways of solving this exercise:
1)
In physics you can find that mmHg is a unit of pressure.
Pressure = Force/area.
If you consider the weight of mercury as your force (mass* acceleration of gravity = density*volume* acceleration of gravity ), then
[tex]P =\frac{\rho Vg}{A} = \frac{\rho Ahg}{A} =\rho g h[/tex]
where h is the height of the mercury column and rho its density.
[tex]\Delta P= P_{atm} - P_{hur} = 21.4\ mm Hg[/tex].
if normal atmospheric pressure is [tex]1.013 \times 10^5\ Pa = 759.81\ mmHg[/tex]
then the pressure in the presence of the hurricane is
[tex]P_{hur} = 759.81 - 21.4 = 738.41\ mmHg = 9.845 \times 10^4\ Pa[/tex]
2)
Considering the definition of pressure
[tex]\Delta P = \rho g h[/tex]
where [tex]\rho = 13.6\ g/cm^3[/tex], [tex]g = 9.8\ m/s^2 =980\ cm/s^2[/tex] and [tex]h = 21.4\ mm = 2.14\ cm[/tex].
[tex]\Delta P = P_{atm} - P_{hur} = 28521,92\ g/cms^2 = 2852,192\ kg/ms^2[/tex], where [tex]kg/ms^2 = Pa[/tex].
if
[tex]P_{atm} = 1.013 \times 10^5\ Pa[/tex],
then
[tex]P_{hur} = 1.013 \times 10^5 - 2852,192 =9.845 \times 10^4\ Pa[/tex].
