Respuesta :
Pressure inside the liquid nitrogen is calculated by Pascal's law
it is given by the equation
[tex]P = P_o + \rho g h[/tex]
given that
at 2 m below the surface pressure is [tex]P = 5 * 10^5 Pa[/tex]
so this equation is given as
[tex]5 * 10^5 = P_o + 2\rho*g[/tex]
also we have [tex]P = 8 * 10^5 Pa[/tex] at 5 m below
[tex]8 * 10^5 = P_o + 5\rho*g[/tex]
in order to find gravity at the planet subtract two equations
[tex](8 - 5)*10^5 = 3 *\rho* g[/tex]
[tex]3* 10^5 = 3 * 808 * g[/tex]
[tex]g = 123.76 m/s^2[/tex]
now to find the atmospheric pressure we can use any one equation
[tex]P_o = 5 * 10^5 - 2* 808 * 123.76[/tex]
[tex]P_o = 3 * 10^5 Pa[/tex]
Answer:
a) [tex](3.00).10^{5}Pa[/tex]
b) [tex]123.762\frac{m}{s^{2}}[/tex]
Explanation:
Let's start defining the following unit of pressure :
[tex]1Pascal=1Pa=1\frac{N}{m^{2}}[/tex]
Where N is newton and m is meter.
Within a mass of liquid and defining the surface of the liquid as depth 0.00 m, we can calculate the absolute pressure in any point of the liquid as :
[tex]Pabs=Pgauge+Patm[/tex]
Where Pgauge is the pressure because of the height of liquid over the point.
Where Patm is the atmospheric pressure.
The gauge pressure in a point of a mass of liquid is define as :
Pgauge = δ.g.h
Where δ is the density of the liquid.
Where g is the acceleration due to gravity.
Where h is the height of liquid over the point.
Using the data from the exercise :
[tex]Pabs=Pgauge+Patm[/tex]
At a depth of 2.00 m :
[tex](5.00).10^{5}\frac{N}{m^{2}}=808\frac{kg}{m^{3}}.g.(2.00)m+Patm[/tex] (I)
At a depth of 5.00 m :
[tex](8.00).10^5\frac{N}{m^{2}}=808\frac{kg}{m^{3}}.g.(5.00)m+Patm[/tex] (II)
In (II) :
[tex]Patm=(8.00)10^{5}\frac{N}{m^{2}}-808\frac{kg}{m^{3}}.g.(5.00m)[/tex] (III)
Putting (III) in (I) :
[tex](5.00).10^{5}\frac{N}{m^{2}}=808\frac{kg}{m^{3}}.g.(2.00)m+(8.00).10^{5}\frac{N}{m^{2}}-808\frac{kg}{m^{3}}.g.(5.00)m[/tex]
[tex]-(3.00).10^{5}\frac{N}{m^{2}}=1616\frac{kg}{m^{2}}.g-4040\frac{kg}{m^{2}}.g[/tex]
[tex]-(3.00).10^{5}\frac{N}{m^{2}}=-2424\frac{kg}{m^{2}}.g[/tex]
[tex]g=\frac{(3.00).10^{5}}{2424}\frac{N.m^{2}}{m^{2}.kg}[/tex]
[tex]g=123.762\frac{m}{s^{2}}[/tex]
This is the gravity for point b)
For a) we need to replace this value in another equation. For example in equation (III) :
[tex](8.00).10^{5}\frac{N}{m^{2}}-808\frac{kg}{m^{3}}.(123.762\frac{m}{s^{2}}).5m=Patm[/tex]
[tex]Patm=(3.00).10^{5}\frac{N}{m^{2}}[/tex]
[tex]Patm=(3.00).10^{5}Pa[/tex]
And that is the value of atmospheric pressure on planet x.
