Answer:
[tex] P_{tot}= 101325 Pa + 13600 \frac{Kg}{m^3} *9.8 \frac{m}{s^2} *0.23m [/tex]
[tex] P_{tot}= 131979.4 Pa *\frac{1Kpa}{1000Pa}= 131.9794 Kpa[/tex]
And using 3 significant digits we got [tex]P_{tot}\approx 132 Kpa[/tex]
Explanation:
For this case we have a cylinder with a height of h =0.23 m
We want to calculate the bottom presure at the bttom of the graduated cylinder, and the formula for the total pressure would be given by:
[tex] P_{tot}= P_{atm} + \rho_{Hg} g h[/tex]
For this case we have the following info:
[tex]\rho_{Hg} = 13600 \frac{Kg}{m^3}[/tex] the density for the mercury
[tex] h =0.23 m[/tex] the heigth of the cylinder
[tex] g = 9.8 \frac{m}{s^2}[/tex] represent the gravity
[tex] P_{atm}= 101325 Pa[/tex] the atmospheric pressure assumed.
And replacing we got:
[tex] P_{tot}= 101325 Pa + 13600 \frac{Kg}{m^3} *9.8 \frac{m}{s^2} *0.23m [/tex]
[tex] P_{tot}= 131979.4 Pa *\frac{1Kpa}{1000Pa}= 131.9794 Kpa[/tex]
And using 3 significant digits we got [tex]P_{tot}\approx 132 Kpa[/tex]
Answer:
[tex]P = 117.791\,kPa[/tex]
Explanation:
The absolute pressure at the bottom of the cylinder is:
[tex]P = P_{atm} + P_{man}[/tex]
[tex]P = 101.325\,Pa + (1000\,\frac{kg}{m^{3}}+13600\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.115\,m)[/tex]
[tex]P = 117790.953\,Pa[/tex]
[tex]P = 117.791\,kPa[/tex]
