Respuesta :
Answer:
0.22 m
Explanation:
[tex]P_{o}[/tex] = Atmospheric pressure = 101325 Pa
[tex]P[/tex] = Pressure at the bottom = tex]2 P_{o}[/tex] = 2 (101325) = 202650 Pa
[tex]h[/tex] = height of the container = 7.59 m
[tex]x[/tex] = depth of the mercury
Pressure at the bottom = Atmospheric pressure + Pressure due to mercury + Pressure due to water
[tex]P = P_{o} + \rho _{w} g (h - x) + \rho _{hg} g x\\2 P_{o} = P_{o} + \rho _{w} g (h - x) + \rho _{hg} g x\\P_{o} = \rho _{w} g (h - x) + \rho _{hg} g x\\101325 = (1000) (9.8) (7.59 - x) + (13600) (9.8) x\\x = 0.22 m[/tex]
Answer:
1.0335 m
Explanation:
Atmospheric pressure, Po = 1.01 x 10^5 Pa
Let the depth of mercury is h
the depth of water is 7.59 - h
density of mercury = 13.6 x 10^3 kg/m^3
density of water = 1000 kg/m^3
Pressure due to mercury + pressure due to water = 2 x Po
height of mercury x density of mercury x g + height of water x density x g
= 2Po
h x 13.6 x 1000 x g + (7.59 - h) x 1000 x g = 2 x 1.01 x 10^5
h x 13.6 x 9.8 + (7.59 - h) x 9.8 = 2 x 1.01 x 100
133.28 h + 74.382 - 9.8 h = 202
123.48 h = 127.618
h = 1.0335 m
Thus, the height of mercury column is 1.0335 m.
