Given: 76cm
A.water = 1000kg/m^3
B. Mercury = 13.6 x 10^3 kg m/3
Solution:
76 x 1/100 = 0.76m
A. P = hdg
=(1000kg/m^3)(.76m)(9.8)
=7448 Pa
B. P = hdg
=(13.6x10^3kg/m^3)(.76)(9.8)
=101293 Pa
find the pressure at the bottom of a vessel 76cm deep when filled with
a) water
b)mercury (sp.g. 13.6)
3 answers
Find the pressure at the bottom of a vessel 76cm deep filled with glycerin 1.26
Given:
Depth (h) = 76 cm = 0.76 m
Density (ρ) of Glycerin = 1.26 g/cm³ = 1260 kg/m³
Solution:
Pressure is given by the formula:
P = ρgh
where:
P = pressure
ρ = density of the liquid
g = acceleration due to gravity
h = height or depth
Substituting the given values:
P = (1260 kg/m³) x 9.8 m/s² x 0.76 m
P = 9286.88 Pa
Therefore, the pressure at the bottom of a vessel 76 cm deep filled with glycerin 1.26 is 9286.88 Pa.
Depth (h) = 76 cm = 0.76 m
Density (ρ) of Glycerin = 1.26 g/cm³ = 1260 kg/m³
Solution:
Pressure is given by the formula:
P = ρgh
where:
P = pressure
ρ = density of the liquid
g = acceleration due to gravity
h = height or depth
Substituting the given values:
P = (1260 kg/m³) x 9.8 m/s² x 0.76 m
P = 9286.88 Pa
Therefore, the pressure at the bottom of a vessel 76 cm deep filled with glycerin 1.26 is 9286.88 Pa.
1 answer