Answer:
[tex]8616.7468 \ kg/m^3[/tex]
Explanation:
Pressure is measured is [tex]p=\rho gh[/tex] here p is pressure [tex]\rho[/tex] is density and h is height
We have given pressure [tex]p=9.891\times 10^4\ Pa[/tex] acceleration due to gravity [tex]g=9.9870\ m/sec^2[/tex] height =1.163 m
[tex]\rho =\frac{p}{gh}=\frac{9.891\times 10^4}{9.870\times 1.163}=8616.7468 \ kg/m^3[/tex]
Answer:
Density is 8669.44kg/m^3
Explanation:
At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at the same location is filled with an unknown liquid. What is the density of the unknown liquid if its height in the barometer is 1.163 m?
Solution
Pressure is the ratio of force per unit area. A barometer can used to derive the atmospheric pressure P in terms of the height h of the unknown liquid column.
Density is the ratio of mass to volume
Pressure is defined as
P=force/unit area
P=m*g/(A).......... 1
Recall that density=mass/volume.............2
Volume =A*h. Area *height
From equation 2
D=m/Ah
m=DAh...............3
Substituting equ 3 into 1
P=DAh *g/A
P=Dgh
9.891 × 10^4=D*1.163*9.81
D=9.891*10^(4)/(1.163*9.81)
D=8669.44kg/m^3
Density is 8669.44kg/m^3
