Answer:
1.553 m
Explanation:
This is a pressure problem.
On one side we have got the pressure from the arm's blood and on the other side we have got the pressure from blood height
[tex]P1=P2\\P1=Patm+Parm'sblood\\P2= Patm+Pbloodheight\\Patm+Parm'sblood=Patm+Pbloodheight[/tex]
Patm is atmospheric pressure.Patm is applied on each side of the equation
Then :
[tex]Parm'sblood=Pbloodheight[/tex]
Pbloodheight =ρ.g.h
where ρ is the blood density, g is the gravity and h is the blood height.
ρ= 1050 kg/m3
[tex]g=9.81\frac{m}{s^{2} }[/tex]
[tex]Parm'sblood= 120 mmHg[/tex]
[tex]1 mmHg=133.322 Pa[/tex]
[tex]120 mmHg.\frac{133.322Pa}{1mmHg} =15998.64 Pa\\Pa=\frac{N}{m^{2} } \\N=kg.\frac{m}{s^{2} }[/tex]
[tex]Parm'sblood=Pbloodheight\\15998.64 Pa=1050\frac{kg}{m^{3} } .9.81\frac{m}{s^{2} } .h\\\\h=\frac{15998.64 Pa}{1050\frac{kg}{m^{3} } .9.81\frac{m}{s^{2} } } \\h=1.553m[/tex]