Respuesta :
Answer:
[tex]\Delta p = 90.7 kPa[/tex]
Explanation:
specific gravity of oil is [tex] = \frac{\rho_{oil}}{\rho_w}[/tex]
[tex]\rho_{oil} = 0.85*1000 = 850 kg/m3[/tex]
we know that
change in pressure for oil is given as
[tex]\Delta p = \rho gh[/tex]
here density and h is for oil
[tex]\Delta p = 850*5 *9.81 = 41,692.5 kPa[/tex]
change in pressure for WATER is given as
[tex]\Delta p = \rho gh[/tex]
here density is for water and h is for water
[tex]\Delta p = 1000*5 *9.81 = 49,050 kPa[/tex]
pressure change due to both is given as
[tex]\Delta p = 41692.3 + 49050 = 90742.5 N/m2[/tex]
[tex]\Delta p = 90.7 kPa[/tex]
