Selena Hardin

2023-02-25

The density of seawater is $1024kg/{m}^{3}$. The pressure at a depth of 5 m below the surface of the sea is ___ $N/{m}^{2}$.

Troy Jimenez

Beginner2023-02-26Added 3 answers

Solution:

Depth of the point h = 5 m

Density of sea water $\rho =1024kg/{m}^{3}$

Acceleration due to gravity $=g=10m/{s}^{2}$

Pressure = ρgh

$=1024\times 10\times 5$

Pressure $=51200N/{m}^{2}$

Depth of the point h = 5 m

Density of sea water $\rho =1024kg/{m}^{3}$

Acceleration due to gravity $=g=10m/{s}^{2}$

Pressure = ρgh

$=1024\times 10\times 5$

Pressure $=51200N/{m}^{2}$

The radius of the aorta is about 1.1 cm , and the blood passing through it has a speed of about 34 cm/s . Coefficient of viscosity for the whole blood (37∘) is η = 4×10−3Pa⋅s Calculate the pressure drop per cm along the aorta.

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