Question

 boy jumps from a cliff and strikes the bodies of water and reaches 40meters down under. If the mass density of the water is 1.03g/cm^3, calculate the pressure he experienced in Newton per square meter.​

Answer 1

Given:-

density of water, $\rho=1.03g/cm^{3}$

height of boy from free surface of liquid, h=40 m

To Find:-

Pressure experienced by the boy.

Explanation:-

we have,

$\rho=1.03g/cm^{3}=1030kg/m^{3}$

From hydrostatic paradox,

Pressure acting on the body submerged in a liquid upto height 'h' is given by:-

$P=\rho\times g\times h\\\\\Rightarrow P=1030\times9.81\times40=404172N/m^{2}$

Answer 2

The pressure experienced by the boy is 403760 N/m²

Explanation:

Given:

The depth below the water = 40 m

Mass density of water = 1.03 g/cc

To find out:

The pressure experienced in N/m²

Solution:

We know that in a liquid of density $\rho$ the pressure at depth h is given by

$P=h\rho g$

Given

$h=40$ m

$\rho=1.03$ g/cc = $1.03\times 10^{-3}/(10^-2)^3$ kg/m³

$\implies \rho=1030$ kg/m³

Thus, the pressure experienced by the boy

$P=h\rho g$

$\implies P=40\times 1030\times 9.8$ N/m²

$\implies P=403760$ N/m²

$\implies P=4\times 10^5$ N/m²

Hope this answer is helpful.

Know More:

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