Respuesta :
Answer:[tex]59.78\times 10^3\ Pa[/tex]
Explanation:
Given
Diver is at a depth of [tex]h=6.1\ m[/tex] inside the water[tex](\rho=10^3\ kg/m^3)[/tex]
As we go down pressure increases due to the weight of liquid column above us
Pressure difference is given by [tex]P=\rho \times g \times h[/tex]
Where [tex]\rho [/tex]=density of liquid
h=depth
g=acceleration due to gravity
So external-internal pressure difference is
[tex]\Delta P=10^3\times 9.8\times 6.1[/tex]
[tex]\Delta P=59.78\times 10^3\ Pa[/tex]
The external–internal pressure difference is 59.78 × 10³ Pa.
- Diver is at a depth of inside the water which is represented as the height h= 6.1 m
- Density = 10³ kg/m³
Pressure difference P = ρgh
where ρ is density of liquid, g is acceleration due to gravity and h is depth
The external-internal pressure difference is
= 10³ × 9.8 × 6.1
= 59.78 × 10³ Pa
ΔP = 59.78 × 10³ Pa
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