Answer :
Pressure under water is called the gauge pressure.
[tex]Pressure= \frac{Force}{Area} \\ F=mg \\ P= \frac{mg}{A}[/tex]
Now the next part is difficult to put into an equation because you need the symbol for density, which is ρ, pronounced rho. I cannot type ρ into the equation so I will use p.
[tex]p= \frac{m}{V} [/tex] (m=mass, V=Volume)
[tex]m=pV \\ P= \frac{pVg}{A} \\ V=Ah[/tex]
A=Area of the base of the object, or cylinder of water above the object, h= height of object
[tex]P= \frac{p(Ah)g}{A} [/tex]
The "A"s cancel, leaving us with
[tex]P=pgh[/tex]
You are only looking for the water pressure, so you would not be concerned with atmospheric pressure on top of that.
[tex]P_{gauge} =p_{liquid} g h_{object} [/tex]
The density of water is [tex]1000 \frac{kg}{ m^{3} } [/tex]
Use 1000 for ρ
Gravity is 9.8
Height is 220
[tex]P=pgh \\ P=1000*9.8*220 \\ P=2.16*10^{6} Pa[/tex]
(Pa is Pascals, the unit for pressure)
[tex]Pressure= \frac{Force}{Area} \\ F=mg \\ P= \frac{mg}{A}[/tex]
Now the next part is difficult to put into an equation because you need the symbol for density, which is ρ, pronounced rho. I cannot type ρ into the equation so I will use p.
[tex]p= \frac{m}{V} [/tex] (m=mass, V=Volume)
[tex]m=pV \\ P= \frac{pVg}{A} \\ V=Ah[/tex]
A=Area of the base of the object, or cylinder of water above the object, h= height of object
[tex]P= \frac{p(Ah)g}{A} [/tex]
The "A"s cancel, leaving us with
[tex]P=pgh[/tex]
You are only looking for the water pressure, so you would not be concerned with atmospheric pressure on top of that.
[tex]P_{gauge} =p_{liquid} g h_{object} [/tex]
The density of water is [tex]1000 \frac{kg}{ m^{3} } [/tex]
Use 1000 for ρ
Gravity is 9.8
Height is 220
[tex]P=pgh \\ P=1000*9.8*220 \\ P=2.16*10^{6} Pa[/tex]
(Pa is Pascals, the unit for pressure)
