Question

I really bad at physics can someone help me with this question ?

The pressure at a depth of (2.74x10^1) m below the surface of the ocean, expressed to 3 significant figures in Pa, is? (The atmospheric Pressure is 101.325 kPa and the density of seawater can be taken as 1035 kg/m3 )

Answer 1

`P=10.2 P_a`

Explanation:

`P_a=101.325 kP_a=0.101325P_a`
`h=2.74*10^1m=27.4m`
`g=9.8 ms^-2`
by the formula of gauge pressure,
i.e. `P-P_a=hrhog`
`P=P_a+hrhog`
`P=1.01325*10^-1+27.4*1.035*9.8`
`P=0.101325+10.134=10.2 P_a`

Answer 2

`P_"Total" = 3.79*10^5 P_a`

Explanation:

Being at a depth of 2.74x10^1 m below the surface of the ocean adds pressure given by

`DeltaP = rho*g*h`

`DeltaP = 1035 (kg)/m^3*9.8 m/s^2*27.4 m = 277,918 (kg*m/s^2)/m^2`

`DeltaP = 277,918 N/m^2 = 2.779 *10^5 P_a`

The total pressure at that depth is the sum of `DeltaP` and atmospheric pressure.

Atmospheric pressure is `101.325 kPa`. That k is a x1000 multiplier, so `101.325 kP_a = 1.01325*10^5 P_a`. So

`P_"Total" = 2.779 *10^5 P_a + 1.01325*10^5 P_a = 3.79*10^5 P_a`

I hope this helps,
Steve