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 )

2 Answers
Shubh
Apr 7, 2018

P=10.2 P_aP=10.2Pa

Explanation:

P_a=101.325 kP_a=0.101325P_aPa=101.325kPa=0.101325Pa
h=2.74*10^1m=27.4mh=2.74⋅101m=27.4m
g=9.8 ms^-2g=9.8ms−2
by the formula of gauge pressure,
i.e. P-P_a=hrhogP−Pa=hρg
P=P_a+hrhogP=Pa+hρg
P=1.01325*10^-1+27.4*1.035*9.8P=1.01325⋅10−1+27.4⋅1.035⋅9.8
P=0.101325+10.134=10.2 P_aP=0.101325+10.134=10.2Pa

Steve J.
Apr 8, 2018

P_"Total" = 3.79*10^5 P_aPTotal=3.79⋅105Pa

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

Related questions
Impact of this question
1572 views around the world
You can reuse this answer
Creative Commons License