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pentagon [3]
3 years ago
11

A container is filled to a depth of 19.0 cm with water. On top of the water floats a 31.0-cm-thick layer of oil with specific gr

avity 0.600. What is the absolute pressure at the bottom of the container?
Physics
1 answer:
IceJOKER [234]3 years ago
6 0

Answer:

Absolute pressure of the oil will be 102822.8 Pa  

Explanation:

We have given height h = 31 cm = 0.31 m

Acceleration due to gravity g=9.8m/sec^2

Specific gravity of oil = 0.600

So density of oil \rho =0.6\times 1000=600kg/m^3

We know that absolute pressure is given by P=P_0+\rho gh, here P_0=1.01\times 10^5Pa

So absolute pressure will be equal to P=1.01\times 10^5+600\times 9.8\times 0.31=102822.8Pa

So absolute pressure of the oil will be 102822.8 Pa

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Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha
mixer [17]

The density of seawater at a depth where the pressure is 500 atm is 1124kg/m^3

Explanation:

The relationship between bulk modulus and pressure is the following:

B=\rho_0 \frac{\Delta p}{\Delta \rho}

where

B is the bulk modulus

\rho_0 is the density at surface

\Delta p is the variation of pressure

\Delta \rho is the variation of density

In this problem, we have:

B=2.3\cdot 10^9 N/m^2 is the bulk modulus

\rho_0 =1100 kg/m^3

\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3

Learn more about pressure in a fluid:

brainly.com/question/9805263

#LearnwithBrainly

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