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sveticcg [70]
3 years ago
6

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

s a bulk
modulus of 2.3 × 10^9 N/m^2.Answer in units of kg/m^3
Physics
1 answer:
mixer [17]3 years ago
7 0

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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a concave lens creates a virtual image at -47.0 cm and a magnification of +1.75. what is the focal length?
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The focal length of given concave lens will be -26.85 cm

The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.

Given  concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.

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a block of wood with a mass of 2.33 kg is at rest on a frictionless pole. A .011 kg bullet is fired 722 m/s and is embedded with
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