Question
Estimate the change in the density of water in ocean at a depth of $400 \mathrm{~m}$ below the surface. The density of water at the surface $=1030 \mathrm{~kg} \mathrm{~m}^{-3}$ and the bulk modulus of water $=2 \times 10^{\circ} \mathrm{N} \mathrm{m}^{-2}$.
Step 1
Step 1: The change in density at a depth h is given by the formula $\Delta \rho = \rho \frac{\Delta P}{B}$, where $\Delta P$ is the change in pressure, $\rho$ is the initial density and B is the bulk modulus of the substance. Show more…
Show all steps
Your feedback will help us improve your experience
Narayan Hari and 88 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Calculate the density of sea water at a depth of $1000 \mathrm{~m}$, where the water pressure is about $1.00 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}$. (The density of sea water is $1.030 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ at the surface.)
The deepest point in the ocean is in the Mariana Trench, about 11 $\mathrm{km}$ deep. The pressure at the ocean floor is huge, about $1.13 \times 10^{8} \mathrm{N} / \mathrm{m}^{2} .(\mathrm{a})$ Calculate the change in volume of 1.00 $\mathrm{m}^{3}$ of water carried from the surface to the bottom of the Pacific. (b) The density of water at the surface is $1.03 \times 10^{3}$ $\mathrm{kg} / \mathrm{m}^{3}$ . Find its density at the bottom.
What is the density of ocean water at a depth, where the pressure is $80.0 \mathrm{~atm}$, given that its density at the surface is $1.03 \times 10^{3} \mathrm{kgm}^{-3}$ (compressiblility of water $=45.8 \times 10^{-11} \mathrm{~Pa}^{-1}$ )?
Solids and Fluids
Section F
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD