Question
A steel ball is tossed into the ocean and comes to rest at a depth of $2.40 \mathrm{~km}$. Find its fractional change in volume, assuming the density of seawater is $1.025 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$.
Step 1
This is represented as: \[ \Delta V/V = -\frac{\Delta P}{B} \] where $\Delta V/V$ is the fractional change in volume, $\Delta P$ is the change in pressure, and $B$ is the bulk modulus. Show more…
Show all steps
Your feedback will help us improve your experience
Prabhu Ramji and 85 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
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 $=2 \times 10^{\circ} \mathrm{N} \mathrm{m}^{-2}$.
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven $\mathrm{km}$ beneath the surface of water. The water pressure at the bottom of the trench is about $1.1 \times 10^{8} \mathrm{~Pa}$. A steel ball of initial volume $0.32 \mathrm{~m}^{3}$ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?
A waterproof ball made of rubber with a bulk modulus of $6.309 \cdot 10^{7} \mathrm{~N} / \mathrm{m}^{2}$ is submerged under water to a depth of $55.93 \mathrm{~m}$. What is the fractional change in the volume of the ball?
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD