Calculate Pressure & Height for Steel Bar on 1L Water

  • Thread starter Thread starter predentalgirl1
  • Start date Start date
predentalgirl1
Messages
61
Reaction score
1
I have tried and tried to figure this out but I'm getting no where. Can anyone help me?

(a) Calculate the gauge pressure (it does not include atmosphereric P) required to compress pure water from 1.000 L to 989 mL. Give an answer in both N/m^2 and psi, lb/in^2.

(b) What height in meters and feet of a 1x1 cm^2 steel bar (density=7.86 g/cm^3) is required to exert the pressure found in part (a)?


(c) How many total pounds of steel would be required to exert this pressure on one cubic face of a liter of water (1 L= 1 cubic decimeter)?
 
Physics news on Phys.org
The answer has to do with the compressibility of water. Look that up. Then start the problem and if you have a problem, post a more specific question.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top