Recent content by Chestermiller

It doesn’t.

Volume per unit mass, For an incompressible fluid,fhe equation of state is degenerate.

Yes, but for a compressible fluid, the equation of state says that the pressure is independent of specific volume.

All the equations you have written are the incompressible case.

If you are considering the liquid water as incompressible. then its specific volume is independent of pressure, and the static pressure (as you call it) is equal to the hydrostatic pressure at all depths. The equation of state for an ideal gas is $$P=\frac{\rho RT}{M}$$where M is the molecular...

Your equation for the hydrostatic pressure is incorrect if you are allowing the water to have a pressure which varies with specific volume (i.e,. your 'static pressure relationship). That is, if the water is considered compressible.

If I understand you correctly, the hydrostatic pressure must match the static pressure at all depths in the glass.

The equation of state of liquid water is approximated by $$\rho=\rho_0e^{(-\alpha(T-T_0)+\frac{P-P_0}{B})}$$where ##\alpha## is the coefficient of thermal expansion and B is the bulk modulus; ##T_0=293 K## and ##P_0=1 atm##. So the hydrostatic equation reads $$\frac{dP}{dh}=\rho...

What is the difference between static pressure and hydrostatic pressure in this system? What is your definition of these terms?

Work is work. It is the integral of the applied force over a displacement. In thermodynamics, you have to very carefully define what you are calling your system, and where your system ends and its surrounding begins. This choice is entirely up to you. You also have to specify what is...

To simplify things, replace the ball with a short elastic rod and neglect gravity. The key to understanding what happens in impact of the rod against the wall is to recognize that the rod is not a rigid body and so it does not have to stop all at once. Different parts of the rod deform at...

The hard part of attacking physical is to articulate in words the physical mechanisms involved and translate these into a set of equations. Solving the equation(s) is supposed to be a gimme.

Are you saying that you don't know how to solve this linear first order ordinary differential equation and that you want. us to solve it for you?

It goes like this: $$MC_p\frac{dT}{dt}=-kS(T-20)$$The initial temperature is 1500 C and, apparently, your final temperature is supposed to be 40C.

In thess equations, the 0.9724 should be 0.8