Recent content by JoeyBob

So it would be rate of change with respect to volume?

So I use final volume and pressure for this? I treid that and it didnt give the right answer for me.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html

Gives rate of change. For instance, if you take the derivative of velocity, you get acceleration, which is the rate of change of velocity.

I know the integral of a P(V) eqn gives an eqn for work. I was wondering if taking the derivative of a P(V) eqn gives an eqn for change in pressure?

Im confused on working backwards so to speak to find adiabatic work. To find work for this adiabatic process, I either need to know the change in temperature OR the initial pressure (I think?). The issue is that I don't know either the initial temperature nor the initial pressure so I am not...

You know what, after thinking about it and reading it a few more times I think I know what it means. The fish 3.5 m above the mirror. The fish is also 3.5 m deep. So the mirror is actually 7m deep. Very poorly worded question. Since the mirror is underneath, I can find the location of the...

For the ice at 273.15 K, its being heated up (dQ/dt is positive). where here dQ/dt is per min (not second) DeltaS=dQ/dt/T=3910000/273.15 = 14314 For the steam at 373.15, its being cooled down (dQ/dt is NEGATIVE) -3910000/373.15 = -10478 Add them up and I get a total change of 3836... close...

The heat absorbed by the ice-water mixture is the heat emitted by the steam-water mixture. I divided it by temperature because that gives the right units and is consistent with the equation Q=T*change in entropy I suppose I need to find the temperature of the interphase maybe instead? Here...

So what I did was find the change in Q per min. Mass melted per min * latent heat capacity = Q per min = 11.5 kg /min * 3.4*10^5 J/kg = 3910000 J/min Now the equilibrium temperature is 100 degrees Celsius or 373.15 degrees kelvin. If I do 3910000 J/min / 373.15 K I get 10478 J/(K*min). This...

So let's pretend there was a person right above the water looking down and I wanted to find how deep the reflection of the fish appeared to be to the person. Is that when I would use n1/p+n2/q=(n2-n1)/R where n2 is 1.29 and p=3.5?

The way I read "3.5m above" was above the water since why else would refractive indexes even be given? If we assume its 3.5 above the fish, then 1/q+1/p=-2/R, where p=3.5m and R=6.2m This does indeed give -1.644 m or 1.644 m behind the mirror. What a dumb question. Who knows what the right...

i think this is where I am confused. The light is coming from the fish. but then it needs to come back to the fish for the fish to see itself. First light comes off fish and hits water. Then light travels through air and hits mirror. Then light goes back through air and hits water. Then goes...

i see, I was using a double slit eqn instead of a single slit. Looking at your link, tanx=x=y/D. So i don't know what y is nor d. d=y/x=y/0.108 = 9.2593y Now I can use y=(m*wavelength*D)/a to find width. m i assume is 2 because second angle measurement... 0.108d=(2*539 nm *d)/a a=9981.4815...

So first I looked at where the image of the fish appeared to be when it went through the water surface. since we can assume the water is flat, R is infinity, so n1/p=-n2/q. plugging in the values (n1=1.29, n2=1, p=3.5) I get q=-0.3686. So the image of the fish appears at 0.369 above the...