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...
The destructive interference equation for small angles is angle=wavelength/(2a), where a is the width. I assume it means destructive interference since its talking about areas where no light is present.
Using the equation after changing degrees into radians I get the answer of 2491 nm when the...
Im confused by the statement of an "absolute certainty" here. Dont all uncertainties have a certain confidence to them?
Ex. A 99% confidence interval will provide a much wider uncertainty than a 95% confidence interval? Isnt the uncertainty also dependent on the confidence interval you pick...
How does it work then? I've determined the x-component of the velocity to be 2.7784E8 using ux'=(ux-v)/(1-(ux*v)/c^2) where I defined the variables in the same manner. i believe my answer is correct, though I have no way of veryfying.
An earlier comment said I needed to find the x-component...
u'=u/(δ(1-(uv)/c^2)
u' is the velocity of the object according to the individual moving at velocity v.
u is the velocity of the object according the the individual in the rest frame.
The question is asking for u`', the velocity of the object according to the individual moving at velocity v...
So for the formula, u'=u/(δ(1-(uv)/c^2)
u=2.06E8 and v=0. I am only looking at the y components here.
Since v=0 it really becomes u'=u/δ or u'= u*sqrt(1-(u^2)/c^2)).
Anyways when I plug this in I am getting 1.49E8 when the answer should be 0.951E8. Am I not using the correct formula here?
So I am not really familiar with lens questions when there's 2 different refraction indexes. I tried using n1/p+n2/q=-(n2-n1)/R but it doesn't seem to work.
p would be the actual location of the fly and q would be the virtual location, what the fish sees if I am understanding correctly. n1...
So first I calculate the final velocity by multiplying the time by the acceleration, 9.8, to get 88.2 m/s.
Now I use the equation. (343/(343-(-88.2))*108.3 = 86.1477.
But the answer should be 88.47. What am I doing wrong here?
So first I tried to find b.
0.454=(0.6)e^((-b/(2*11.6)*50)
Anyways with some natural log algebra etc. I get b = 0.129378
But when I plug this into the same equation only changing mass to 17.7 kg I get 0.4998 or 50% when the answer should be 59.6%?
So first I find the energy using the eqn (1/2)kA^2. Since there are two springs with the same k I multiply it by two to get kA^2. Energy I get is 2.0475,
Now I use E=(1/2)m(wA)^2 to find mass. Again since there are two springs I use E=m(wA)^2.
m=E/(wA)^2. w=(2(pi))/T btw.
I get the answer of...
I mean I thought I understood it since I was able to use momentum or energy to solve the problem you cite in that thread. The issue here is I don't have the final wavelength so nothing seems to work.
Units are what I use as a last resort and don't always help. Trying to cancel them out units to...
So I can find the initial momentum using p=h/wave = 4.98 x 10-23. Now my problem is that I don't know the final momentum of the photon nor electron, I just know the photon is scattered at an angle of 34 degrees.
I know how to solve this problem if I was given the final wavelength of the light...
So either I need to find the angle the electron is ejected or the y component of its momentum. The y component of the photons momentum is sin(89.422)*4.872 x 10-23 = 4.872 x 10-23. The initial y component of the lights momentum is 0. Therefore the y component of the electrons momentum is -4.872...
so
0.59161 x 10-22 = cos(89.422) * magnitude light momentum + x component of electron momentum
So I really now just need to find the magnitude of light momentum after scattering occurs.
The scattered light wavelength ss 13.6 x 10-12, p=h/wavelength gives 4.8721 x 10-23.
Using the above...
Ok so I used the formula
change in wavelength = (h/mc)(1-cos(angle) to get an angle of 89.422 degrees, but isn't this giving the angle the photon is scattered and not the angle the electron is scattered?
I could understand how I could use this vector to for lights momentum, but i don't see...
So the initial wavelength gives the total momentum, p=h/11.2p. Which is 59.161y.
Then I tried to substract the momentum from the scattered light to get the momentum of the electron.
59.161y-h/13.6p, which ends up being 0.4872 as the final answer, but the answer is supposed to be 0.77?
Qin=W/e so
e=1-Qout/(W/e)
e=1-eQout/W
1-e=eQout/W
(1-e)W/e = Qout
(1-0.35481)*3560/0.35481
Youre right, must have messed up my algebra somewhere because this gives the right answer.
Thanks.
The answer is 6470 J.
So since I have the two temperatures I could calculate the efficiency. First I convert to kelvin then get an efficiency of 0.35481. Now I can use e=W/Qin to get Qin. I get a value of 10033.54J.
Now I can use e=(Qin-Qout)/Qout to get Qout, the waste heat. I get 7405.9 J...
Okay I finally figured out how to do but would really appreciate if you could explain why it works and why my earlier method was wrong (I hate this chapter so much).
As you said, I calculated Q using nc(T change). Did this for each step. Some Qs were negative, some were positive. Now I then...
It would be the sum of the work minus the sum of the heat. Idk if its because I am calculating it wrong or not, but I get 0. But this seems to be wrong because if the internal energy didnt change, doesn't that mean the efficiency was 100%?
Okay so I think I've gotten somewhere but I am not sure how to use heat to get work.
What I did for each temperature changes was use the equation Q=nc(T change). the c is either (3/2)R or (5/2) R depending on if its isochoric or isobaric. Now I have all the Qs, some positive, some negative. The...
What do you mean by set them equal?
I could try to simplify my equation above to get (R(change T1 + change T2))/(-cv(change T1) - cp (change T2)) but i don't think its right.
I think the bottom signs need a negative? Just looking at the graph I know that Pi can't be Pf (they have different height).
The problem with PV=nRT is that I don't know n. Or P. Or V. But let's say I substitute change in T into the equation everywhere.
Pi(nR(change T)/P) + Pf(nR(change T)/P)...
So efficiency is W/Qin.
W= 0 for isochoric processes and for the isobaric, P(change in V). So W=Pi(Vi-Vf)+Pf(Vi-Vf)
Qin is negative Qs.This would happen at step 2 and 3. For the isobaric, Q=ncv(change in T) and for isochoric, Q=ncp(change in T).
Now if I put everything in the equation I get...