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1. . Growth in salmon smolt

urgent. Growth in salmon smolt Homework Statement smolt weith is 35g we shall find the weight after 61 days. Homework Equations specific growth rate:; K' = Ln (N2 / N1) / (t2 - t1) growth rate is 1,9 % The Attempt at a Solution I find 110g when I use this forumula...
2. Final Temperature After Phase Change

Yeah actually :) Only I did not put them up that way

why nobody ?

any?
5. Final Temperature After Phase Change

This should be a easy task for you :p

Help :)
7. Thermodynamic ( quanity of heat)

:D so thats the answer then 3859000J Thanks for the help :)
8. Thermodynamic ( quanity of heat)

Ok :) thanks for the advice. But what do I now Is it just to add the heat from the container 55000J too the 4 parts added = 3804000J = 3804000J + 55000J = 3859000J ?
9. Thermodynamic ( quanity of heat)

hmm Starting temprature of the container must be -10 Celsius and ending temprature of 100Celsius ? So then the heat from container is 500j/k *110K = 55000J ?
10. Final Temperature After Phase Change

did I do something right here?
11. Thermodynamic ( quanity of heat)

Ok I think what I dont seems to find now is the Also set your equations equal to one another. Energy lost = energy gained ect...I had the same problem awhile back and thats what I was leaving out. hmm is the energy lost in the phase changes ?
12. Final Temperature After Phase Change

What I tryed is: Water Q=mt = 0,3*4190 = 1257J Ice to liquid: Q=mt = 0,15*334000 = 50100J 1) The melted Ice is now water of 0C = (0,15kg)*(4190)j/kg*K)(T-0) 2) The water of 50c = (0,3kg)*(4190)j/kg*K)(T-50) solving that (0,15)*(4190)= 628,6T...
13. Thermodynamic ( quanity of heat)

hmm. So what I do is to find the temprature out of thoose 4 parts and * it with the 500J/K ?
14. Thermodynamic ( quanity of heat)

Aii yes it is 500J/K not Kg :) I tryed to find the heat gained from ice to vapor. 1: Ice heats up to melting point Q=mcdeltaT = 2kg*2100j/kg*K*10K = 42000J 2: Ice melts at C0 Q=+-ml = 2kg* 334000J = 668000J 3: liquid water of C0 heats up to 100C Q=mcdeltaT = 2kg*4190j/kg*K*100K...
15. Thermodynamic ( quanity of heat)

Hmm =| Off course the container is inportnant. I mixed up. It has a heat capacity of 500J/Kg But it doesnt have a mass. So what I do with it ?
16. Thermodynamic ( quanity of heat)

ok I will try to solve it this way then.
17. Thermodynamic ( quanity of heat)

The container is not inportnant in this task it says its uninportant. So it must be solved another way then.
18. Final Temperature After Phase Change

Homework Statement find the final temperature of A 150g (0,15kg) cube of ice at 0.0 degrees Celsius is added to 300g(0.3 kg) of water at 50.0 degrees Celsius. Specific heat capacity of ice: 2.09 x 10^3 J/(kg x degrees Celsius) Specific heat capacity of water: 4.186 x 10^3 J/(kg x...
19. Thermodynamic ( quanity of heat)

[SOLVED] Thermodynamic ( quanity of heat) Homework Statement Calculate he quantity of heat that needed to transfer 2,0 kg ice of -10C to steam/vapor of 100C The process in this case take place in a good heat-isolated container with heat capacity of 500J/Kg Homework Equations Q=mcT Q+-ml...
20. Centre of gravity and force

Yeah I should =| But I belive I learned a bit from it. Thanks for your time:smile:
21. Centre of gravity and force

I dont see why I cant:confused: And I dont know about that but I need to get the task done so I need to do the way I can :wink: But I would like to see your ways of doing it caus its probably a better and "safer" way.
22. Centre of gravity and force

Forces. 4+1 = 5F 4/5 in the thick end 1/5 in the thin end :blushing: It is correct isnt it ?
23. Centre of gravity and force

Found another (easy) way to solve it :) thin end 1/5* 10= 2 thick end 4/5*10= 8 So the gravity of force is 2m from the thick end and 8m from the thin end. So now you can reveal your way that Iv tryed to understand for 4 days now :)
24. Centre of gravity and force

So the equations are correct now ? Its Wx we need to find ? Do i solve it by using both equations ? never been good at equations either.
25. Centre of gravity and force

That make sense :) 1: 0 + xW -10F 2: (10-x)W + 40F then ?
26. Centre of gravity and force

I dont know how to explain I dont think I understand some basics here. why is the 1 one 0 + wX and the second one 10 - x ? Is there another way to solve it ? =s http://www2.ucdsb.on.ca/tiss/stretton/PhysicsFilm/ForceTorque/sld011.html Found this. Could this problem be solved in a...
27. Centre of gravity and force

Im struggling with this task =| 1: 0 + xW -10F thats right but the way I see the 2 one is either 2: 0 +xW-40F or 2: -10F + xW+-40F
28. Pressure and resultant force

"The resultant force agains the bottom of the tub" I would say it also have info about a hatch but I think that is only relevant to the next question who is what the resultant force is agains the hatch..
29. Pressure and resultant force

Dont ask me You are the smart one:wink: But the resultant force to the bottom of the "tub" I dont know but we got the information about lenght and width and so far i havent used that info, maybe i should :confused:
30. Centre of gravity and force

Im thinking..