# Search results

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..