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To likninger to ukjente

Posted: 07/05-2012 14:56
by Razzy
[tex]$${\rm I}.\;\;\;\;\;\;{1 \over 2}{M_A}{V_1}^2 = {1 \over 2}{M_A}{V_2}^2 + {1 \over 2}{M_B}{U_2}^2$$[/tex]

[tex]$$\Pi .\;\;\;\;{M_A}{V_1} = {M_A}{V_2} + {M_B}{U_2}$$[/tex]

[tex]$$der\;{M_A} = 1\;kg\;\;og\;\;{M_B} = 2\;kg$$[/tex]


[tex]$${Loser\;{\rm I}\;m.h.p\;{U_2}{\rm{:}}\;$$[/tex]

[tex]$${U_2} = {{{M_A}{V_1} - {M_A}{V_2}} \over {{M_B}}}$$[/tex]

[tex]$${U_2} = {{{M_A}} \over {{M_B}}}\left( {{V_1} - {V_2}} \right)$$[/tex]


[tex]$${\rm{Onsker}}\;{\aa}\;lose\;\Pi \;m.h.p\;{V_2}\;og\;sette\;inn\;for\;{U_2}{\rm{:}}$$[/tex]

[tex]$$Fasit: \;\;{{\rm{V}}_2} = {V_1}{{{{{M_A}} \over {{M_B}}} - 1} \over {1 + {{{M_A}} \over {{M_B}}}}}$$[/tex]


Får dere dette til? NB: Siste del av fasiten i nevner skal være: [tex]$${{{{M_A}} \over {{M_B}}}}$$[/tex]

Posted: 07/05-2012 19:25
by 2357
Du mangler en kvadratrot i uttrykket ditt for [tex]U_2[/tex].