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Mechanics11/Seite11/Loesung4.6

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c)

geg.:

  • m = 4000 m3
  • 1,0 m/s

ges.:

  • Wa

W_a = \frac{1}{2} * m * v^2

W_a = \frac{1}{2} * 4.000.000 kg * (1,0 \frac{m}{s})^2

W_a = 2.000.000 J \frac{}{}

W_a = 2,0 MJ \frac{}{}


--FatF 12:12, 15 February 2008 (CET)


Lösungen zu Aufgabe 4.6:

a) m/2*v²=E =192901,2346J ~ 193kJ

b) mgh=E =1962000J ~ 2MJ

c) m/2*v²=E =25920J ~ 2.6kJ

--Alexander m11 11:21, 16 February 2009 (UTC)

4.6.

Lösungen: 

         a)= 192901,2346 ~ 192kJ
         b)= 1962000J    ~ 2,0MJ
         c)= 25920J      ~ 2,0MJ

Benötigte Formeln: 

                  1) 0.5*m*v² (KinetischeEnergie)
                  2) m*g*h    (HöhenEnergie)


Amageddon m11

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