dwa ciala ktorych stosunek mas jest rowny 3 maja jednakowe pedy. oblicz stosunek ich energii kinetycznej
p1=p2
m1= m m2=3m
p= m*V
p1=m1*V1=m*V1
V1= p1/m
p2=m2*V2=3m*V2
V2=p2/3m
Ek=mV^2/2
Ek1= m1*V1^2/2 = (m*p1^2/m^2) = (p1^2/m)/2
Ek2= m2*V2^2/2= (3m*p2^2/9m^2)/2 = (p2^2/3m)/2
Ek2/Ek1= [(p1^2/m)/2]/[(p2^2/3m)/2]= (p1^2/2m)/(p2^2/6m)
Ek2/Ek1 = (P1^2/2m)*(6m/p2^2) = 3
stosunek ich energii kinetycznej rowniez wynosi 3
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p1=p2
m1= m m2=3m
p= m*V
p1=m1*V1=m*V1
V1= p1/m
p2=m2*V2=3m*V2
V2=p2/3m
Ek=mV^2/2
Ek1= m1*V1^2/2 = (m*p1^2/m^2) = (p1^2/m)/2
Ek2= m2*V2^2/2= (3m*p2^2/9m^2)/2 = (p2^2/3m)/2
p1=p2
Ek2/Ek1= [(p1^2/m)/2]/[(p2^2/3m)/2]= (p1^2/2m)/(p2^2/6m)
Ek2/Ek1 = (P1^2/2m)*(6m/p2^2) = 3
stosunek ich energii kinetycznej rowniez wynosi 3