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m=M*I*t/(z*F)
masa miedzi
m= (64g/mol*1000A*150*3600s)/(2*96500C/mol) = 179067g
objętość otrzymanej miedzi
V=m/d =179067g/8,93g/cm³ =20052,34cm³
objętość walca
V = πr²h
h= V/(π r²)
h=20052,34cm³/(3,14*(0,1)²cm² = 638609,4cm = 6386,094m
Można otrzymać 6386,094m drutu