Jawaban:
Persamaan asas kontinuitas
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Q subscript 1 end cell equals cell Q subscript 2 end cell row cell v subscript 1 times A subscript 1 end cell equals cell v subscript 2 times A subscript 2 end cell end table end style
luas penampangnya sama dengan luas lingkaran begin mathsize 14px style open parentheses 1 fourth straight pi d squared close parentheses end style, maka
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 times open parentheses 1 fourth straight pi d subscript 1 squared close parentheses 2 end cell equals cell blank v subscript 2 times open parentheses 1 fourth straight pi d subscript 2 squared close parentheses end cell row cell v subscript 1 times d subscript 1 squared end cell equals cell v subscript 2 times d subscript 2 squared end cell row cell 5 times open parentheses 4 cross times 10 to the power of negative 2 end exponent close parentheses squared end cell equals cell 20 times d subscript 2 squared end cell row cell 5 times open parentheses 16 cross times 10 blank to the power of negative 4 end exponent close parentheses end cell equals cell 20 cross times d subscript 2 squared end cell row cell d subscript 2 squared end cell equals cell fraction numerator 5 cross times 16 cross times 10 to the power of negative 4 end exponent over denominator 20 end fraction end cell row cell d subscript 2 squared end cell equals cell 4 cross times 10 to the power of negative 4 end exponent end cell row cell d subscript 2 end cell equals cell 2 cross times 10 to the power of negative 2 end exponent space straight m end cell row cell d subscript 2 end cell equals cell 2 blank cm end cell end table end style
Jadi, jawaban yang tepat adalah A.
Penjelasan:
2cm
