bantuin pakai caranya
Jawaban:
∑F[tex]=ma[/tex]
[tex]Fcos(53)=ma\\a=\frac{Fcos(53)}{m} \\a=\frac{20 . 0,6}{4} \\\\a=3[/tex]
Penjelasan:
[answer.2.content]Jawaban:
∑F[tex]=ma[/tex]
[tex]Fcos(53)=ma\\a=\frac{Fcos(53)}{m} \\a=\frac{20 . 0,6}{4} \\\\a=3[/tex]
Penjelasan:
[answer.2.content]