seutas kawat berdiameter 2mm dan panjang 18m,dihubungkan dengan sumber tegangan220v,jika hambatan jenis 3,14×10 pangkat 6=3,14 tentukan arus listrik yang mengalir pada kawat tersebut

5 g + 4 ( − 5 + 3 g ) = 1 − g 5 g − 20 + 12 g = 1 − g 17 g − 20 = 1 − g 17 g + g = 1 + 20 18 g = 21 g = 18 21 ​ = 6 7 ​ ​

Answered by mariamikayla | 2024-06-10

5 g + 4 ( − 5 + 3 g ) = 1 − g 5 g − 20 + 12 g = 1 − g 17 g − 20 = 1 − g ∣ A dd g t o e a c h s i d e 18 g − 20 = 1 ∣ A dd 20 t o e a c h s i d e 18 g = 21 ∣ D i v i d e b y 18 g = 18 21 ​ ​

Answered by luana | 2024-06-10

To solve the equation 5 g + 4 ( − 5 + 3 g ) = 1 − g , we simplify it step by step to isolate g . The final answer is g = 6 7 ​ .
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Answered by mariamikayla | 2024-12-26

hambat jenisnya 3,14 x 10^6 ya. kalau menurut saya harusnya 3,14 x 10^-6

Answered by lrntrn17 | 2025-07-16