Question

Fisika.
Vektor.
Mohon di bantu.​

Answer 1

Jawaban A

[tex] |F1| = |F2| = F[/tex]

[tex] \frac{ |F1 -F2| }{ |F1 +F2| } = \sqrt{3} [/tex]

[tex] \frac{ \sqrt{ {F1}^{2} + {F2}^{2} - 2F1F2 \cos\alpha} }{ \sqrt{ {F1}^{2} + {F2}^{2} + 2F1F2 \cos \alpha } } = \sqrt{3} [/tex]

[tex] \frac{ {F1}^{2} + {F2}^{2} - 2F1F2 \cos \alpha }{ {F1}^{2} + {F2}^{2} + 2F1F2 \cos \alpha } = 3[/tex]

[tex] {F1}^{2} + {F2}^{2} - 2F1F2 \cos \alpha = 3( {F1}^{2} + {F2}^{2} + 2F1F2 \cos \alpha )[/tex]

[tex] {F1}^{2} + {F2}^{2} - 2F1F2 \cos \alpha = {3F1}^{2} + {3F2}^{2} + 6F1F2 \cos \alpha [/tex]

[tex] - 2F1F2 \cos \alpha - 6F1F2 \cos \alpha = - {F1}^{2} - {F2}^{2} + {3F1}^{2} + {3F2}^{2}[/tex]

[tex]- 8F1F2 \cos \alpha = {2F1}^{2} + {2F2}^{2}[/tex]

[tex]- 8 {F}^{2} \cos \alpha = {2F}^{2} + {2F}^{2}[/tex]

[tex]- 8 {F}^{2} \cos \alpha = {4F}^{2} [/tex]

[tex]\cos \alpha = \frac{4 {F}^{2} }{- 8 {F}^{2} } [/tex]

[tex] \cos \alpha = - \frac{1}{2} [/tex]

[tex] \alpha = 120°[/tex]

Jadi, sudut apit kedua vektor adalah 120°.