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muhammad-azhar-hadi
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pembahasan soal fisika vektor
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Bx = B . Cos θ
-6 = B. Cos θ
B = -6 / cos θ
By = B. Sin θ
-6 = B. Sin θ
B= -6 / sin θ
−6
cos θ=
−6
sin θ
sin θ
cos θ=
−6
−6
sin θ
cos θ= 1
tan 𝜃 = 1
Θ= arc tan 1 = 45o
Untuk cari nilai B, maka
Bx = B . Cos θ
-6 = B. Cos 45
-6 = B . 1
2√2
B = −6.2
√2 = - 6√2