(a) 12.8 m, 150°;
(b) 3.30cm, 60.0°;
(c) 22.0 in., 215°.
Answer:
a) 12.8 m, 150° = (-11.09 i + 6.4 j) m
b)3.30 cm, 60.0° = (1.65 i + 2.86 j) cm
c)22.0 in., 215° = (-18.02 i + 9.30 j) in.
Explanation:
Consider vector having magnitude X and inclined θ degree to positive horizontal axis, its horizontal component = X cosθ and vertical component = X sinθ.
So X = X cosθ i + X sinθ j
a)12.8 m, 150°
= 12.8 cos 150 i + 12.8 sin 150 j = (-11.09 i + 6.4 j) m
b)3.30 cm, 60.0°
= 3.30 cos 60 i + 3.30 sin 60 j = (1.65 i + 2.86 j) cm
c)22.0 in., 215°
= 22.0 cos 215 i + 22.0 sin 215 j = (-18.02 i + 9.30 j) in.
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