A vector is 253m long and points in a 55.8degree direction. Find the y-component of the vector.
2 answers:
Answer:
The answer to your question is: 209 m
Explanation:
Data
length = 253 m
d = 55.8°
y - component = ?
Formula
To solve this problem we need to use a right triangle a the trigonometric functions (sine, cosine, tangent, etc)
Now we have to choose among the trigonometric functions which one to use that relates the opposite side and the hypotenuse.
And that one is sineФ = os/h
we clear os = sineФ x h
os = sine 55.8 x253
os = 209 m
soh, cah, toa
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Answer:
The length of the flagpole is approximately 87.43 m
Explanation:
The given parameters of the cable attached to the flagpole are;
The point along the flagpole at which the cable is attached = 32.0 m
The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°
The downward force exerted by the cable, = 1.233 × 10⁴ N
The force exerted by the cable to the left = 1.233 × 10⁴ N
Let 'W' represent the weight of the flagpole, at equilibrium, we have;
The sum of vertical forces = 0
Therefore;
+ W - R = 0
W - R = -1.233 × 10⁴ N
Taking moment about the support at the base of the pole, we get;
1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0
∴ W × d/2 ×cos(60.0°) ≈ 4513.093·d
W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N
R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373
R ≈ 30,382.373 N
Taking moment about the point of attachment of the cable to the ground, we have;
W × ((d/2) × cos(60.0°) + 32) = R × 32
∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281
d = 2 × 43.71281 ≈ 87.43
The length of the flagpole, d ≈ 87.43 m