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jame asked in Science & MathematicsMathematics · 10 months ago

Vectors Math Problem Help?

a) Determine the magnitude of the resultant vector, u + v, to the nearest tenth of a newton.

b) Determine the direction of the resultant vector, u + v, to the nearest degree.

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  • TomV
    Lv 7
    10 months ago
    Favorite Answer

    Resultant Vector:

    <R> = <30 + 50cos42°, 50sin42°>

    a) |R| = √[(30 + 50cos42°)² + (50sin42°)²]

    = √(900 + 3000cos42° + 2500)

    = √(3400 + 3000cos42°)

    ≈ 75.0 N (to the nearest 10th)

    b) Arg<R> = arctan[50sin42°/(30 + 50cos42°)]

    ≈ arctan(33.45653/67.15724)

    ≈ 26° (to the nearest degree)

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  • 10 months ago

    Sum the horizontal components:

    30+50*cos42 = 67.2

    Sum the vertical components:

    50*sin42 = 33.5

    arctan 33.5/67.2 = 26.5° <<<<<

    67.2²+33.5² = 75²

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  • Mike G
    Lv 7
    10 months ago

    s = resultant vector magnitude

    θ = direction

    Cosine Rule

    s^2 = 30^2+50^2-2*30*50cos138

    s = 75 N

    sinθ/50 = sin138/75.03

    θ = 26.5°

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  • Ian H
    Lv 7
    10 months ago

    M^2 = x^2 + y^2 = (30 + 50c)^2 + (50s)^2

    M^2 = 2500(s^2 + c^2) + 900 + 3000c = 3400 + 3000cos42

    M ~ √[5629.43] ~ 72. 0296 or 72.0 to nearest tenth

    tan(θ) = 50 sin(42)/[30 + 50 cos(42)]

    θ = artan(0.498182) ~ 26.48 degrees or 26 to nearest degree.

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  • ted s
    Lv 7
    10 months ago

    U + V = < - 30 + 50 cos 42° , 50 sin 42° >...| U + V | = √ ( 2590 - 600 cos 42° )

    tan Θ = [ 50 sin 42° / (- 30 +50 cos 42°) ]....Θ ≈102 °

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  • 10 months ago

    It's already drawn for you...?

    Magnitude sqrt[ (50sin42°)^2 + (30+50cos42°)^2 ]

    Direction arctan[ 50sin42° / (30+50cos42°) ] counterclockwise from horizontal

    • ted s
      Lv 7
      10 months agoReport

      should be ' - 30 '

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  • david
    Lv 7
    10 months ago

    180 - 42 = 138

    c^2 = 30^2 + 50^2 - 2*30*50*cos138 = 5629.43

    c = 75.03 N <<< mag. or resultant

    sin X / 50 = sin 138 / 75.03

    x = 26.48 degrees <<< angle for the resultant

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