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10 February, 07:40

Wind resistance is jointly proportional to velocity and surface area of the object. If an object is traveling 45 mph and has a surface area of 49 square feet it will experience a wind resistance of 496.125 Newtons. How fast must an object be moving to experience a wind resistance of 164.25 Newtons if it has a surface area of 10 square feet

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Answers (2)
  1. 10 February, 07:44
    0
    Given:

    Velocity, v1 = 45 mph

    Converting from mph to m/s,

    45 mi/h * 1609.3 m/1 mi * 1 h/3600 s

    = 20.12 m/s

    surface area, A1 = 49 square feet

    Converting ft^2 to m^2,

    49 ft^2 * (0.305 m) ^2 / (1 ft) ^2

    = 4.56 m^2

    Wind resistance, R1 = 496.125 Newtons.

    Wind resistance, R2 = 164.25 Newtons

    Surface area, A2 = 10 square feet

    Converting ft^2 to m^2,

    10 ft^2 * (0.305 m) ^2 / (1 ft) ^2

    = 0.93 m^2

    Wind resistance, R is jointly proportional to velocity, v and surface area of the object, A

    R = k (v * A)

    R1 / (v1 * A1) = R2 / (v2 * A2)

    v2 = R2 * (v1 * A1) / (R1 * A2)

    = 164.25 * (20.12 * 4.56) / (496.125 * 0.93)

    = 32.66 m/s

    Converting from m/s to mph,

    32.66 m/s * 1 mi/1609.3 m * 3600 s/1 h

    = 73.06 mph
  2. 10 February, 07:55
    0
    an object will be moving 73 mph to experience a wind resistance of 164.25 Newtons if it has a surface area of 10 square feet

    Explanation:

    given information:

    object speed, v₁ = 45 mph

    surface area, s₁ = 49 ft²

    wind resistance, R₁ = 496.125 N

    if

    R₂ = 164.25 N

    s₂ = 10 ft²

    v₂ = ?

    first we need to find the k from the first case. we know that

    R₁ = k v₁ s₁

    k = R / (v s)

    = 496.125 / (45 x 49)

    = 0.225

    now we can find the v₂

    R₂ = k v₂ s₂

    v₂ = R₂ / k s₂

    = 164.25 / (0.225 x 10)

    = 73 mph
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