If the x-component of a vector is 17, and the angle between the vector and the x-axis is 46 degrees, what is the magnitude of the vector? Ro

If the x-component of a vector is 17, and the angle between the vector and the x-axis is 46 degrees, what is the magnitude of the vector? Round your answer to the nearest hundredth.

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  1. Answer:

    24.47

    Explanation:

    cosθ = adjacent/hypotenuse = x-component/vector magnitude, so cos46° = 17/ vector magnitude. the vector magnitude is, therefore, 24.47

    FROM CK-12

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  2. Answer:

    17.00 N

    Explanation:

    Given that the x-component of a vector is 17, and the angle between the vector and the x-axis is 46 degrees

    The magnitude of the vector will be calculated by first resolving the vector into x component and y component.

    X – component

    17cos46 = 11.809

    Y component

    17sin46 = 12.229

    We will find the resultant vector by using pythagorean theorem

    R = sqrt ( X^2 + Y^2 )

    R = sqrt ( 11.809^2 + 12.229^2 )

    R = sqrt ( 288.995 )

    R = 16.999

    R = 17.00 N

    Therefore, the magnitude of the vector is 17 .00N

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