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Answer:</h2><h2>
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The x- and y- components of vector A are -12.99cm and 7.5cm respectively.
The x- and y- components of vector B are 7.71m and 9.20m respectively.
The x- and y- components of vector C are -21.45m/s and -18.00m/s respectively.
The x- and y- components of vector D are 22.94m/s² and -16.06m/s² respectively.
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Explanation:</h2><h2>
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Given an arbitrary vector M (written in bold face), of magnitude M (not written in bold face), which makes an angle θ to the positive x axis, the x-component of M (written as ) is given by;
= M cos θ
And its y-component (written as ) is given by;
= M sin θ
Therefore, taking due east as positive x axis, for the vector:
<em>(i) A = 15.0 cm; the vector is 30.00 north of west; </em>
the x-component of A (written as ) is given by;
= A cos θ --------------------(i)
And its y-component (written as ) is given by;
= A sin θ --------------------(ii)
<em>Where;</em>
A = 15.0cm
θ = 30.00 north of west = 180° - 30.00° [measured due east] = 150°
<em>Substitute these values into equations (i) and (ii) as follows;</em>
= 15.0 cos 150° = 15.0 x -0.8660 = -12.99cm
= 15.0 sin 150° = 15.0 x 0.5 = 7.5cm
Therefore, the x- and y- components of vector A are -12.99cm and 7.5cm respectively.
<em>(ii) B = 12.0 m; the vector is 55.00 north of east; </em>
the x-component of B (written as ) is given by;
= B cos θ --------------------(iii)
And its y-component (written as ) is given by;
= B sin θ --------------------(iv)
<em>Where;</em>
B = 12.0m
θ = 55.0 north of east = 50° [measured due east]
<em>Substitute these values into equations (iii) and (iv) as follows;</em>
= 12.0 cos 50° = 12.0 x 0.6428 = 7.71m
= 12.0 sin 50° = 12.0 x 0.7660 = 9.20m
Therefore, the x- and y- components of vector B are 7.71m and 9.20m respectively.
<em>(iii) C = 28.0 m/s; the vector is 40.00 south of west; </em>
the x-component of C (written as ) is given by;
= B cos θ --------------------(v)
And its y-component (written as ) is given by;
= B sin θ --------------------(vi)
<em>Where;</em>
C = 28.0m/s
θ = 40.00 south of west = 180° + 40.00° [measured due east] = 220°
<em>Substitute these values into equations (v) and (vi) as follows;</em>
= 28.0 cos 220° = 28.0 x -0.7660 = -21.45m/s
= 28.0 sin 220° = 28.0 x -0.6428 = -18.00m/s
Therefore, the x- and y- components of vector C are -21.45m/s and -18.00m/s respectively.
<em>(iv) D = 55.5 m/s²; the vector is 35.00 south of east; </em>
the x-component of D (written as ) is given by;
= D cos θ --------------------(vii)
And its y-component (written as ) is given by;
= D sin θ --------------------(viii)
<em>Where;</em>
D = 55.5m/s²
θ = 35.00 south of east = 360° - 35.00° [measured due east] = 325°
<em>Substitute these values into equations (vii) and (viii) as follows;</em>
= 55.5 cos 325° = 28.0 x 0.8192 = 22.94m/s²
= 55.5 sin 325° = 28.0 x -0.5736= -16.06m/s²
Therefore, the x- and y- components of vector D are 22.94m/s² and -16.06m/s² respectively.