Respuesta :
Answer:
A = 15.0 cm; the vector is 30.00 north of west; = 15 cm N30W
Y - component = 7.5cm and X - component = -12.99 cm
B = 12.0 m; the vector is 55.00 north of east; = 12.0 m N55E
Y - component = 9.83 m and X - component = 6.88 m
C = 28.0 m/s; the vector is 40.00 south of west; = 28.0 m/s S40W
Y - component = - 17.998 m/s and X - component = - 21.448 m/s
D = 55.5 m/s2; the vector is 35.00 south of east; = 55.5 m/s² S35E
Y - component = - 31.835 m/s² and X - component = 45.466 m/s²
Explanation:
1. A = 15.0 cm; the vector is 30.00 north of west; = 15 cm N30W
Y - component = 15cm × sin30 = 7.5cm
X - component = -15cm × cos30 = -12.99 cm
2. B = 12.0 m; the vector is 55.00 north of east; = 12.0 m N55E
Y - component = 12.0 m × sin55 = 9.83 m
X - component =12.0 m × cos55 = 6.88 m
3. C = 28.0 m/s; the vector is 40.00 south of west; = 28.0 m/s S40W
Y - component = - 28.0 m/s × sin40 = - 17.998 m/s
X - component = - 28.0 m/s × cos40 = - 21.448 m/s
4. D = 55.5 m/s2; the vector is 35.00 south of east; = 55.5 m/s² S35E
Y - component = - 55.5 m/s² × sin35 = - 31.835 m/s²
X - component = 55.5 m/s² × cos35 = 45.466 m/s²
Answer:
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.
Explanation:
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 [tex]M_{X}[/tex]) is given by;
[tex]M_{X}[/tex] = M cos θ
And its y-component (written as [tex]M_{Y}[/tex]) is given by;
[tex]M_{Y}[/tex] = M sin θ
Therefore, taking due east as positive x axis, for the vector:
(i) A = 15.0 cm; the vector is 30.00 north of west;
the x-component of A (written as [tex]A_{X}[/tex]) is given by;
[tex]A_{X}[/tex] = A cos θ --------------------(i)
And its y-component (written as [tex]A_{Y}[/tex]) is given by;
[tex]A_{Y}[/tex] = A sin θ --------------------(ii)
Where;
A = 15.0cm
θ = 30.00 north of west = 180° - 30.00° [measured due east] = 150°
Substitute these values into equations (i) and (ii) as follows;
[tex]A_{X}[/tex] = 15.0 cos 150° = 15.0 x -0.8660 = -12.99cm
[tex]A_{Y}[/tex] = 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.
(ii) B = 12.0 m; the vector is 55.00 north of east;
the x-component of B (written as [tex]B_{X}[/tex]) is given by;
[tex]B_{X}[/tex] = B cos θ --------------------(iii)
And its y-component (written as [tex]B_{Y}[/tex]) is given by;
[tex]B_{Y}[/tex] = B sin θ --------------------(iv)
Where;
B = 12.0m
θ = 55.0 north of east = 50° [measured due east]
Substitute these values into equations (iii) and (iv) as follows;
[tex]B_{X}[/tex] = 12.0 cos 50° = 12.0 x 0.6428 = 7.71m
[tex]B_{Y}[/tex] = 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.
(iii) C = 28.0 m/s; the vector is 40.00 south of west;
the x-component of C (written as [tex]C_{X}[/tex]) is given by;
[tex]C_{X}[/tex] = B cos θ --------------------(v)
And its y-component (written as [tex]C_{Y}[/tex]) is given by;
[tex]C_{Y}[/tex] = B sin θ --------------------(vi)
Where;
C = 28.0m/s
θ = 40.00 south of west = 180° + 40.00° [measured due east] = 220°
Substitute these values into equations (v) and (vi) as follows;
[tex]C_{X}[/tex] = 28.0 cos 220° = 28.0 x -0.7660 = -21.45m/s
[tex]C_{Y}[/tex] = 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.
(iv) D = 55.5 m/s²; the vector is 35.00 south of east;
the x-component of D (written as [tex]D_{X}[/tex]) is given by;
[tex]D_{X}[/tex] = D cos θ --------------------(vii)
And its y-component (written as [tex]D_{Y}[/tex]) is given by;
[tex]D_{Y}[/tex] = D sin θ --------------------(viii)
Where;
D = 55.5m/s²
θ = 35.00 south of east = 360° - 35.00° [measured due east] = 325°
Substitute these values into equations (vii) and (viii) as follows;
[tex]D_{X}[/tex] = 55.5 cos 325° = 28.0 x 0.8192 = 22.94m/s²
[tex]D_{Y}[/tex] = 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.
