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3 years ago

Find out the x-component and y-component of the following vectors:

Physics

2 answers:

Nina [5.8K]3 years ago

8 0

<h2>Answer:</h2><h2></h2>

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.

<h2>Explanation:</h2><h2></h2>

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 $M_{X}$ ) is given by;

$M_{X}$ = M cos θ

And its y-component (written as $M_{Y}$ ) is given by;

$M_{Y}$ = 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 $A_{X}$ ) is given by;

$A_{X}$ = A cos θ --------------------(i)

And its y-component (written as $A_{Y}$ ) is given by;

$A_{Y}$ = 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>

$A_{X}$ = 15.0 cos 150° = 15.0 x -0.8660 = -12.99cm

$A_{Y}$ = 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 $B_{X}$ ) is given by;

$B_{X}$ = B cos θ --------------------(iii)

And its y-component (written as $B_{Y}$ ) is given by;

$B_{Y}$ = 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>

$B_{X}$ = 12.0 cos 50° = 12.0 x 0.6428 = 7.71m

$B_{Y}$ = 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 $C_{X}$ ) is given by;

$C_{X}$ = B cos θ --------------------(v)

And its y-component (written as $C_{Y}$ ) is given by;

$C_{Y}$ = 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>

$C_{X}$ = 28.0 cos 220° = 28.0 x -0.7660 = -21.45m/s

$C_{Y}$ = 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 $D_{X}$ ) is given by;

$D_{X}$ = D cos θ --------------------(vii)

And its y-component (written as $D_{Y}$ ) is given by;

$D_{Y}$ = 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>

$D_{X}$ = 55.5 cos 325° = 28.0 x 0.8192 = 22.94m/s²

$D_{Y}$ = 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.

Leno4ka [110]3 years ago

6 0

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

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