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Tanya [424]
3 years ago
8

What are (a) the x component and (b) the y component of a vector in the xy plane if its direction is 259° counterclockwise from

the positive direction of the x axis and its magnitude is 5.4 m?
Physics
2 answers:
kozerog [31]3 years ago
5 0
<h2>Answer:</h2>

(a) -1.030m

(b) -5.301m

<h2>Explanation:</h2>

Given a vector F in the xy plane, of magnitude F and in a direction θ counterclockwise from the positive direction of the x-axis;

The x-component (F_{X}) of vector F is given by;

F_{X} = F cos θ    ---------------------(i)

And;

The y-component (F_{Y}) of vector F is given by;

F_{Y} = F sin θ            -----------------------(ii)

Now to the question;

Let the vector be A

Therefore;

The magnitude of vector A is A = 5.4m

The direction θ of A counterclockwise from the positive direction of the x-axis = 259°

(a) The x-component (A_{X}) of the vector A is therefore given by;

A_{X} = A cos θ       ------------------------(iii)

Substitute the values of θ and A into equation (iii) as follows;

A_{X} = 5.4 cos 259°

A_{X} = 5.4 x (-0.1908)

A_{X} = -1.030

Therefore, the x-component of the vector is -1.030m

(b) The y-component (A_{Y}) of the vector A is therefore given by;

A_{Y} = A sin θ       ------------------------(iv)

Substitute the values of θ and A into equation (iv) as follows;

A_{Y} = 5.4 sin 259°

A_{Y} = 5.4 x (-0.9816)

A_{Y} = -5.301m

Therefore, the y-component of the vector is -5.301m

garik1379 [7]3 years ago
3 0

Answer:

|Ax| =1.03 m  (directed towards negative x-axis)

|Ay|= 5.30 (directed towards negative y-axis)

Explanation:

Let A is a vector = 5.4 m

θ = 259°

to Find Ax, Ay

Sol:

according the condition it lies in 3rd quadrant

we know that Horizontal Component Ax = A cos θ

Ax = 5.4 Cos 259°

Ax = - 1.03 m

|Ax| =1.03 m  (directed towards negative x-axis)

Now Ay = A sin θ

Ay = 5.4 Sin 259°

Ay = -5.30

|Ay|= 5.30 (directed towards negative y-axis)

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