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A displacement vector points in a direction of θ = 23° left of the positive y-axis. The magnitude of this vector is D = 155 m. Refer to the figure. Enter an expression for the x-component vector, Dx, in terms of D, θ, and the unit vectors i and j.

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stigawithfun

Answer:

Dₓ = -155 sin 23° i + 0 j

Explanation:

The diagram showing the vector has been attached to this response.

As shown in the diagram,

The vector D has an x-component (also called horizontal component) of -D sinθ i. i.e

Dₓ = -D sin θ i   [The negative sign shows that D lies in the negative x direction]

Where;

D = magnitude of D = 155m

θ = direction of D = 23°

Therefore;

Dₓ = -155 sin 23° i

Since Dₓ represents the x component, its unit vector, j component has a value of 0.

Therefore, Dₓ can be written in terms of D, θ and the unit vectors i and j as follows;

Dₓ = -155 sin 23° i + 0 j

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onyebuchinnaji

The x-component of the vector is written as [tex]D_x = (-60.56 \ i + 0j)\ m[/tex].

The given parameters;

  • direction of the vector, θ = 23°
  • magnitude of the vector, D = 155 m

The x-component of the vector is calculated as;

[tex]D_x = D \ \times sin\theta \\\\D_x = 155 \times sin(23)\\\\D_x = 60.56 \ m[/tex]

Since the direction of the vector is in negative x-axis, the x-component of the vectors is written as;

[tex]D_x = (-60.56 \ i + 0j)\ m[/tex]

Learn more here:https://brainly.com/question/13219489

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