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sladkih
3 months ago
12

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. R

efer to the figure. Enter an expression for the x-component vector, Dx, in terms of D, θ, and the unit vectors i and j.

Physics
1 answer:
Keith_Richards [3.2K]3 months ago
3 0

Answer:

Dₓ = -155 sin 23° i + 0 j

Explanation:

The accompanying diagram illustrates the vector.

According to the diagram,

The vector D has an x-component (or horizontal component) expressed as -D sinθ i. Specifically,

Dₓ = -D sin θ i   [The negative sign indicates that D is directed along the negative x-axis]

Where;

D = magnitude of D = 155m

θ = angle of D = 23°

Consequently;

Dₓ = -155 sin 23° i

Since Dₓ signifies the x component, its unit vector, j component holds a value of 0.

<pthus d="" can="" be="" formulated="" in="" terms="" of="" and="" the="" unit="" vectors="" i="" j="" as="" follows="">

Dₓ = -155 sin 23° i + 0 j

</pthus>
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