A displacement vector, has magnitude $$ d=197 \mathrm{~m} $$ and points in a direction $$ heta=21.6^{\circ} $$ measured counterclockwise from the positive $$ y $$ axis Enter an expression for the horizontal component of the vector d using the symbols provided.

Question

A displacement vector, has magnitude $$ d=197 \mathrm{~m} $$ and points in a direction $$ 	heta=21.6^{\circ} $$ measured counterclockwise from the positive $$ y $$ axis Enter an expression for the horizontal component of the vector d using the symbols provided.

UpStudy AI · Accepted Answer

The horizontal component ($$ d_x $$) of the displacement vector can be expressed using the magnitude $$ d $$ and the angle $$ \theta $$ as follows:

$$
d_x = d \cdot \sin(\theta)
$$

**Explanation:**

- The angle $$ \theta = 21.6^\circ $$ is measured counterclockwise from the positive $$ y $$-axis.
- To find the horizontal ( $$ x $$-axis) component, you use the sine of the angle:
  $$
  d_x = d \sin(\theta)
  $$
  
So, plugging in the values:
$$
d_x = 197 \, \text{m} \cdot \sin(21.6^\circ)
$$

This expression gives you the horizontal component of the displacement vector.

**Final Answer:**
$$
d_x = d \times \sin{\theta}
$$
The horizontal component of the vector is $$ d_x = 197 \, \text{m} \times \sin(21.6^\circ) $$.

A displacement vector, has magnitude \( d=197 \mathrm{~m} \) and points in a direction \( \theta=21.6^{\circ} \) measured counterclockwise from the positive \( y \) axis Enter an expression for the horizontal component of the vector d using the symbols provided.

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