QuestionJuly 16, 2025

A displacement vector is 28.0 m long and makes an angle of 60.0^circ counterclockwise from the y-axis (on the side of the -x axis). What is the x-component of this vector? (Include a minus sign if necessary.) square lm

A displacement vector is 28.0 m long and makes an angle of 60.0^circ counterclockwise from the y-axis (on the side of the -x axis). What is the x-component of this vector? (Include a minus sign if necessary.) square lm
A displacement vector is 28.0 m long and makes an angle of 60.0^circ  counterclockwise from the y-axis (on the side of the -x
axis).
What is the x-component of this vector? (Include a minus sign if necessary.)
square  lm

Solution
4.5(184 votes)

Answer

x = -28.0 \cdot \cos(30.0^{\circ}) = -24.2 \, \text{m} Explanation 1. Identify the angle with the x-axis The vector makes an angle of 60.0^{\circ} with the y-axis, so it makes an angle of 30.0^{\circ} with the x-axis. 2. Calculate the x-component using cosine Use the formula for the x-component: **x = r \cdot \cos(\theta)**, where r = 28.0 \, \text{m} and \theta = 30.0^{\circ}. 3. Apply the sign based on direction Since the vector is on the side of the -x axis, the x-component will be negative.

Explanation

1. Identify the angle with the x-axis<br /> The vector makes an angle of $60.0^{\circ}$ with the y-axis, so it makes an angle of $30.0^{\circ}$ with the x-axis.<br />2. Calculate the x-component using cosine<br /> Use the formula for the x-component: **$x = r \cdot \cos(\theta)$**, where $r = 28.0 \, \text{m}$ and $\theta = 30.0^{\circ}$.<br />3. Apply the sign based on direction<br /> Since the vector is on the side of the $-x$ axis, the x-component will be negative.
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