Question

consider a vector 5 units long at an angle 53° to the horizontal.
a. determine the horizontal component of the vector.
vx = v cosθ v = 5 θ = 53° = 0.6
v = 5 cosθ = 0.6 vx = 5×0.6 = 3
b. determine the vertical component of the vector.
53° = 0.8 vy = v sin θ
v = 5 sinθ = 0.8 vy = 5×0.8 = 4
consider a vector 12.0 centimeters long at an angle 150° to the horizontal.
a. determine the horizontal component of the vector. provide your answer as a decimal.
v = 12 cm cosθ = vx / v v cosθ = vx vx=-10.4 cm
θ = 150 vx = 12 cos(150°)
b. determine the vertical component of the vector.

Explanation:

Step1: Recall vector - component formula

The horizontal component of a vector $\vec{V}$ with magnitude $V$ and angle $\theta$ with the horizontal is given by $V_x = V\cos\theta$, and the vertical component is given by $V_y=V\sin\theta$.

Step2: Calculate horizontal component for $V = 12$ cm and $\theta = 150^{\circ}$

We know that $V = 12$ cm and $\theta = 150^{\circ}$. Using the formula $V_x=V\cos\theta$, and $\cos(150^{\circ})=-\frac{\sqrt{3}}{2}\approx - 0.866$. So $V_x = 12\times\cos(150^{\circ})=12\times(- 0.866)=-10.392\approx - 10.4$ cm.

Step3: Calculate vertical component for $V = 12$ cm and $\theta = 150^{\circ}$

Using the formula $V_y = V\sin\theta$, and $\sin(150^{\circ})=\frac{1}{2}=0.5$. So $V_y=12\times\sin(150^{\circ})=12\times0.5 = 6$ cm.

Answer:

a. - 10.4 cm
b. 6 cm