Question

a vector a has components a x equals -5.00 m in a y equals 9.00 meters find the magnitude and the direction of the vector

Answer 1

Answer:

The magnitude = 10.30 m

The direction of the vector proceeds at angle of 119.05°

Explanation:

Given that:

A vector $\bar A$ has component $A_x$ = -5 m and $A_y$ = 9 m

The magnitude of vector  $\bar A$ can be represented as:

$\bar A$  = $\sqrt{A_x^2 + A_y^2}$

$\bar A$  = $\sqrt{(-5)^2 + (9)^2}$

$\bar A$  = $\sqrt{25 + 81}$

$\bar A$  = $\sqrt{106}$

$\bar A$  = 10.30 m

If we make $\bar A$  an angle $\theta$ with y- axis:

Then;   tan $\theta$  = $\frac{A_x}{A_y}$

tan $\theta$  = $\frac{5}{9}$

tan $\theta$  = 0.555

$\theta$  = tan⁻¹ (0.555)

$\theta$  = 29.05°

Angle with positive x-axis = 90 + $\theta$  

= 90° + 29.05°

= 119.05°