#### Answer

(a) The magnitude of the vector is 5.7 and the direction is an angle of $45^{\circ}$ above the negative x-axis.
(b) The magnitude of the vector is 2.2 cm and the direction is an angle of $26.6^{\circ}$ below the negative x-axis.
(c) The magnitude of the vector is 100.5 m/s and the direction is an angle of $84^{\circ}$ below the negative x-axis.
(d) The magnitude of the vector is $22~m/s^2$ and the direction is an angle of $26.6^{\circ}$ above the positive x-axis.

#### Work Step by Step

(a) We can find the magnitude of the vector.
$B = \sqrt{B_x^2+B_y^2}$
$B = \sqrt{(-4.0)^2+(4.0)^2}$
$B = 5.7$
We can find the angle above the negative x-axis.
$tan(\theta) = \frac{4.0}{4.0}$
$\theta = tan^{-1}(\frac{4.0}{4.0}) = 45^{\circ}$
The magnitude of the vector is 5.7 and the direction is an angle of $45^{\circ}$ above the negative x-axis.
(b) We can find the magnitude of the vector.
$r = \sqrt{r_x^2+r_y^2}$
$r = \sqrt{(-2.0~cm)^2+(-1.0~cm)^2}$
$r = 2.2~cm$
We can find the angle below the negative x-axis.
$tan(\theta) = \frac{1.0}{2.0}$
$\theta = tan^{-1}(\frac{1.0}{2.0}) = 26.6^{\circ}$
The magnitude of the vector is 2.2 cm and the direction is an angle of $26.6^{\circ}$ below the negative x-axis.
(c) We can find the magnitude of the vector.
$v = \sqrt{v_x^2+v_y^2}$
$v = \sqrt{(-10~m/s)^2+(-100~m/s)^2}$
$v = 100.5~m/s$
We can find the angle below the negative x-axis.
$tan(\theta) = \frac{100}{10}$
$\theta = tan^{-1}(\frac{100}{10}) = 84^{\circ}$
The magnitude of the vector is 100.5 m/s and the direction is an angle of $84^{\circ}$ below the negative x-axis.
(d) We can find the magnitude of the vector.
$a = \sqrt{a_x^2+a_y^2}$
$a = \sqrt{(20~m/s^2)^2+(10~m/s^2)^2}$
$a = 22~m/s^2$
We can find the angle above the positive x-axis.
$tan(\theta) = \frac{10}{20}$
$\theta = tan^{-1}(\frac{10}{20}) = 26.6^{\circ}$
The magnitude of the vector is $22~m/s^2$ and the direction is an angle of $26.6^{\circ}$ above the positive x-axis.