Question

Find the magnitude of the sumof these two vectors:5.60 m A,30.0°B 6.00 mmagnitude (m)

Answer 1

In order to find the magnitude of the vectors given, let's find first the components

The component in the x-axis

$C_X=5\cos (0)+6\cos (180+30)$$C_X=-0.196$

then the component in the y-axis

$C_Y=5\sin (0)+6\sin (180+30)=-3$

Then the magnitude can be calculated with the next formula

$|M|=\sqrt[]{Cx^2+C_Y^2}$

we substitute the values

$|M|=\sqrt[]{(-0.196)^2+(-3)^2}=3.01m$

The magnitude is 3.01m

then for the direction, we can use the next formula

$\theta=\tan ^{-1}(\frac{C_Y}{C_x})$

we substitute the values

$\theta=\tan ^{-1}(\frac{-3}{-0.196})=86.26\text{\degree}$

then because both components are negative the direction will be

$\text{direction}=86.26+180=266.26\text{\degree}$