Question

Vector G is 40.3 m long in a

Answer 1

Answer:

Approximately 92.51.

Not sure what the desired rounding is since it isn't listed.

Step-by-step explanation:

So the first vector G is 40.3 m long in a -35 degree direction.

Lat's find the components of G.

$G_x=40.3\cos(-35)=33.0118$.

$G_y=40.3\sin(-35)=-23.1151$.

The second vector H is 63.3 m long in a 270 degree direction.

$H_x=63.3\cos(270)=0$.

$H_y=63.3\sin(270)=-63.3$.

The resultant vector can be found by adding the corresponding components:

$R_x=G_x+H_x=33.0118+0=33.0118$

$R_y=G_y+H_y=-23.1151+(-63.3)=-86.4151$

Now we are asked to find the magnitude of $(R_x,R_y)$ which is given by the formula $\sqrt{R_x^2+R_y^2}$.

Since $(R_x,R_y)=(33.0118,-86.4151)$ then the magnitude is $\sqrt{(33.0118)^2+(-86.4151)^2}=\sqrt{8557.34844}=92.51$.