Practice Questions

(V=sqrt{(-5)^2+(-12)^2}=13) N.

Taking east as positive gives (R=4-3=1) km east.

The displacements are perpendicular. (R=sqrt{6^2+8^2}=10) m.

(V_x=40(0.6)=24) N and (V_y=40(0.8)=32) N.

(V_y=Vsintheta). It increases from zero at 0° to V at 90°.

(V_x=Vcostheta). Cosine decreases from 1 to 0 over this range.

The new components are 8 N and 3 N. Thus (R=sqrt{8^2+3^2}=sqrt{73}) N.

For perpendicular vectors (R=sqrt{10^2+10^2}=10sqrt2) N.

(V=sqrt{18^2+24^2}=30) N.

(F_x=20cos120^circ=-10) N and (F_y=20sin120^circ=17.32) N.