Taking east as positive gives (R=4-3=1) km east.
(V=sqrt{(-5)^2+(-12)^2}=13) N.
The first force contributes (100cos30^circ=86.6) N horizontally. The vertical force has no x-component.
(F_x=50cos37^circ=50(0.8)=40) N.
(V=sqrt{8^2+(-6)^2}=10) m.
(R_x=8-3+5=10) N.
(R_y=6-10+4=0) N.
(V_y=sqrt{25^2-7^2}=sqrt{576}=24) m.
(F_x=20cos120^circ=-10) N and (F_y=20sin120^circ=17.32) N.
(V=sqrt{18^2+24^2}=30) N.
For perpendicular vectors (R=sqrt{10^2+10^2}=10sqrt2) N.
The new components are 8 N and 3 N. Thus (R=sqrt{8^2+3^2}=sqrt{73}) N.
(V_x=Vcostheta). Cosine decreases from 1 to 0 over this range.
(V_y=Vsintheta). It increases from zero at 0° to V at 90°.
(V_y=Vsin30^circ=10(0.5)=5) N.
(V_x=40(0.6)=24) N and (V_y=40(0.8)=32) N.
Parallel vectors pointing in the same direction can be added directly.
The displacements are perpendicular. (R=sqrt{6^2+8^2}=10) m.
Negative x and positive y components place the vector in quadrant II.
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