Respuesta :
Answer:
Explanation:
Given
mass of duck=2.40 kg
[tex]F_1[/tex]=0.150 N
[tex]F_2[/tex]=0.160 N in a direction south of east
Net force in x- direction
[tex]F_x=0.150+0.160\cos 45=0.263 N[/tex]
[tex]a_x=\frac{0.263}{2.4}=0.109 N[/tex]
net force in Y direction
[tex]F_y=0.160\times \sin 45=0.113 N[/tex]
[tex]a_y=\frac{0.113}{2.4}=0.047 m/s^2[/tex]
Thus displacement in x direction
[tex]x=ut+\frac{at^2}{2}[/tex]
[tex]x=0.15\times 2.8+\frac{0.109\times 2.8^2}{2}=0.847 m[/tex]
Displacement in Y direction
[tex]Y=ut+\frac{at^2}{2}[/tex]
here u=0
[tex]Y=0+\frac{0.047\times 2.8^2}{2}=0.184 m[/tex]
net displacement [tex]\sqrt{0.847^2+0.184^2}=\sqrt{0.7513}[/tex]
=0.866 m
Direction of displacement
[tex]tan\theta =\frac{0.184}{0.847}[/tex]
[tex]\theta =12.25^{\circ}[/tex] south of east
[tex]\theta =-12.25^{\circ}[/tex]
