A..........................................
Answer:
Option (2)
Explanation:
From the figure attached,
Horizontal component, 
![A_x=12[\text{Sin}(37)]](https://tex.z-dn.net/?f=A_x%3D12%5B%5Ctext%7BSin%7D%2837%29%5D)
= 7.22 m
Vertical component, ![A_y=A[\text{Cos}(37)]](https://tex.z-dn.net/?f=A_y%3DA%5B%5Ctext%7BCos%7D%2837%29%5D)
= 9.58 m
Similarly, Horizontal component of vector C,
= C[Cos(60)]
= 6[Cos(60)]
= 
= 3 m
![C_y=6[\text{Sin}(60)]](https://tex.z-dn.net/?f=C_y%3D6%5B%5Ctext%7BSin%7D%2860%29%5D)
= 5.20 m
Resultant Horizontal component of the vectors A + C,
m
= 4.38 m
Now magnitude of the resultant will be,
From ΔOBC,

= 
= 
= 6.1 m
Direction of the resultant will be towards vector A.
tan(∠COB) = 
= 
= 
m∠COB = 
= 46°
Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.
Option (2) will be the answer.
Vo= 331+0.6T
360=331+0.6T
360-331=0.6T
29=0.6T
0.6T/29
T=6/290 so change it to simplest form and us formulas good luck
<span>work =V*Q
=12*50*10^-6
The total work done will be equal to
work = V.Q
which means
w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>