Two forces act on a 16 kg object. the first force has a magnitude of 68 n and is directed 24 degrees north of east. the second force is 32n , 48 degrees of west. what is the acceleration of the object resulting from the application of these forces? a. 1.6 m/s^2, 5.5 degrees north of west b. 1.9 m/s^2, 18 degrees north of west c. 2.4 m/s^2 , 34 degrees north of east d. 3.6 m/s^2 , 5.5 degrees north of west e.4.1 m/s^2, 52 degrees north of east

e. 4.1 m/s^2, 52 degrees north of east

Explanation:

Two forces act on a 16 kg object. The first force has a magnitude of 68 N and is directed 24 degrees north of east. The second force is 32N , 48 degrees of west. What is the acceleration of the object resulting from the application of these forces? a. 1.6 m/s^2, 5.5 degrees north of west b. 1.9 m/s^2, 18 degrees north of west c. 2.4 m/s^2 , 34 degrees north of east d. 3.6 m/s^2 , 5.5 degrees north of west e.4.1 m/s^2, 52 degrees north of east

There are two components of force

finding the sum of the horizontal component of the forces

Efx

 68 cos 24 - 32cos48 = 40.7088 N

Resolve the sum of the vertical components  of the forces

    68 sin 24   +  32 Sin 48 = 51.438  

now we find the resultant of the two components

R=(Fy^2+Fx^2)^0.5

      resultant force =   (51.438 ^ 2  +  40.7088 ^ 2)^ 1/2  

    = 65.597

recall that force is the product of mass and acceleration

ma = 65.597

m=16 kg

 a = 65.597/16 = 4.09 m/s/s

tan=EFy/EFx

    tan  = 51.438/ 40.7088 =

1.263

taking the tan inverse of both sides

 = 51.64 degree north of east

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

your answer is c! hope this : )