Only two horizontal forces act on a 9.1 kg body. One force is 5.3 N, acting due east, and the other is 1.6 N, acting 56° north of west. What is the magnitude of the body's acceleration?

figure the acceleration in each direction.

a1=5.3/9.1 E

a2=1.6/9.1 in direction 53g NofW

You know the angle between, 180-56=124

Law of cosines:
magnitude acceleration^2=a1^2+a2^2-a1*a2*cos124

To find the magnitude of the body's acceleration, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

In this case, we have two horizontal forces acting on the body: one due east with a magnitude of 5.3 N, and the other 56° north of west with a magnitude of 1.6 N.

To find the net force acting on the body, we need to resolve the force of 1.6 N into its east and north components. We can do this using trigonometry.

The north component of the force is given by: 1.6 N * sin(56°) = 1.6 N * 0.829 = 1.327 N (approx.)

The east component of the force is given by: 1.6 N * cos(56°) = 1.6 N * 0.559 = 0.894 N (approx.)

Now we can calculate the net force acting on the body in the horizontal direction.

Net force = 5.3 N (east) - 0.894 N (east) = 4.406 N (east)

Now we can use Newton's second law to find the acceleration of the body.

Net force = mass * acceleration

Rearranging the equation, we get:

Acceleration = Net force / mass

Acceleration = 4.406 N / 9.1 kg = 0.484 m/s² (approx.)

Therefore, the magnitude of the body's acceleration is approximately 0.484 m/s².