Forces of 10.6 N north, 20.8 N east, and 14.3 N south are simultaneously applied to a 3.76 kg mass as it rests on an air table. What is the magnitude of its acceleration?

Find the magnitude of the resultant force

F=√(10.6²+20.8²)
a=F/m

To find the magnitude of acceleration, you can use Newton's second law of motion:

F = ma

Where F is the net force applied to the object, m is the mass of the object, and a is the acceleration of the object.

In this case, you have three forces acting on the mass simultaneously: 10.6 N north, 20.8 N east, and 14.3 N south.

First, determine the net force by adding up the forces in each direction. For the north and south forces, since they are in opposite directions, you subtract the smaller force from the larger one:

Net force in the north direction = 10.6 N - 14.3 N = -3.7 N

Forces in the east and west directions don't cancel each other out, so you add them up:

Net force in the east direction = 20.8 N

Now you have the net force in both the north and east directions. To find the magnitude of the net force, you can use the Pythagorean theorem:

Net force = √(net force in north direction)^2 + (net force in east direction)^2

Net force = √((-3.7 N)^2 + (20.8 N)^2)

Net force = √(13.69 N^2 + 432.64 N^2)

Net force = √446.33 N^2

Net force ≈ 21.14 N

Now that you have the net force, you can use Newton's second law to find the acceleration:

a = F/m

Where F is the net force and m is the mass of the object.

a = 21.14 N / 3.76 kg

a ≈ 5.63 m/s^2

Therefore, the magnitude of the acceleration of the 3.76 kg mass is approximately 5.63 m/s^2.