Only two forces act on an object (mass=4.00 kg), as in the drawing. Find (a) the magnitude and (b) the direction (relative to the x axis) of the acceleration of the object.

It matters how the forces are applied. Do a vector addition, then Net force=mass*acceleration where force is a vector, as is acceleration.

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To find the magnitude and direction of the acceleration of the object, we first need to determine the net force acting on it.

In the drawing, we can see that there are two forces acting on the object. Let's call the force along the x-axis as F1 and the force along the y-axis as F2.

To find the net force in the x-axis direction, we need to find the x-components of each force. Looking at the drawing, we can see that F1 is acting horizontally along the x-axis. Therefore, the x-component of F1 is F1x = F1 * cosθ, where θ is the angle between F1 and the x-axis.

Since we are not given any specific values or the angle θ, we cannot determine the value of F1x.

Moving on to the y-component of the forces, we have F2 acting vertically along the y-axis. Therefore, the y-component of F2 is F2y = F2 * sinθ.

Similarly, without any specific values or the angle θ, we cannot determine the value of F2y.

To find the net force, we need to sum up the x-components and y-components of the forces. The net force in the x-axis direction is the sum of all x-components, and the net force in the y-axis direction is the sum of all y-components.

Since we don't have any specific values or angles, we cannot compute the net forces. Therefore, we cannot determine the magnitude or direction of the acceleration of the object.

To find the magnitude and direction of the acceleration of the object, we need to analyze the forces acting on it.

First, let's label the forces in the given drawing:
- The force F1 with a magnitude of 5.00 N is acting at an angle of 30 degrees above the positive x-axis.
- The force F2 with a magnitude of 8.00 N is acting at an angle of 45 degrees below the positive x-axis.

To find the net force acting on the object, we need to break down the forces into their x and y components.

For Force F1:
- The x-component (F1x) is given by F1 * cos(theta), where theta is the angle above the x-axis.
- The y-component (F1y) is given by F1 * sin(theta).

For Force F2:
- The x-component (F2x) is given by F2 * cos(theta), where theta is the angle below the x-axis.
- The y-component (F2y) is given by -F2 * sin(theta) since the force is acting in the negative y-direction.

Now, let's calculate the x and y components of the forces:
F1x = 5.00 N * cos(30°) = 4.33 N
F1y = 5.00 N * sin(30°) = 2.50 N

F2x = 8.00 N * cos(45°) = 5.66 N
F2y = -8.00 N * sin(45°) = -5.66 N

Next, let's calculate the net forces in the x and y directions by adding the corresponding components:
Net Force in the x-direction (Fnetx) = F1x + F2x = 4.33 N + 5.66 N = 9.99 N
Net Force in the y-direction (Fnety) = F1y + F2y = 2.50 N - 5.66 N = -3.16 N

Now, to find the magnitude of the net force:
Magnitude of the net force (|Fnet|) = sqrt(Fnetx^2 + Fnety^2)

|Fnet| = sqrt((9.99 N)^2 + (-3.16 N)^2) = sqrt(99.8 N^2 + 9.99 N^2) = sqrt(109.8 N^2) = 10.48 N

So, the magnitude of the net force is 10.48 N.

To find the direction relative to the x-axis, we can use trigonometric functions:

Direction (θ) = arctan(Fnety / Fnetx)

θ = arctan((-3.16 N) / (9.99 N)) = -17.6°

Therefore, the direction of the acceleration is 17.6° below the negative x-axis.

Hence, the answers are:
(a) The magnitude of the acceleration is 10.48 N.
(b) The direction of the acceleration is 17.6° below the negative x-axis.