1 - To find the horizontal component of the 20 N force, we can use the equation F_horizontal = F * cos(theta), where F is the force and theta is the angle from the horizontal.

F_horizontal = 20 N * cos(30 degrees)
F_horizontal = 20 N * (sqrt(3)/2) (cosine of 30 degrees is sqrt(3)/2)
F_horizontal = 10 * sqrt(3) N ≈ 17.32 N

The horizontal component of the force is approximately 17.32 N.

2 - To find the net force acting on the object, we first need to determine the horizontal and vertical components of each force.

Force 1:
F1_horizontal = 5 N * cos(30 degrees) = 5 N * (sqrt(3)/2) ≈ 4.33 N
F1_vertical = 5 N * sin(30 degrees) = 5 N * (1/2) = 2.5 N

Force 2:
F2_horizontal = 8 N * cos(60 degrees) = 8 N * (1/2) = 4 N
F2_vertical = 8 N * sin(60 degrees) = 8 N * (sqrt(3)/2) ≈ 6.93 N

Now, add the horizontal and vertical components of each force:

Net_horizontal = F1_horizontal + F2_horizontal = 4.33 N + 4 N ≈ 8.33 N
Net_vertical = F1_vertical + F2_vertical = 2.5 N + 6.93 N ≈ 9.43 N

Finally, we can use the Pythagorean theorem to find the net force:

Net_force = sqrt(Net_horizontal^2 + Net_vertical^2) ≈ sqrt((8.33 N)^2 + (9.43 N)^2) ≈ sqrt(69.44 N^2 + 88.96 N^2) ≈ sqrt(158.4 N^2) ≈ 12.58 N

The net force acting on the object is approximately 12.58 N.

3 - To find the component of the 500 N force that acts in the upward direction, we can use the equation F_vertical = F * sin(theta).

F_vertical = 500 N * sin(40 degrees)
F_vertical ≈ 500 N * 0.6428 (sine of 40 degrees is approximately 0.6428)
F_vertical ≈ 321.4 N

The component of the force that acts in the upward direction is approximately 321.4 N.

4 - Since the object is moving with a constant velocity of 10 m/s to the right, it means that there is no net force acting on the object. When a force of 5 N is applied in the opposite direction, there will be a net force, and the object will start to decelerate. However, we can't directly find the resulting velocity without knowing the mass of the object or the time the force is applied.

5 - To find the component of the weight force acting perpendicular to the slope, we can use the equation F_perpendicular = F * cos(theta).

F_perpendicular = 100 N * cos(30 degrees)
F_perpendicular = 100 N * (sqrt(3)/2)
F_perpendicular = 50 * sqrt(3) N ≈ 86.6 N

The component of the weight force acting perpendicular to the slope is approximately 86.6 N.
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