yes, the Y component of the net forces are opposite in direction, so they cancel out. And the X component of the net forces are opposite in direction so they cancel out.

No, the net forces add up to 14 N. $\Sigma F\ =\ 14N$

yes, if it was originally in motion when these forces were applied. Since the net force equals zero, there would be no change in motion. 

no, a net force in equilibrium means 0, therefore, it is at rest. 

No, the net forces are balanced, therefore in equilibrium. The box remains at rest

The forces are not balanced, so the Net force is not in equilibrium, so the box will accelerate to the left.

The forces are not balanced, so the Net force is not in equilibrium, so the box will accelerate to the right.

The forces are not balanced, so the Net force is not in equilibrium, so the box will accelerate up.

The forces are not balanced, so the Net force is not in equilibrium, so the box will accelerate down.

$\Sigma F\ =\ 65N\ North$

$\Sigma F\ =\ 15N\ North$

$\Sigma F\ =\ 15N\ South$

$\Sigma F\ =\ 0N\ \ @\ Equilibrium$

$\Sigma F\ =\ 0N$ The box is at equilibrium. It will be at rest.

$\Sigma F\ =\ 0N\$ It is at equilibrium. The box will remain traveling at 25 m/s in the Z axis direction out of the screen.

$\Sigma Fy\ =\ 6N,\ \ \Sigma Fx=\ 10N$ The Net force is 16N.

 $\Sigma Fx\ =\ -8300\ N$  

 $\Sigma Fx\ =\ 2300\ N$  

 $\Sigma Fx\ =6000N$  

 $\Sigma Fx\ =-6000N$  

 $\Sigma Fx\ =\ 10,6000\ N$  

 $Fnormal\ =\ -9500\ N$  

 $Fnormal\ =\ 9500\ N$  

 $Fnormal\ =\ 1000kg$  

Fg=mg

Fg=ma

-9800 N

9800 N

10500 N

8500 N

 $-6\ \frac{m}{s^2}$  

-6000  $\frac{m}{s^2}$  

7000  $\frac{m}{s^2}$