A565 n rightward force pulls a large box across the floor with a constant velocity of 0.75 m/s. if the coefficient of friction is 0.8, what is the weight of the box?

The first thing you should know is that the friction force is equal to the coefficient of friction due to normal force.
 Therefore, clearing the normal force we have:
 The friction is 565N.
 (565 / 0.8) = 706.25N. weight.

To use anything transparent

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.