Aprofessor's office door is 0.89 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg ; and pivots on frictionless hinges. a "door closer" is attached to the door and the top of the door frame. when the door is open and at rest, the door closer exerts a torque of 4.9 n⋅m . what is the least force that you need to apply to the door to hold it open?

In order to answer this question ... strange as it may seem ...
we only need one of those measurements that you gave us
that describe the door.

The door is hanging on frictionless hinges, and there's a torque
being applied to it that's trying to close it.  All we need to do is apply
an equal torque in the opposite direction, and the door doesn't move.

Obviously, in order for our force to have the most effect, we want
to hold the door at the outer edge, farthest from the hinges.  That
distance from the hinges is the width of the door ... 0.89 m.

We need to come up with 4.9 N-m of torque,
applied against the mechanical door-closer.

Torque is (force) x (distance from the hinge).

                                    4.9 N-m  =  (force) x (0.89 m) 

Divide each side by 0.89m:    Force = (4.9 N-m) / (0.89 m)

                                                             =  5.506 N .

It equals just about 6,300 km per hour each day an week

Answerstatic electricity shock;