[Gary Dillard]

1) Yes, the resultant ideal torque on shaft B would be 20Nm, and it would turn 1/4 as fast as shaft A. Input power x efficiency (1.0 for ideal) = output power, so Ta x Na = Tb x Nb. The tooth stresses are the same in both gears since the forces are equal - the torque is different because the distance at which it is applied is different (the radius).
2) The Wheel doesn't apply a "torque" on the ground, it applies a force on the ground, equal to the torque on the shaft divided by the distance (the wheel radius). This would be 20 Nm /[5 in x (.0254 m/in)] = 157.48 N. The linear speed would be (Nb rpm) x (Pi x 10 in/rev) x (1 min/60 sec) x (1 ft/12 in) = Nb x Pi/72 fps.

Hope that helps and I hope I got my units right.