When the saw slices wood, the wood exerts a 104-n force on the blade, 0.128 m from the blade’s axis of rotation. if that force is at right angles to the lever arm, how much torque does the wood exert on the blade? does this torque make the blade turn faster or slower?

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

= force x perpendicular distance from the axis of rotation

= 104 x .128

=13.312 Nm.

Since this torque opposes the movement of blade , it turns the blade slower.

the area between the graph and the t axis

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.