Answer:
The displacemnent of the nut will be 0.3 m.
Explanation:
As we know the torque ([tex]\large{\overrightarrow{\tau}}[/tex]) is given by
[tex]\large{\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}}[/tex]
where [tex]\large{\overrightarrow{r}}[/tex] is the displacement and [tex]\large{\overrightarrow{F}}[/tex] is the applied force.
Given, [tex]\large{|\overrightarrow{\tau}|} = 40 N-m[/tex] and [tex]\large{|\overrightarrow{F}|} = 133 N[/tex]
So the magnitude of the displacement of the nut is given by
[tex]\large{r} = \dfrac{\tau}{F} = \dfrac{40}{133} = 0.3 m[/tex]