Answer: [tex]4.8(10)^{-5} m[/tex]
Explanation:
We can solve this problem by the following equation:
[tex]\eta=\frac{F.h}{A \Delta x}[/tex]
Where:
[tex]\eta=3.5(10)^{10}Pa[/tex] is the shear modulus for brass
[tex]F=4.2(10)^{4}N[/tex] is the applied force
[tex]h=2.5 cm=0.025 m[/tex] is the height of the cube
[tex]A=h^{2}=(0.025 m)^{2}=625(10)^{-6} m^{2}[/tex] is the area of each surface of the cube
[tex]\Delta x[/tex] is the shear displacement
Finding [tex]\Delta x[/tex]:
[tex]\Delta x=\frac{F.h}{A \eta}[/tex]
[tex]\Delta x=\frac{(4.2(10)^{4}N)(0.025 m)}{(625(10)^{-6} m^{2})(3.5(10)^{10}Pa)}[/tex]
Finally:
[tex]\Delta x= 4.8(10)^{-5} m[/tex]
