PHYSICS HIGH SCHOOL

A shear force of 400 N is applied to one face of an aluminum cube with sides of 30 cm. What is the resulting relative displacement? (the shear modulus for aluminum is 2.5 X 1010N/m2).a) 4.4 X 10-8 m
b) 8.2 X 10-8 m
c) 1.9 X 10-8 m
d) 5.3 X 10-8 m

Answers

Answer 1
Answer:

Answer:

\Delta x=5.33* 10^(-8)m

So option (d) will be correct option

Explanation:

We have given shear force F = 400 N

Size of the cube L_0=30cm=0.3m

Shear modulus of aluminium S=2.5* 10^(10)N/m^2

We have to find the resulting relative displacement

Area of the cube A=L_0^2=0.3^2=0.09m^2

We know that shear force is given by

F=S* (\Delta x)/(L_0)* A

So 400=2.5* 10^(10)* (\Delta x)/(0.3)* 0.09

\Delta x=5.33* 10^(-8)m

So option (d) will be correct option

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Piston is the part in the combustion engine which moves up and down.
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