Solution)
Part a)
Amount of force applied tangentially , F=0.245 N
Area of the displaced plane, A= (40 mm)(40 mm) =16x10-4 m^2
We know, shearing stress = F/A
=(0.245 N)/(16x10-4 m2 )
=153.125 Pa (Ans)
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Part b)
Strain=tanθ=d/h =5.5 mm/20 mm = 0.275 (Ans)
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Part c)
Shear modulus, η =shearing stress /Strain
=(153.125 Pa)/(0.275)
=556.818 N/m2 (Ans)
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Good luck!:)
5.50 mm-시-. F.. 0.245 N 1 point) 20.0 mm 40.0 mm- A block of gelatin has...
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