A disk between vertebrae in the spine is subjected to a shearing force of 375 N. Find its shear deformation, taking it to have a shear modulus of 1.60×109 N/m2. The disk is equivalent to a solid cylinder 0.750 cm high and 6.50 cm in diameter.

5.29×10^-7

Explanation:

shear stress τ = F/ A

shear deformation δ = (VL)/ (AG)

= (τL)/ G

V=shear force

L=height of disk=6.50×10^-2

A=cross sectional area

G= shear modulus= (1.60x10^9N/m^2)

A=πd^2/4

Then substitute the values we have

4×(375N)(0.00750m)

= δ

(π*0.00650^2)(1.60x10^9N/m^2)

= 5.29×10^-7

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

not too sure thinking all accelerations are pointed down