Suppose a disk is 0.50 cm thick and 4.0 cm in diameter. if this disk supports half the weight of a 95 kg person, by what fraction of its thickness does the disk compress?

37.04 %

Explanation:

Young modulus of the disk = stress / strain = F t/ AΔt

where t = thickness in meters = 0.5 cm = 0.5 / 100 = 0.005 m

Δt = the change in the thickness of the disk in meters

F = 95 × 9.8 ( mass × acceleration due to gravity) = 931 N

half of the weight of the man = 931 / 2 = 465.5

young modulus of the disk = 1.0 × 10⁶ N/m²

radius = diameter / 2 = 4 /2 = 2 cm = 2/100 = 0.02 m

Area of the disk = πr² = 3.142 × ( 0.02)²= 0.0012568 m²

1.0 × 10⁶ N/m²  = (  465.5 × 0.005) / ( 0.0012568 × Δt)

0.0012568 Δt = 465.5 × 0.005) / (1.0 × 10⁶) = 2.3275 × 10⁻⁶

Δt = 2.3275 × 10⁻⁶ / 0.0012568 = 0.001852 m

the fraction the disk was compress = 0.001852 / 0.005 × 100 = 37.04 %

found in the nucleus

has mass of one amu