A basketball player dunks the ball and momentarily hangs from the rim of the basket. Assume that the player can be considered as a 95.0 kg...

From the information given, we have a player of mass 95 kg
hanging from the rim of a basket. The behavior of the basket rim can be simplified to that of a
spring with a spring constant of 7.4*10^3 N/m.

The displacement of
the basket rim can be determined using Hooke's law. According to Hooke's law, F = -kx, where F is
the force causing the displacement, k is the spring constant and x is the displacement from the
equilibrium position. The negative sign indicates the tendency of a spring to return to the
equilibrium length when displaced.

The player of weight 95 kg exerts
a force given by m*g where m is the mass of the player and g is the acceleration due to gravity
equal to 9.8 m.s^2. The force is 95*9.8 = 931 N

The displacement due
the force x is given by F/k. Here F = 931 N and k = 7.4*10^3 N/m

x =
931/7.4*10^3 = 0.125 m

The displacement of the basket's rim from its
horizontal position due to the player hanging from it is 0.125 m.