Final answer:
If the radius of a sphere is increased by a factor of 1.4 and the difference between the sphere's density and the liquid's density is reduced by a factor of 2, the new terminal velocity of the sphere will be 4.9 cm/s, slightly less than its original terminal velocity of 5 cm/s.
Step-by-step explanation:
When a sphere is suspended in a liquid and reaches terminal velocity, the gravitational force on the sphere is balanced by the net force of the drag force and buoyant force acting on it. The terminal velocity V of the sphere can be given by the equation:
V = (2/9) * (R²(g(ρs - ρ1))/η)
Given that the velocity of the sphere initially is 5cm/s, when the radius of the sphere is increased by a factor of 1.4, its new radius R_new = 1.4R. Also, if the difference in density (ρs - ρ1) is reduced by a factor of two, the new difference in density is (ρs - ρ1)/2.
Substituting these into the equation, the new terminal velocity V_new is:
V_new = (2/9) * ((1.4R)²(g((ρs - ρ1)/2η))
Comparing V_new with the initial V, keeping other factors constant:
V_new = V * ((1.4)² / 2)
V_new = V * (1.96 / 2)
V_new = V * 0.98
With the initial velocity of the sphere at 5cm/s, the new terminal velocity becomes:
V_new = 0.98 * 5 cm/s
V_new = 4.9 cm/s
The new terminal velocity of the sphere after the given changes is 4.9 cm/s.