Aforce of 6 pounds is required to hold a spring stretched 0.1 feet beyond its natural length. how much work (in foot-pounds) is done in stretching the spring from its natural length to 1.1 feet beyond its natural length?

36.3 lb-ft

Explanation:

F = force required to hold the spring stretched = 6 lb

x = stretch in the spring = 0.1 ft

k = spring constant

Using the equation

F = k x

6 = k (0.1)

k = 60 lb/ft

U = work done in stretching the spring

x' = 1.1 ft

Work done in stretching the spring is given as

U = (0.5) k x'²

U = (0.5) (60) (1.1)²

U = 36.3 lb-ft

c. rb

explanation:

the atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.

answer; /// i believe that the correct answer is (c)