9. A 9,000 lb load is suspended from the roof in a shopping mall with a 16 ft long solid aluminum rod. The modulus of elasticity of the aluminum is 10,000,000 psi. If the maximum rod elongation must be limited to 0.50 in. and the maximum stress must be limited to 30,000 psi, determine the minimum diameter that may be used for the rod. (precision to 0.00)

Step-by-step explanation:

Given:

Force, f = 9000 lb

Young modulus, E = 10,000,000 psi

Stress, S = 30000 psi

Elongation, x = 0.50 in

Length, L = 16 ft

Converting ft to in,

12 in = 1 ft

= 192 in

Young modulus, E = stress/strain

Stress = force/area, A

Strain = elongation, x/Length, L

Maximum stress = 30,000 psi

Stress = force/area

Area = 9000/30000

= 0.3 in^2

Using minimum area of 0.3 in^2,

A = (pi × d^2)/4

0.3 = (pi × d^2)/4

= sqrt(1.2/pi)

diameter, d = 0.62 inches

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms