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:

elongation, x = 0.50 in

Force, f = 9000 lb

Young modulus, E = 10,000,000 psi

Maximum Stress, Sm = 30000 psi

Length, L = 16 ft

Converting ft to in,

12 in = 1 ft

=16 × 12 = 192 in

Young modulus, E = stress/strain

Stress = force/area, A

Strain = elongation, x/Length, L

E = f × L/A × E

1 × 10^7 = stress/(0.5/16)

= 26041.7 psi

Minimum stress = 26041.7 psi

Maximum stress = 30,000 psi

Stress = force/area

Area = 9000/26041.7

= 0.3456 in^2

Stress = force/area

Area = 9000/30000

= 0.3 in^2

Using minimum area of 0.3 in^2,

A = (pi/4)(d^2)

0.3 in^2 = (pi/4)(d^2)

d = 0.618 inches

diameter, d = 0.618 inches

the answer is c. 23/3

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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