the fuel tanks for airplanes are in the wings, cross section below. the tank must hold 6000 lb of fuel with…

the fuel tanks for airplanes are in the wings, cross section below. the tank must hold 6000 lb of fuel with density 42 lb/ft³. estimate the length of the tank using simpsons rule\nhorizontal spacing = 1.8 ft\ny₀ = 1.3 ft, y₁ = 1.5 ft, y₂ = 1.8 ft, y₃ = 2 ft, y₄ = 2.1 ft, y₅ = 2.3 ft, y₆ = 2.1 ft\nthe length of the tank is □ ft.\n(round to the nearest tenth as needed.)

the fuel tanks for airplanes are in the wings, cross section below. the tank must hold 6000 lb of fuel with density 42 lb/ft³. estimate the length of the tank using simpsons rule\nhorizontal spacing = 1.8 ft\ny₀ = 1.3 ft, y₁ = 1.5 ft, y₂ = 1.8 ft, y₃ = 2 ft, y₄ = 2.1 ft, y₅ = 2.3 ft, y₆ = 2.1 ft\nthe length of the tank is □ ft.\n(round to the nearest tenth as needed.)

Answer

Explanation:

Step1: Calculate the area using Simpson's Rule

Simpson's Rule formula for (n = 6) (even) is (A=\frac{\Delta x}{3}(y_0 + 4y_1+2y_2 + 4y_3+2y_4+4y_5 + y_6)), where (\Delta x = 1.8) ft. Substitute (y_0 = 1.3), (y_1 = 1.5), (y_2 = 1.8), (y_3 = 2), (y_4 = 2.1), (y_5 = 2.3), (y_6 = 2.1) into the formula: [ \begin{align*} A&=\frac{1.8}{3}(1.3 + 4\times1.5+2\times1.8 + 4\times2+2\times2.1+4\times2.3 + 2.1)\ &= 0.6(1.3+6 + 3.6+8+4.2+9.2 + 2.1)\ &=0.6\times34.4\ &=20.64 \end{align*} ]

Step2: Find the volume and then the length

The weight of the fuel (W=\text{density}\times\text{Volume}), so (\text{Volume}=\frac{W}{\text{density}}). Given (W = 6000) lb and (\text{density}=42) lb/ft³, (\text{Volume}=\frac{6000}{42}\approx142.86) ft³. Also, (\text{Volume}=A\times L) (where (L) is the length of the tank). Then (L=\frac{\text{Volume}}{A}). Substitute (A = 20.64) and (\text{Volume}\approx142.86) into the formula: (L=\frac{142.86}{20.64}\approx6.9)

Answer:

(6.9)