Asked by Fabian
I need help
The design of a new airplane requires a gasoline tank of constant cross-sectional area in each wing. A scale drawing of a cross section is shown here. The tank must hold 5000lb of gasoline, which has a density of 42 lb/ft3. Estimate the length of the tank.
Y0 = 1.5 ft, Y1 = 1.6 ft, Y2 = 1.8 ft, Y3 = 1.9 ft, Y4 = 2.0 ft, Y5 = Y6 = 2.1 ft
And the horizontal spacing is 1 ft.
The design of a new airplane requires a gasoline tank of constant cross-sectional area in each wing. A scale drawing of a cross section is shown here. The tank must hold 5000lb of gasoline, which has a density of 42 lb/ft3. Estimate the length of the tank.
Y0 = 1.5 ft, Y1 = 1.6 ft, Y2 = 1.8 ft, Y3 = 1.9 ft, Y4 = 2.0 ft, Y5 = Y6 = 2.1 ft
And the horizontal spacing is 1 ft.
Answers
Answered by
drwls
Without the drawing or a better verbal description of the wind geometry, we cannot help you. The volume of the tank must be
V = (mass)/(density)
= (5000 lb)/(42 lb/ft^3) = 119.0 ft^3
The volume V equals the cross sectional area times the length in this case. Figure out the area and use it to calculate the length
V = (mass)/(density)
= (5000 lb)/(42 lb/ft^3) = 119.0 ft^3
The volume V equals the cross sectional area times the length in this case. Figure out the area and use it to calculate the length
Answered by
katy
use the trapezoidal rule with the scale so its
height =1, so
(1/2) (1.5+ 2*1.6+ 2*1.8+ 2*1.9+ 2*2.0+ 2*2.1+ 2.1)
which equals 11.2
so 119/11.2= 10.625 or 10.63
and that is the answer
height =1, so
(1/2) (1.5+ 2*1.6+ 2*1.8+ 2*1.9+ 2*2.0+ 2*2.1+ 2.1)
which equals 11.2
so 119/11.2= 10.625 or 10.63
and that is the answer
Answered by
Joseph President
Satan
Answered by
14 years later
Thank you for the answer!!! I'm tired
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