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alexgriva [62]
3 years ago
9

Using an "AND" and an "OR", list all information (Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per

Gallon) on aircraft that have a seat capacity less than 250, or aircraft that have a miles per gallon greater than 4.0 miles per gallon and fuel capacity less than 2500.

Engineering
1 answer:
Tomtit [17]3 years ago
5 0

Answer:

Explanation :

The given  information to be listed can are Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per Gallon.

Check the attached document for the solution.

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Technician A says that if fuel pump pressure is correct, fuel pump volume will be correct as well. Technician B says that a fuel
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Explanation:

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2 years ago
Given : f(x) = x³- 7x²+ 36 Draw the graph of f neatly on F graph paper. Clearly indicate an Intercepts and coordinates of turnin
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Answer:

Explanation:

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3 years ago
A vertical pole consisting of a circular tube of outer diameter 127 mm and inner diameter 115 mm is loaded by a linearly varying
Anna [14]

Maximum shear stress in the pole is 0.

<u>Explanation:</u>

Given-

Outer diameter = 127 mm

Outer radius,r_{2} = 127/2 = 63.5 mm

Inner diameter = 115 mm

Inner radius, r_{1} = 115/2 = 57.5 mm

Force, q = 0

Maximum shear stress, τmax = ?

 τmax  = \frac{4q}{3\pi } (\frac{r2^2 + r2r1 + r1^2}{r2^4 - r1^4} )

If force, q is 0 then τmax is also equal to 0.

Therefore, maximum shear stress in the pole is 0.

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3 years ago
Problem definition
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Answer:

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Explanation:

8 0
3 years ago
A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
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