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Alinara [238K]
2 years ago
10

An A-26C aircraft has 2 piston engines driving propellers. It is climbing at a flight-path angle of 10 degrees. The aircraft wei

ghs 32,000 pounds (142,343 Newtons). For a steady climb, EACH engine must overcome how much weight in addition to the drag of the airplane?
Engineering
1 answer:
8090 [49]2 years ago
5 0

Answer:

320lbs

Explanation:

Weight = T–D / %gradient x 100

Where T = thrust and D = Drag

Due to engine out case, thrust will be from 2 engines piston i.e. 32,000x 2 = 64,000

Weight = (64,000-)/2x 100

Weight = 320lbs

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An aircraft component is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 40 MPa1m (36.4 ksi1in.)
sergiy2304 [10]

Answer:

Y = \frac{40 Mpa \sqrt{m}}{300 Mpa \sqrt{\pi 0.002m}}= 1.682

Now we can solve for the fracture level with this formula:

\sigma_F = \frac{K_{lc}}{Y \sqrt{\pi d_x}}

We can use the parameter of Y from the previous equation since this value not changes.

d_x = 6/2 = 3 mm =0.003m

And replacing we have:

\sigma_F= \frac{40 Mpa \sqrt{m}}{1.682 \sqrt{\pi 0.003m}}= 244.96 Mpa

And for this case we see that this value not correspond to 260 Mpa the value reported, so then the correct answer for this case would be 244.96 Mpa for the new crack length of 6mm = 0.006 m

Explanation:

We need to solve for the parameter Y given by this expression:

Y = \frac{K_{lc}}{\sigma \sqrt{\pi d}}

Where:

K_{lc} = 40 MPa \sqrt{m} represent the fracture toughness

\sigma = 300 MPa represent the stress level

d = \frac{4 mm}{2}= 2mm = 0.002 m

And if we replace we got:

Y = \frac{40 Mpa \sqrt{m}}{300 Mpa \sqrt{\pi 0.002m}}= 1.682

Now we can solve for the fracture level with this formula:

\sigma_F = \frac{K_{lc}}{Y \sqrt{\pi d_x}}

We can use the parameter of Y from the previous equation since this value not changes.

d_x = 6/2 = 3 mm =0.003m

And replacing we have:

\sigma_F= \frac{40 Mpa \sqrt{m}}{1.682 \sqrt{\pi 0.003m}}= 244.96 Mpa

And for this case we see that this value not correspond to 260 Mpa the value reported, so then the correct answer for this case would be 244.96 Mpa for the new crack length of 6mm = 0.006 m

6 0
2 years ago
Fly thermostat, an automatic temperature controller for homes, learns the patterns for raising and lowering the temperature in a
Lubov Fominskaja [6]

Answer:

The Internet of things

Explanation:

When devices embedded with a network connectivity, software or sensors, are able to exchange data with the manufacturer through the network of physical things, then we say the Internet of things (IoT) has been defined.

Internet of things is when devices like smartphones, sensors and other smart devices are connected together which allows them to exchange data through the network of physical things. By exchanging data, these devices learn new patterns of adaptation.

In this case, the fly thermostat, learns the patterns for raising and lowering the temperature in a house through the Internet of things beacause the sensor has made it possible by automatically observing the temperature pattern of the house.

4 0
2 years ago
Read 2 more answers
A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process t
iren2701 [21]

Work done = -19.7 KJ

Heat transferred = 17.4 KJ

Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, P_{1}= 1 bar

v_{2} = 4 bar

pV¹°¹ = constant

From superheated propane table, at  P_{1}= 1 bar andT_{1}  = 27⁰C

v_{1} = 0.557 m³/kg

v_{2} = 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

V2 = (\frac{P1}{P2})^\frac{1}{1.1} * V1  \\\\V2 = \frac{1}{4}^\frac{1}{1.1} * 0.557  \\\\V2 = 0.158 m^3/kg

At  = 4 bar and v = 0.158 m³/kg

u2 = 548.45K J/kg

To find work done in the process:

W = \frac{P2V2 - P1V1}{1-n} \\\\W = \frac{m(P2V2 - P1V1)}{1-n} \\\\W = \frac{v}{u} * \frac{P2V2 - P1V1}{1-n}\\  \\W = \frac{0.2}{0.5571} * \frac{4 X 0.158 - 1 X 0.577}{1-1.1} X 10^5 \frac{Pa}{Bar} \frac{1KJ}{10^3Nm} \\   \\W = -19.75KJ

(b) Heat transfer = ?

Q = m(u2 - u1) + W\\\\Q = \frac{0.2}{0.5571} * (548.45 - 473.73) + (-19.7)\\\\Q = 17.4KJ

8 0
2 years ago
Why is it so dangerous to use a ground lift on a metal cased power tool
MaRussiya [10]
Answer:

Removal of the safety ground connection on equipment can expose users to an increased danger of electric shock and may contradict wiring regulations. ... If a fault develops in any line-operated equipment, cable shields and equipment enclosures may become energized, creating an electric shock hazard.
7 0
1 year ago
Holmes owns two suits: one black and one tweed. He always wears either a tweed suit or sandals. Whenever he wears his tweed suit
vazorg [7]

Answer:

He wore his black suit, another color of shirt (not purple) and shoes

Explanation:

Holmes owns two suits: one black and one tweed.

Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.

So, if he wore a bow tie yesterday, it means he wore his black suit, another color of shirt (not purple) and shoes because the shirt color is not purple

4 0
2 years ago
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