All categories

2 years ago

What is the single largest contributor to increasing takeoff or landing distance?

Engineering

1 answer:

hodyreva [135]2 years ago

3 0

An uphill slope improves the take-off ground run, and a downhill slope increases the landing ground run.

<h3>What is uphill slope?</h3>

Driving uphill suggests climbing a “positive” six percent slope . Driving downhill, the “rise” exists as a drop, so there is a “negative,” or downhill, slope (Figure B). When dealing with slope, a positive slope simply indicates uphill and a negative slope indicates downhill. An uphill slope improves the take-off ground run, and a downhill slope increases the landing ground run.

If something or someone lives moving downhill or is downhill, they exist moving down a slope or are located toward the bottom of a hill. He headed downhill toward the river. adverb. If you communicate that something exists going downhill, you mean that it is becoming worse or less prosperous.

To learn more about uphill slope refer to:

brainly.com/question/13361896

#SPJ4

You might be interested in

You are hired as the investigators to identify the root cause and describe what should have occurred based on the following info

creativ13 [48]

Answer:

The mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.

Explanation:

Since density ρ = m/v where m = mass of fuel and v = volume of fuel, we need to find the mass of each volume of fuel.

So, m = ρv now ρ = specific gravity × density of water = 0.803 × 1000 kg/m³ = 803 kg/m³.

To find the mass of the 7,682 L of fuel, its volume is 7,682 dm³ = 7,682 dm³ × 1 m³/1000 dm³ = 7.682 m³.

It's mass, m = 803 kg/m³ × 7.682 m³ = 6168.646 kg

To find the mass of the extra 4,916 L of fuel added, we have

m' = ρv' where v' = 4,916 L = 4,916 dm³ = 4916 dm³ × 1 m³/1000 dm³ = 4.916 m³

m' = 803 kg/m³ × 4.916 m³ = 3947.548 kg

So, the total mass of the fuel is m" = m + m' = 6168.646 kg + 3947.548 kg = 10116.194 kg ≅ 10,166.2 kg

<u>Since this mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.</u>

4 0

3 years ago

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0

3 years ago

The chart shows the bids provided by four engineers to test a prototype.

klasskru [66]

Answer:

Explanation:

To know which is most or least cost-effective, it's not enough to look at only the per day rate, or only the time to complete. You have to multiply them to get the total cost of the project.

$\left[\begin{array}{ccccc}&Cost\ per\ day\ (\$)&Time\ to\ complete\ (days)&Total\ cost\ (\$)\\Zoe&500&8&4000\\Greg&650&10&6500\\Orion&400&12&4800\\Jin&700&5&3500\end{array}\right]$

As you can see, Greg is the least cost-effective because he charges the most for the project.

8 0

3 years ago

An air conditioning system operating on reversed carnot cycle is required to remove heat from the house at a rate of 32kj/s to m

Brilliant_brown [7]

Answer:

(e) 1.64 kW

Explanation:

The Coefficient of Performance of the Reverse Carnot's Cycle is:

$COP = \frac{T_{L}}{T_{H}-T_{L}}$

$COP = \frac{293.15\,K}{308.15\,K-293.15\,K}$

$COP = 19.543$

Lastly, the power required to operate the air conditioning system is:

$\dot W = \frac{\dot Q_{L}}{COP}$

$\dot W = \frac{32\,kW}{19.543}$

$\dot W = 1.637\,kW$

Hence, the answer is E.

3 0

3 years ago

Paint can shaker mechanisms are common in paint and hardware stores. While they do a good job of mixing the paint, they are also

Ymorist [56]

Answer:

A good design for a portable device to mix paint minimizing the shaking forces and vibrations while still effectively mixing the paint. Is:

The best design is one with centripetal movement. Instead of vertical or horizontal movement. With a container and system of holding structures made of materials that could absorb the vibration effectively.

Explanation:

First of all centripetal movement would be friendlier to our objective as it would not shake the can or the machine itself with disruptive vibrations. Also, we would have to use materials with a good grade of force absorption to eradicate the transmission of the movement to the rest of the structure. Allowing the reduction of the shaking forces while maintaining it effective in the process of mixing.

6 0

3 years ago