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Dennis_Churaev [7]
2 years ago
5

If my current directory is ‘AR’ write the path for my current directory

Engineering
1 answer:
salantis [7]2 years ago
6 0

Answer:

sorry not having idea about answer

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Technician A says that a voltage drop of 0.8 volts on the starter ground circuit is within specifications. Technician B says tha
Romashka-Z-Leto [24]

Answer:

Technician A is wrong

Technician B is right

Explanation:

voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.

A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.

8 0
3 years ago
A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A
OLEGan [10]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos \frac{28.655*S}{R} )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos \frac{28.65 *s }{678}  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + \frac{u^2}{30(\frac{a}{3.2} )-G1}  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + \frac{u^2}{30[(11.2/32.2)-0 ]}

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

5 0
2 years ago
a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What
Flura [38]

Answer:

Temperature on the inside ofthe box

Explanation:

The power of the light bulb is the rate of heat conduction of the bulb, dq/dt = 100 W

The thickness of the wall, L = 1.2 cm = 0.012m

Length of the cube's side, x = 20cm = 0.2 m

The area of the cubical box, A = 6x²

A = 6 * 0.2² = 6 * 0.04

A = 0.24 m²

Temperature of the surrounding, T_0 = 20^0 C = 273 + 20 = 293 K

Temperature of the inside of the box, T_{in} = ?

Coefficient of thermal conductivity, k = 0.8 W/m-K

The formula for the rate of heat conduction is given by:

dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C

5 0
3 years ago
Steam enters a turbine operating at steady state with a mass flow of 10 kg/min, a specific enthalpy of 3100 kJ/kg, and a velocit
Natali [406]

Answer:

\dot W_{out} = 133.327\,kW

Explanation:

The model for the turbine can be derived by means of the First Law of Thermodynamics:

-\dot Q_{out}-\dot W_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right] =0

The work produced by the turbine is:

\dot W_{out}=-\dot Q_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right]

The mass flow and heat transfer rates are, respectively:

\dot m = (10\frac{kg}{min})\cdot (\frac{1\,min}{60\,s} )

\dot m = 0.167\,\frac{kg}{s}

\dot Q_{out} = (0.167\,\frac{kg}{s} )\cdot (1.1\times 10^{3}\,\frac{J}{kg} )

\dot Q_{out} = 183.7\,W

Finally:

\dot W_{out} = -183.7\,W + (0.167\,\frac{kg}{s} )\cdot \left(8\times 10^{5}\,\frac{J}{kg} -562,5\,\frac{J}{kg} +29.43\,\frac{J}{kg} \right)

\dot W_{out} = 133.327\,kW

3 0
3 years ago
What are some aircraft aging problems? How can you as an Aviation Maintenance Manager monitor problems that relate to aircraft f
julia-pushkina [17]

Answer:

Answered

Explanation:

The two key processes that lead to aircraft ageing are fatigue and corrosion. These processes generally affect the aircraft structure, but can also affect wiring, flight controls, power plants, and other components. Fatigue and corrosion can work independently from one another, or they can interact. The interaction between fatigue and corrosion can increase the rate of ageing to a greater extent than that due to either process alone.

Fatigue predominately takes place in metal components, but it can also affect non-metallic materials. Fatigue occurs through cyclic loading patterns, where a component is repeatedly loaded. Bending a metal paper clip backwards and forwards is an example of fatigue; the paper clip will not break if only bent once, however, if it is repeatedly loaded, it will eventually break. Fatigue failures will often take place at loads much lower than the materials ultimate strength.

Generally, the initiation point for fatigue will be a microscopic crack that forms at a location of high stress, such as a hole, notch, or material imperfection. The crack will then grow as loads are repeatedly applied. If not detected and treated, the crack will eventually grow to a critical size and failure will occur at loads well below the original strength of the material.

The relationship between repetitive loading and fatigue crack growth, creates a link between fatigue related ageing, the number of flight cycles, and the number of flight hours that an aircraft has accumulated.

Aircraft components that are susceptible to fatigue include most structural components such as the wings, the fuselage, and the engine.

During initial manufacturing the research department is responsible for designing the aircraft to withstand fatigue. During operations the fatigue risk management of aviation maintenance will try to rectify the problems due to fatigue.

4 0
3 years ago
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