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Oxana [17]
2 years ago
11

Is it possible to pass in a course if I passed in its prerequisites without understanding most about them?​

Engineering
1 answer:
MAXImum [283]2 years ago
4 0

Answer:

I don't think so but you could  remember  little bit and  you could pass.

Explanation:

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Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
Nina [5.8K]

dutile is the correct answer

6 0
3 years ago
Match the types of AI to the correct automation tasks.
Maru [420]

Answer:

Theory of Mind : A robotic head has a face that recognizes and simulates emotions.

Self aware : A robot tries to protect itself from harm

Purely Reactive : A door automatically opens when a person steps in front of it.

Limited Memory : A personal assistant software tracks a persons travel routes and suggests shorter routes.

Explanation:

Artificial intelligence is simply a technology which enables automation. It enables system perform task without being explicitly controlled. Purely reactive systems do not store Data in memory, it simply observes what going on at the moment which is what it was programmed to do and takes a step. One the machine detects someone approaching up to a certain distance, it opens.

Limited Memory systems store information about the past and this enhances its Decison making, prediction engines, self driving cars use this kind of artificial intelligence.

Theory of Mind : Here, systems are trained to detect, understand and replicate what is understood. Once the robot identifies an emotion, it replicates it.

Self - Aware : An advanced level of AI, where systems will not only be able to replicate what they see, but also make conscious decisions as to which action to take in different circumstances.

5 0
3 years ago
What are the general rules for press fit allowances
Keith_Richards [23]

Explanation:

As a general rule of thumb, the large the diameter of a bearing, bushing or pin, the larger the tolerance range,” Brieschke points out. “The inverse is true for smaller-diameter pieces.”

Mike Brieschke, vice president of sales at Aries Engineering, says a 0.25-inch-diameter metal dowel that is press-fit into a mild steel hole usually has an interference of ±0.0015 inch. Parts in noncritical assemblies tend to have looser tolerances

please rate brainliest if helps and follow

4 0
1 year ago
Las dos torres que sostienen un puente colgante tienen una separación de 240m y una altura de 110m a partir de la carretera, si
Sveta_85 [38]

La altura es de 169.4 metros.

Dado que las dos torres que sostienen un puente colgante tienen una separación de 240m y una altura de 110m a partir de la carretera, si el cable tensor más corto mide 10m, para determinar cuál es altura de un cable que se encuentra a 100m de distancia del centro se debe realizar los siguientes cálculos, aplicando la ecuación parabólica:  

  • (240)² = 4P x (110-10)
  • 57600 = 4P x 100
  • 57600 = 400P
  • 57600/400 = P
  • 144 = P
  • 200 x 200 = 4 x 144 x (Altura - 100)
  • 40000 = 576Altura - 57600
  • 40000 + 57600 / 576 = Altura
  • 169.4 metros = Altura

Por lo tanto, la altura es de 169.4 metros.

Aprende más en brainly.com/question/20333463

8 0
2 years ago
1 A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is expo
Alexeev081 [22]

Answer:

a) q' = 351.22 W/m

b) q'_total = 1845.56 W / m

c) q'_loss = 254.12 W/m

Explanation:

Given:-

- The diameter of the steam line, d = 100 mm

- The surface emissivity of steam line, ε = 0.8

- The temperature of the steam, Th = 150°C

- The ambient air temperature, T∞ = 20°C

Find:-

(a) Calculate the rate of heat loss per unit length for a calm day.

Solution:-

- Assuming a calm day the heat loss per unit length from the steam line ( q ' ) is only due to the net radiation of the heat from the steam line to the surroundings.

- We will assume that the thickness "t" of the pipe is significantly small and temperature gradients in the wall thickness are negligible. Hence, the temperature of the outside surface Ts = Th = 150°C.

- The net heat loss per unit length due to radiation is given by:

                     q' = ε*σ*( π*d )* [ Ts^4 - T∞^4 ]      

Where,

          σ: the stefan boltzmann constant = 5.6703 10-8 (W/m2K4)

          Ts: The absolute pipe surface temperature = 150 + 273 = 423 K

          T∞:The absolute ambient air temperature = 20 + 273 = 293 K

Therefore,

                    q' = 0.8*(5.6703 10-8)*( π*0.1 )* [ 423^4 - 293^4 ]    

                    q' = (1.4251*10^-8)* [ 24645536240 ]    

                    q' = 351.22 W / m   ... Answer

Find:-

(b) Calculate the rate of heat loss on a breezy day when the wind speed is 8 m/s.

Solution:-

- We have an added heat loss due to the convection current of air with free stream velocity of U∞ = 8 m/s.

- We will first evaluate the following properties of air at T∞ = 20°C = 293 K

                  Kinematic viscosity ( v ) = 1.5111*10^-5 m^2/s

                  Thermal conductivity ( k ) = 0.025596

                  Prandtl number ( Pr ) = 0.71559

- Determine the flow conditions by evaluating the Reynold's number:

                 Re = U∞*d / v

                      = ( 8 ) * ( 0.1 ) / ( 1.5111*10^-5 )

                      = 52941.56574   ... ( Turbulent conditions )

- We will use Churchill - Bernstein equation to determine the surface averaged Nusselt number ( Nu_D ):

           Nu_D = 0.3 + \frac{0.62*Re_D^\frac{1}{2}*Pr^\frac{1}{3}  }{[ 1 + (\frac{0.4}{Pr})^\frac{2}{3} ]^\frac{1}{4}  }*[ 1 + (\frac{Re_D}{282,000})^\frac{5}{8} ]^\frac{4}{5}    \\\\Nu_D = 0.3 + \frac{0.62*(52941.56574)^\frac{1}{2}*(0.71559)^\frac{1}{3}  }{[ 1 + (\frac{0.4}{0.71559})^\frac{2}{3} ]^\frac{1}{4}  }*[ 1 + (\frac{52941.56574}{282,000})^\frac{5}{8} ]^\frac{4}{5}  \\\\

           Nu_D = 0.3 + \frac{127.59828 }{ 1.13824  }*1.27251  = 142.95013

- The averaged heat transfer coefficient ( h ) for the flow of air would be:

            h = Nu_D*\frac{k}{d} \\\\h = 143*\frac{0.025596}{0.1} \\\\h = 36.58951 W/m^2K

- The heat loss per unit length due to convection heat transfer is given by:

           q'_convec = h*( π*d )* [ Ts - T∞ ]

           q'_convec = 36.58951*( π*0.1 )* [ 150 - 20 ]

           q'_convec = 11.49493* 130

           q'_convec = 1494.3409 W / m

- The total heat loss per unit length ( q'_total ) owes to both radiation heat loss calculated in part a and convection heat loss ( q_convec ):

           q'_total = q_a + q_convec

           q'_total = 351.22 + 1494.34009

           q'_total = 1845.56 W / m  ... Answer

Find:-

For the conditions of part (a), calculate the rate of heat loss with a 20-mm-thick layer of insulation (k = 0.08 W/m ⋅ K)

Solution:-

- To reduce the heat loss from steam line an insulation is wrapped around the line which contains a proportion of lost heat within.

- A material with thermal conductivity ( km = 0.08 W/m.K of thickness t = 20 mm ) was wrapped along the steam line.

- The heat loss through the lamination would be due to conduction " q'_t " and radiation " q_rad":

             q'_t = 2*\pi*k \frac{T_h - T_o}{Ln ( \frac{r_2}{r_1} )}  

             q' = ε*σ*( π*( d + 2t) )* [ Ts^4 - T∞^4 ]

             

Where,

             T_o = T∞ = 20°C

            T_s = Film temperature = ( Th + T∞ ) / 2 = ( 150 + 20 ) / 2 = 85°C

             r_2 = d/2 + t = 0.1 / 2 + 0.02 = 0.07 m

             r_1 = d/2 = 0.1 / 2 = 0.05 m

- The heat loss per unit length would be:

            q'_loss = q'_rad - q'_cond

- Compute the individual heat losses:

            q'_t = 2*\pi*0.08 \frac{150 - 85}{Ln ( \frac{0.07}{0.05} )}\\\\q'_t = 0.50265* \frac{65}{0.33647}\\\\q'_t = 97.10 W/m

Therefore,

             q'_loss = 351.22 - 97.10

            q'_loss = 254.12 W / m   .... Answer

- If the wind speed is appreciable the heat loss ( q'_loss ) would increase and the insulation would become ineffective.

6 0
3 years ago
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