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3 years ago

The complexity of bfs and dfs

Engineering

1 answer:

Lelechka [254]3 years ago

3 0

Answer:

BFS uses Queue to find the shortest path. DFS uses Stack to find the shortest path. ... Time Complexity of BFS = O(V+E) where V is vertices and E is edges. Time Complexity of DFS is also O(V+E) where V is vertices and E is edges.

Explanation:

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You are about to perform PMCS on your M1114? What resource should you use for the procedures and instructions for performing PMC

aniked [119]

The resources and instructions that should be used for the procedures of performing PMCS are:

Operator's manuals
Safety cautions and warnings.
Fording kit
Heating and cooling systems.

PMCS is an acronym for preventive maintenance checks and services and it can be defined as the maintenance, checks, and services that are typically performed before, during, and after the use of any type of military equipment such as:

M1114

M1151

M1123

Basically, the resources and instructions that should be used for the procedures of performing PMCS are:

Operator's manuals
Safety cautions and warnings.
Fording kit
Heating and cooling systems.

Read more on PMCS here: brainly.com/question/15720250

#SPJ1

4 0

2 years ago

A glass bottle washing facility uses a well agitated hot water bath at 50°C with an open top that is placed on the ground. The b

DochEvi [55]

Answer:

do the wam wam

Explanation:

7 0

3 years ago

Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large

salantis [7]

Answer:

$Q_{cv} = -1007.86kJ$

Explanation:

Our values are,

State 1

$V=3m^3\\P_1=1bar\\T_1 = 295K$

We know moreover for the tables A-15 that

$u_1 = 210.49kJ/kg\\h_i = 295.17kJkg$

State 2

$P_2 =6bar\\T_2 = 296K\\T_f = 320K$

For tables we know at T=320K

$u_2 = 228.42kJ/kg$

We need to use the ideal gas equation to estimate the mass, so

$m_1 = \frac{p_1V}{RT_1}$

$m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}$

$m_1 = 3.54kg$

Using now for the final mass:

$m_2 = \frac{p_2V}{RT_2}$

$m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}$

$m_2 = 19.59kg$

We only need to apply a energy balance equation:

$Q_{cv}+m_ih_i = m_2u_2-m_1u_1$

$Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i$

$Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)$

$Q_{cv} = -1007.86kJ$

The negative value indidicates heat ransfer from the system

7 0

3 years ago

Considering the analogy between electrical circuit and thermal circuit, show your approach to derive an expression for the therm

Simora [160]

Answer:

Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.

Explanation:

Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow. You can also see that $R_{th}$ represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:

Wall Thickness. More thickness, more thermal resistance.
Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
Cross-section Area. More cross-section area, less thermal resistance.

A expression to define the thermal resistance for the wall is as follows: $R_{th} =\frac{l}{Ak}$ , where l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.