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Afina-wow [57]
3 years ago
10

What are the specifications for a mine shaft headgear ​

Computers and Technology
1 answer:
Licemer1 [7]3 years ago
4 0

Answer: THERE ARE LOTS BUT I ONLY KNOW THIS

Explanation:

the depth of the shaft,

the carrying load of the skip and the mass of the counterweight,

the approximate size of the winding drum,

the approximate height of the headgear and the sheave wheel

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i just took this test on brainly

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Con respecto a la bicicleta y a la escuela, identifique un sistema​
dimulka [17.4K]

Answer:

Cycling, also called bicycling or biking, is the use of bicycles for transport, recreation, exercise or sport.[1] People engaged in cycling are referred to as "cyclists",[2] "bicyclists",[3] or "bikers".[4] Apart from two-wheeled bicycles, "cycling" also includes the riding of unicycles, tricycles, quadricycles, recumbent and similar human-powered vehicles (HPVs).

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Bicycles were introduced in the 19th century and now number approximately one billion worldwide.[5] They are the principal means of transportation in many parts of the world.

Cycling is widely regarded as an effective and efficient mode of transportation[6][7] optimal for short to moderate distances.

Bicycles provide numerous possible benefits in comparison with motor vehicles, including the sustained physical exercise involved in cycling, easier parking, increased maneuverability, and access to roads, bike paths and rural trails. Cycling also offers a reduced consumption of fossil fuels, less air or noise pollution, reduced greenhouse gas emissions,[8] and greatly reduced traffic congestion.[9] These have a lower financial cost for users as well as for society at large (negligible damage to roads, less road area required). By fitting bicycle racks on the front of buses, transit agencies can significantly increase the areas they can serve.[10]

In addition, cycling provides a variety of health benefits. The World Health Organization (WHO) states that cycling can reduce the risk of cancers, heart disease, and diabetes that are prevalent in sedentary lifestyles.[11][12] Cycling on stationary bikes have also been used as part of rehabilitation for lower limb injuries, particularly after hip surgery.[13] Individuals who cycle regularly have also reported mental health improvements, including less perceived stress and better vitality.[14]

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6 0
3 years ago
This is a broad category of applications and technologies for gathering, storing, analyzing, and providing access to data to hel
igomit [66]

Answer:

Business intelligence.

Explanation:

That concept that applies to that of the selection, incorporation, evaluation, and analysis of companies or business information systems, software, and behaviors. It is direct at promoting good business decision-making.

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6 0
3 years ago
There are n poor college students who are renting two houses. For every pair of students pi and pj , the function d(pi , pj ) ou
Nuetrik [128]

Answer:

Here the given problem is modeled as a Graph problem.

Explanation:

Input:-  n, k and the function d(pi,pj) which outputs an integer between 1 and n2

Algorithm:-We model each student as a node. So, there would be n nodes. We make a foothold between two nodes u and v (where u and v denote the scholars pu and pv respectively) iff d(pu,pv) > k. Now, Let's call the graph G(V, E) where V is that the vertex set of the graph ( total vertices = n which is that the number of students), and E is that the edge set of the graph ( where two nodes have edges between them if and only the drama between them is bigger than k).

We now need to partition the nodes of the graph into two sets S1 and S2 such each node belongs to precisely one set and there's no edge between the nodes within the same set (if there's a foothold between any two nodes within the same set then meaning that the drama between them exceeds k which isn't allowed). S1 and S2 correspond to the partition of scholars into two buses.

The above formulation is akin to finding out if the graph G(V,E) is a bipartite graph. If the Graph G(V, E) is bipartite then we have a partition of the students into sets such that the total drama <= k else such a partition doesn't exist.

Now, finding whether a graph is bipartite or not is often found using BFS (Breadth First algorithm) in O(V+E) time. Since V = n and E = O(n2) , the worst-case time complexity of the BFS algorithm is O(n2). The pseudo-code is given as

PseudoCode:

// Input = n,k and a function d(pi,pj)

// Edges of a graph are represented as an adjacency list

1. Make V as a vertex set of n nodes.

2. for each vertex  u ∈ V

\rightarrow  for each vertex v ∈ V

\rightarrow\rightarrowif( d(pu, pj) > k )

\rightarrow\rightarrow\rightarrow add vertex u to Adj[v]   // Adj[v] represents adjacency list of v

\rightarrow\rightarrow\rightarrow add vertex v to Adj[u] // Adj[u] represents adjacency list of u

3.  bool visited[n] // visited[i] = true if the vertex i has been visited during BFS else false

4. for each vertex u ∈ V

\rightarrowvisited[u] = false

5. color[n] // color[i] is binary number used for 2-coloring the graph  

6. for each vertex u ∈ V  

\rightarrow if ( visited[u] == false)

\rightarrow\rightarrow color[u] = 0;

\rightarrow\rightarrow isbipartite = BFS(G,u,color,visited)  // if the vertices reachable from u form a bipartite graph, it returns true

\rightarrow\rightarrow if (isbipartite == false)

\rightarrow\rightarrow\rightarrow print " No solution exists "

\rightarrow\rightarrow\rightarrow exit(0)

7.  for each vertex u ∈V

\rightarrow if (color[u] == 0 )

\rightarrow\rightarrow print " Student u is assigned Bus 1"

\rightarrowelse

\rightarrow\rightarrow print " Student v is assigned Bus 2"

BFS(G,s,color,visited)  

1. color[s] = 0

2. visited[s] = true

3. Q = Ф // Q is a priority Queue

4. Q.push(s)

5. while Q != Ф {

\rightarrow u = Q.pop()

\rightarrow for each vertex v ∈ Adj[u]

\rightarrow\rightarrow if (visited[v] == false)

\rightarrow\rightarrow\rightarrow color[v] = (color[u] + 1) % 2

\rightarrow\rightarrow\rightarrow visited[v] = true

\rightarrow\rightarrow\rightarrow Q.push(v)

\rightarrow\rightarrow else

\rightarrow\rightarrow\rightarrow if (color[u] == color[v])

\rightarrow\rightarrow\rightarrow\rightarrow return false // vertex u and v had been assigned the same color so the graph is not bipartite

}

6. return true

3 0
3 years ago
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