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malfutka [58]
1 year ago
11

Write the design brief (identify the problem) of mine headgear​

Engineering
1 answer:
grin007 [14]1 year ago
3 0

The design brief that identified the problem of mine headgear is:

The structural structure above an underground mine shaft that facilitates the hoisting of machinery, persons, or supplies is known as a headframe (also called as a gallows frame, winding tower, hoist frame, pit frame, shafthead frame, headgear, headstock, or poppethead).

Mine headgear supports wheel systems that suspend winding cables that convey employees and ore up and down deep level shafts. These weird humanoid constructions have become the mining industry's defining emblem.

A miner's helmet consists of four major components:

Part 1: The hoist or winch is in a winding house. This component of the system is responsible for winding and unwinding the steel cable.

A motor and a control system are connected to the hoist.

When a steel cable unwinds from the winch, the mine cage and skips are lowered into the mine.

When the steel cable is wound up again, the mine cage and skips are elevated.

The sheave wheel is a pulley wheel that stands above the mining shaft in Part 2. The hoist rope travels over the sheave wheel and down the mine shaft.

The sheave wheel minimizes the mine cable's sliding friction.

Part 3: The head frame is the framework that holds the sheave wheel in place. When lifting the heavy mine cage, it must be robust enough to maintain the sheave wheel in place.

The head frame's left "legs" slope towards the hoist. This is due to the cable's strain dragging the entire frame in that direction. The sloping legs keep the head frame from tipping or collapsing.

Part 4: The cage and the jumps. Miners and equipment are transported up and down the mine in the cage. Skips are attached beside or beneath the cage.

Skips are used to transport ore and waste materials from mines.

<h3>What is a design brief?</h3>

A design brief, also called as a creative brief, is a program management document that identifies the scope, scale, and key aspects of your impending design project.

Learn more about design brief:

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This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1
Ulleksa [173]

Answer:

<em>L is not a regular language with formal proofs  </em>

Explanation:

<em>(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails  that L is a regular language. Then by the Pumping Lemma for Regular Languages, </em>

<em>there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, </em>

<em>s = xyz subject to the following conditions: </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L</em>

<em />

<em>(b) To determine that L is not a regular language, we mke use of proof by contradiction.  lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,</em>

<em>The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject  to the condtions as follows : </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L. </em>

<em>Choose s = 0p10p </em>

<em>. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.</em>

<em>for some  k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus,  xy0 </em>

<em>z should be in L. xy0 </em>

<em>z = xz = 0(p−k)10p </em>

<em>It is shown that is is  not in L. This is a  contraption with the pumping lemma.  our assumption that L is regular is  incorrect, and L is not a regular language</em>

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3 years ago
An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 Ω) are connected in series. A 44.0-Hz AC generator
MakcuM [25]

Answer:

(A) Maximum voltage will be equal to 333.194 volt

(B) Current will be leading by an angle 54.70

Explanation:

We have given maximum current in the circuit i_m=385mA=385\times 10^{-3}A=0.385A

Inductance of the inductor L=400mH=400\times 10^{-3}h=0.4H

Capacitance C=4.43\mu F=4.43\times 10^{-3}F

Frequency is given f = 44 Hz

Resistance R = 500 ohm

Inductive reactance will be x_l=\omega L=2\times 3.14\times 44\times 0.4=110.528ohm

Capacitive reactance will be equal to X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 44\times 4.43\times 10^{-6}}=816.82ohm

Impedance of the circuit will be Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{500^2+(816.92-110.52)^2}=865.44ohm

So maximum voltage will be \Delta V_{max}=0.385\times 865.44=333.194volt

(B) Phase difference will be given as \Phi =tan^{-1}\frac{X_C-X_L}{R}=\frac{816.92-110.52}{500}=54.70

So current will be leading by an angle 54.70

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