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nirvana33 [79]
4 years ago
12

This is an essential safety procedure that protects workers from injury while working on or near electrical circuits and equipme

nt by locking the device and/or power source to prevent anyone from turning on the hazardous power sources while someone is performing maintenance or servicing work: Select the best option a. Shutdown b. Lockout/Tagout c. Closeout
Engineering
1 answer:
Mice21 [21]4 years ago
8 0

Answer:

Option B

Lockout/ Tagout

Explanation:

lockout and tagout is a safety procedure used in most industries and work places. This ensures that during any maintenance work, that dangerous machines are properly shut off and not able to be started up again prior to the completion of maintenance or repair work.

The procedure includes:

1. Turns off and disconnects the machinery or equipment from its energy source(s) before performing service or maintenance.

2. The authorized employee(s) either lock or tag the energy-isolating device(s) to prevent the release of hazardous energy.

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Describe the steps, tools, and technology needed in detail and
Mumz [18]

Answer:

Mass production methods are based on two general principles: (1) the division and specialization of human labour and (2) the use of tools, machinery, and other equipment, usually automated, in the production of standard, interchangeable parts and products.

Explanation:

Step 1: Product Concept. This is where you begin to flesh out your basic idea. ...

Step 2: Research. ...

Step 3: Product Design Development. ...

Step 4: Research and development of the final design. ...

Step 5: CAD. ...

Step 6: CAM. ...

Step 7: Prototype Testing. ...

Step 8: Manufacturing.

8 0
3 years ago
Water flows through a horizontal 60 mm diameter galvanized iron pipe at a rate of 0.02 m3/s. If the pressure drop is 135 kPa per
maksim [4K]

Answer:

pipe is old one with increased roughness

Explanation:

discharge is given as

V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}

V = 7.07  m/s

from bernou;ii's theorem we have

\frac{p_1}{\gamma}  +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma}  +\frac{V_2^2}{2g} + z_2 + h_l

as we know pipe is horizontal and with constant velocity so we have

\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}

P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma

135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81

solving for friction factor f

f = 0.0324

fro galvanized iron pipe we have \epsilon  = 0.15 mm

\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025

reynold number is

Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}

Re = 378750

from moody chart

For Re = 378750 and \frac{\epsilon}{d} = 0.0025

f_{new} = 0.025

therefore new friction factor is less than old friction factoer hence pipe is not new one

now for Re = 378750 and f = 0.0324

from moody chart

we have \frac{\epsilon}{d} =0.006

\epsilon = 0.006 \times 60

\epsilon = 0.36 mm

thus pipe is old one with increased roughness

5 0
3 years ago
You are designing a 200mm long x 100mm wide x 50mm deep rectangular housing, with a wall width of 1.5 mm, and 1 degree draft. Yo
Nimfa-mama [501]
365gpa I don’t even know what it is but this is with 10 point and I need it
6 0
2 years ago
Let's model this housing price data! Before we can do this, however, we need to split the data into training and test sets. Reme
Lilit [14]

The program reads in a dataset into a pandas dataframe, and uses the train_test_split function in the sklearn library to split the data into <em>training and test sets</em>. The code goes thus :

import pandas as pd

<em>#import</em><em> </em><em>the</em><em> </em><em>pandas</em><em> </em><em>dataframe</em><em> </em><em>and</em><em> </em><em>alias</em><em> </em><em>it</em><em> </em><em>as</em><em> </em><em>pd</em>

from sklearn.model_selection import train_test_split

<em>#import</em><em> </em><em>the</em><em> </em><em>train_test_split</em><em> </em><em>function</em><em> </em>

housing_df = pd.read_csv('housing price.csv')

<em>#read</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>housing</em><em> </em><em>data</em><em> </em>

features_df = df.iloc[:,1:]

<em>#seperate</em><em> </em><em>the</em><em> </em><em>features</em><em> </em><em>from</em><em> </em><em>the</em><em> </em><em>label</em><em> </em><em>;</em>

target_df = df.iloc[:,0]

<em>#put</em><em> </em><em>the</em><em> </em><em>label</em><em> </em><em>into</em><em> </em><em>a</em><em> </em><em>seperate</em><em> </em><em>dataframe</em><em> </em><em>as</em><em> </em><em>well</em><em>.</em><em> </em>

X_train, X_test, Y_train, Y_test = train_test_split(features_df, target_df, test_size = 0.1, random_state = 1)

<em>#uses</em><em> </em><em>tuple</em><em> </em><em>unpacking</em><em> </em><em>to</em><em> </em><em>randomly</em><em> </em><em>assign</em><em> </em><em>the</em><em> </em><em>data</em><em> </em><em>each</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>4</em><em> </em><em>variables</em><em>.</em><em> </em>

<em>#</em><em>Test</em><em> </em><em>size</em><em> </em><em>is</em><em> </em><em>test</em><em> </em><em>percent</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>entire</em><em> </em><em>dataset</em><em> </em>

Learn more :brainly.com/question/4257657?referrer=searchResults

3 0
3 years ago
A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co
NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx  32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

V_{hem}=\frac{2}{3}\pi r^3

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

V_{cyl}=\pi r^2\cdot h

3) Conical bottom

Volume of conical bottom of radius'r' and angle \theta is

V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}

Applying the given values we obtain the volume of the container up to cylinder is

V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}

Hence the capacity in liters is V=32.355\times 1000=32355Liters

3 0
4 years ago
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