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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.

You might be interested in
the water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of
Brilliant_brown [7]

The maximum height of the water stream that could rise is 40.65 m. It can be calculated using Principle of Bernoulli.

Principle of Bernoulli can be described an increase in the speed of a fluid happen simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. Equation of Bernoulli is shown as a conservation of energy law for a flowing fluid. Bernoulli equation is shown below:

P_1/ρg + v_1²/2g + h_1  = P_2/ρg +v_2²/2g + h_2

Where:

ρ = fluid density

g = acceleration due to gravity

P_1 = pressure at elevation 1

v_1 = velocity at elevation 1

h_1 = height of elevation 1

P_2 = pressure at elevation 2

v_2 = velocity at elevation 2

h_2 = height at elevation 2

Assumed that the water at free surface, so v_1 = v_2 = 0

So, the formula will be

P_1/ρg + h_1  = Patm/ρg + h_2

Based on the scenario, we know that:

h_1 = 20 m

P_1gage = 2 atm ≈ 20265 N/m

ρ = 1000 kg/m²

From the scenario, we will determine the maximum height by using the equation of bernoulli and it will be

h_2 = (P_1 - Patm)/ρg + Z_1

h_2 = \frac{2 atm}{1000(9.81)} x (\frac{101325N/m^{2} }{1 atm} )(\frac{1 kg.m/s^{2} }{1 N} ) + 20 = 40.65 m

So, from that, we can conclude the maximum height of the water stream that could rise is 40.65 m.

Learn more about bernoulli principle at brainly.com/question/15415820

#SPJ4

4 0
1 year ago
Is there anyone who can help me with welding?
Gnoma [55]
what is it ill know your question??!!
5 0
3 years ago
Your manager has asked you to research and recommend a writing guide that examiners in your digital forensics company can use fo
vivado [14]

Answer:

Kindly Check the explanation

Explanation:

Report

A Report is a way to present the forensic examination result in front of judge. A report contains warrants, all affidavits issued for arrest, expenses occur during forensic test and lists of evidence. Apart from evidence a report also contains expert opinions.  

Guidelines  

These are the set of rules and written statement which are used to present the important information in a structured and clear format. The main motive behind to define the guidelines is to achieve standardization in the reports. It also improves the readability of the report.  

Guidelines for report  

Abstract:  

Each report must contain an abstract of the report which gives the brief review about the report. It came at the starting of the report. The main motive behind to give abstract is to give brief about report to someone in a very less time. The size of abstract must not be more than one page.  

Index:  

Each report must contain an index which display's which content is available at which page of report. The content of index must be appropriate and clear so that the reader will not get confuse.

Format of report:

The formatting of report is also one of the essential parts of the guidelines. It makes the report attractive and also due to this reader not gets bored while reading the report. The guidelines regarding layout of report are as follow:  

1. Use time roman font and 12 font size as a default font to explain anything in the report.  

2. The explanation must be justified

3. As possible divide the long paragraph. if possible explain in points.

4. The page must have border on all sides

5. Each pages must have numbering and footer

Rules for grammar:

1. The language of the report must be simple and clear. So that everyone can understand it easily.

2. Avoid repeating same sentence or word again and again.

3. Use active instead of passive voice, in order to give importance to doer.

4. Write expressions between commas.

5. Avoid spelling mistakes in report which also make wrong impact.

6. As possible give heading or subheading to material or explanation which must be accurate.

7. Explain, new or difficult words used in the report.

8. What ever written in the report must be through the point and accurate on bases of fact.

Body of report:

The body of the report referred to as a content written inside the report. In order to write an impressive report, it must include all these things:

1. Avoid use of hypothetical question, if still hypothetical question arises it must be bases on factual evidence.  

2. Add all the affidavits and warrants in the report issued for search and arrest.

3. The cost money expenses while solving the case.

4. The expert opinion which must be bases on knowledge or previously declare case results.

5. The list of people and evidence which are collected and interrogative during the investigation.

6. If investigation is further going on than write the area in which the investigation further goes on and also mention when the investigation get complete.  

Forensic software used:  

A short must be given about the software used in the investigation. The result generated by the software must also be pasted in that report. It is more good if attach the snap shot of the result. Add any previous case result in which the same software result is taken as evidence, in order to justify that the result generated by the software is true. Some of forensic softwares are FTK, ProDiscover, Hexworkshop. The result of these softwares must be heighted so that everyone can notice it.  

Conclusion:  

The conclusion must be given at the end of report to tell according him what result came after investigation. In conclusion also write on which basis the result concludes. There must be strong behind the conclusion.  

References:  

In Reference write all the cases, thesis, books which an investigator followed to solve the current case. In addition if they referred any internet article. They can also write the website name where  that article lies. In short can say that write everything where they got an idea which is helpful in solving the case.

 

8 0
3 years ago
Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain
fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

N_{A}  = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1}  } ln(\frac{P_{T} -P_{A2}  }{P_{T} -P_{A1} })

Where

D_{AB} = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

P_{A1} = Pressure at first boundary

P_{A2} = Pressure at the destination boundary

T = System temperature

P_{T} = System pressure

Where P_{T} = 101.3 kPa P_{A2} =0, P_{A1} =y_{A}, P_{T} = 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R =  \frac{kJ}{(kmol)(K)} ,    T = 298 K   and  D_{AB} = 1.18 \frac{cm^{2} }{s} = 1.8×10⁻⁵\frac{m^{2} }{s}

N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65}) = 5.153×10⁻⁴\frac{kmol}{m^{2}s }

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

5 0
3 years ago
Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 5208C.
algol [13]

This question is incomplete, the complete question is;

Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520°C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500°C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa.

Determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency.

Answer:

- the quality of the steam exiting the second stage of the turbine is 0.9329  

- the thermal efficiency is 36.05%  

Explanation:

get the properties of steam at pressure p1 = 28 MPa and temperature T2 = 520°C .

Specific enthalpy h1= 3192.3 kJ/kg

Specific entropy s1 = 5.9566 kJ/kg.K  

Process 1 to 2s is isentropic expansion process in the turbine

S1 = S2s

get the enthalpy at state 2s at pressure p2 = 6 MPa and S2s = 5.9566 kJ/kg.K

h2s = 2822.2 kJ/kg

get the enthalpy at state 2 using isentropic turbine efficiency of the turbine. nT1 = (h1 - h2) / (h1 - h2s)

0.78 = (3192.3 - h2) / (3192.3 - 2822.2)

h2 = 2903.6 kJ/kg

get the enthalpy at state 3 at pressure p2 = p3 = 6 MPa and T3 = 500°C

h3 = 3422.2 kJ/kg

s3 = 6.8803 kJ/kg.K

Process 3 to 4s is isentropic expansion process in the turbine

S3 = S4s

get the enthalpy at state 4s at pressure p4s = p4 = 6 kPa and S4s = 6.8803 kJ/kg.K

h4s = 2118.8 kJ/kg

get the enthalpy at state 4 using isentropic turbine efficiency of the turbine. nT2 = (h3 - h4) / (h3 - h4s)

0.78 = (3422.2 - h4) / ( 3422.2 - 2118.8 )

h4 = 2405.5 kJ/kg

get the properties at pressure, p5 = 6 kPa

h5 = hf

= 151.53 kJ/kg

v5 = Vf  

= 0.0010064 m³/kg  

get the enthalpy at state 6 using isentropic pump efficiency of the turbine, at

p6 = p1 = 28 MPa

np = v5( p6 - p5) / (h6 - h5)

0.82 =  ((0.0010064)( 28000 - 6)) / (h6 - 151.53)

h6 = 185.89 kJ/kg  

Now to find the quality of the steam at the exit of the second stage of the turbine

At stat4, p4 = 6kPa  

h4f = 151.53 kJ/kg

h4fg = 2415.9 kJ/kg  

h4 = h4f + x4h4fg

2405.5 = 151.53 + (x4 (2415.9))

x4 = 0.9329  

the quality of the steam exiting the second stage of the turbine is 0.9329  

Also to find the efficiency of the power plant, we use the following equation;

n = Wnet / Qin  

= (Wt1 + Wt2 - Wp) / (Q61 + Q23)

=  [(h1 - h2) + (h3 - h4) - (h6 - h5)] / [(h1 - h6) + (h3 - h2)]

[(3192.3 - 2903.6) + (3422.2 - 2405.5) - (185.89 - 151.53)] / [(3192.3 - 185.89) + (3422.2 - 2903.6)]

= 0.3605

n = 36.05%  

therefore the thermal efficiency is 36.05%  

3 0
3 years ago
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