A journeyman electrician with 16 years experience on-the-job was removing metal fish
tape from a hole at the base of a metal light pole. It was raining during the course of the
work. The fish tape became energized, electrocuting him. What could have been done to
prevent this accident? Select all that apply.
Explanation:
1. Ensure all circuits are de-energized before beginning work (29 CFR 1926.416(a)(3)).
2. Controls to be deactivated during the course of work on energized or de-energized
equipment or circuits must be tagged (29 CFR 1926.417(a)).
3. Employees must be instructed to recognize and avoid unsafe conditions associated with
their work (29 CFR 1926.21(b)(2)).
Which term refers to an application of biotechnology to forensic science?
Cloning
Transgenesis
Bioprocessing
DNA profiling
Biomanufacturing
Answer:
DNA profiling
Explanation:
took the test
5. (20 points) Identify and resolve (via forwarding) all the data hazards in the MIPS pipeline for the following sequence of MIPS instructions. Insert pipeline bubbles as necessary if forwarding cannot completely solve the data hazards. You may use multi-cycle pipeline representation to show the forwarding and bubbles. sub $4, $1, $3 sw $4, 0($2) lw $1, 0($2) add $1, $3, $4
Answer:
a
Explanation:
Ideal gas helium flows through the inlet of an isentropic nozzle with a velocity of 25 m/s, and the exit flow is at 100 kPa, 300 K, and has a velocity of 250 m/s. a. Determine the inlet temperature. b. Determine the inlet pressure. c. Determine the area ratio between inlet and exit
Answer:
a. The inlet temperature is approximately 305.232 K
b. The inlet pressure is approximately 452.0108 kPa
c. The area ratio between the inlet and exit is approximately 2.2509
Explanation:
a. From the energy equation related to the question, we have;
[tex]C_p \cdot (T_i - T_e) = \dfrac{1}{2} \cdot \left (v_e^2 - v_1^2 \right)[/tex]
Where;
[tex]C_p[/tex] = The specific heat capacity for helium = 5.913 kJ/(kg·K)
[tex]T_i[/tex] = The inlet temperature
[tex]T_e[/tex] = The exit temperature = 300 K
[tex]v_i[/tex] = The inlet velocity = 25 m/s
[tex]v_e[/tex] = The exit velocity = 250 m/s
Therefore, we have;
[tex]T_i= \dfrac{ \dfrac{1}{2} \cdot \left (v_e^2 - v_1^2 \right)}{C_p} + T_e = \dfrac{ \dfrac{1}{2} \times \left (250^2 - 25^2 \right)}{5.913 \times 1000} + 300 \approx 305.232[/tex]
The inlet temperature = [tex]T_i[/tex] ≈ 305.232 K
b. From the following equation for the critical pressure, for helium, we have;
[tex]\dfrac{P_c}{P_i} = \left (\dfrac{2}{n + 1} \right ) ^{\dfrac{n}{n - 1} } = 0.487[/tex]
Where;
[tex]P_c[/tex] = The critical pressure = 2.26 atm for helim
[tex]P_i[/tex] = The inlet pressure
n = The polytropic constant
We have;
[tex]\dfrac{2.26 \ atm}{P_i} = 0.487[/tex]
[tex]\therefore P_i = \dfrac{2.26 \ atm}{0.487} \approx 4.641 \ atm[/tex]
The inlet pressure, [tex]P_i[/tex] ≈ 4.641 atm ≈ 452.0108 kPa
c. The inlet to exit pressure ratio is given as follows;
[tex]P_e = \dfrac{A_i \times T_e \times v_i}{A_e \times T_i \times v_e} \times P_i[/tex]
Therefore, we have;
[tex]\dfrac{A_i}{A_e} = \dfrac{P_e \times T_i \times v_e}{ T_e \times v_i \times P_i} = \dfrac{100 \times 305.232 \times 250}{ 300 \times 25 \times 452.0108} = 2.2509[/tex]
The area ratio between the inlet and exit, [tex]A_i/A_e[/tex] ≈ 2.2509.
Type the correct answer in the box. Spell all words correctly.
Who focuses on planning a long-term business?
focus on planning a long-term business.
Reset
Next
Explanation:
A business man who really focus on victory achieved through a right procedure will focus on long-term planning.
Let us understand what a short-term and long-term planning is.
Short-term will plan only for two-three years. But a long-term plan will look for future five years income projection, plan of expansion, bigger goals, etc.
A business man is the person who take risks and achieve more. A victory can be achieved in many ways one is taking bigger risks, next is focusing on long-term plans
If this is wrong, give me the answer choices so I know what's right or wrong. I'll edit the question if given to me.
Given these observed times (in minutes) for five elements of a job, determine the observed time (OT) for each element. Note: Some of the elements occur only periodically. (Round your answers to 2 decimal places.) CYCLE Element 1 2 3 4 5 6 1 1.2 1.2 1.2 1.2 1.2 - 2 - .7 - .7 - 0.7 3 2.2 2.2 2.2 3.2 2.2 2.2 4 3.2 - - 4.2 - - 5 0.7 .7 1.2 0.7 0.7 0.7
Answer:
Following are the solution to this question:
Explanation:
[tex]CYCLE[/tex]
[tex]Element \ \ \ \ \ \ 1\ \ \ \ \ \ 2 \ \ \ \ \ \ 3 \ \ \ \ \ \ 4 \ \ \ \ \ \ 5 \ \ \ \ \ \ 6 \ \ \ \ \ \ Average[/tex]
[tex]1 \ \ \ \ \ \ 1.2 \ \ \ \ \ \ 1.2 \ \ \ \ \ \ 1.2 \ \ \ \ \ \ 1.2 1.2 \ \ \ \ \ \ - \ \ \ \ \ \ 1.2\\\\[/tex]
[tex]2 \ \ \ \ \ \ - \ \ \ \ \ \ .7 \ \ \ \ \ \ - \ \ \ \ \ \ .7 \ \ \ \ \ \ - \ \ \ \ \ \ 0.7 \ \ \ \ \ \ 0.7 \\\\3 \ \ \ \ \ \ 2.2 \ \ \ \ \ \ 2.2 \ \ \ \ \ \ 2.2 \ \ \ \ \ \ 3.2 \ \ \ \ \ \ 2.2 \ \ \ \ \ \ 2.2 \ \ \ \ \ \ 2.36\\\\[/tex]
[tex]4 \ \ \ \ \ \ 3.2 \ \ \ \ \ \ - \ \ \ \ \ \ - \ \ \ \ \ \ 4.2 \ \ \ \ \ \ - \ \ \ \ \ \ - \ \ \ \ \ \ 3.7\\\\5 \ \ \ \ \ \ 0.7 \ \ \ \ \ \ .7 \ \ \ \ \ \ 1.2 \ \ \ \ \ \ 0.7 \ \ \ \ \ \ 0.7\ \ \ \ \ \ 0.7\ \ \ \ \ \ 0.78[/tex]
[tex]Element \ \ \ \ \ \ \ \ \ \ \ \ Observed \ time \\\\1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.2\\\\2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.7\\\\3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.36\\\\4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.7\\\\5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.78[/tex]
A piston–cylinder device contains 15 kg of saturated refrigerant-134a vapor at 280kPa. A resistance heater inside the cylinder with a value of 110V is used to transfer heat of 20.67kJ/kg to the refrigerant at constant pressure for 6min. Determine the current supplied if the final temperature is 75℃
Answer:
18.77 A
Explanation:
To solve this we use the energy balance equation, that is:
[tex]E_{in}-E_{out}=\Delta E_{sys}\\\\Q_{in}+W_{in}+W_{out}=\Delta U\\\\Q_{in}+W_{in}=\Delta U\\\\But\ W_{in}=Voltage(V)*Current(I)*change\ in\ time(\Delta t)=VI\Delta t,\Delta U=m(h_2-h_1)\\\\Given\ that\ m=15kg,V=110\ V,\Delta t=6\ min = (6*60\ s)=360\ s,Q_{in}=20.67\ kJ/kg*15\ kg=310.05\ kJ=310050\ J\\\\From\ table: At\ P_1=280kPa, h_1=249.71\ kJ/kg=249710\ J/kg;At\ P_2=280kPa \ and\ T_1=75^oC,P_2=319.95\ kJ/kg=319950\ J/kg\\\\Substituting:\\\\310500+(110*360*I)=15(319950-249710)\\\\[/tex]
[tex]39600I=1053600-310500\\\\39600I=743100\\\\I=18.77\ A[/tex]
Which process made making copies of technical drawings easier?
In the nineteenth century, the process of ___________
helped professionals quickly create copies of technical drawings.
Answer:
The Process of printing
Explanation:
hope this helps :)
A fruit juice is being heated in an indirect heat exchanger using steam as a heating medium. The product flows through the heat exchanger at a rate of 1000 kg/h and the inlet temperature is 30 oC. Determine the quantity of steam required to heat the product to 100 oC when only latent heat of vaporization (2230.2 kJ/kg) from steam at 110oC is used for heating. The specific heat of the product is 4 kJ/(kg oC).
Answer:
The appropriate response will be "125.55 kg/hr".
Explanation:
The given values are:
Rate of heat exchanger,
m = 1000 kg/h
Inlet temperature,
T = 30°C
Specific heat,
[tex]c_p[/tex] = 4 kJ/(kg°C)
Now,
The heat taken by juice from steam will be:
⇒ [tex]Q=mc_p(\Delta T)[/tex]
On substituting the values, we get
⇒ [tex]=1000\times 4(100-30)[/tex]
⇒ [tex]=280000 \ kJ/hr[/tex]
The heat given by the steam to juice will be:
⇒ [tex]Q=m\times L[/tex]
On substituting the values, we get
⇒ [tex]=m\times 125.55[/tex]
On equating both the heat values, we get
⇒ [tex]280000=m\times 2230.2[/tex]
⇒ [tex]m=\frac{280000}{2230.2}[/tex]
⇒ [tex]m=125.55 kg/hr[/tex]
A light commercial building located in Philadelphia, Pa has construction and use characteristics much like a residence and design heating load of 120,000 Btu/h and cooling of 60,000 Btu/h. Determine the following: (a) Heating energy requirement, Btu(b) Gallons of No. 2 fuel oil/year, assuming efficiency of the furnace is 75 %(c) Cubic feet of natural gas/year(d) kWh of electricity for cooling for air conditioner with SEER of 8.5 using degree day method.(6 points)
Answer:
b
Explanation:
15 kg of ice at 0°C is heated to water of temperature 20°C. What is the amount of heat required if the specific latent heat of ice fusion is 335 kJ/kg and the specific heat capacity of water is 4.19kJ/kg/°C?
Answer:
6282 kJ
Explanation:
Given that:
The mass (m) = 15 kg
The specific latent heat of ice fusion [tex]h_{fg}_{ice}[/tex] = 335 kJ/kg
The specific heat capacity of water = 4.19 kJ/kg.c
The initial temperature of the ice [tex]T_i[/tex]= 0° C
The final temperature of the water [tex]T_f[/tex] = 20° C
To find the needed amount of heat to convert 0° C ice to 20° C of water.
To do that, we need to find the latent heat required for the phase change from 0° C ice to 0° C water, then the heat required to convert 0° C water to 20° C water.
Heat required = [tex]m \times h_{fg}_{ice}+ m \times c_{water} \times \Delta T[/tex]
Heat required = (15 × 335) + (15 × 4.19) × (20 - 0)
Heat required = 5025 + 62.85 × 20
Heat required = 5025 + 1257
Heat required = 6282 kJ
A gas stream (A) of 15.0 mol% H2, and the balance N2, is to be mixed with another gas stream
(B) containing 40.0 mol% H2, and the balance N2, to produce 100 kg/h of a 25 mol% H2, and the
balance N2 gas stream (C).
(a) Draw and fully label a flowchart of the mixing process.
(b) Calculate the average molecular weight of the product stream (C).
(c) Calculate the molar flow rates of the product stream (C) in kmol/h.
(d) Calculate the required flow rates of the feed mixtures A and B in kmol/h.
Answer:
A gas stream and other points are discussed below in details.
Explanation:
The method flow chart is displayed above.
B)Mole fraction of H2 in stream C=25%=0.25
Mole fraction of N2 in steams C=1-Mole fraction of H2=1-0.25=0.75
Average molecular weight of stream C=Mole fraction of H2*Molecular weight of H2+ Mole fraction of N2*Molecular weight of N2=0.25*2 kg/kmol+0.75*28 kg/kmol=21.5 kg/kmol
C) Mass flow rate of product C=100 kg/h
Molar flow rate of product C=Mass flow rate /Molecular weight=100 kg/h/21.5 kg/kmol=4.6512 kmol/h
D) Apply overall mole balance
Mole in=Mole out
Mole in=Mole of A+Mole of B=A+B
Let mole of A=A kmol/h
Mole of B=B kmol/h
Mole out=Mole of C stream=4.6512 kmol/h
A+B=4.6512 eq.1
Apply hydrogen mole balance
Mole of H2 in=Mole of H2 in A stream+ Mole of H2 in B stream=0.15 A+0.4 B
Mole of H2 out=Mole of H2 in C stream=0.25*4.6512=1.1628 kmol/h
1.1628 =0.15 A+0.4 B eq.2
Multiply eq.1 by 0.15
0.69768=0.15 A+0.15 B eq.3
Subtract eq.3 from eq.2
0.25 B=0.46512
B=1.86048 kmol/h
Substitute in eq1.
A=4.6512-1.86048=2.79042 kmol/h
Molar flow rate of A=2.79042 kmol/h
A molar flow rate of B=1.86048 kmol/h
the term applied to the chemistry of the body
Answer:
Biochemistry
Explanation:
Hope this helps :)
Answer:
Biochemistry
Have an amazing day!
A heat exchanger is used to recover waste heat from an industrial furnace and use it to heat pressurized water. The heat exchanger consists of a tube bank with a square 10 x 10 array of thin-walled smooth tubes (total of 100 tubes). The tubes are aligned in-line with a transverse pitch, ST = 50 mm. Each tube is 25 mm in diameter and 4 m long. Water passes through each tube at a flow rate of 0.025 kg s-1, so the total flow rate through all 100 tubes is 2.5 kg s-1. The exhaust gas is in cross-flow over the tubes, with an upstream velocity of 5.0 m s-1 and total flow rate of 2.25 kg s-1. The exhaust gas has properties similar to air. The inlet temperatures of the water and exhaust gas are 300 K and 800 K, respectively. (a) What is the overall heat transfer coefficient between the water and the gas? (b) What are fluid outlet temperatures? As a first approximation, assume the water flow through the tubes is fully developed and turbulent, with an estimated bulk mean temperature of 350 K. Use a mean temperature of 600 K and properties of air to represent the exhaust gases. Please comment on the accuracy of these assumptions after you obtain your results, but you do not need to repeat any calculations.
Answer:
Explanation:
ok this is as test dont cheat i d
a 1-w, 350-ω resistor is connected to 24 v. Is this resistor operating within its power rating?
Answer:
No.
Explanation:
[tex]P_r[/tex] = Power rating = 1 W
R = Resistance = [tex]350\ \Omega[/tex]
V = Voltage = [tex]24\ \text{V}[/tex]
Power is given by
[tex]P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}[/tex]
[tex]1.65\ \text{W}>1\ \text{W}[/tex]
So
[tex]P>P_r[/tex]
Hence, the resistor is not operating within its power rating.
What is the base unit that is divided in an engineering scale?
An engineering scale is based on unit increments of a/n ____.
Explanation:
The most often used architect's scale is 1/4" = 1'-0". The basic unit of measure for metric scales is the millimeter (mm). Drafting scales are either open-divided or full-divided.
I really have no clue if this is right or not because, well, I'm not an expert.
If this is wrong, don't hesitate to report my answer if it's wrong.
discuss 7 habits of highly effective people and how important are ethics in today's society
Answer:
Explanation:
The 7 Habits of Highly Effective People, is a book written and first published in 1989. It is a business and self-help book that was written by Stephen Covey. The seven habits include
Being proactive
Starting anything with the end in mind
First things first
Always thinking towards a win-win situation
Seeking initially to understand, then going on to want to be understood
Synergize, and lastly
Growing
Type the correct answer in the box. Spell all words correctly.
A genetically engineered hormone, , can treat Mary’s child for growth hormone deficiencies by stimulating body growth and increasing muscle mass.
Answer:
recombinant human growth
Explanation:
Answer:
recombinant human growth
Explanation:
yes
Determine the design moment strength for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb
This question is incomplete, the complete question is;
Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.
Use Fy=50 ksi and assume Cb=1.0 (if needed).
Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft
Explanation:
Given that;
section W 21 x 73 steel beam;
now from the steel table table for this section;
Zx = Sx = 151 in³
also given that; fy = 50 ksi and Cb = 1.0
QMn = 0.9 × Fy × Zx
so we substitute
QMn = 0.9 × 50 × 151
QMn = 6795 k-inch
we know that;
12inch equals 1 foot
so
QMn = 6795 k-inch / 12
QMn = 566.25 f-ft
Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft
In order to avoid slipping in the shop, your footwear should ___________.
A) Be brand-new
B) Have steel toes
C) Not have shoestrings
D) Have proper tread
(This is for my Automotive class by the way)
In order to avoid slipping in the shop, your footwear should have proper tread. The correct option is D.
What is slipping?Slipping is an action of falling or sliding down due to a lack of friction between the surfaces. This occurs when both surfaces coming into contact are smooth or frictionless.
During rainy seasons, due to rain everywhere the chances of slipping increase, the slippers slip on mud or water. Vehicles also slip down on roads.
To avoid slipping, wear proper shoes. The shoes that have a good grip are less to slip. The tires are given with tread, which increases the grip on the road. Tread-on shoes or slippers are also good. Wear tread full shoes. It can avoid slipping and accidents.
Thus, the correct option is D. Have proper tread.
To learn more about slipping, refer to the link:
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Which is a better hydraulic cross section for an open channel: one with a small or a large hydraulic radius?
For a soil, cohesion is 15 kN/sq.M, unit weight is 20 kN/cu.M and the factor of safety is 1.5 along with stability number of 0.05. The safe maximum height of slope is *
Answer:
10 m
Explanation:
The Soil Cohesion is the force that holds together molecules or like particles within a soil.
The unit weight of a soil mass is the ratio of the total weight of soil to the total volume of soil.
The stability number (Sn) is given by the formula:
[tex]S_n=\frac{C_m}{\gamma *H}[/tex]
Given that:
Cohesion = C = 15 kN/m², Factor of safety = FOS = 1.5, unit weight = γ = 20 kN/m³, H = maximum height, Sn = 0.05. Hence:
[tex]C_m=\frac{C}{FOS}\\\\S_n=\frac{C}{\gamma *FOS *H}\\\\H= \frac{C}{\gamma *FOS *S_n}\\\\Substituting:\\\\H=\frac{15}{20*1.5*0.05} \\\\H=10\ m[/tex]
In addition to bandages and gauze, it's recommended that a first aid kit also contain ___________.
A) Portable eyewash bottles
B) Tweezers
C) A finger splint
D) Scissors
In addition to bandages and gauze, it's recommended that a first aid kit also contain scissors. The correct option is D.
What is First aid?First aid is the initial and urgent care given to anyone who has either a small or significant disease or injury, with the goal of preserving life, preventing the condition from getting worse, or assisting in the healing process.
First aid is the initial and urgent care given to anyone who has either a small or significant disease or injury, with the goal of preserving life, preventing the condition from getting worse, or assisting in the healing process.
A first aid kit normally contains :
1. plasters
2. scissors
3. bandages
4. antiseptic ointments or lotions
5. basic medicines
6. gauzes
Anyone who has been hurt in an accident or has been injured can receive quick and basic medical care with the help of these materials.
To know more about the First-Aid box follow
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A(94,0,14) B(52,56,94) C(10,6,48) D(128,64,10)
The goal of the assignment is to apply methods of transforming orthogonal projections to solving metric problems.
The coordinates of four points ABCD are given (see next table). Determine the following values by transforming projection planes:
• 1. The size of the dihedral angle at the edge AB.
• 2. The shortest distance between ribs DA and BC.
• 3. Distance from vertex D to plane ABC.
• 4. Natural size of triangle ABC.
An example how to complete the task is shown after the table
9514 1404 393
Answer:
see below
Explanation:
We presume the drawing you want is contained in the figure you attached. The true sizes are identified with corresponding numbers.
Without your example, or your given methods of transforming projections, it is difficult to tell exactly how you want to compute the various distances or sizes.
Which one of the following skills is a software engineer specializing in system support expected to possess?
ability to correct system maintenance issues
ability to direct traffic between devices
ability to develop user programming
ability to recommend system upgrades
Answer:
A. Ability to correct system maintenance issues
Explanation:
As a systems engineer, you will be responsible for a range of tasks, including management of IT systems, working with software developers to design and document new system architecture, and testing and implementing processes to improve the efficacy of existing systems.
A lake is fed by a polluted stream and a sewage outfall. The stream and sewage wastes have a decay rate coefficient (k) of 0.5/day (1st order units). Assuming complete mixing and no other water losses or gains, what is the steady-state pollutant concentration as mg/L in the lake? Incoming Stream: C = 10 mg/L, Q = 40 m^3/s Sewage Outfall: C = 100 ppm, Q = 0.5 m^3/s Lake: V= 200 m^3
Solution :
Given :
k = 0.5 per day
[tex]$C_s = 10 \ mg/L \ ; \ \ Q_s= 40 \ m^3/s$[/tex]
[tex]$C_{sw} = 100 \ ppm \ ; \ \ Q_{sw}= 0.5 \ m^3/s$[/tex]
Volume, V [tex]$= 200 \ m^3$[/tex]
Now, input rate = output rate + KCV ------------- (1)
Input rate [tex]$= Q_s C_s + Q_{sw}C_{sw}$[/tex]
[tex]$=(40 \times 10) + (0.5\times 100)$[/tex]
[tex]$= 2 \times 10^5 \ mg/s$[/tex]
The output rate [tex]$= Q_m C_{m}$[/tex]
= ( 40 + 0.5 ) x C x 1000
[tex]$=40.5 \times 10^3 \ C \ mg/s$[/tex]
Decay rate = KCV
∴[tex]$KCV =\frac{0.5/d \times C \ \times 200 \times 1000}{24 \times 3600}$[/tex]
= 1.16 C mg/s
Substituting all values in (1)
[tex]$2 \times 10^5 = 40.5 \times 10^3 \ C+ 1.16 C$[/tex]
C = 4.93 mg/L
What is a shop air compressor
Answer:
need more information about the question