Answer:
a) [tex] \dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW[/tex]
b) T_2 = 36.4 degree C
c) [tex]\Delta S_{gen} = 0.006512 KW/K[/tex]
Explanation:
Given data:
[tex]P_1 = 160 kPa[/tex]
volumetric flow [tex]V_1 = 0.03 m^3/s[/tex]
[tex]P_2 = 800 kPa[/tex]
power input [tex]W_{in} = 10 kW[/tex]
[tex]T_{surr} = 20 degree C[/tex]
entropy = 0.008 kW/K
from refrigerant table for P_1 = 160 kPa and x_1 = 1.0
[tex] v_1 = 0.12355 m^3/kg[/tex]
[tex]h_1 = 241.14 kJ/kg[/tex]
[tex]s_1 = 0.94202 kJ/kg K[/tex]
a) mass flow rate [tex] \dot m = \frac{V_1}{v_1}[/tex]
[tex]\dot m = \frac{0.03}{0.12355} = 0.2428 kg/s[/tex]
heat loss[tex] = T_{surr} \times entropy[/tex]
heat loss[tex] \dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW[/tex]
b) from energy balance equation
[tex]W_{in} 0 \dot Q_{out} = \dot m (h_2 -h_1}[/tex]
[tex]10 - 2.344 = 0.2428 (h_2 - 241.14}[/tex]
[tex]h_2 = 272.67 kJ/kg[/tex]
from refrigerant table, for P_2 = 800 kPa and h_2 = 272.67 kJ/kg
T_2 = 36.4 degree C
c) from refrigerant table P_2 = 800 kPa and h_2 = 272.67 kJ /kg
[tex]s_2 = 0.93589 kJ/kg K[/tex]
rate of entropy
[tex]\Delta S_R = \dot m =(s_2 -s_1)[/tex]
[tex]\Delta S_R = 0.2428 \times (0.93589 -0.94202) = - 0.0014884 kW/K[/tex]
rate of entropy for entire process
[tex]\Delta S_{gen} = \Delta _S_R + \Delta_{surr}[/tex]
[tex]\Delta S_{gen} = 0.0014884 + 0.008 = 0.006512 KW/K[/tex]
What is the correct statement regarding the stress over the section of a shaft in torsion?
a) A normal stress that varies linearly over the section
b) A normal stress that is uniform over the section.
c) A shear stress that is uniform over the section.
d) A shear stress that varies linearly over the section
Answer:
d) A shear stress that varies linearly over the section.
Explanation:
Given that shaft is under pure torsion
As we know that relationship between shear stress in the shaft and radius given as
[tex]\dfrac{T}{J}=\dfrac{\tau}{r}[/tex]
For solid shaft
[tex]J=\dfrac{\pi}{64}d^4[/tex]
Therefore shear stress given as
[tex]\tau=\dfrac{T}{J}\times r[/tex]
T=Applied torque on the shaft
τ=Shear stress at any radius r
From the above equation we can say that shear stress is vaying linearly with the radius of the shaft
Therefore the answer will be d.
A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at
a completely reversed stress amplitude of 70 kpsi.
Answer:
104,576 cycles
Explanation:
Step 1: identify given parameters
Ultimate strength of steel ([tex]S_{ut}[/tex])= 120 Kpsi
stress amplitude ([tex]\alpha_{a}[/tex])= 70 kpsi
life of the specimen (N) = ?
[tex]N = (\frac{\alpha_{a}}{a})^\frac{1}{b}[/tex]
where a and b are coefficient of fatigue cycle
Step 2: calculate the the endurance limit of specimen
[tex]S_{e} = 0.5*S_{ut}[/tex]
[tex]S_{e}[/tex] = 0.5*120 = 60 kpsi
Step 3: calculate coefficient 'a'
[tex]a=\frac {(0.8XS_{ut})^2}{S_{e}}[/tex]
[tex]a=\frac {(0.8X120)^2}{60}[/tex]
[tex]a= 153.6 kpsi
Step 4: calculate the coefficient 'b'
[tex]b =-\frac{1}{3}log(\frac{f*S_{ut} }{S_{e}})[/tex]
[tex]b =-\frac{1}{3}log(\frac{0.8*120}{60})[/tex]
[tex]b =-0.0680
Step 5: calculate the life of the specimen
[tex]N=(\frac{\alpha_{a}}{a})^\frac{1}{b}[/tex]
[tex]N=(\frac{70}{153.6})^\frac{1}{-0.068}[/tex]
[tex]N=104,576 cycles [/tex]
∴ the life (N) of the steel specimen is 104,576 cycles
Explain the difference, on the basis of the test results, between the ultimate strength and the "true" stress at fracture.
Answer / Explanation:
On the basis of the test result, Ultimate strength which is mostly known as the ultimate tensile strength is the strength attached to the ability or capacity of a structural element or material used in the test to withstand elongation forces or pull force applied to it.
WHILE,
True stress at fracture can be classified as stress or load associated to the point where yielding or fracture occurred divided by the cross-sectional area at the yield point.
A world class runner can run long distances at a pace of 15 km/hour. That runner expends 800 kilocalories of energy per hour. Find the power (in Watts [W]) the runner is exerting while running. b) Find the total energy (in Joules [J]) exerted by the runner in a 15 km run.c) How many Milky Way (Original Single 52.2g) chocolate bars does the runner need to buy to supply the amount of energy to complete a half-marathon (13.1 miles)?
Answer:
a) power = 929.78W
b) energy = 3,347,200J
c) 5 milkyway bars
Explanation:
a) Power is the rate of use of energy per unit time. This is given in the question as 800 kilocalories per hour. Part a) requires the power in Watts (W) which is equivalent to Joules per second. Thus, kilocalories per hour need to be converted to Joules per second:
[tex]1kcal = 4184J[/tex]
[tex]1hour = 3600s[/tex]
[tex]power=\frac{energy}{time}[/tex]
[tex]power=\frac{800kcal*\frac{4184J}{kcal}}{1hour*\frac{3600s}{hour}}[/tex]
[tex]power=\frac{3347200}{3600}[/tex]
[tex]power=929.78J/s=929.78W[/tex]
b) Total time required for a 15km run can be calculated by the speed of the runner (15km/h):
[tex]time=\frac{distance}{speed}[/tex]
[tex]time=\frac{15km}{15km/h}[/tex]
[tex]time=1hour=3600s[/tex]
The energy exerted over this time can be found by:
[tex]energy=power*time[/tex]
[tex]energy=929.78J/s*3600s[/tex]
[tex]energy=3,347,200J[/tex]
c) Total time required for a 13.1mile run can be calculated by the speed of the runner (15km/h):
[tex]1mile=1.6km[/tex]
[tex]time=\frac{distance}{speed}[/tex]
[tex]time=\frac{13.1mile*\frac{1.6km}{mile}}{15km/h}[/tex]
[tex]time=\frac{20.96}{15}[/tex]
[tex]time=1.40h=5030.4s[/tex]
The energy exerted over this time can be found by:
[tex]energy=power*time[/tex]
[tex]energy=929.78J/s*5030.4s[/tex]
[tex]energy=4,677,165J[/tex]
Assuming one Milky Way (Original Single 52.2g) has 240 kilocalories
[tex]number=\frac{energy}{energy_{milkyway}}[/tex]
[tex]number=\frac{4677165}{240kcal*\frac{4184J}{kcal}}[/tex]
[tex]number=\frac{4677165}{1004160}[/tex]
[tex]number=4.65[/tex]
5 milkyway bars are needed
An air-conditioning system operates at a total pressure of 95kPa and consists of a heating section and a humidifier that supplies wet steam (saturated water vapor) at 100°C. Air enters the heating section at 10°C and 70% relative humidity at a rate of 35 m^3/min, and it leaves the humidifying section at 20°C and 60 percent relative humidity. Determine:
(a) the temperature and relative humidity of air when it leaves the heating section.
(b) the rate of heat transfer in the heating section.
(c) the rate at which water is added to the air in the humidifying section.
Answer:
The temperature and relative humidity when it leaves the heating section = T2 = 19° C and ∅2 = 38%
Heat transfer to the air in the heating section = Qin = 420 KJ/min
Amount of water added = 0.15 KG/min
Explanation:
The Property of air can be calculated at different states from the psychometric chart.
At T1 = 10° C and ∅ = 70%
h1 = 87 KJ/KG of dry air
w1 = 0.0053 kg of moist air/ kg of dry air
v1 = 0.81 m^3/kg
AT T3 = 20° C , 3 ∅ = 60%
h3 = 98 KJ/KG of dry air
w3 = 0.0087 kg of moist air/ kg of dry air
The moisture in the heating system remains the same when flowing through the heating section hence, (w1 = w2)
The mass flow rate of dry air,
m1 = V'1/V1 = 35/0.81
m1 = 43.21 kg/min
By balancing the energy in heating section we get:
mwhw + ma2h2 = mah3
(w3 -w2)hw + h2 = h3
h2 = h3 - (w3 -w2)hw @ 100 C
Hence, hw = hg @ 100 C and w2 = w1
h2 = h3 - (w3 -w2) hg @ 100 C
h2 = 98 - ( 0.0087 - 0.0053) * 2676
h2 = 33.2 KJ/KG
The exit temperature and humidity will be,
T2 = 19.5° C and 2 ∅ = 37.8%
(b) Calculating the transfer of heat in the heating section
Qin = ma(h2 -h1) = 43.21(33.2 - 23.5)
Qin = 420 KJ/min
(c) Rate at which water is added to the air in the humidifying section,
mw = ma(w3 - w2) = (43.2)(0.0087 - 0.0053)
mw = 0.15 KG/min
Make two lists of applications of matrices, one for those that require jagged matrices and one for those that require rectangular matrices. Now argue whether just jagged, just rectangular, or both should be included in a programming language.
Matrices, particularly jagged and rectangular ones, serve different functions in programming. Jagged matrices handle non-uniform data and graphs, while rectangular matrices are necessary for graphics and linear algebra. Both should be included in programming languages to provide flexibility and cater to various applications.
Explanation:Applications of Matrices in ProgrammingMatrices are a fundamental aspect of both mathematics and computer science. They are primarily used for representing and manipulating linear transformations, data, and relations. Matrices come in various forms, with two common types being jagged matrices and rectangular matrices.
When considering whether to include jagged or rectangular matrices, or both, in a programming language, it's essential to recognize the different use cases each type serves. Inclusion of both types allows for greater flexibility and functionality. A programming language that supports both can cater to a wider range of applications and user needs. Nullable matrices can be used when necessary to mimic jaggedness within rectangular matrix structures, offering a middle ground.
Including both jagged and rectangular matrices in a programming language offers flexibility for diverse applications, optimizing memory usage and performance across various data structures and operations.
Applications of Matrices
Applications that Require Jagged Matrices
1. Sparse Data Representation: Efficient storage of data where the majority of elements are zero, such as social network adjacency matrices.
2.Graph Representation: Storing graphs where the number of edges varies widely between nodes.
3. Image Processing: Non-uniform image sampling or varying resolution images.
4. Variable-Length Data: Managing records of varying lengths in databases or documents.
5. Numerical Solutions: Adaptive mesh refinement in computational fluid dynamics.
Applications that Require Rectangular Matrices
1. Linear Algebra: Solving systems of linear equations, eigenvalue problems.
2. Data Analysis: Handling datasets in machine learning, where data points have the same number of features.
3. Computer Graphics: Transformations and projections in 2D and 3D graphics.
4. Control Systems: State-space representations of dynamic systems.
5. Signal Processing: Filtering, convolution operations in audio and image processing.
Arguments for Including Matrix Types in a Programming Language
Just Jagged Matrices
- Pros:
- Flexibility in handling datasets with variable row lengths.
- Efficient memory usage for sparse or highly irregular data.
- Cons:
- Increased complexity in implementation and usage, as many linear algebra operations assume rectangular structure.
- Can be less performant for operations that benefit from contiguous memory storage.
Just Rectangular Matrices
- Pros:
- Simplicity in usage and implementation for most standard applications.
- Better support for linear algebra operations and optimizations in libraries.
- Cons:
- Inefficient memory usage for sparse or variable-length data, leading to wasted space.
- Lack of flexibility for applications requiring variable row lengths.
Both Jagged and Rectangular Matrices
- Pros:
- Flexibility to choose the most appropriate type based on the specific application requirements.
- Optimal memory usage and performance by using jagged matrices for sparse data and rectangular matrices for linear algebra operations.
- Cons:
- Increased complexity in the programming language and potential confusion for users in choosing the correct type.
- Possible performance overhead in managing two types of matrices.
Including both jagged and rectangular matrices in a programming language provides the greatest flexibility and efficiency across a wide range of applications. While it adds some complexity, the benefits of being able to choose the optimal structure for specific tasks outweigh the drawbacks. This approach ensures that the language can handle diverse data structures and operations effectively, catering to both irregular and uniform datasets.
Discuss how, as a safety professional, you would respond to the overlap in Occupational Safety and Health Administration (OSHA) standards and state or local building, electrical, and life safety codes. What would be the most significant challenges?
Answer:
Explanation:
As a security professional, I will respond positively to the OSHA requirements overlap. OSHA guidelines are meant to provide general guidance to all members of various entities throughout the country, while local or state codes also ensure compliance with laws unique to their areas, taking into account workplace safety and security.
OSHA accepts the security codes of the state building To the degree that such codes comply with OSHA regulations, such as BOCA. All the codes and regulations for local, state-owned construction, electrical and life protection are under the same umbrella. Generally, all security protocols and specifications are in accordance with OSHA guidelines. Nonetheless, certain points will overlap, while localized codes will also be addressed to a particular community or state that may
2.73 A fan within an insulated duct delivers moist air at the duct exit at 358C, 50% relative humidity, and a volumetric flow rate of 0.4 m3 /s. At steady state, the power input to the fan is 1.7 kW. The pressure in the duct is nearly 1 atm throughout. Using the psychrometric chart, determine the temperature, in 8C, and relative humidity at the duct inlet.
Answer:
Temperature at 8C = 32 °C × 8 = 256
Relative humidity is = 60% × 8 of 256°C accordingly.
Explanation:
Considering the volumetric flow rate = 0.4 m³/s
Moist air delivered (Temperature T1 = 358 C)
With Relative humidity at duct outlet = 50%
Power input at steady rate = 1.7 KW
Pressure in the duct = 1 atmosphere
Temperature at 8C = ?
Relative humidity at the duct inlet = ?
Recalling that the value for specific enthalpy and specific volume is
Specific Enthalpy = h1 = 81 kj/kilogram of dry air
and Specific Volume = v1 = 0.9 m³/kg
Now, recalling the formula for mass flow rate,
We have, m = Volumetric flow rate / Specific volume
Therefore, 0.4 m³/s ÷ 0.9 m³/kg
= 0.44 kilogram / second
Recalling the enthalpy at inlet,
we have, h2 = h1 - p/m
Where h1 = Specific enthalpy
p = power input at steady rate
m = calculated mass flow rate
Now, if we substitute the values into the equation,
we have h2 = 81 kj/kilogram of dry air - 1.7 KW / 0.44 kilogram / second
h2 = 77.175 kj / kilogram of dry air.
Therefore, the properties of air at constant absolute humidity and specific enthalpy is 77.175 kj/kg.
Temperature at 8C = 32 °C × 8 = 256
Relative humidity is = 60% × 8 of 256°C accordingly.
True/False
If a contractor chooses to use their own quality system, they must demonstrate compliance with the established military standard through formal third party certification. [Identify DoD policy regarding Basic Quality Systems and the role of ISO 9001.
Answer:
false jdbebheuwowjwjsisidhhdd
Employers are not required to keep a record of an employee who has the flu.
a. True
b. False
Employers aren't required to keep records specifically for an employee with the flu, unless under certain conditions related to workplace illnesses. Reasonable exceptions to the FOIA include the protection of sensitive personal information such as government employees' medical records.
Explanation:Employers are not generally required to keep records of an employee who has simply contracted the flu. However, certain regulations may apply if the illness could be work-related or if it pertains to a larger public health concern that requires tracking. In the context of the Freedom of Information Act (FOIA), the question of keeping medical records would fall under exemptions related to personal privacy. For instance, medical records for government employees would be a reasonable exception to FOIA, as they contain sensitive personal information that is protected from public disclosure.
One kilogram of ammonia initially at 8.0 bar and 50°C undergoes a process to 4.5 bar, 20°C while being rapidly expanded in a piston–cylinder assembly. Heat transfer between the ammonia and its surroundings occurs at an average temperature of 40°C. The work done by the ammonia is 40 kJ. Kinetic and potential energy effects can be ignored. Determine the heat transfer, in kJ, and the entropy production, in kJ/K.
The heat transfer for the process is -20.9 kJ, and the entropy production is 1 kJ/K.
Explanation:The heat transfer for the given ammonia process is -20.9 kJ. To calculate the entropy production, we can use the equation: ΔS = Q/T, where temperature T is in Kelvin. Given Q = 40 kJ, and the average temperature is 40°C, the entropy production is 1 kJ/K.
Show the ERD with relational notation with crowfoot. Your ERD must show PK, FKs, min and max cardinality, and correct line types.
Drexel University Financial Office has made contracts with several local banks to help students to easily obtain small loans up to $10,000 for qualified students for each year. A student can apply for loans up to 3 banks each year. For each application, student must indicate bank name, loan amount, and requested date. Obviously, a bank approves some student loans and also decline some loans depending on the student status (which is beyond the scope of this database). For each approved loan, there is a loan#, interest rate, approved amount, monthly payment amount, and the beginning date of loan payments. For each loan, the office also keeps tracks of history of payments to the loan, including payment date and payment amount. The office will record student ID, name, major, and year. For banks, we just keep bank number and name.
You may add other unstated, but common-sense oriented, facts to the ERD, but you must represent the facts stated above.
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
What is the stress concentration factor of a shaft in torsion, where D=1.25 in. and d=1 in. and the fillet radius is, r=0.2 in.a. K=1.4b. K=1.5c. K=1.2d. K=1.3
Answer:
Concentration factor will be 1.2
So option (C) will be correct answer
Explanation:
We have given outer diameter D = 1.25 in
And inner diameter d = 1 in and fillet ratio r = 0.2 in
So [tex]\frac{r}{d}[/tex] ratio will be [tex]=\frac{0.2}{1}=0.2[/tex]
And [tex]\frac{D}{d}[/tex] ratio will be [tex]=\frac{1.25}{1}=1.25[/tex]
Now from the graph in shaft vs torsion the value of concentration factor will be 1.2
So concentration factor will be 1.2
So option (C) will be correct answer.
The nozzle bends the flow from vertically upward to 30 degrees with the horizontal as it discharges the water (at 20 degrees C) to the atmosphere at V = 125 ft/s.
The volume of water within the nozzle itself (above the flange) is 100 lb.
Find the horizontal and vertical forces that must be applied to the flange (by the pipe below it) to hold it in place.Area of flange = 1.0 ft^2Area of nozzle = 0.50 ft^2Volume of area above flange = 1.8 ft^3Vertical height from flange to nozzle = 2 ft
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
Fy = -11267.294 lbf
Explanation:
given data
nozzle flow = 30 degrees
discharges the water = 20 degrees C
volume of water = 100 lb
Area of flange = 1.0 ft²
Area of nozzle = 0.50 ft²
Volume of area flange = 1.8 ft³
Vertical height flange to nozzle = 2 ft
solution
we will apply here continuity equation that is
A1 × V1 = A2 × V2 .............1
put here value and we get volume V1 that is
V1 = [tex]\frac{0.5\times 125}{1}[/tex]
V1 = 62.5 ft/s
and
now we will apply here Bernoulli equation that is
[tex]\frac{p1}{\gamma 1} + \frac{V1^2}{2g} + z1 = \frac{p2}{\gamma 2} + \frac{V2^2}{2g} + z2[/tex] .............................2
put here value and we will get
p1 = 0 + [tex]\frac{62.4}{2\times 32.2}(125^2 - 62.5^2) + 62.4 (2)[/tex]
p1 = 11479.614 psf
so here moment in y will be
∑ Fy = m [ (Vo)y - (Vi)x ]
so here we get
p1 ×A1 + Fy - Wn - Ww = [tex]\rho[/tex] Q [ V2 × sin30 - V1 ]
put here value and we get Fy
1147.614 × 1 + Fy - 100 - (62.4 × 1.8) = (1.94) × (0.5 ×130) × (125sin30 - 62.5)
solve it we get
Fy = -11267.294 lbf
A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequality of Clausius. Which process is possible. What is the maximum power available?
a) Pout = 3 kW (power out)
b) Pout = 2 kW
c) Carnot Cycle.
Answer:
Explanation:
Given
[tex]T_h=250^{\circ}C\approx 523\ K[/tex]
[tex]T_L=30^{\circ}C\approx 303\ K[/tex]
[tex]Q_1=6 kW[/tex]
From Clausius inequality
[tex]\oint \frac{dQ}{T}=0[/tex] =Reversible cycle
[tex]\oint \frac{dQ}{T}<0[/tex] =Irreversible cycle
[tex]\oint \frac{dQ}{T}>0[/tex] =Impossible
(a)For [tex]P_{out}=3 kW[/tex]
Rejected heat [tex]Q_2=6-3=3\ kW[/tex]
[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]
[tex]=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K[/tex]
thus it is Impossible cycle
(b)[tex]P_{out}=2 kW[/tex]
[tex]Q_2=6-2=4 kW[/tex]
[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]
[tex]=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K[/tex]
Possible
(c)Carnot cycle
[tex]\frac{Q_2}{Q_1}=\frac{T_1}{T_2}[/tex]
[tex]Q_2=3.47\ kW[/tex]
[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]
[tex]=\frac{6}{523}-\frac{3.47}{303}=0[/tex]
and maximum Work is obtained for reversible cycle when operate between same temperature limits
[tex]P_{out}=Q_1-Q_2=6-3.47=2.53\ kW[/tex]
Thus it is possible
Water at 20 bar, 400°C enters a turbine operating at steady state and exits at 1.5 bar. Stray heat transfer and kinetic and potential energy effects are negligible. A hard-to-read data sheet indicates that the quality at the turbine exit is 98%. Can this quality value be correct? If no, explain. If yes, determine the power developed by the turbine, in kJ per kg of water flowing.
Answer:
quality value is not 98%
Explanation:
given data
inlet pressure p1 = 20 bar
outlet pressure p2 = 1.5 bar
temperature t1 = 400°C
solution
as assume here assume isentropic process so equation that states stray heat transfer is negligible so Q = 0 and S1 = S2
S1 = Sf2 + x Sfg2 ........................1
here x is quantity of steam and we get all other value by steam table
so at pressure 20 bar and 400°C and at pressure 1.5 bar
S1 = 7.127 kJ/kg K and Sf2 = 1.4336 kJ/kg K
and Sfg2 = 5.7898 kJ/kg K
so put all value in equation 1 we get x that is
x = [tex]\frac{7.127-1.4336}{5.7898}[/tex]
x = 0.9833
x = 98.33 %
so here we can say quality value is not 98%
The minimum requirements for engineering documents are enumerated in
a. The Florida Building Code.
b. Chapter 471, F.S.
c. Engineer's Responsibwty Role
d. Role 61615-21, F.A.C, relating to 'Seals.
Answer:
The answer will be Rule 61G15-23 F.A.C, relating to Seals.
Explanation:
According to the description given by: Florida administrative code&Florida administrative register the Minimum requirements for engineering documents are in the section 'Final 61G15-23' from 11/3/2015. This document provides specifications of materials required for the safe operation of the system that is the result of engineering calculations, knowledge and experience.
When implementing a safety and health program, management leadership does not need employee participation a True b) False Effective management of worker safety and health programs has O a) Stopped all on-site injuries o b) Improve employee productivity and morale in the workplace c)Cost companies more money O d)Become a waste of time Nearly 3 of all serious occupational injuries and illnesses stem from overexertion of repetitive motion e a1/2 O b) 1/4 c)All o d) 1/3 Training is a way for employers to provide 4 to enable employees to protect themselves and others from injuries O a) Ideas b) Tools o cInteractions O d) Money Under OSHA, employees are not protected from discrimination when reporting a work-related injury, illness, or fatality 5 a True b) False
Answer: I have answered the questions in rephrased sentences as below;
When implementing a safety and health program, management leadership need employee participation. Effective management of worker safety and health programs has improved employee productivity and morale in the workplace.
Nearly a third of all serious occupational injuries and illnesses stem from overexertion of repetitive motion.
Training is a way for employers to provide tools to enable employees to protect themselves and others from injuries.
Under OSHA, employees are protected from discrimination when reporting a work-related injury, illness, or fatality.
Explanation: All personnel including management & employees must be directly involved when workplace HSE policies are being made & reviewed. This is because everyone in the work environment is impacted one way or the other when incidents occur.
Training & Reporting are key responsibilities of managers, employers & supervisors, so it is mandatory to be done without discrimination so as to foster employees happiness which ultimately lead to zero incidents & increased productivity & profit.
Forward Error Correction (FEC) is a popular method of error control in networks where bandwidth utilization is a secondary issue. (FEC transmits information using a redundant coding that allows full recovery of the data even in cases where some parts of the transmission were not delivered). Can FEC fully replace the standard timeout/acknowledgment-based error control mechanisms? Explain your answer.
Answer: NO
Explanation:
FEC transmits information using a redundant coding that allows full recovery of the data even in cases where some parts of the transmission were not delivered.FEC cannot be always the alternative for standard timeout/acknowledgment-based error control mechanisms. One of the reasons is that it can recover full data only when some part of transmission is not delivered.
Marco is a franchisee with Daggies, a chain of sandwich shops. His business was doing well until several Daggies franchisees got in trouble and were forced to close their shops. Soon afterward, Marco's business deteriorated and he too was forced to close. This is an example of:
Answer:
The coattail effect
Explanation:
Clearly, the Daggies franchise were a huge part of Marco's success and attracted many customers for the business. This is why Marco deteriorated as soon as many Daggies franchisees closed.
The coattail effect is the phenomenon where an influencing member in a party (franchisee in this case) contributes largely to the success of another, which is the case with Marco and Daggies
A horizontal 2-m-diameter conduit is half filled with a liquid (SG=1.6 ) and is capped at both ends with plane vertical surfaces. The air pressure in the conduit above the liquid surface is 200kPa. Determine the resultant force of the fluid acting on one of the ends caps, and locate this force relative to the bottom of the conduit.
Answer:
Resultant force = 639 kN and it acts at 0.99m from the bottom of the conduit
Explanation:
The pressure is given as 200 KPa and the specific gravity of the liquid is 1.6.
The resultant force acting on the vertical plate, Ft, is equivalent to the sum of the resultant force as a result of pressurized air and resultant force due to oil, which will be taken as F1 and F2 respectively.
Therefore,
Ft = F1 + F2
According to Pascal's law which states that a change in pressure at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere, the air pressure will act on the whole cap surface.
To get F1,
F1 = p x A
= p x (πr²)
Substituting values,
F1 = 200 x π x 1²
F1 = 628.32 kN
This resultant force acts at the center of the plate.
To get F2,
F2 = Π x hc x A
F2 = Π x (4r/3π) x (πr²/2)
Π - weight density of oil,
A - area on which oil pressure is acting,
hc - the distance between the axis of the conduit and the centroid of the semicircular area
Π = Specific gravity x 9.81 x 1000
Therefore
F2 = 1.6 x 9.81 x 1000 x (4(1)/3π) x (π(1)²/2)
F2 = 10.464 kN
Ft = F1 + F2
Ft = 628.32 + 10.464
Ft = 638.784 kN
The resultant force on the surface is 639 kN
Taking moments of the forces F1 and F2 about the centre,
Mo = Ft x y
Ft x y = (F1 x r) + F2(1 - 4r/3π)
Making y the subject,
y = (628.32 + 10.464(1 - 4/3π)/ 638.784
y = 0.993m
Chairs and bolsters are used to:
a. keep formwork from moving during concrete placement.
b. maintain proper spacing between members during installation of precast concrete members.
c. help during posttensioning of concrete slabs.
d. splice reinforcing bars that require greater length.
e. support reinforcing bars in beams and slabs, prior to concrete placement.
Answer:e. support reinforcing bars in beams and slabs, prior to concrete placement.
Explanation: Chairs are used for construction of foundations,large steel supports,deck constructions and for underground works. Chairs can be for rebar support ( rebar support chairs),post tension chairs.
Bolsters are usually long made of metallic materials mainly used as support for construction of different infrastructures like roads it helps to ensure the concretes and other construction materials stay firmly connected. Image 1 is a metal be chair
Image 2 is a metal bolster
A 0.5m diameter sphere containing pollution monitoring equipment is dragged through the Charles River at a relative velocity of 10m/s. The sphere has a specific gravity of 0.5, is fully submerged, and tethered to the towing device by a 2m cable. What is the angle the towing cable makes with the horizontal? Assume the water is at 10°C.
Answer:
[tex]\phi = 155.57[/tex]
Explanation:
from figure
taking summation of force in x direction be zero
[tex]\sum x = 0 [/tex]
[tex]F_D = Tsin \theta[/tex] .....1
[tex]\frac{c_d \rho v^2 A}{2} =Tsin \theta[/tex]
taking summation of force in Y direction be zero
[tex]F_B - W- Tcos \theta[/tex]
[tex]T = \frac{F_B -W}{cos \theta}[/tex] .........2
putting T value in equation 1
[tex]F_D - \frac{F_B -W}{cos \theta} sin\theta[/tex]
[tex]F_D = \rho g V ( 1 -Sg) tan \theta [/tex].........3
[tex]F_D = \rho g [\frac{\pi d^3}{6}] ( 1 -Sg) tan \theta [/tex]
[tex]tan \theta = \frac{6 c_D \rho v&2 A}{ 2 \rho g V \pi D^3 (1- Sg)}[/tex]
Water at 10 degree C has kinetic viscosity v = 1.3 \times 10^{-6} m^2/s
Reynold number
[tex]Re = \frac{ VD}{\nu} = \frac{10\times 0.5}{1.3 \times 10^{-6}} = 3.84 \times 10^6[/tex]
so for Re =[tex] 3.84 \times 10^6 [/tex] cd is 0.072
[tex]tan \theta = \frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}[/tex]
[tex]\theta = tan^{-1} [\frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}][/tex]
[tex]\theta = - 65.57 degree[/tex]
[tex]\phi = 90 - (-65.57) = 1557.57[/tex] degree
Lot ABCD between two parallel street lines is 350.00 ft deep and has a 220.00-ft frontage (AB) on one street and a 260.00-ft frontage (CD) on the other. Interior angles at A and B are equal, as are those at C and D. What distances AE and BF should be laid off by a surveyor to divide the lot into two equal areas by means of a line EF parallel to AB?
Answer:
240.83 ft
Explanation:
The distances AE and BF will be equal = 182.58 ft
The area of the lot will be the product of the depth and the average of the two frontages
= ( 350 * (220 + 260)/2) = 84000 ft
Half of the area becomes 42000 ft
<A = arctan (20/350) = 3. 3°
Hence 42000=h/2*(220 + 220 + 2h*tan3.3)
Solving, we obtain h= 182.28ft
EF = 220 + (2*182.28*tan 3.3)
= 240.83ft
A mitochondrial membrane complex consisting of ATP synthase, adenine nucleotide translocase (ATP–ADP translocase), and phosphate translocase functions in oxidative phosphorylation. Adenine nucleotide translocase, an antiporter located in the inner mitochondrial membrane, moves ADP into and ATP out of the matrix. Phosphate translocase is also located in the inner mitochondrial membrane. It transports H + ions and phosphate H 2 PO − 4 ions into the matrix. The energy derived from the movement of H + ions down an electrochemical gradient from the intermembrane space into the matrix is used to drive the synthesis of ATP. How many H + ions must be moved into the matrix for the synthesis of 1 ATP? number of H + ions:'
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Consider a 3-phase, 4 Pole, AC induction motor being driven at 75 Hz. Initially, the motor is connected to a mechanical load that requires 1,816 W of power. While connected to this load, the motor spins at 1,996 rpm. Now a new load is connected that requires 2,334 W of power. What is the new slip on the motor for the new load assuming that the motor is operating in its normal operating range?
Answer:
New slip = 0.145
Explanation:
Revolutions per minute rpm = 1996
Synchronous speed Ns= 120 * f / p
= 120 * 75 / 4 = 2250
Hence initial slip S1 = 2250 - 1996 / 2250 = 0.113
From output power = [tex] 3*S*E^2*R2 [/tex]
Hence at constant E and R, output power depends on slip
So for new power and new slip to new initial power and initial slip,
Hence 2334/1816 = S2/ S1
New slip S2 = 2334 * 0.113 / 1816 = 0.145
Hence, new speed = Ns ( 1 - S2 )
= 2250*(1 - 0.145)
= 1924 rpm
In an aligned and continuous carbon fiber-reinforced nylon 6,6 composite, the fibers are to carry 97% of a load applied in the longitudinal direction. (a) Using the data provided, determine the volume fraction of fibers required.
Answer:
a) 0.26
b) 1077 MPa
Explanation:
a) The following equation can be used to determine the volume fraction:
[tex]\frac{F_f}{F_m} =\frac{E_fV_f}{E_m(1-V_f)}[/tex]
[tex]\frac{0.97}{1-0.97} =\frac{260V_f}{2.8(1-V_f)}[/tex]
[tex]32.3 = \frac{260V_f}{2.8-2.8V_f}[/tex]
[tex]V_f = 0.26[/tex]
b) Tensile strength can be found by using the following equation:
[tex]\sigma_{cl} = \sigma_m(1-V_f)+\sigma_fV_f = 50*(1-0.26)+4000*0.26 = 1077[/tex] MPa
A steam turbine in a power plant receives 5 kg/s steam at 3000 kPa, 500°C. Twenty percent of the flow is extracted at 1000 kPa to a feed water heater, and the remainder flows out at 200 kPa. Find the two exit temperatures and the turbine power output.
Answer:
The temperature of the first exit (feed to water heater) is at 330.15ºC. The second exit (exit of the turbine) is at 141ºC. The turbine Power output (if efficiency is %100) is 3165.46 KW
Explanation:
If we are talking of a steam turbine, the work done by the steam is done in an adiabatic process. To determine the temperature of the 2 exits, we have to find at which temperature of the steam with 1000KPa and 200KPa we have the same entropy of the steam entrance.
In this case for steam at 3000 kPa, 500°C, s= 7.2345Kj/kg K. i=3456.18 KJ/Kg
For steam at 1000 kPa and s= 7.2345Kj/kg K → T= 330.15ºC i=3116.48KJ/Kg
For steam at 200 kPa and s= 7.2345Kj/kg K → T= 141ºC i=2749.74KJ/Kg
For the power output, we have to multiply the steam flow with the enthalpic jump.
The addition of the 2 jumps is the total power output.
For a two-source network, if current produced by one source is in one direction, while the current produced by the other source is in the opposite direction through the same resistor, the resulting current is? a. The product of the two and the direction of the smaller b. The difference between the two and has the same direction. c. The sum of the two and the direction of either d. The average of the two and the direction of the largest
Answer:
c. The sum of the two and the direction of either.
Explanation:
If both sources are linear, we can apply the superposition theorem, which in this case, states simply that the resulting current is just the algebraic sum of both currents.
This is due to any of them, is independent from the other, so we can calculate her influence without any other source present.
Assuming that both sources are ideal (no shunt resistances) , we can apply superposition, removing all sources but one each time, just replacing the sources by an open circuit.
Let's suppose that we have a 5A source flowing to the rigth, and a 3A source in the opposite direction.
Applying superposition, we have:
I₁ = 5A (the another source is removed)
I₂ = -3A (The minus sign indicates that is flowing in the opposite direction, 5A source is removed)
I = I₁ + I₂ = 5A + (-3A) = 2 A
As it can be seen , the resulting current is the sum (algebraic) of both current sources, and has the direction of either of the sources, depending on the absolute value of the current sources.
(If the directions of I₁ and I₂ were inverted, with the same absolute values, the direction would be the opposite).
We could have arrived to the same result applying KCL.
Most economists believe that real economic variables and nominal economic variables behave independently of each other in the long run. For example, an increase in the money supply, a variable, will cause the price level, a variable, to increase but will have no long-run effect on the quantity of goods and services the economy can produce, a variable. The distinction between real variables and nominal variables is known as .
Answer:
The distinction between real variables and nominal variables is known as inflation rate.
Explanation:
The inflation rate is what distinguishes real variables (such as increase or decrease in prices/price level of goods or services) from nominal variables (such as the quantity of available money: high or low money supply). Real variables, which are affected by nominal variables, are actually nominal variables that have been adjusted for inflation.