A work cell is currently operated 2000 hr/yr by a human worker who is paid an hourly rate of $23.00, which includes applicable overhead costs. One work unit is produced in a cycle time of 4.8 min. Management would like to increase output to meet increasing demand and a robot cell is being considered as a replacement for the present manual cell. The cycle time of the proposed cell would be reduced to 4.0 min. The installed cost of the robot plus supporting equipment is $120,000. Power and other utilities to operate the robot will be $0.30/hr, and annual maintenance costs are $2500. Determine:
a. the number of parts produced annually by the manually operated cell.
b. cost per part produced.
c. how does the cost per part of the robot cell compare with your answer in part (b), given a 4-year service life, 10% rate of return, and no salvage value.

Answer:

Symbols in The Secret Garden:

The secret garden: The garden seems to be a symbol for Mary’s transformation. She changes from a sickly child to a healthy one. It is a contextual symbol.  

The robin: The robin symbolizes the kindness and compassion of nature. The robin is a contextual symbol.

Spring: The spring season is a symbol for change and new life. It represents these ideas in each of the characters at the end of the story. It is a universal symbol.

Symbols in The Hunger Games:

Katniss’s mockingjay pin: It might stand for rebellion and resistance against the Capitol. It is a contextual symbol.

The mockingjay: The bird itself is a symbol for resistance against the Capitol. It is a contextual symbol.

Bread: It is a symbol for the connection between Katniss and Peeta. It is a contextual symbol.

Symbols in Hatchet:

The hatchet: The hatchet symbolizes survival and self-reliance. The hatchet is a contextual symbol.

The moose and the tornado: Both symbolize confusion and the power nature has over people. They are contextual symbols. However, a tornado can also be a universal symbol for destruction.

The plane: The plane symbolizes the changes in Brian’s life. It represents how Brian moves from one stage of his life to another. It is a contextual symbol.

Symbols in The Red Badge of Courage:

The dead soldier: The dead soldier symbolizes how all things will eventually end. It expresses the idea that accomplishments may mean very little. It is a universal symbol for an end.

Henry’s wound: For Henry, a wound symbolizes a soldier’s bravery at war. It is a contextual symbol.

Flag: A flag symbolizes the values and identity of a state or nation. It also represents the reason why two opposing sides of a war are fighting. A flag is a universal symbol.

Explanation:

These are the answers for Plato,  these are the answers from each novel.

I hope this helps you!

Answer:

The correct option is B) Balance Sheet

Explanation:

A Balance Sheet offers a description of a company's obligations, assets, and investments as well as net income over a given span of time such as a period of 6 months or 12 months, for instance.

Also known as the Statement of Financial Position, it contains sufficient information for investors and business owners to determine the company's financial performance in that period as well as to compare the performance of that company with industry norms or competition.

Cheers

Answer:

Mechanic: a person who repairs and maintains machinery

Mechanical engineers: design power-producing machines

Explanation:

Answer:

Design engineer

Explanation:

Answer:

The wind turbine generates [tex]19297.222[/tex] kilowatt-hours of electricity daily.

The wind turbine makes a daily revenue of 1736.75 US dollars.

Explanation:

First, we have to determine the stored energy of wind ([tex]E_{wind}[/tex]), measured in Joules, by means of definition of Kinetic Energy:

[tex]E_{wind} = \frac{1}{2}\cdot \dot m_{wind}\cdot \Delta t \cdot v_{wind}^{2}[/tex] (Eq. 1)

Where:

[tex]\dot m_{wind}[/tex] - Mass flow of wind, measured in kilograms per second.

[tex]\Delta t[/tex] - Time in which wind acts in a day, measured in seconds.

[tex]v_{wind}[/tex] - Steady wind speed, measured in meters per second.

By assuming constant mass flow and volume flows and using definitions of mass and volume flows, we expand the expression above:

[tex]E_{wind} = \frac{1}{2}\cdot \rho_{air}\cdot \dot V_{air} \cdot \Delta t \cdot v_{wind}^{2}[/tex] (Eq. 1b)

Where:

[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.

[tex]\dot V_{air}[/tex] - Volume flow of air through wind turbine, measured in cubic meters per second.

[tex]E_{wind} = \frac{1}{2}\cdot \rho_{air}\cdot A_{c}\cdot \Delta t\cdot v_{wind}^{3}[/tex] (Eq. 2)

Where [tex]A_{c}[/tex] is the area of the wind flow crossing the turbine, measured in square meters. This area is determined by the following equation:

[tex]A_{c} = \frac{\pi}{4}\cdot D^{2}[/tex] (Eq. 3)

Where [tex]D[/tex] is the diameter of the wind turbine blade, measured in meters.

If we know that [tex]\rho_{air} = 1.25\,\frac{kg}{m^{3}}[/tex], [tex]D = 100\,m[/tex], [tex]\Delta t = 86400\,s[/tex] and [tex]v_{wind} = 8\,\frac{m}{s}[/tex], the stored energy of the wind in a day is:

[tex]A_{c} = \frac{\pi}{4}\cdot (100\,m)^{2}[/tex]

[tex]A_{c} \approx 7853.982\,m^{2}[/tex]

[tex]E_{wind} = \frac{1}{2}\cdot \left(1.25\,\frac{kg}{m^{3}} \right) \cdot (7853.982\,m^{2})\cdot (86400\,s)\cdot \left(8\,\frac{m}{s} \right)^{3}[/tex]

[tex]E_{wind} = 2.171\times 10^{11}\,J[/tex]

Now, we proceed to determine the quantity of energy from wind being used by the wind turbine in a day ([tex]E_{turbine}[/tex]), measured in joules, with the help of the definition of efficiency:

[tex]E_{turbine} = \eta\cdot E_{wind}[/tex] (Eq. 4)

Where [tex]\eta[/tex] is the overall efficiency of the wind turbine, dimensionless.

If we get that [tex]E_{wind} = 2.171\times 10^{11}\,J[/tex] and [tex]\eta = 0.32[/tex], then the energy is:

[tex]E_{turbine} = 0.32\cdot (2.171\times 10^{11}\,J)[/tex]

[tex]E_{turbine} = 6.947\times 10^{10}\,J[/tex]

The wind turbine generates [tex]6.947\times 10^{10}[/tex] joules of electricity daily.

A kilowatt-hours equals 3.6 million joules. We calculate the equivalent amount of energy generated by wind turbine in kilowatt-hours:

[tex]E_{turbine} = 6.947\times 10^{10}\,J\times\frac{1\,kWh}{3.6\times 10^{6}\,J}[/tex]

[tex]E_{turbine} = 19297.222\,kWh[/tex]

The wind turbine generates [tex]19297.222[/tex] kilowatt-hours of electricity daily.

Lastly, the revenue generated per day can be found by employing the following:

[tex]C_{rev} = c\cdot E_{turbine}[/tex] (Eq. 5)

Where:

[tex]c[/tex] - Unit price, measured in US dollars per kilowatt-hour.

[tex]C_{rev}[/tex] - Revenue generated by the wind turbine in a day, measured in US dollars.

If we know that [tex]c = 0.09\,\frac{USD}{kWh}[/tex] and [tex]E_{turbine} = 19297.222\,kWh[/tex], then the revenue is:

[tex]C_{rev} = \left(0.09\,\frac{USD}{kWh} \right)\cdot (19297.222\,kWh)[/tex]

[tex]C_{rev} = 1736.75\,USD[/tex]

The wind turbine makes a daily revenue of 1736.75 US dollars.

Answer:

3. have known

4. wrote

5. not sure

6.have changed, has only

7. answered

8. have missed

9.has never seen

10.never saw

11.became,have changed

12. have changed, have grown

13.was,found

14. was never

am not sure but hope it helps

Answer:

  4800 watts

Explanation:

Power is the product of voltage and current:

  P = VI = (240 V)(20 A) = 4800 W

The wattage is 4800 watts.

The phase identifying the need involves research to determine exactly what the client expects. The correct option is B.

Research is a systematic inquiry process that includes data gathering, documentation of important information, analysis, and interpretation of that data and information.

These all are in accordance with appropriate procedures established by particular academic and professional disciplines.

Action-informing research is its goal. As a result, your study should attempt to place its findings in the perspective of the wider body of knowledge. In order to develop knowledge that is usable outside of the research setting, research must constantly be of the highest calibre.

Research is done during the step of determining the need to ascertain the precise expectations of the client.

Thus, the correct option is B.

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Answer:

Not sure

Explanation:

Looking now

Answer:

  technician A is correct

Explanation:

Technician B has circuit topologies confused. In a series circuit, there is only one path for electrical current to take. In a parallel circuit, the current will divide between paths in proportion to the inverse of their resistance. The least resistance path will have the most current.

Technician A is mostly correct.

Answer:

  its requirement of labor or materials from far-flung sources

Explanation:

"Globalization" generally means having or creating sources of supply in foreign countries, usually those with lower labor or material costs. If some industry sector is dependent upon that, then it's probably because it depends on labor or materials from low-cost sources.

Answer:

The answer is below

Explanation:

Given that:

[tex]k_t=torque\ constant=0.075\ Nm/A\\\\k_v=voltage\ constant=0.12\ V/(rad/sec)\\\\R_a=armature \ resistance=2.5 \Omega\\\\V_t=terminal\ voltage=24\ V\\\\a)The \ starting\ current\ I_a\ is\ given \ as:\\\\I_a=\frac{V_t}{R_a} =\frac{24}{2.5} =9.6\ A\\\\The \ starting\ torque(T)\ is:\\\\T=k_tI_a=0.075*9.6=0.72\ N.m[/tex]

b) The maximum speed occurs when the terminal voltage and back emf are equal to each other i.e.

[tex]V_t=e_b=k_v\omega\\\\\omega=\frac{V_t}{k_v}=\frac{24}{0.12} =200\ rad/s[/tex]

c) The load torque is given as:

[tex]T_L=0.0125\Omega\\\\The\ motor\ torque \ is:\\\\T=k_t(\frac{V_t-k_v\omega}{R_a} )\\\\but\ T = T_L,hence:\\\\0.0125\omega=0.075(\frac{24-0.12\omega}{2.5} )\\\\0.03125\omega=1.8-0.009\omega\\\\0.04025\omega=1.8\\\\\omega=44.72\ rad/sec\\\\N=\frac{60\omega}{2\pi} =\frac{60*44.72}{2\pi} =427\ rpm[/tex]

Answer:

Usually it can be changed by flipping a switch. If your machine does not have one, then you just need to exchange the cables to the electrode holder and ground clamp.

Explanation:

The things that must be done before you change the polarity on a welding machine are coarse and fine adjustment.

The heated tool assembly, with two exposed surfaces, and two fixtures for holding the component parts to be welded. The tooling is for bringing the component parts into contact with the heated tool and bringing the molten joint surfaces together to form the weld. And displacement stops on the platen and holding fixtures make up a heated tool welding machine.

Polarity refers to the fact that the electrical circuit formed when you turn on the welder contains a negative and a positive pole. When welding, polarity is crucial, since the weld's strength and quality are affected by the choice of polarity.

Therefore, before changing the polarity of welding equipment, coarse and fine adjustments must be made.

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Answer: a drafting engineer

Explanation:

Answer: the total chronic daily intake is  0.03296 mg/ kg.d  

Explanation:

First of  we have to determine the CDI due to ingestion of drinking water

so

CDI = (C x IR x EF x ED) / (BW x AT)     let leave this as equation 1

where C is the concentration of the life time risk of drinking water(1.0mg/L)

ED is the risk dying per year((70),

EF is the number of days per year(365 day/year)

BW is the weight of body(65.4)kg

AT is the average tenure time of life(75years)

IR = 2.3 L/days

so we substitute

CDI₁ = (1(mg/L) x 2.3(L/day) x 365 x 70) / (65.4 x 365 x 75)

= 3.29 x 10⁻² mg/kg.d  

next we determine the CDI due to dermal contact with water.  

find the ingestion rate which is equal to the exposure time in hour/day.

IR = ET

(20 min/day) /  (60 min/hour)

= 0.333 h/day

now lets substitute for CD1₂

1.0 mg/L for C , 0.333 h/day for m , 365 day/year for EF, 70 year for ED, 65.4 kg for Bw , and 75 year for AT

0.8 for submergence and 1.69m² for area of skin of adult female in our equation 1

CDI₂ = [(1 mg/1) × (1.69 m²) × (9.0 x 10⁻⁶(m/h)) × 0.333(h/day) x 365 x 70)) / (65.4 x 365 x 75)]  × ( 0.8 x 10³ L/m³)

CDI₂ = 5.41 x 10⁻⁵ mg/kg.d  

now we find the CDI due to inhalation during bath

we substitute 1.0 mg/L for C, 11.3m³/day for IR , 365 day/year for EF , 70 year for ED. 65.4 kg for BW , and 75 year for AT in our equation 1

CDI₃ = [( 1ug/m³ × 1mg/10³ug) x (11.3m³/day x 1day/24 hr) x 365 x 70)] / (65.4 kg x 365 x 75 )

= 6.71 x 10⁻⁶ mg/kg.d

finally we Calculate the total chronic daily intake value

CDI_total = CDI₁ + CDI₂ + CDI₃  

we substitute

CDI_total = 3.29 x 10⁻² + 5.41 x 10⁻⁵ + 6.71 x 10⁻⁶

= 0.03296 mg/kg.d  

so the total chronic daily intake is  0.03296 mg/ kg.d  

Answer:

Beam of 25" depth and 12" width is sufficient.

I've attached a detailed section of the beam.

Explanation:

We are given;

Beam Span; L = 20 ft

Dead load; DL = 0.50 k/ft

Live load; LL = 0.65 k/ft.

Beam width; b = 12 inches

From ACI code, ultimate load is given as;

W_u = 1.2DL + 1.6LL

Thus;

W_u = 1.2(0.5) + 1.6(0.65)

W_u = 1.64 k/ft

Now, ultimate moment is given by the formula;

M_u = (W_u × L²)/8

M_u = (1.64 × 20²)/8

M_u = 82 k-ft

Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.

Effective depth of a beam is given by the formula;

d_eff = d - clear cover - stirrup diameter - ½Main bar diameter

Now, let's adopt the following;

Clear cover = 1.5"

Stirrup diameter = 0.5"

Main bar diameter = 1"

Thus;

d_eff = 25" - 1.5" - 0.5" - ½(1")

d_eff = 22.5"

Now, let's find steel ratio(ρ) ;

ρ = Total A_s/(b × d_eff)

Now, A_s = ½ × area of main diameter bar

Thus, A_s = ½ × π × 1² = 0.785 in²

Let's use Nominal number of 3 bars as our main diameter bars.

Thus, total A_s = 3 × 0.785

Total A_s = 2.355 in²

Hence;

ρ = 2.355/(22.5 × 12)

ρ = 0.008722

Design moment Capacity is given;

M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12

Φ is 0.9

f’c = 4,000 psi = 4 kpsi

fy = 60,000 psi = 60 kpsi

M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12

M_n = 220.03 k-ft

Thus: M_n > M_u

Thus, the beam of 25" depth and 12" width is sufficient.

Answer: 11470.4 cubic yards

Explanation:

first we calculate the volume of the site;

V1 = Area × depth

V1 = 2.5 acre ×  (43560 ft² / 1 acre )×3ft

V1 = 326700 ft

next we is the relative compaction

RC = [γd(field) / γdmax(laboratory)] × 100

so we substitute

93 =  [γd(field) / 122 lb/ft³)] × 100

γd(field) = 113.46 lb/ft³

then the dry unit weight of the site

γday1 = γavg / ( 1 + w)

= 118 lb/ft³ / ( 1 + (9.6/100))

= 118 lb/ft³ /1.096

= 107.664 lb/ft³

finally we find the fill volume of the site

V2/V1 = γd / γd(field)

we substitute

V2/326700 = 107.664 / 113.46

V2 = 310010.83 ft³

we convert to cubic yards

= 310010.83 ft³ × (0.037 cubic yard / 1 ft³)

= 11470.4 cubic yards

Answer:

The relation between the discharge volume flow rate and the inlet volume flow rate is [tex]\frac{1}{5}[/tex].

Explanation:

No matter if fluid is compressible or not, mass throughout compressor, a device that works at steady state, must be conserved according to Principle of Mass Conservation:

[tex]\dot m_{in}-\dot m_{out} = 0[/tex] (Eq. 1)

Where [tex]\dot m_{in}[/tex] and [tex]\dot m_{out}[/tex] are mass flows at inlet and outlet, measured in kilograms per second.

After applying Dimensional analysis, we expand the equation above as follows:

[tex]\rho_{in}\cdot \dot V_{in} - \rho_{out}\cdot \dot V_{out} = 0[/tex] (Eq. 2)

Where:

[tex]\rho_{in}[/tex], [tex]\rho_{out}[/tex] - Fluid densities at inlet and outlet, measured in kilograms per cubic meter.

[tex]\dot V_{in}[/tex], [tex]\dot V_{out}[/tex] - Volume flow rates at inlet and outlet, measured in cubic meters per second.

After some algebraic handling, we find the following relationship:

[tex]\rho_{out}\cdot \dot V_{out} = \rho_{in}\cdot \dot V_{in}[/tex]

[tex]\frac{\dot V_{out}}{V_{in}} = \frac{\rho_{in}}{\rho_{out}}[/tex] (Eq. 3)

If we know that [tex]\rho_{in} = 1\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{out} = 5\,\frac{kg}{m^{3}}[/tex], then the relation between the discharge volume flow rate and the inlet volume flow rate is:

[tex]\frac{\dot V_{out}}{\dot V_{in}} = \frac{1\,\frac{kg}{m^{2}} }{5\,\frac{kg}{m^{3}} }[/tex]

[tex]\frac{\dot V_{out}}{\dot V_{in}} = \frac{1}{5}[/tex]

The relation between the discharge volume flow rate and the inlet volume flow rate is [tex]\frac{1}{5}[/tex].

Answer:

False

Explanation:

False, though they could do it.

It's most likely and Industrial Engineer, it could also be a Chemical Engineer.

Answer:

Aluminium, Timber, Fibreglass

Explanation:

Most ladders are made of aluminium to have resistance against oxidisation and avoid rust; aluminium also doesnt spark either and is non magnetic. Some ladders are made out of timber like Douglas Fir, Oak, and Hickory. Some ladders are made out of fibreglass because it is strong and sturdy. They are also very to carry and are electrically resistant and heat tolerant.

Answer:D

Explanation:

Option D is correct. Balancing the crankshaft is NOT one of the functions of a flywheel.

A flywheel is part of an automobile. It is a heavy wheel that is attached to a rotating shaft of an automobile. Since there may be the presence of large forces within an automobile structure, the presence of a flywheel helps to balance the force evenly across it.  

Some of the functions of the flywheel are but are not limited to help in transferring power to the transmission, engaging the starter, and smoothing out of power pulses.

Hence based on the explanation above, balancing the crankshaft is NOT one of the function of a flywheel.

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Answer:

a

Explanation:

Answer:

The answer is "[tex]\bold{ 259.2 \times 10^{11} }[/tex]".

Explanation:

The amount of kilograms, which travel in a thick sheet of hydrogen:

[tex]M= -DAt \frac{\Delta C}{ \Delta x} \\\\[/tex]

[tex]D =1.0 \times 10^{8} \ \ \ \frac{m^2}{s} \\\\ A = 0.20 \ m^2\\\\t = 1\ \ h = 3600 \ \ sec \\\\[/tex]

calculating the value of [tex]\Delta C:[/tex]

[tex]\Delta C =C_A -C_B[/tex]

  [tex]= 2.4 - 0.6 \\\\ = 1.8 \ \ \frac{kg}{m^3}[/tex]

calculating the value of [tex]\Delta X:[/tex]

[tex]\Delta x = x_{A} -x_{B}[/tex]

     [tex]= 0 - (5\ mm) \\\\ = - 5 \ \ mm\\\\= - 5 \times 10^{-3} \ m[/tex]

[tex]M = -(1.0 \times 10^{8} \times 0.20 \times 3600 \times (\frac{1.8}{-5 \times 10^{-3}})) \\\\[/tex]

    [tex]= -(1.0 \times 10^{8} \times 720 \times (\frac{1.8}{-5 \times 10^{-3}})) \\\\= -(1.0 \times 10^{8} \times \frac{ 1296}{-5 \times 10^{-3}})) \\\\= (1.0 \times 10^{8} \times 259.2 \times 10^3)) \\\\= 259.2 \times 10^{11} \\\\[/tex]

Answer:

k-1000

Explanation:

Answer:

k = 1;

total = 0;

do {

total = total + (k*k);

k++;

} while (k <= 50);

Explanation:

answer using Java

Answer:

Quick Breads

Explanation:

I think this is correct. I am not for sure i just looked it up and it says word for word that this is the answer.

Answer:

Quick Breads

Explanation:

I got it right on my Plato test

Answer:

The answer "K = 0.0075"

Explanation:

If we try to measure up to 69 kPa of air, find mercury or fluid for gauge.  

While mercury was its largest liquid with a density of 13600 kg / m3 at normal room temperature.  

Let's all measure for 69 kPa that height of the  mercury liquid column.

[tex]\to P = 69 \ kPa[/tex]

       [tex]= 69000 Pa \\\\[/tex]

[tex]\to \rho = 13600 \ \ \frac{kg}{m^3} \\\\\\to g = 9.81 \ \ \frac{m}{s^2} \\\\[/tex]

Formula:

[tex]\to P=\rho \ gh[/tex]

[tex]\to 69000 = 13600\times9.81 \times h\\\\\to h= \frac{69000}{13600\times9.81} \\\\\to h= \frac{69000}{133416} \\\\\to h= 0.517179349 \\\\ \to h= 517 \ mm \\\\[/tex]

The right choice for pressure measurements up to 69 kPa is mercury.  

Atmospheric Mercury up to 69 kPa Air 517 mm  

The relationship of Hg to Pa is = 134.22 Pa 1 mm Hg  

Static sensitivity to Pa of mm hg = change of mercury height to Pa:

[tex]= \frac{\Delta Hg }{ \Delta P }\\\\= \frac{1 }{ 133.3 }\\\\= 0.0075[/tex]

Answer:

Colloids are very low diameter particles which are responsible for the turbidity or the color of surface water. Because of their very low sedimentation speed the best way to eliminate them is the coagulation-flocculation processes.

Answer:

Check  the 2nd, 3rd and 4th statements.

Explanation:

The router provides two basic functions traffic management between these networks through the transmission of data packets to the designated IP address, and the use of the same Internet connection by many devices.

Therefore, the final answer is "the second choice".

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