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barxatty [35]
4 years ago
14

Two production methods are being compared. One manual and the other automated. The manual method produces 10 pc per hour and req

uires one worker at $ 15.00 per hour. Fixed cost of the manual method is $ 5,000 per year. The automated method produces 25 pc per hour, has a fixed cost of $ 55,000 per year, and a variable cost of $ 4.50 per hour. Determine the Break – Even Point (BEP) for the two methods; that is, determine the annual production quantity at which the two methods have the same annual cost. Ignore the cost of material used in the two methods

Engineering
1 answer:
Genrish500 [490]4 years ago
6 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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What is the power system?
Inessa [10]

Answer: An electric power system is a network of electrical components deployed to supply, transfer, and use electric power.

Explanation:

4 0
4 years ago
A good visual lead is .... seconds from the front of the vehicle, focusing in the center of the path of travel. Searching 20 to
Tomtit [17]

A good visual lead is 20-30 seconds from the front of the vehicle, focusing in the center of the path of travel. Searching 20 to 30 seconds ahead, gives you time to assess within the next ..12-15 seconds, actions you may need to take to control an approaching risk.

Discussion:

Keeping eye focus centered in a path of travel at an interval of 20 to 30 seconds away from the vehicle is critical to gaining enough info. as possible in the driving scene. Good targeting sets up good sight lines for referencing and good peripheral fields for observing changes.

  • It is important to look ahead 12-15 seconds into your target area as one drives. Compromise. space by giving as much space to the greater of two hazards.

Read more on driving visual leads:

brainly.com/question/7067386

3 0
3 years ago
A receptacle, plug, or any other electrical device whose design limits the ability of an electrician to come in contact with any
o-na [289]

Answer:

Recessed  safe

Explanation:

A recessed, receptacle or plug is one in which the metal interfacing parts have been retracted back in such a way that common contact between a person handling the receptacle or outlet will not ordinarily lead to exposure to electric shock hazard and as such improving the safety of use of such plug, receptacle or electrical device either at home or in industrial setting

A recessed outlet is also more ideal than standard outlets where there are hazards such as potential contact with water.

4 0
3 years ago
In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri
ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

Here

  • \sigma_x is the normal stress which is 1750 psf
  • \sigma_y is 0
  • \tau_{xy} is the shear stress which is 800 psf

So the formula becomes

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}

Similarly, the minimum normal stress is given as

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}

The maximum shear stress is given as

\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}

5 0
3 years ago
A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

3 0
4 years ago
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