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N76 [4]
3 years ago
7

A concrete block making company is developing an aggregate capacity plan from the following sales forecast for its 6” and 8” con

crete blocks shown in the table below. Product Sales Forecasts (Concrete Blocks) 1st Quarter 2nd Quarter 3rd Quarter 4th Quarter 6” Block 2,000 1,500 1,600 1,800 8” Block 1,200 1,000 800 1,000 Ample machine capacity exists to produce the forecast. Each 6” concrete block takes an average of 20 labour-hours while each 8” concrete block takes an average of 15 labour-hours. a) Compute the aggregate number of labour-hours in each quarter. b) If the plant works the same number of days in each quarter, 13 five-day weeks on a single shift, how many workers will be required in each quarter? c) It costs $1,000 to hire a worker and $500 to lay off a worker, and inventory carrying cost is $100 for each 6” concrete block and $100 for each 8” concrete block (this means that if one 6” concrete block were held in inventory for a year, it would cost $100 for finance charge, insurance, warehousing expense, etc.). The plant works the same number of days in each quarter, 13 five-day weeks. Assuming that the quarterly demand pattern repeats from year to year, compute the cost of each of the following two aggregate plans for next year: i. Chase strategy plan. Beginning inventory in this plan is zero for both types of concrete blocks. ii. Level strategy plan. Beginning inventory in this plan is 275 for 6” concrete blocks and 200 for 8” concrete blocks.
Engineering
1 answer:
Fiesta28 [93]3 years ago
4 0

Answer:

nah ask ur mom then

Explanation:

ur mom is smarter than u then ask her dont be rude

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What is clearance? What is backlash? What is interference? Explain briefly.
Anton [14]

Explanation:

Clearance:

For easy matching  and dis matching  of  hole and  shaft we use size of hole little bit more than the size of shaft and this difference in size is called clearance.

Backlash:

  It is the clearance between the two mating gear to avoids failure of gears.Actually when temperature of gears increases then at the same time the size of gear also increases ,due this there is a possibility foe jamming of gears so to avoids this backlash is provides.

Interference:

  When two gears are matting then addendum of one gear inters into the deddendum of another gear and due to this gears get jam .This phenomenon is called interference.

5 0
3 years ago
Consider a single crystal oriented such that the slip direction and normal to the slip plane are at angles 42.7° and 48.3°, resp
zlopas [31]

Answer:

Applied Stress > 58.29 MPa

Explanation:

  • Resolved shear stress should be greater than critically resolved shear stress in order to cause the single crystal to yield

Given angles are

∅ = 42.7 degree

Ф = 48.3 degree

Critically resolved shear stress = 28.5 MPa

If we consider

Critically resolved shear stress = resolved shear stress

Applied stress can be found by

Z_{R} = applied stress X cos\phi X cos\theta    (1)

Applied Stress = \frac{Z_{R} }{Cos\phi XCos\theta}

Applied Stress = \frac{28.5}{Cos(48.3)XCos(42.7)}

Applied Stress = 58.29 MPa

We got reference

  • By putting applied stress values of greater than 58.29 MPa in equation 1 we get

        Resolved Shear Stress = 60 x Cos(48.3) x Cos(42.7)

        Resolved Shear Stress = 29.33 MPa

Therefore, by the above calculation we conclude that applied stress should be greater than 58.29 MPa, In order to make resolved shear stress to be greater than critically resolved shear stress that is essential for single crystal to yield.

8 0
3 years ago
What are the two types of furnaces used in steel production?
Kobotan [32]

Explanation:

The two types of furnaces used in steel production are:

<u>Basic oxygen furnace </u>

In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.

<u>Electric arc furnace</u>

Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.

3 0
4 years ago
Info security:
il63 [147K]

Answer:

True

Explanation:

Dual home host - it is referred to as the firewall that is incorporated with two or more networks. out of these two networks, one is assigned to the internal network and the other is for the network. The main purpose of the dual-homed host is to ensure that no Internet protocol traffic is induced between both the network.

The most simple example of a dual-homed host is a computing motherboard that is provided with two network interfaces.

7 0
4 years ago
In an apartment the interior air temperature is 20°C and exterior air temperatures is 5°C. The wall has inner and outer surface
ELEN [110]

Answer:

20 W/m², 20 W/m², -20 W/m²

Yes, the wall is under steady-state conditions.

Explanation:

Air temperature in room = 20°C

Air temperature outside = 5°C

Wall inner temperature = 16°C

Wall outer temperature = 6°C

Inner heat transfer coefficient = 5 W/m²K

Outer heat transfer coefficient = 20 W/m²K

Heat flux = Concerned heat transfer coefficient × (Difference of the temperatures of the concerned bodies)

q = hΔT

Heat flux from the interior air to the wall = heat transfer coefficient of interior air × (Temperature difference between interior air and exterior wall)

⇒ Heat flux from the interior air to the wall = 5 (20-6) = 20 W/m²

Heat flux from the wall to the exterior air = heat transfer coefficient of exterior air × (Temperature difference between wall and exterior air)

⇒Heat flux from the wall to the exterior air = 20 (6-5) = 20 W/m²

Heat flux from the wall to the interior air = heat transfer coefficient of interior air × (Temperature difference between wall and interior air)

⇒Heat flux from the wall and interior air = 5 (16-20) = -20 W/m²

Here the magnitude of the heat flux are same so the wall is under steady-state conditions.

7 0
3 years ago
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