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Lelu [443]
3 years ago
13

According to the eNotes, a program that eliminates sales and promotions in an effort to minimize the bullwhip effect would be ca

lled...
CRMWRONGEDLP
Engineering
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0

Options: True or false

Answer:True, it will be called CRMWRONGEDLP.

Explanation:eNotes was founded in 1998 by Brad Satoris and Alexander Bloomingdale, to enable and enhance the ability of students to solve assignments as given and other home works,eNote is also used by students to prepare effectively and efficiently for examinations.

Since its launch eNote has been a great source of educational materials for students who have gained a lot in the use of its materials.

bullwhip effect is a concept in supply chain, where forcasts causes the supply chain to be inefficient, it usually starts from retailers who raise concern of high demands.

CRMWRONGEDLP, is the program that minimizes the bullwhip effects according to eNote.

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Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pres
LekaFEV [45]

Answer:

44.59°c

Explanation:

Given data :

Total pressure = 105 kpa

complete combustion

A) Determine air-fuel ratio

A-F = \frac{N_{air} }{N_{fuel} }  = \frac{(Nm)_{air} }{(Nm)_{c} (Nm)_{n} }

N = number of mole

m = molar mass

A-F = \frac{(6.75*4.76)kmole * ( 29kg/mol)}{(3kmole)* 12kg/mol + (6kmol)*(1kg/mol)}  =  22.2 kg air/fuel

hence the ratio of Fuel-air = 1 : 22.2

B) Determine the temperature at which water vapor in the products start condensing

First we determine the partial  pressure of water vapor before using the steam table to determine the corresponding saturation temp

partial pressure of water vapor

Pv = \frac{(N_{water vapor}) }{N_{pro} } * ( P_{ro} )

N watervapor ( number of mole of water vapor ) = 3

N pro ( total number of mole of product = 3 + 3 + 2.25 + 25.28 = 33.53 kmol

Pro = 105

hence Pv = ( 3/33.53 ) * 105 =  9.39kPa

from the steam pressure table the corresponding saturation temperature to 9.39kPa =  44.59° c

Temperature at which condensing will start = 44.59°c

An equation showing the products of propylene with their mole numbers is attached below

8 0
3 years ago
6. At a construction site, cement, sand, and gravel are used to make concrete. The ratio of cement to sand to gravel is 1 to 2.4
S_A_V [24]

Answer:

Mass of cement used is 62.5 lb

Mass of gravel used is 225 lb

Explanation:

The ratio given here is cement to sand to gravel = 1 : 2.4 : 3.6

So, for 150 lb of sand

C : S : G = 1 : 2.4 : 3.6

\frac{C}{S}=\frac{1}{2.4}\\\Rightarrow C=S\frac{1}{2.4}\\\Rightarrow C=150\frac{1}{2.4}\\\Rightarrow C=62.5\ lb

Mass of cement used is 62.5 lb

\frac{S}{G}=\frac{2.4}{3.6}\\\Rightarrow G=S\frac{3.6}{2.4}\\\Rightarrow C=150\frac{3.6}{2.4}\\\Rightarrow C=225\ lb

Mass of gravel used is 225 lb

7 0
3 years ago
A hydraulic lift is to be used to lift a 2500 kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Deter
Lilit [14]

Answer:here in Hydraulic system the pressure on both sides will remain the same

P_1 = P_2

\frac{F_1}{A_1} = \frac{F_2}{A_2}

\frac{mg}{\pi*\frac{10^2}{4}} = \frac{Mg}{\pi \frac{d^2}{4}}

\frac{25}{10^2} = \frac{1900}{d^2}

d = 87.2 cm

so its diameter is 87.2 cm

Explanation:

8 0
3 years ago
Tech a says that the weight of the flywheel smoothest out the engines power pulses. Tech B says that the flexplate and torque co
lakkis [162]

Answer:

both statement is correct

Explanation:

Flywheel engine uses to reduce fluctuations.

And                                                                

FlexPlate is a metal disk that connects the output from the engine to the input of the torque converter. This will replace the flywheel

so that both statement is correct

4 0
3 years ago
Another common unit for viscosity is the centipoise, which is 10-2 poise (just like a centimeter is 10-2 meters). (1 centipoise
Oduvanchick [21]

Answer and Explanation

Picking room temperature to be 20°C, the viscosity of water (μ) obtained from literature is:

1 centipoise = 0.01 poise = 0.001 Pa.s = 0.001 N.s/m² (This viscosity is the dynamic viscosity of water at 20°C)

Kinematic viscosity of water, η = μ/ρ

At 20°C, μ = 0.001 Pa.s, ρ = 998.23 kg/m³

η = 0.001/998.23 = 1.0 × 10⁻⁶ m²/s

Picking the room temperature to be 25°C, the viscosityof water (μ) is

0.89 centipoise = 0.0089 poise = 0.00089 Pa.s = 8.9 × 10⁻⁴ N.s/m²

Kinematic viscosity of water, η = μ/ρ

At 25°C, μ = 0.00089 Pa.s, ρ = 997 kg/m³

η = 0.00089/997 = 8.9 × 10⁻⁷ m²/s

4 0
4 years ago
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