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Aleksandr-060686 [28]
2 years ago
15

Merchandise without an expiration date like electronics, tools and home goods typically have a longer

Engineering
1 answer:
Aleks [24]2 years ago
5 0

Answer:

Shelf life.

Explanation:

In Business, an inventory is a term used to describe a list of finished goods, goods still in the production line and raw materials that would be used for the manufacturing of more goods in a bid to meet the unending consumer demands.

Simply stated, an inventory can be classified into three (3) main categories; finished goods, work in progress, and raw materials.

An inventory is recorded as a current asset on the balance sheet because it's primarily the most important source of revenue for a business entity.

Also, the three (3) main cost concept associated with an inventory are;

1. First In First Out (FIFO).

2. Last In First Out (LIFO).

3. Weighted average cost.

Shelf life can be defined as a measure of the length of time that a particular product could be kept or stored without it getting bad or becoming unsuitable for use by the consumers. Thus, the shelf life of a product is largely dependent on its expiration date.

Hence, merchandise without an expiration date like electronics, tools and home goods typically have a longer shelf life.

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A centimeter is Viooth of a meter, while a kilo-
tatuchka [14]

Answer:

b. False

Explanation:

A centimeter is a hundredth of a meter. This means 1/100 ,1 cm = 0.01 m

A kilo is 1000 grams.

The first answer is false because a hundredth isnot written well.

The second statement is false because the comparison given is of different units of measure. For distance is meters where as for weight is kilograms.

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A steel plate has a hole drilled through it. The plate is put into a furnace and heated. What happens to the size of the inside
djverab [1.8K]

Answer:

The diameter increases

Explanation:

The expansion in the metal is uniform in every dimension

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Basil is installing a system in Toronto, Ontario where they are wide temperature variations. What is the TD of the system he’s i
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3 years ago
Which of these fuel injection systems operates with fuel injectors located only in the intake manifold near each intake
Stella [2.4K]

Answer:

C. Multipoint fuel injection

Explanation:

A fuel injection system can be defined as a system found in the engine of most automobile cars, used for the supply of a precise amount of fuel or fuel-air mixture to the cylinders in an internal combustion engine through the use of an injector.

There are different types of fuel injection system and these includes;

I. Central-point injection.

II. Throttle (single point) body injection.

III. Gasoline direct injection.

IV. Multipoint (port) fuel injection.

Multipoint fuel injection is a type of fuel injection system that operates with fuel injectors located only in the intake manifold near each intake valve and sprays fuel toward the valve. As a result, it allows for the supply of a precise amount of fuel and as such creating a better air-fuel ratio for automobile cars.

8 0
2 years ago
Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin
Alborosie

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress  sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy =  tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

6 0
2 years ago
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