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Lesechka [4]
3 years ago
13

Under which of the following conditions would material flood damage be considered an extraordinary item for financial reporting

purposes?
a. Only if floods in the geographical area are unusual in nature and occur infrequently.
b. Only if the flood damage is material in amount and could have been reduced by prudent management.
c. Under any circumstances as an extraordinary item.
d. Flood damage should never be classified as an extraordinary item.
Engineering
1 answer:
dedylja [7]3 years ago
6 0

Answer: A.Only if floods in the geographical area are unusual in nature and occur infrequently.

Explanation: Extraordinary items are special losses or gains that don't occur regularly and are unusual on nature. Since the materials damaged as a result of floods in the area are unusual and infrequently occurs they are to be considered as extraordinary items for financial statements purposes.

Business entities usually prepare the financial statements for Extraordinary items separately as they only occur on a one time basis.

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Answer:

correct option is c. a testable prediction leading to design of an experiment

Explanation:

The results of raising tadpoles were estimated to be water with an atrazine level of 0.1 ppb compared to those grown in pure water. So, this is not the question. If this assumption can now be tested, an experiment can be made in which a certain number of tadpoles can be raised in pure water and the same number of tadpoles can be raised in water with a 0.1ppb atrazine level can. The difference between the two populations can be estimated or compared.

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3 years ago
_______ is a material property that pertains to local resistance to plastic deformation, such as scratching or denting. It is of
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Answer: hardness

Explanation:

Hardness is a measure of a material's ability to resist plastic deformation. In other words, it is a measure of how resistant material is to denting or scratching. Diamond, for example, is a very hard material. It is extremely difficult to dent or scratch a diamond. In contrast, it is very easy to scratch or dent most plastics.

7 0
3 years ago
Assume a function requires 20 lines of machine code and will be called 10 times in the main program. You can choose to implement
Volgvan

Answer:

"Macro Instruction"

Explanation:

A macro definition is a rule or pattern that specifies how a certain input sequence should be mapped to a replacement output sequence according to a defined procedure. The mapping process that instantiates a macro use into a specific sequence is known as macro expansion.

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3 years ago
A constant-head permeability test gives the following information: - Water flows horizontally through the soil sample. - The hei
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5 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

7 0
4 years ago
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