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3 months ago

In the first paragraph, the words "disrobed," "unveiling" and "deconstructed" primarily serve to

2 answers:

8 0

The terms "disrobed," "unveiling," and "deconstructed" in the initial paragraph primarily aim to (a) underline the negative implications associated with laser technology, (b) stress the broad impact of laser technology, (c) showcase the innate unknowable traits of objects, regardless of laser technology, or (d) implicitly draw comparisons between lasers and other technologies.

Anarel [2.9K]3 months ago

4 0

The correct option is (b) which highlights the broad impact of laser technology. The authors of the passage, "These Are the Days of Lasers in the Jungle" by Joseph Mascaro and others, discuss how laser technology evolves and revolutionizes processes. The expressions "disrobed," "unveiling," and "deconstructed" convey an idea of revelation, suggesting the stripping away of obscuring elements that conceal the true nature. Thus, their usage in context signifies the vast potential reach of laser technology.

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"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92

lorasvet [2795]

Answer: The percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

From the information provided:

Let the fractional abundance for $_{14}^{28}\textrm{Si}$ isotope be 'x'

For $_{14}^{28}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 27.9769 amu

Percentage abundance of $_{14}^{28}\textrm{Si}$ isotope = 92.22 %

Fractional abundance for $_{14}^{28}\textrm{Si}$ isotope = 0.9222

For $_{14}^{29}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 28.9764 amu

Percentage abundance for $_{14}^{28}\textrm{Si}$ isotope = 4.68%

Fractional abundance of $_{14}^{28}\textrm{Si}$ isotope = 0.0468

For $_{14}^{30}\textrm{Si}$ isotope:

Mass of $_{14}^{30}\textrm{Si}$ isotope = 29.9737 amu

Fractional abundance for $_{14}^{30}\textrm{Si}$ isotope = x

The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

$28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309$

To convert this fractional abundance into a percentage, multiply by 100:

$\Rightarrow 0.0309\times 100=3.09\%$

This shows that the percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

6 0

3 months ago

One tank of gold fish is fed the normal amount of food once a day. A second tank is fed twice a day. A third tank is fed four ti

Tems11 [2777]

The inquiry is incomplete; here is the full question:

One tank of goldfish receives the standard amount of feeding once daily, a second tank is given two feedings a day, and a third tank is fed four times daily throughout a six-week experiment. The body fat of the fish is recorded every day.

Independent Variable-

Dependent Variable-

Constants

Control Group-

Answer:

A) The quantity of food given to the goldfish

B) The body fat of the goldfish

C) -Type of fish in the experiment (goldfish)

Time period for feeding the fish (six weeks)

Shape and size of the tanks

D) group of goldfish receiving the standard feeding amount

Explanation:

The objective of the experiment is to assess how the quantity of food affects the body fat of goldfish. Consequently, the amount of food serves as the independent variable while the body fat acts as the dependent variable.

The control group is the one given the standard feeding amount (once daily). All subjects are goldfish, fed over a six-week duration, with all tanks being the same shape and size, establishing the constants in the research.

4 0

4 months ago