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m_a_m_a [10]
3 years ago
14

The fertile soil present in most parts of the state of Mississippi can be attributed to

Biology
2 answers:
Ostrovityanka [42]3 years ago
7 0
The water makes the soul more fertile idk it is probobly wrong
fiasKO [112]3 years ago
4 0

The answer is ( A ) the alluvium deposited by the Mississippi river

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1. What is the total magnification of an object if the ocular lens magnification is 20x and the objective lens magnification is
V125BC [204]

1. Answer: 900x

A typical microscope has two lens: ocular lens which located near the eye, and objective lens which located near the object. The image will undergo magnification of both lens. The total magnification of the object would be the product of both lens multiplication, not the sum. The multiplication will be: 20x * 45x= 900x

total mag= ocular * object

total mag= 20x * 45x

total mag= 900x


2.  Answer: 100x

The total magnification of the object would be the product of both lens multiplication. If the total magnification 1000x, that mean it was the product of ocular lens and objective lens. If ocular lens magnification is 10x, objective lens magnification would be:

total mag= ocular * object

1000x= 10x*object

object=1000x/ 10x=

object=100x

3. Answer: 0.2 mm

The area of the microscope that can be viewed by the eye should be the same. But since the magnification is different, the actual area that it represent will be different. Microscope with bigger magnification will have smaller diameter of the field of view(DFV).  Remember that micrometer(μm) is 1/1000 of millimeter(mm). The DFV would be:500 μm/ ( 1000x/400x)= 200μm= 0.2 mm


4. Answer: 750μm

This question is similar to number 3 question. Remember that 1 millimeter(mm) equal to 1000 micrometer(μm) .

The DFV of a 10x objective lens is 3 mm. Then the DFV of 40x lens would be:

DFV2= DFV1/ (mag1/mag2)

DFV2= 3mm/ ( 40x/10x)= 0.75 mm= 750μm

5. Answer: 125μm x 37.5μm

The lens DFV of microscope from question 4 is 750μm, so the width and length of the area would be 750μm.

If 6 organisms could fit across the DFV if they were laid end-to-end, the length would be: 750μm/ 6 organism= 125μm

If 20 organism could fit is stacked side by-side, then the width would be: 750μm/ 20 organism= 37.5μm

6. Answer: 6mm

This question is similar to number 3 and number 4 question. Remember that 1 millimeter(mm) equal to 1000 micrometer(μm).

The DFV of a 100x objective lens is 1.5 mm. Then the DFV of 25x lens would be:

DFV2= DFV1/ (mag1/mag2)

DFV2= 1.5mm/ (100x/25x)= 6 mm


7.Using the 100x objective lens from question 6, you estimate 12 organisms could fit across the DFV if they were laid end-to-end and 30 could fit is stacked side-by-side. What is the length and width of this organism (in microns)?  

Answer: 0.5mm x 0.2mm or 500μm x 200μm

The lens DFV of microscope from question 6 is 6 mm, so the width and length of the area would be 6 mm.

If 12 organisms could fit across the DFV if they were laid end-to-end, the length would be: 6mm/ 12 organism=0.5mm

If 30 organism could fit is stacked side by-side, then the width would be: 6mm/ 30 organism= 0.2mm

6 0
3 years ago
The body fluid of sharks has a much lower concentration of sodium chloride than that of the surrounding seawater, and yet they a
enyata [817]
A. Sharks store enough urea to match the total solute concentration of the surrounding seawater
4 0
3 years ago
What is the importance of manipulation and control in the experimental method?
Vitek1552 [10]
This is an experiment where the researcher manipulates one variable, and control/randomizes the rest of the variables. It has a control group, the subjects have been randomly assigned between the groups, and the researcher only tests one effect at a time.
6 0
3 years ago
Cellular respiration and photosynthesis are similar in that both require A. carbon dioxide molecules B. energy-carrier molecules
yulyashka [42]

Answer:

A. carbon dioxide molecules  C. glucose molecules

Explanation:

4 0
3 years ago
Because they alter the reading frame of all base-pair triplets, base-pair additions and deletions are collectively referred to a
Rom4ik [11]

Answer:

The correct answer is option c. "frameshift mutations".

Explanation:

The reading frame of a gene is based on base-pair triplets, starting from the start codon until the ribosome encounters with the end codon. Base-pair additions and deletions are collectively referred to as frameshift mutations because they alter the reading frame of the gene. Base-pair additions and deletions break down the original sequence of the gene triplets, which alters the open reading frame and usually results in the production of non active proteins.

7 0
3 years ago
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