Presence of tissues in a multicellular organism ensures

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

after digestion and absorption, an amino acid not used to build protein will first be subjected to ____.

yulyashka [42]

<span>After digestion, an amino acid not absorbed by the body and not used to build proteins will first be subjected to removal from its amino group. If they are not used for protein synthesis, amino acids becomes part of the metabolism process. The body has the capability to create a subset of amino acids, also known as non-essential amino acids because we don’t have requirement for them in our diet. </span>

3 0

4 years ago

In 1985 a biologist counted 750 pine trees in a 250 hectare forest. Using similar counting techniques, the biologist counted 1,2

Alisiya [41]

Answer:

Suppose that we have a given function f(x)

The average rate of change of the function between two values x₁ and x₂ is given by:

$r = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

a) We want to find the average (rate) of change on the size of population from 1985 to 1995.

We have that:

f(1985) = 750

f(1995) = 1500

Then we have:

$r = \frac{1500 - 750}{1995 - 1985} = 750/10 = 75$

This means that the population of trees increases, in average, at a rate of 75 trees per year.

b) What is the density of trees each year that they were counted?

This will be equal to the quotient between the number of trees and the area.

1985: number of trees = 750 pines

area = 250 ha

Then the density is:

D(1985) = (750 pines)/(250 ha) = 3 pines/ha

So 1985, there were 3 pines per hectare.

1990: number of trees = 1250 pines

area = 250 ha

Then the density is:

D(1990) = (1250 pines)/(250 ha) = 5 pines/ha

1995: number of trees = 1500 pines

area = 250 ha

The density is:

D(1995) = (1500 pines)/(250 ha) = 6 pines/ha

3) now we want to get the average change between 1985 and 1995 in the density, this will be:

$r = \frac{D(1995) - D(1885)}{1995 - 1985} = \frac{6 pines/ha - 3pines/ha}{10} = 0.3 pines/ha$

So, on average, each year the number of pines per hectare increases by 0.3

7 0

3 years ago

Because of biochemical cycling

shepuryov [24]

Nutrients are circulated throughout the biosphere because of biochemical cycling.

Answer: Option C

<u>Explanation:</u>

Apart from vitality, water and a few other chemical components spin through biological systems and impact the rates at which life forms develop and duplicate. Around 10 noteworthy supplements and six minor nutrients are basic to all creatures and plants, while others assume significant jobs for chosen species.

The most significant biochemical cycles influencing environment well being of the water, carbon, nitrogen, and phosphorus cycles. Thus it is due to biochemical cycling, nutrients are circulated through out the biosphere.

7 0

3 years ago

Discuss three ways to sustain the carbon cycle

Savatey [412]

We can maintain the carbon cycle by

4 0

3 years ago