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enyata [817]
2 years ago
5

Once a slide is prepared and placed onto the microscope, the magnification and focus need to be altered. Describe how to change

the magnification and bring the image into focus using a light microscope like the one in the simulation.
Chemistry
2 answers:
Harlamova29_29 [7]2 years ago
5 0

Answer:

我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈

Explanation:

我實際上不知道答案,我只是為了點我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈數而這樣做,哈哈,祝你好運哈哈

Evgen [1.6K]2 years ago
5 0

Answer:

They said: I don’t actually know the answer, I just do it for points, haha, good luck haha

Explanation:

I actually don’t know the answer, I’m just for the point I don’t actually know the answer, I just do it for the point, haha, good luck haha ​​and do this, haha, good luck haha

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PLEASE HELP I WILL GIVE BRAINLIEST!!! Its due tonight at 12:00
Dmitry [639]

Answer:

c

Explanation:

we have only distance vs time graph so c is the right answer that cart is moving with constant speed of 0.5m/s

confirmed

first calculate distance which is from 2 to 5.5

5.5-2=3.5

apply the formula of velocity

v=s/t

v=3.5/7=0.5

8 0
2 years ago
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When an object falls to the ground, potential energy is most likely transformed into
pentagon [3]

Answer:

kinetic energy

Explanation:

When the object is released, the gravitational potential energy is gradually converted into kinetic energy as it picks up speed.

8 0
2 years ago
The diagram shows a model of the nitrogen cycle. which role do plants play in the nitrogen cycle?
kenny6666 [7]

Answer

D

Explanation:

They take up usable forms of nitrogen found in soil

3 0
2 years ago
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3. The diet of white-tailed deer consists primarily of shrubs. Sika are
olga_2 [115]
Sika have more food choices because they eat both grasses and shrubs, compared to the white-tailed dear who only eats shrubs.
3 0
2 years ago
Suppose a system returns to its original overall, internal energy ((∆U) after the following changes. In step 1, 25 J of work is
Brums [2.3K]

Answer:

Heat transfer in step 2 = 47.75 J

Explanation:

Internal energy = heat + work done

U = Q + W

In a cyclic process the total internal energy change of the system = 0.

In the process there are two steps. The total heat exchange in the process is the sum of heat exchanges in the two processes.

We have to find the heat exchange in step 2.

In step 1,

W = 1.25 J          Q = -37 J

U_{1} = -37 + 1.25 = -35.75 J

In step 2, the internal energy change will be negative of that in step 1.

U = 35.75 J

W = -12 J

U = Q + W

35.75 = Q -12

Q = 47.75 J

Heat transfer in step 2 = 47.75 J

6 0
3 years ago
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