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Mamont248 [21]
2 years ago
11

Plz someone help me on this problem 77 ×667

Mathematics
2 answers:
7nadin3 [17]2 years ago
7 0
51,359 uwuuuuuuuuuuuuuu :)))
Scilla [17]2 years ago
5 0
Ggggggggggggggggggg 51,359
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2 years ago
The focal lengths of the objective lens and the eyepiece of a microscope are 0.50 cm and 2.0 cm, respectively, and their separat
VladimirAG [237]

Answer:

- 103.7

Step-by-step explanation:

Given:

Focal length of the eyepiece, f = 2.0 cm

Focal length of the objective lens, f' = 0.50 cm

Separation for minimum eyestrain = 6,0 cm

Image distance, v = - 25 cm

Now, from the lens formula,

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

here, u is the object distance

on substituting the respective values, we get

\frac{1}{2}=\frac{1}{u}+\frac{1}{-25}

or

u = 1.852 cm

also, the separation is adjusted for minimum eyestrain,

therefore, image distance for the objective lens, v' = 6 - 1.852 = 4.148 cm

Now, for the objective lens

using the lens formula, we get

\frac{1}{0.5}=\frac{1}{u'}+\frac{1}{4.418}

Here, u' is the distance between the physical object and objective lens

or

u' = 0.568 cm

Thus,

Magnification, m = -\frac{vv'}{ff'}

or

m =-\frac{25\times4.418}{0.5\times2}

or

m = - 103.7

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R(a) is the change in the population of rats based on the whole number of acres(a) of pasture. What is an appropriate domain for
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Its all positive integers

Step-by-step explanation:


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