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Len [333]
3 years ago
9

A person's prescription for bifocals is -0.25 diopter for distant vision and +2.50 diopters for near vision, the near vision bei

ng accomplished by inserts into the distant vision lens. What is the net power of the near vision correction?
Physics
1 answer:
Darina [25.2K]3 years ago
6 0

Answer:

net power is + 2.25 D

Explanation:

Given data

distance vision = -0.25 D

near vision = + 2.50 D

to find out

net power

solution

we have given a person lens power for near is - 0.25 diopter and lens power for near power is  +2.50 diopter so

net power is sum of both the power vision

so

net power = distance + near power

put both value we get net power

net power = ( -0.25 D) + ( + 2.50 D)

net power = + 2.25 D

so net power is + 2.25 D

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2 years ago
A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o
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Explanation:

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    \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

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                 = 0.945 \times 10^{-6} F

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Now, we will calculate the charge across each capacitance as follows.

              Q = CV

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