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Lemur [1.5K]
2 years ago
9

To correct myopia (nearsightedness) requires ___________ lenses; to correct hyperopia (farsightedness) requires _____________ le

nses to correct.
Physics
2 answers:
zysi [14]2 years ago
4 0

By using concave lens, we can correct Myopia and Hyperopia can be corrected by convex lenses.

To find the answer, we need to know about Myopia and Hyperopia.

<h3>What is Myopia and Hyperopia?</h3>
  • Myopia is an optical defect, in which a person can see nearby objects clearly and cannot see far off objects clearly.
  • Here the eyeball becomes too longer, and the focal length of eye lens become too short.
  • It can be corrected by using concave lens.
  • For Hyperopia, the person can see far off objects clearly and cannot see nearby objects clearly.
  • This can be corrected by using convex lens.

Thus, we can conclude that, by using concave lens, we can correct Myopia and Hyperopia can be corrected by convex lenses.

Learn more about Myopia and Hyperopia here:

brainly.com/question/28052359

#SPJ1

natta225 [31]2 years ago
3 0

Myopia can be corrected using concave lenses, while the Hyperopia is corrected by convex lenses.

We need to be aware of myopia and hyperopia in order to locate the solution.

<h3>Myopia and hyperopia: what are they?</h3>
  • Myopia is an optical defect in which a person can see clearly up close but not clearly far away.
  • Here, the eyeball lengthens excessively and the eye lens focal length contracts.
  • Concave lenses can be used to rectify it.
  • A person with hyperopia can clearly see distant objects but cannot see nearby objects clearly.
  • Convex lenses can be used to fix this.

Thus, we can deduce that myopia and hyperopia can be corrected using concave lenses and convex lenses, respectively.

Learn more about the myopia and hyperopia here:

brainly.com/question/28052359

#SPJ1

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Answer:

T_{f}  = 17º C

Explanation:

This is a calorimetry problem, where heat is yielded by liquid water, this heat is used first to melt all ice, let's look for the necessary heat (Q1)  

Let's reduce the magnitudes to the SI system  

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    Q₁ = m L  

     Q₁ = 0.080 3.33 10⁵

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Now let's see what this liquid water temperature is when this heat is released  

      Q = M c_{e} ΔT = M c_{e} (T₀₁ -T_{f1})  

      Q₁ = Q  

     T_{f1} = T₀₁ - Q / M ce  

     T_{f1} = 26.0 - 2,664 10⁴ / (0.860 4186)  

     T_{f1} = 26.0 - 7.40  

     T_{f1} = 18.6 ° C  

The initial temperature of water that has just melted is T₀₂ = 0ª  

The initial temperature of the liquid water is T₀₁= 18.6  

     m c_{e} T_{f} + M c_{e} T_{1} = M c_{e} T₀₁ - m c_{e} T₀₂o2  

         T_{f} = (M To1 - m To2) / (m + M)  

         T_{f} = (0.860 18.6 - 0.080 0) / (0.080 + 0.860)  

T_{f} = 17º C

 

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