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vampirchik [111]
3 years ago
6

An elderly sailor is shipwrecked on a desert island but manages to save his eyeglasses. The lens for one eye has a power of 1.16

diopters, and the other lens has a power of 9.37 diopters. What is the magnifying power of the telescope he can construct with these lenses
Physics
1 answer:
igomit [66]3 years ago
8 0

Answer:

m = 8

Explanation:

A telescope is a device that allows us to see objects that were very far from us, it is built by the combination of two lenses, the one with the lowest focal length near the eye and that is the one or the one with the greatest focal length, the most eye-flounder . The magnification of the telescope is

            m = - f₀ / f_{e}

Where f₀ is the focal length of the lens and f_{e} is the false distance of the eyepiece.

It is this problem that gives us the diopter of each lens, these are related to the focal length in meters

           D = 1 / f

Let's find the focal length

       f₁ = 1 / D₁

       f₁ = 1 / 1.16

       f₁ = 0.862 m

     

        f₂ = 1 / 9.37

        f₂ = 0.1067 m

Therefore, the lens with f₂ is the eyepiece and the slow one with the  

distance focal  f₁ is the objective.

Let's calculate

       m = - f₂ / f₁

      m = - 0.862 / 0.1067

      m = 8

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Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final
Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B =  -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s  x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A  after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

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