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Sholpan [36]
4 years ago
7

A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely far away. T

here exist, however, certain conditions under which the range of vision is not so extended. For example, a nearsighted person cannot focus on objects farther than a certain point (the far point), while a farsighted person cannot focus on objects closer than a certain point (the near point). Note that even though the presence of a near point is common to everyone, a farsighted person has a near point that is much farther from the eye than the near point of a person with normal vision.
Both nearsightedness and farsightedness can be corrected with the use of glasses or contact lenses. In this case, the eye converges the light coming from the image formed by the corrective lens rather than from the object itself.

Required:
a. If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that the person would need to see an object at infinity clearly?
b. If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes?
Physics
1 answer:
NeTakaya4 years ago
3 0

Answer:

a)   f₁ = 3.50 m ,  b)     f₂ = 0.84 m  

Explanation:

For this exercise we must use the constructor equation

          1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image

a) the distance where the image should be placed is q = 3.50 m and the object is located at infinity p = ∞

           1 / f₁ = 1 /∞ + 1 / 3.50

           f₁ = 3.50 m

b) in this case the image is at q = -0.600 m and the object p = 0.350 m

           1 / f₂ = 1 / 0.350 -1 / 0.600

the negative sign, is because the image is in front of the object

           1 / f₂ = 1,1905

            f₂ = 1 / 1,1905

            f₂ = 0.84 m

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Answer:

0.144 kg of water

Explanation:

From Raoult's law,

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 423 mmHg ÷ 528.8 mmHg = 0.8

Let the moles of solvent (water) be y

Moles of solute (C3H8O3) = 2 mole

Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.8 = y/(y + 2)

y = 0.8(y + 2)

y = 0.8y + 1.6

y - 0.8y = 1.6

0.2y = 1.6

y = 1.6/0.2 = 8

Moles of solvent (water) = 8 mol

Mass of water = moles of water × MW = 8 mol × 18 g/mol = 144 g = 144/1000 = 0.144 kg

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3 years ago
Explain why the bright spectral lines of light emitted from the atoms of an element are unique to that element
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3 years ago
The only force acting on a 2.9 kg canister that is moving in an xy plane has a magnitude of 7.5 N. The canister initially has a
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Answer:15.66 J

Explanation:

mass of block \left ( m\right )=2.9 kg

Force magnitude=7.5 N

Initial velocity =3.9\hat{i} m/s

Final velocity=5.1 \hat{j} m/s

Initial Kinetic Energy=\frac{1}{2}mv^2

=\frac{1}{2}\times 2.9\times 3.9^2=22.05 J

Final Kinetic Energy=\frac{1}{2}mv^2

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I got it off a Quizlet hope this helps!!

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3 years ago
Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 4.50 m above
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8.98×10^6\:\text{m}

Explanation:

First we need to find the acceleration due to gravity on the planet. The wrench took 0.809 s to fall from a height of 4.50 m so we can use the equation

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Solving for g, we get

g = -\dfrac{2y}{t^2} = -\dfrac{2(-4.50\:\text{m})}{(0.809\:\text{s})} = 13.8\:\text{m/s}^2

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The expression for g then becomes

g = \dfrac{G}{R_p^2} \left(\dfrac{4\pi}{3}\rho R_p^3\right) = \dfrac{4\pi G}{3}\rho R_p

Solving for R_p, we get

R_p = \dfrac{3g}{4\pi G\rho}

\:\:\:\:\:\:\:= \left[\dfrac{3(13.8\:\text{m/s}^2)}{4\pi (6.674×10^{-11}\:\text{Nm}^2\text{/kg}^2)(5500\:\text{kg/m}^3)}\right]

\:\:\:\:\:\:\:= 8.98×10^6\:\text{m}

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3 years ago
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