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vlabodo [156]
1 year ago
10

He lens of the eye loses its ________ with age, which impairs the ability to focus.

Physics
1 answer:
Rom4ik [11]1 year ago
7 0

The lens of the eye loses its <u>flexibility</u> with age, which impairs the ability of the eyes to focus. This process is called presbyopia.

Presbyopia happens when the lens of the eye loses its flexibility due to age. The lens of the eye works similar to the lens of a digital camera. The digital camera focuses on objects that are far or near by by adjusting the placement of the lens, by slightly moving it forward or backward, constantly changing its focal length. The human eye does the same, but instead of moving the lens, the lens changes its shape to focus on far or near objects, thereby also changing its focal length. This is also called accommodation.

To focus on far objects, certain muscles around the lens relax, making the lens rounder in shape, while for close objects, the muscles contract, making the lens flatter in shape.

As a person ages, the lens thickens and gradually loses its flexibility, and around the age of 40, the lens is no longer flexible. Therefore, it becomes difficult to focus on close objects. This condition is therefore known as presbyopia.

To learn more about how the human eye functions, check out;

brainly.com/question/16834488

#SPJ4

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As mass increases what happens to the kinetic energy
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As mass increases kinetic energy also increases; kinetic energy is directly proportional to mass so whatever is done to either affects the other one the same. i hope this helps :)
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When you balance a redox equation using the half-reaction method, what do you do after you write the oxidation and reduction hal
Ann [662]
For each half reaction:
>Balance all except O and H
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3 years ago
Read 2 more answers
1. Driving at slower speeds than traffic flow _____________ .
r-ruslan [8.4K]

<u>The correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge. </u>

<u> </u>

Further Explanation:

The electric field intensity at a point is the measure of the force exerted by a charge particle on another charge particle in the particular area of its strength.

The electric field intensity at a distance d due to a static charge having charge q is directly proportional to the amount of charge and inversely proportional to the square of the distance between them.

The Electric field intensity due to a charge is given as:

E = \dfrac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}

Here, E is the electric field intensity, q is the amount of charge and r is the distance of the charge from the point.

The above expression of electric field shows that the electric field intensity at a point depends on the amount of charge as well as the distance of the point from the charge.

<u>Thus, the correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge. </u>

<u> </u>

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Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Electrostatics

Keywords:  Strength, electric field, charge, distance, electric field intensity, magnitude of charge, electrostatic, test charge, kq/r^2.

7 0
3 years ago
Read 2 more answers
A bungee jumper of mass m jumps off a bridge. Assume that the bungee cord behaves like am ideal spring of spring constant k. Whe
DanielleElmas [232]

Answer:

b) √[(kx²/m) - 2gx]

Explanation:

The energy at the lowest point is equal to:

E_{elas}=\frac{1}{2} *k*x^{2}

where:

Eelas = elastic energy [J]

k = spring constant [N/m]

x = extension of the spring [m]

We consider the lowest point, as the point where the potential energy is zero. At the moment when the person goes back through the point of the normal length of the elastic cord, it is this point that the person will have potential energy and kinetic energy.

E_{elas}=E_{pot}+E_{kine}\

\frac{1}{2}*k*x^{2}=m*g*x +\frac{1}{2} *m*v^{2}  \\v^{2} = \frac{k*x^{2} }{m}-2*g*x\\ v=\sqrt{\frac{k*x^{2} }{m}-2*g*x}

8 0
3 years ago
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