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1 year ago

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

1 answer:

7 0

The lens of the eye loses its flexibility 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

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Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani

Dahasolnce [82]

Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

$n = \frac{t}{t_{1/2}}$

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

$t_{1/2}$ = half-life = 700 million years

Therefore,

$n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3$

Now, we will calculate the number of uranium nuclei left ( $n_u$ ):

$n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u = 800\ million$

and the rest of the uranium nuclei will become thorium nuclei ( $u_{th}$ )

$n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million$

dividing both:

$\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u$

Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.

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2 years ago

It takes 180 kj of work to accelerate a car from 21.0 m/s to 27.0 m/s. what is the car's mass?

beks73 [17]

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3 years ago

A change in kinetic of an object is equal to the ___

podryga [215]

Answer:C..net work done on the object.

Explanation:

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3 years ago

The primary job of a(n) ____ is to increase the power of a modified radio wave.

amm1812

The primary job of a(n) ____

is to increase the power of a modified radio wave.

Answer:

<h2>amplifier</h2>

Explanation:

Hope it helps:)

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2 years ago

Read 2 more answers

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

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3 years ago