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vlabodo [156]
2 years ago
10

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

Physics
1 answer:
Rom4ik [11]2 years 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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An electric pump rated 1.5 KW lifts 200kg of water through a vertical height of 6m in 10 secs: way is the efficiency of the pump
ruslelena [56]

Answer:

80%

Explanation:

Efficiency = Power output / Power input × 100 %

To calculate efficiency we need to find power output of electric pump.

We can use,

Work done = Energy change

Work done per second = Energy change per second

Work done per second = Power

Therefore, Power = Energy change per second

                              = Change in potential energy of water per second

                              =mgh / t

                              = 200× 10×6 / 10

                              = 1200 W = 1.2 kW

Now use the first equation to find efficiency,

Efficiency = \frac{1.2}{1.5} × 100%

                = 80 %

8 0
3 years ago
A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia
aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

E_{t} = KE_{i} + PE_{i}

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

   

2) The <em>potential energy</em> at point A is:

PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

KE_{A} + PE_{A} = KE_{B} + PE_{B}

KE_{B} = KE_{A} + PE_{A} - PE_{B}

Since

KE_{A} + PE_{A} = KE_{i} + PE_{i}

we have:

KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J

5) The <em>kinetic energy</em> of the roller coaster at point C is:

KE_{i} + PE_{i} = KE_{C} + PE_{C}            

KE_{C} = KE_{i} + PE_{i} = 19820 J      

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

KE_{C} = \frac{1}{2}mv_{C}^{2}

v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

3 0
3 years ago
What is energy converted to if it is not used to do work?
Romashka [77]

Answer:

the human body isn't very efficient at converting food into useful work. The human body is less than 5% efficient most of the time. The rest of the energy is converted to heat, which may or may not be useful, depending on how cool or warm a person wants to be.

Explanation:

3 0
2 years ago
What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa
dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}

E=3.315\times10^{-17}\ J

(b).  We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}

E=6.709\times10^{-21}\ J

(c). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}

E=3.315\times10^{-11}\ J

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}

E=6.709\times10^{-9}\ J

Hence, This is the required solution.

6 0
4 years ago
A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal
Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, a=g=-9.8\ m/s^2(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

S=ut+\frac{1}{2}at^2

Plug in the given values and solve for 'u'. This gives,

0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0
4 years ago
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