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3 years ago

What condition occurs when the eyeball is too long?

2 answers:

7 0

Hello!

Myopia, also known as nearsightedness, occurs when the eyeball is too long. This condition is very common in the U.S and throughout the world. It can be corrected by wearing glasses or contact lenses.

Contact [7]3 years ago

6 0

Explanation :

There can be many eye defects in the human eye. Some of them are Myopia, Hypermetropia, Astigmatism etc.

In Myopia the size of the eyeball is too long and the image is formed in front of the retina. A person suffering from this defect is not able to see far objects clearly but can see nearby objects normal.

Concave lenses are used to correct this defect so that the image is focused on the retina.

You might be interested in

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

3 years ago

If the force applied to an object is not greater than the starting friction, what will happen to the object? Which answer is cor

AleksandrR [38]

Answer:

The object will move in the opposite direction of the force applied. - 2.

8 0

3 years ago

A car changes its speed by 2 meters per second each second. What is its acceleration?

KatRina [158]

Answer:

2 m/s²

Explanation:

If changes speed by 2 meters per second each second means:

2 m/s²

Because it changes constantly it veloctity.

Remember the aceleration changes the velocity.

4 0

3 years ago

Read 2 more answers

Order from biggest to smallest?<br> Milky Way, universe, planets, clusters, and stars

Zinaida [17]

Hello,

The answer is "universe, Milky Way, clusters, stars, planets".

Reason:

The universe would be the biggest because it has all the galaxy's, starts, clusters, and planets into one. Then it would be Milky Way because this is a galaxy that contains: stars, planets, and clusters. Then it would be clusters because that contains stars, or planets in one group. Then be stars because stars are bigger than planets. Then it would be planets. Therefore the order should go like this: <span>Milky Way, universe, planets, clusters, and stars.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>

5 0

3 years ago

Read 2 more answers

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

75 kmh⁻¹

Explanation:

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago