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PSYCHO15rus [73]
3 years ago
12

What condition occurs when the eyeball is too long?

Physics
2 answers:
Diano4ka-milaya [45]3 years ago
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.

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Which statement accurately describes science?
slava [35]
Answer: We don’t really ever need to know how to dissect a frog but hey, one day maybe I’ll need to give a frog a discectomy
5 0
3 years ago
What is the weight of a rock with a mass of 3.6 kilograms
Marysya12 [62]
3.85 pounds is the answer
4 0
3 years ago
A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can
Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

\mu mg = m\frac{v^2}{r}

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

\mu is the coefficient of friction between the tires and the road

m is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

v=85 mph \cdot \frac{1609}{3600}=38.0 m/s is the speed

And solving for \mu, we find the coefficient of friction required to keep the car in circular motion:

\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196

Learn more about circular motion:

brainly.com/question/2562955  

brainly.com/question/6372960  

#LearnwithBrainly

8 0
3 years ago
Please help.. I'm desperate.
rosijanka [135]

Answer:

Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry.

Explanation:

7 0
3 years ago
Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di
jeyben [28]

<u>Answer :</u>

(a) d = 0.25 m

(b) d = 0.5 m

<u>Explanation :</u>

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at 20^0\ C is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

d=\dfrac{\lambda}{2}........(1)

Velocity of sound wave is given by :

v=f\times \lambda

d=\dfrac{v}{2f}

d=\dfrac{343\ m/s}{2\times 686\ Hz}

d=0.25\ m

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

d=n\lambda  ( n = integers )

Let n = 1

So, d=\dfrac{v}{f}

d=\dfrac{343\ m/s}{686\ Hz}

d=0.5\ m

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

4 0
3 years ago
Read 2 more answers
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