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andrezito [222]
3 years ago
15

Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be ______

____ the distance between the front and back of his eye. View Available Hint(s) Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be __________ the distance between the front and back of his eye. greater than less than equal to
Physics
1 answer:
kotykmax [81]3 years ago
5 0

Answer: The focal length of the cornea-lens system in his eye must be LESS THAN the distance between the front and back of his eye.

Explanation:

The human eye the front part of the eye is the CORNEA. This is the tough white transparent part of the eye that helps in the refraction of light rays. While the backside of the eye is the RETINA. This is the part of the eye when images are focused.

When a normal eye is at rest, parallel rays from a distant object are focused on the retina. The ability of the eye - lens to focus points at different distances on the retina is known as accomodation. The adjustment of the eye lens to focus objects of varying distances is brought about by the ciliary muscles. The have the ability to change the shape of the eye which leads to change in focal length.

When a person with normal vision looks at a distant object at infinity, the lens brings parallel rays to focus on the retina. Thus, the furthest point which the eye can see distinctly is called the far point of the eye and it's infinity for a normal eye. But Joe was able to focus his eye on the tree, meaning that the tree was within his near point. This is the nearest point at which an object is clearly seen. Therefore, when the effective focal length of the cornea-lens system changes, it changes the location of the image of any object in one's field of view.

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A ball is dropped from rest at the top of a 6.10 m
natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0
3 years ago
Technician A says that a tire will increase or decrease approximately 1 psi for each 10°F change of temperature. Technician B sa
bearhunter [10]

Answer:

Technician A

Explanation:

It is seen that a tire pressure will increase or decrease 1 psi for each 10^{\circ} F  change in temperature.

For Technician B vehicle pressure should not be adjusted after tire has been warmed as the warm air may increase the pressure but it will be auto adjusted as  the temperature falls to normal .          

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3 years ago
It is an employer's responsibility to provide training for special equipment. true false
Naddik [55]

Answer:

True I hope you like it

Give me a brainliest answer

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3 years ago
A man does 500 J of work pushing a car a distance of 2 m. How much force does he apply? Assume there is no friction.
Dmitry [639]

The correct answer is A. 250N

Work is a product of force and distance.

That is, work done=force×distance

Therefore substituting for the values in the question:

500J=force×2m

force= 500Nm/2m=250N

another unit for work done is Nm as force as the SI unit of force is newtons and distance in meters

6 0
3 years ago
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