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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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4 years ago
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
Pavel [41]

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

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3 years ago
A wind instrument that produces only odd-numbered standing wave modes has what configuration of its ends?
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Explanation:

It has Open-closed configuration of its end

4 0
3 years ago
What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original
DIA [1.3K]

Answer:

F' = 64 F

Explanation:

The electric force between charges is given by :

F=\dfrac{kq_1q_2}{r^2}

Where

q₁ and q₂ are charges

r is the distance between charges

When  each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

F'=\dfrac{kq_1'q_2'}{r'^2}

Apply new values,

F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F

So, the new force becomes 64 times the initial force.

7 0
3 years ago
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