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Harlamova29_29 [7]
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.
a. greater than
b. less than
c. equal to

Physics
2 answers:
Grace [21]3 years ago
6 0

Answer:

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.

The answer is option B (less than)

Explanation:

The cornea which is the strongest part of the refracting power of the eye, is the front part of the eye, and the backside of the eye is called retina.

Light enters the eye at the curved front surface of the cornea, passes through the lens, and then strikes the retina and fovea on the back of the eye. For the eye to receive light, light travels in air, gets refracted majorly at the cornea with some other refraction done by the Aqueous humor, the lens and the vitreous humour. Focal length of the lens must vary to ensure objects at various distance can be brought to a focus at the retina.

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. The fovea which is a small area near the center of the retina, is responsible for the most acute vision. Speed of light changes considerably while traveling from air into cornea.

When muscles in the eye contract and squeeze the lens, the center of the lens bulges, causing the focal length to shorten, and allowing the eye to focus on closer objects. The image distance depends on both the object distance and the focal length of the lens. For objects to be at infinity the focal point must be situated farthest from the lens, while near objects have focal point closer to the lens.

nexus9112 [7]3 years ago
4 0

Answer:

The correct answer is letter b. less than

Explanation:

In the attached schematic figure, the lens (lens) is initially adjusted for a given distance from the object. If the object gets closer, the image loses its sharpness. To recover it, the lens settles down, increasing convergence, that is, decreasing the focal length.

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Answer:

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Explanation:

Applying

q = cm(t₂-t₁)............... Equation 2

Where q = heat energy, c = specific heat of ethanol, m = mass of ethanol, t₁ = initial temperature, t₂ = Final temperature.

Given: c = 2.44 J/g.°C,  m = 300 g, q = 14640 J, t₂ = 30°C

Substitute into equation 2 and solve for t₁

14640 = 2.44×300(30-t₁)

14640 = 732(30-t₁)

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Answer:

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Part c)

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Part f)

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Explanation:

As we know that ball is projected with speed

v = 8.10 m/s at an angle 18 degree below the horizontal

so we will have

v_x = 8.10 cos18

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Part a)

Since it took t =4 s to reach the ground

so its initial y coordinate is given as

y = v_y t + \frac{1}{2}a_y t^2

y = 2.5(4) + \frac{1}{2}(9.81)(4^2)

y = 88.5 m

Part b)

components of the velocity is given as

v_x = 8.10 cos18

v_x = 7.7 m/s

v_y = 8.10 sin18

v_y = 2.5 m/s

Part c)

equation of position in x direction is given as

x = v_x t

equation of position in y direction is given as

y = v_y t + \frac{1}{2}gt^2

Part d)

distance where it will strike the floor is given as

x = v_x t

x = 7.7 \times 4

x = 30.8 m

Part e)

Height from which it is thrown is same as initial y coordinate of the ball

so it is given as

H = 88.5 m

Part f)

time taken by ball to reach 10 m below is given as

y = v_y t + \frac{1}{2}gt^2

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when two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of
Ahat [919]

<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q​</h2>

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Q= m x c x t

In above question , it is given :

For Q

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X=m x C₁ X T₁

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mass of P =m/2

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Heat supplied is same that is : X

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