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Sedbober [7]
3 years ago
5

Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a

squirrel sitting on a rock next to him, a tree a few meters away, and a distant mountain. As Joe is taking in the view, he thinks back to what he learned in his physics class about how the human eye works.
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. The cornea and lens together form a compond lens system. The large difference between the index of refraction of air and that of the aqueous humor behind the cornea is responsible for most of the bending of the light rays that enter the eye, but it is the lens that allows our eyes to focus. The ciliary muscles surrounding the lens can be expanded and contracted to change the curvature of the lens, which in turn changes the effective focal length of the cornea-lens system. This in turn changes the location of the image of any object in one's field of view. Images formed on the fovea appear in focus. Images formed between the lens and the fovea appear blurry, as do images formed behind the fovea. Therefore, to focus on some object, you adjust your ciliary muscles until the image of that object is located on the fovea.
A) 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
B) Joe's eyes are focused on the tree, so the squirrel and the mountain appear out of focus. This is because the image of the squirrel is formed ______ and the image of the mountain is formed _____.
a. between the lens and fovea / between the lens and fovea
b. between the lens and fovea / behind the fovea
c. behind the fovea / between the lens and fovea
d. behind the fovea / behind the fovea
C) Joe now shifts his focus from the tree to the squirrel. To do this, the ciliary muscles in his eyes must have _____ the curvature of the lens, resulting in a(n) _______ focal length for the cornea-lens system. Note that curvature is different from radius of curvature.
a. increased / increased
b. increased / decreased
c. decreased / increased
d. decreased / decreased
Physics
1 answer:
Novosadov [1.4K]3 years ago
6 0

Answer:

A) correct answer is C,   B)   correct answer is b  and C) The correct answer is b

Explanation:

In the exercises of geometric optics, the equation of the constructor tells us the location of the image.

        1 / f = 1 / p + 1 / q

where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.

In this case, the distance to the image on the retina is constant, about 3 cm. Therefore depending on the distance to the object) p = the focal length must change

        1 / q = 1 / f 1 / p

let's apply this expression to our case

A) indicates that the tree is at a medium distance

so that the image is formed on the retina THE SAME AS

correct answer is C

B) The squirrel is at a smaller distance (p ') than the tree (p), therefore if we substitute in the equation above we find that q must decrease. Consequently the image is in front of the retina

The mountain is very far, suppose in infinity, so the image is BEHIND THE RETINA

therefore the correct answer is b

C) The squirrel is very close so the curvature of the lens INCREASES, resulting in a DECREASE in the focal length

The correct answer is b

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1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

v_f = 24.5 m/s

Explanation:

Part 1)

initial speed of the ball upwards

v_i = 24.5 m/s

so maximum height of the ball is given by

H = \frac{v_i^2}{2g}

H = \frac{24.5^2}{2(9.80)}

H = 30.6 m

Part 2)

As we know that final speed will be zero at maximum height

so we will have

v_f - v_i = at

0 - 24.5 = (-9.8)t

t = 2.5 s

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

v_f = 24.5 m/s

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

y = \frac{1}{2}at^2

4.88 = \frac{1}{2}a(1^2)

a = 9.76 m/s^2

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

a = 90 m/s^2

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

v_f = v_i + at

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v_f = 29.7 m/s

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

y = \frac{1}{2}gt^2

y = \frac{1}{2}(9.81)(2.3^2)

y = 25.95 m

Part 2)

Distance traveled by it in horizontal direction is given as

d = v_x t

d = 2.92 \times 2.3

d = 6.72 m

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3 years ago
What are three physical properties of gases
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Answer:   Gases have three characteristic properties: (1) they are easy to compress, (2) they expand to fill their containers, and (3) they occupy far more space than the liquids or solids from which they form. An internal combustion engine provides a good example of the ease with which gases can be compressed.

Explanation:

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Answer:

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Explanation:

The important concept of mass center defined by

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where M is the total mass, x_i and m_i are the position and masses of each body

Let's apply this expression to our case.

Let's set a reference frame where the axis points from the center of the Earth to the Moon,

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the total mass is

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the distance from the Earth is zero because all mass can be considered to be at its gravimetric center

let's calculate

      M = 5.98 10²⁴ + 7.35 10²²

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we substitute

      R_cm = 1 / 6.0535 10²⁴ (0 + 7.35 10²² 3.84 )

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Answer:

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K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

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For circular fin

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m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}

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By solving above equation we get

L= 36.18 mm

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\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}

ε = 23.52

5 0
3 years ago
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