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andrezito [222]
2 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]2 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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How can you increase the momentum of an object
Olin [163]

Answer:

You can change the momentum of an object by giving the object more force or less force.

Explanation:

Think about a ball. It is going slow, you push it and you give it more momentum.

5 0
3 years ago
An ideal refrigerator does 130. 0 j of work to remove 780. 0 j of heat from its cold compartment during each cycle. what is the
zheka24 [161]

The refrigerator's coefficient of performance is 6.

The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.

As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.

The heat produced in the cold compartment, H = 780.0 J

Work done in ideal refrigerator, W = 130.0 J

Refrigerator's coefficient of performance = H/W

                                                                     = 780/130

                                                                     = 6

Therefore, the refrigerator's coefficient of performance is 6.

Energy conservation requires the exhaust heat to be = 780 + 130

                                                                                          = 910 J

Learn more about  coefficient here:

brainly.com/question/18915846

#SPJ4

5 0
2 years ago
If a sound wave traveled from a solid to a liquid its speed would:.
JulijaS [17]

Answer:

increase

Please Give Brainliest!

5 0
2 years ago
Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir
yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) T=2\pi \sqrt{\frac{r^{3}}{GM}}

M=\frac{4\pi ^{2}r^{3}}{GT^{2}}

M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

v=\frac{2\pi r}{T}

v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

E = \frac{GM}{r^{2}}

E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}

E = 0.28 N/kg

6 0
2 years ago
HELP ASAP!! i’ll mark you the brainliest!!
Triss [41]

Answer:

yes

Explanation:

6 0
2 years ago
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