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andrey2020 [161]
1 year ago
15

Measurements on the cornea of a person's eye reveal that the magnitude of the front surface radius of curvature is while the mag

nitude of the rear surface radius of curvature is (see Figure 25.18 ), and that the index of refraction of the cornea is 1.38 . If the cornea were simply a thin lens in air, what would be its focal length and its power in diopters? What type of lens would it be?
Physics
1 answer:
34kurt1 year ago
5 0

(a) If the cornea were simply thin lens then power will be 43 diopters.

(b) This is a concave lens

The cornea is the transparent front part of the eye that covers the iris, pupil, and anterior chamber. Despite injury or disease, the cornea can still repair itself quickly. However, there are situations where damage is too severe for the cornea to heal on its own – such as with a deep injury to the cornea. The following symptoms may indicate that the cornea has sustained a substantial infection, injury or disease: Blurred vision Pain Redness.

Along with the anterior chamber and lens, the cornea refracts light, accounting for approximately two-thirds of the eye's total optical power. In humans, the refractive power of the cornea is approximately 43 diopters.

There are two types of lenses: converging and diverging and here if the cornea was simply thin then the diverging or concave lens is used in the eyes which is thin in the center than their edges.

To know more about cornea, refer: brainly.com/question/13866057

#SPJ4

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8 0
3 years ago
A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is
jasenka [17]

Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

Explanation:

The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, F_{net} = The weight of the student + The inertia force of the slowing elevator

∴ The net force, F_{net} = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

3 0
3 years ago
An electron is placed 5.40 cm due north of a charged sphere and experienced a force of 8.30 x 10^-13 N due north. What is the el
ira [324]

Answer:

The electric field is E = 5.1*10^6N/C.

Explanation:

The force F on a charge q in an electric field E is given by

F = qE,

which can be rearranged to give

E = \dfrac{F}{q }

Now, the force on the electron is F = 8.30*10^{-13}N, and its charge is

q = 1.6*10^{-19}C; therefore,

E = \dfrac{8.30*10^{-13}N}{1.6*10^{-19}C}

\boxed{E = 5.1*10^6N/C}

4 0
3 years ago
And object which is 3cm high is placed vertically 10cm in front of a concave mirror. If this object produces an image 40cm from
Stolb23 [73]

Answer:

25cm is the answer I think

8 0
2 years ago
A small 0.14 kg metal ball is tied to a very light (essentially massless) string that is 0.9 m long. The string is attached to t
Vsevolod [243]

Answer:

a) v=2.743m/s

b) a_c = 8.363m/s^2

c) T=2.543N

Explanation:

First, calculate the height of the ball at the starting point:

y' = 0.9cos(55)

y' = 0.516

At this point, just in the moment the ball is released, all the energy of the system is potencial gravitational energy. When it is at the bottom all the potencial energy is transformed into kinetic energy:

E_p=E_k\\mgh=\frac{mv^2}{2}

Solving for v:

v=\sqrt{2gh}

if h is the height loss: (l-y')

v=2.743m/s

The centripetal acceleration is the acceleration caused by the tension force exercised by the string, and is pointing outside of the trayectory path (at the lowest point, directly dawn):

a_c=\frac{v^2}{r}

a_c = 8.363m/s^2

To calculate tension, just make the free body diagram of forces in the ball, noticing the existence of the centripetal acceleration:

\sum{F_y}=ma_c=T-W\\T=ma_c+W\\T=m(a_c+g)\\T=0.14(8.363+9.8)\\T=2.543N

4 0
3 years ago
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