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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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Answer:

The astronaut can throw the hammer in a direction away from the space station. While he is holding the hammer, the total momentum of the astronaut and hammer is 0 kg • m/s. According to the law of conservation of momentum, the total momentum after he throws the hammer must still be 0 kg • m/s. In order for momentum to be conserved, the astronaut will have to move in the opposite direction of the hammer, which will be toward the space station.

Explanation:

6 0
2 years ago
An electric Kettle is rated at 25 W. Calculate the quantity of heat generated in 2s
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Answer:

Energy consumed by the electric kettle in 9.5 min =Pt=(2.5×10

3

)×(9.5×60)=14.25×10

5

J

Energy usefully consumed =msΔT=3×(4.2×10

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)×(100−15)=10.71×10

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where s=4.2J/g

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4 0
3 years ago
Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both fila
Lemur [1.5K]

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

or

\frac{A_1}{A_2} =\frac{T_{2}^{4}}{T_{1}^{4}}

substituting the values in the above question we get

\frac{A_1}{A_2} =\frac{2100_{2}^{4}}{2700_{1}^{4}}

or

\frac{A_1}{A_2} }=0.3659

6 0
3 years ago
Two teams of nine members each engage in tug-of-war. Each of the first team's members has an average mass of 68 kg and exerts an
diamong [38]

Answer:

(a) Acceleration  = 0.1063 m/s^2      (Second team wins)

(b) Tension in rope = 65.106 N

Explanation:

Total mass of first team = 68 * 9 = 612 kg

Total force of first team = 1350 * 9 = 12150 N

Total mass of second team = 73 * 9 = 657 kg

Total force of seconds team = 1365 * 9 = 12285 N

Difference in force = 12285 - 12150 = 135 N   (towards the second team as it has more force)

(a) For acceleration we get:

F = m * a

135 = (mass of both teams) * a

a = 135 / (612 + 657)

acceleration  = 0.1063 m/s^2      (Second team wins)

(b) Since we know the acceleration of the first team (pulling being pulled towards the second team at an acceleration of 0.1063 m/s^2) , we can find out the force required to move them:

Force required for first team = mass of first team * acceleration

Force required = 612 * 0.1063

Force required = 65.106 N

This is the force exerted on the first team through the rope, so the tension in the rope will also be 65.106 N.

7 0
3 years ago
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