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SOVA2 [1]
3 years ago
8

Contact lenses are designed to be gas permeable in order to allow oxygen to diffuse to the eye. If the edge of a contact lens is

double the thickness of the central portion of the lens, how will the diffusion time differ through the different parts of the contact lens?
a) The diffusion time through the edge is double the diffusion time through the center.
b) The diffusion time through the edge is quadruple the diffusion time through the center
c) The diffusion time through the edge is half the diffusion time through the center
d) The diffusion time is the same for both.
Physics
1 answer:
guapka [62]3 years ago
6 0

Answer: A

The diffusion time through the edge is double the diffusion time through the center.

Explanation:

Only the diffusion coefficient, thickness, surface area, distance affects the time it takes a gas such as oxygen to diffuse a given in a medium. The Diffusion time increases with the square of diffusion distance  Also, the Increased surface area increases the rate of diffusion or the diffusion time, whereas a thicker membrane reduces the rate of diffusion or time it takes the gas to be permeable. Also, The greater the distance that a substance must travel, the slower the rate of diffusion The diffusion coefficient determines the time it takes a gas to diffuse. Also the Diffusion time is inversely proportional to the diffusion coefficient.

Now, If the edge of a contact lens is double the thickness of the central portion of the lens, then it will take more time for the gas to diffuse through the thicker portion. since its edge is double the central position, then the diffusion time will also be doubled. So Option A is the best answer-

The diffusion time through the edge is double the diffusion time through the center.

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Igoryamba

Answer:

300,000,030m/s

Explanation:

Using the relationship between frequency and speed of a wave;

F ∝ V

F = kV

k = F/V

F1/V1 = F2/V2 = k

Let F1 be the frequency of the radio wave = 2,000,000.0Hz

V1 be the speed of light = 300,000,000m/s

F2 = frequency produced by the car = 2,000,000.2Hz

V2 be the velocity of the moving car

Substituting this values in the equation above;

2,000,000/300,000,000 = 2,000,000.2/V2

Cross multiplying

2,000,000V2 = 300,000,000×2,000,000.2

2V2 = 300×2,000,000.2

V2 = 150×2,000,000.2

V2 = 300,000,030m/s

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3 years ago
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3 0
3 years ago
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Oksana_A [137]

Answer:

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F_{in}=10 N is the force in input applied by the worker

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Substituting into the equation, we find

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