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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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Explanation:

Given that,

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We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

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masya89 [10]

Answer:

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