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SOVA2 [1]
4 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]4 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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The higher you go the more potential energy there is, and the lower it is the more kinetic energy there is, so the more kinetic energy there is the higher the ball will bounce.
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the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path
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Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:  

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5 0
3 years ago
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levacccp [35]
Answer
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Explanation
4 0
3 years ago
Read 2 more answers
PLEASE HELP WILL GIVE BRAINLIEST!!!
Elan Coil [88]

The spring constant is 181.0 N/m

Explanation:

We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:

\frac{1}{2}kx^2 = mgh

where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where

k is the spring constant of the spring

x = 2.08 cm = 0.0208 m is the compression of the spring

m = 12.3 g = 0.00123 kg is the mass of the dart

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h = 3.25 m is the maximum height of the dart

Solving for k, we find:

k=\frac{2mgh}{x^2}=\frac{2(0.00123)(9.8)(3.25)}{(0.0208)^2}=181.0 N/m

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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