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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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tekilochka [14]

Answer:

changing the direction in which a force is exerted

8 0
3 years ago
g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th
kati45 [8]

The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

3 0
3 years ago
Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the
joja [24]

As it is given that the air bag deploy in time

t = 10 ms = 0.010 s

total distance moved by the front face of the bag

d = 20 cm = 0.20 m

Now we will use kinematics to find the acceleration

d = v_i*t + \frac{1}{2}at^2

0.20 = 0 + \frac{1}{2}a*0.010^2

0.20 = 5 * 10^{-5}* a

a = 4000 m/s^2

now as we know that

g = 10 m/s^2

so we have

a = 400g

so the acceleration is 400g for the front surface of balloon

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kotykmax [81]

Answer: The electrons flowing through the wire are referred to as a quantity of electricity, and the flow of electricity is referred to as “an electric current.”

Explanation: Hope it Helps have a blessed day

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A sportscar has a mass of 1500 kg and accelerates at 5 meters per second squared. What is the magnitude of the force acting on t
Hatshy [7]

Answer:

7500 Newtons

Explanation:

Mass of the sportscar= 1500 kg

Acceleration of the sportscar= 5m/s^2

Hence, let the Force acting on it be F

We\ know\ that,\\Force=Mass*Acceleration\\F=ma\\\\Here,\\F=1500*5\\=7500 kg m/s^2\ or\ 7500\ Newtons

4 0
3 years ago
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