Answer:

Step-by-step explanation:
Answer:
144
Step-by-step explanation:
If you multiply 4 by 12, then add 100, you get 144 which is the area.
The diffusion time of the red blood cell to take oxygen through the membrane is 25uS.
The small capillaries in the lungs are in close contact and it takes a red blood cell 0.5 seconds to squeeze through the capillary.
The diffusion time for the oxygen across the 1 um thick membrane separating the air is given by,
t = x²/2D
Where,
t is the diffusion time,
x is the thickness of the membrane,
D is the diffusion coefficient.
Putting values,
t = (1 x 10⁻⁹)²/2 x 2 x 10⁻¹¹
t = 10⁻⁷/4
t = 25 x 10⁻⁹ seconds.
t = 25 uS.
so, the diffusion time is 25uS.
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Answer:
T = 47.1875°C
Step-by-step explanation:
Given:
Surrounding temp, Ts = 100°C
Initial Temperature ,T0= 20°C
Increase in temperature = 15°C
Final temperature, T = 20 + 15 = 35°C
Time, t = 9 seconds
Let's take Newton's law of cooling:

We'll solve for k

Divide both sides by -5


Let's now find the temperature of the ball after 18 seconds in boiling water.
Use the Newton's equation again:




T = 47.1875°C
Temperature of the ball after 18 seconds in boiling water is 47.1875°C