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ioda
3 years ago
5

The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a

nd a length of 0.25 cm. What would be the diffusion rate mlt in a channel with a cross- sectional area of 0.30 cm^2 and a length of 0.10 cm?
Physics
1 answer:
zlopas [31]3 years ago
7 0

Answer:

s = 9.6\times 10^{-12}kg/s

Explanation:

Given:

Solute Diffusion rate  = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel  =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel  =0.10 cm

let the Solute Diffusion rate  of new channel = s

now equating the diffusion rate per unit volume for both the channels

\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}

thus,

s = 9.6\times 10^{-12}kg/s

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alexdok [17]

Answer:

20.25 m

Explanation:

  • <u>Centripetal acceleration </u>is given by; the square of the velocity, divided by the radius of the circular path.

That is;

         <em><u>ac = v²/r</u></em>

<em>         </em><em><u> Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the  radius, m</u></em>

Therefore;

r = v²/ac

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Hence the radius is 20.25 meters

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3 years ago
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What is your theory on frame of reference?

Answer:

It is the idea of seeing something from a different perspective

Explanation:

Hope this helps!

8 0
3 years ago
A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

P = \ro g h = 1000 * 10 * 20 = 200000 Pa

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

P_1V_1 = P_2V_2

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

V_1 = Ah

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

P_1Ah_1 = P_2Ah_2

h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m

b) If the temperatures changes, we can still reuse the ideal gas equation above:

\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}

h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m

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3 years ago
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AnnyKZ [126]

Answer:

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What happens to the Force due to gravity between 2 masses if one
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