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ioda
4 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]4 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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You can write the equation in 3 different ways, depending on which quantity you want to be the dependent variable.  Any one of the three forms can be derived from either of the other two with a simple algebra operation.  They're all the same relationship, described by "Ohm's Law".

==> Current = (potential difference) / (resistance)

==> Potential difference = (current) x (resistance)

==> Resistance = (potential difference) / (resistance)

4 0
3 years ago
Determine whether the statement is true or false. If f '(c) = 0, then f has a local maximum or minimum at c.
ankoles [38]

It is False that, If f '(c) = 0, then f has a local maximum or minimum at c.

Local maximum and minimum points are very distinctive on the graph of a function and are thus, helpful in grasping the shape of the graph. Either a local minimum or a local maximum can be considered a local extremum.

A counterexample can be used for:

f(x) = x³

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and,

f'(0) = 2×0² = 0

However, the assertion is false because x = 0 is actually an inflection point rather than a maximum or minimum in f(x) = x³.

Know more about f(x) here: brainly.com/question/28058540

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4 0
2 years ago
A negatively charged object has _______________ electrons than an object that is neutral.
Free_Kalibri [48]

Answer:

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Explanation:

4 0
3 years ago
Lenses are often coated with magnesium fluoride (refractive index n = 1.38) to reduce reflections. How thick should the layer of
Valentin [98]

Answer: 99.64 nm (≅ 100 nm)

Explanation: In order to explain this problem we have to obtain destructive inteference from two waves that reflect in the film and lens surface so both waves have a λ/2 shift then we have to get a difference path (2L) equal to an odd number of the half wavelegth.

Then we have the following expression:

L=(m+1/2)* λ/(2*n2);  where n2 is the refractive index of the coating

for m=0 we have the minimum thickness for the coating

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3 years ago
A car, with an initial velocity of 22 m/s west, travels for 345 seconds. If it has a final velocity of 44 m/s west what is its’
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Answer: 0.064 m/s²

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= 22/345

= 0.0637681

= 0.064 m/s²

Therefore, the acceleration is 0.064 m/s².

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3 years ago
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