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2 years ago

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One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d

istance of each from the line are measured, and the sum of the squares of the distance divided by 100 is 36 in2.
a. Assuming only one-dimensional movement away from the line, calculate the diffusion coefficient of the jumping beans.
b. If the mean jump distance of a bean is equal to 0.1 inches, estimate the mean jump frequency of a bean.

1 answer:

stira [4]2 years ago

8 0

Answer:

a) Diffusion coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

$^{2} = 2Dt$ ..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

$36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr$

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

$^{2} = 2Dt$

$0.1^{2} = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds$

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

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