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kati45 [8]
3 years ago
6

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea

ding of 400. counts has diminished to 100. counts after 66.7 minutes , what is the half-life of this substance? Express your answer with the appropriate units.
Engineering
1 answer:
vovangra [49]3 years ago
3 0

Answer: 33.35 minutes

Explanation:

A(t) = A(o) *(.5)^[t/(t1/2)]....equ1

Where

A(t) = geiger count after time t = 100

A(o) = initial geiger count = 400

(t1/2) = the half life of decay

t = time between geiger count = 66.7 minutes

Sub into equ 1

100=400(.5)^[66.7/(t1/2)

Equ becomes

.25= (.5)^[66.7/(t1/2)]

Take log of both sides

Log 0.25 = [66.7/(t1/2)] * log 0.5

66.7/(t1/2) = 2

(t1/2) = (66.7/2 ) = 33.35 minutes

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slamgirl [31]

Answer:

d. 90%

Explanation:

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So option d is correct.

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Answer:

V = 0.30787 m³/s

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diameter = 28 cm

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temperature = 20°C

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solution

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m = ρ V

flow rate = V = \frac{m}{\rho}

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V = 0.30787 m³/s

and

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m = ρ A v

m = ρ V

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m = 2.6963 kg/s

and

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velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

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v2 = \frac{0.3705}{\frac{\pi}{4} * 0.28^2}

v2 = 6.017 m/s

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Answer:

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