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Alexus [3.1K]
3 years ago
8

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:
Reika [66]3 years ago
4 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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For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value
OleMash [197]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

Y = 1 - e^{-kt^n}

here Y is percentage of completion  of reaction = 50%

t  is duration of reaction = 146 sec

so,

0.50 = 1 - e^{-k^146^2.1}

0.50 = e^{-k306.6}

taking natural log on both side

ln(0.5) = -k(306.6)

k = 2.26\times 10^{-3}

for 86 % completion

0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}

e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14

-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)

t^{2.1} = 869.96

t = 25.10 sec

5 0
3 years ago
Knowing that v = –8 m/s when t = 0 and v = 8 m/s when t = 2 s, determine the constant k. (Round the final answer to the nearest
docker41 [41]

Answer:

a)We know that acceleration a=dv/dt

So dv/dt=kt^2

dv=kt^2dt

Integrating we get

v(t)=kt^3/3+C

Puttin t=0

-8=C

Putting t=2

8=8k/3-8

k=48/8

k=6

5 0
3 years ago
About ceramics: Only can be optically opaque or semi-transparent. a) True b)-False
julia-pushkina [17]

Answer: True

Explanation: Ceramics have the property that when the band gap present between the atoms are larger than the light energy then the tend to become opaque because the light scattering is caused . They also show the property of being translucent when there are chances of the light to get a path through the surface of ceramic so they get the light at some parts e.g.porcelain .Therefore the statement given is true that ceramics can be optically opaque or semi-transparent(translucent).

6 0
3 years ago
جائت فكرة ربط الحواسيب لغرض نقل البيانات و مشاركتها و بعدها بفتره قصيره جائت إمكانية مشاركة الموارد بين الحواسيب صح ام خطأ​
melamori03 [73]

Answer:

بدلاً من ذلك يُشار إليه باسم مشاركة أو مشاركة شبكة ، الدليل المشترك هو دليل أو مجلد يمكن الوصول إليه من قبل العديد من المستخدمين على الشبكة. هذه هي الطريقة الأكثر شيوعًا للوصول إلى المعلومات ومشاركتها على شبكة محلية

Explanation:

5 0
3 years ago
Read 2 more answers
Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo
AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

     = ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

           

7 0
3 years ago
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