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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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Answer:

Code in MATLab is given as  below:

Explanation:

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Answer:b

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The function below takes two numeric parameters. The first parameter specifies the number of hours a person worked and the secon
Mekhanik [1.2K]

def calculate_pay(total_worked_hours, rate_per_hour):

  if total_hours_worked > 40:

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Explanation:

  • First we create the calculate_pay function which will takes 2 parameters.
  • Secondly ,inside the function we check if  the total_worked_hours is greater than 40 and then return the pay by calculating with the help of formula for work over 40 hours.
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3 years ago
A 10-mm-diameter Brinell hardness indenter produced an indentation 1.55 mm in diameter in a steel alloy when a load of 500 kg wa
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Answer:

HB = 3.22

Explanation:

The formula to calculate the Brinell Hardness is given as follows:

HB = \frac{2P}{\pi D\sqrt{D^{2}- d^{2} } }

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HB = Brinell Hardness = ?

P = Applied Load in kg = 500 kg

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d = Diameter of the indentation in mm = 1.55 mm

Therefore, using these values, we get:

HB = \frac{(2)(500)}{\pi (10)\sqrt{10^{2}- 1.55^{2} } }

<u>HB = 3.22 </u>

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3 years ago
when electric power is transmitted over long distance, line losses can be reduced by generating AC rather than DC voltage and by
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When electric power is transferred over a long distance then, line loss can be reduced by high voltage and low current. Thus, option A is correct.

<h3>What is electric power?</h3>

The complete question is attached to the image below.

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The AC transmitted is of high voltage, and the power is low. The equation for current is given,

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