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satela [25.4K]
3 years ago
12

The count rate of a radioactive source decreases from 1600 counts per minute to 400 counts per minute in 12 hours. What is the h

alf-life of the source?
Physics
1 answer:
kirill115 [55]3 years ago
4 0

Answer:

t_{1/2}=6 h

Explanation:

Let's use the decay equation.

A=A_{0}e^{-\lambda t}

Where:

  • A is the activity at t time
  • A₀ is the initial activity
  • λ is the decay constant

We know that \lambda=\frac{ln(2)}{t_{1/2}}

So we have:

\lambda=\frac{ln(A/A_{0})}{t}

\frac{ln(2)}{t_{1/2}}=\frac{ln(A/A_{0})}{t}

t_{1/2}=\frac{t*ln(2)}{ln(A/A_{0})}

t_{1/2}=6 h

Therefore, the half-life of the source is 6 hours.

I hope it helps you!

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To obtain the power, we first need to find the work made by the force.

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Q|C (a) You need a 45-ω resistor, but the stockroom has only 20-ω and 50-ω resistors. How can the desired resistance be achieved
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  1. In order to achieve the desired resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with the 20 Ohms resistors.
  2. In order to get a 35 Ohms resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with two 20 Ohms resistors that are connected in parallel.

<h3>How to achieve the desired resistance under these circumstances?</h3>

In order to achieve the desired resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with the 20 Ohms resistors.

Mathematically, the total equivalence resistance of two resistors that are connected in parallel is given by:

1/Rt = 1/R₁ + 1/R₂

1/Rt = 1/50 + 1/50

1/Rt = 2/50

1/Rt = 1/25

Rt = 25 Ohms.

Next, we would connect this 25 Ohms resistor in series with the 20 Ohms resistor:

R₃ = 20 + Rt

R₃ = 20 + 25

R₃ = 45 Ohms.

<h3>Part B.</h3>

In order to get a 35 Ohms resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with two 20 Ohms resistors that are connected in parallel.

1/Rt = 1/R₁ + 1/R₂

1/Rt = 1/50 + 1/50

1/Rt = 2/50

1/Rt = 1/25

Rt = 25 Ohms.

1/R't = 1/R₁ + 1/R₂

1/R't = 1/20 + 1/20

1/R't = 2/20

1/R't = 1/10

R't = 10 Ohms.

Next, we would connect the 25 Ohms resistor in series with the 10 Ohms resistor:

R₃ = 10 + Rt

R₃ = 10 + 25

R₃ = 35 Ohms.

Read more on resistors in parallel here: brainly.com/question/15121871

#SPJ4

Complete Question:

You need a 45-ω resistor, but the stockroom has only 20-ω and 50-ω resistors.

(a) How can the desired resistance be achieved under these circumstances?

(b) What can you do if you need a 35-ω resistor?

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The current in the second diode is 400mA

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<h3>Current In a Series</h3>

The current in the first diode is equal to 400mA. In a series circuit, the current passing the diodes are equal. This implies that the current in the series are equal.

Diodes connected in series will be the equal.

I_1 = I_2

Since I1 is 400mA, I2 will be equal to 400mA

I_1 = I_2\\I_1 = 400mA\\I_2 = 400mA

The current in the second diode is 400mA

Learn more on current in a diode here;

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