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tino4ka555 [31]
3 years ago
8

Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin

al number of nuclei remain. find an expression for this time, thalf.
Mathematics
1 answer:
Airida [17]3 years ago
4 0
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

2 =   e^{ \alpha t} 

 \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




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dalvyx [7]

Answer:

13) (5x)^{-\frac{5}{4} ⇒ \frac{1}{\sqrt[4]{(5x)^5}}

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Step-by-step explanation:

Given expression:

13) (5x)^{-\frac{5}{4}

15) (10n)^{\frac{3}{2}

Write the expressions in radical form.

Solution:

For an expression with exponents as fraction like

(x)^{\frac{m}{n}

the numerator m represents the power it is raised to and the denominator n represents the nth root of the expression.

For an expression with exponents as negative  fraction like

(x)^{-\frac{m}{n}

We take the reciprocal of the term by rule for negative exponents.

So it is written as:

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using the above properties we can write the given expressions in radical form.

13) (5x)^{-\frac{5}{4}

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3 years ago
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irakobra [83]

We have that the total length of paper needed to cover all the tables is

X=288.5ft

From the question we are told that:

Number of tables n=36

Length of tablel=8

Extension e=3 inch =>0.25feet

Generally, the equation for the Total length  tables is mathematically given by

l_t=n*l\\\\l_t=36*8

l_t=288ft

Therefore with an extension of 0.25ft at both end we have the Total length of  tablecloths to be

X=l_t+2(e)\\\\X=288ft+2(0.25)

X=288.5ft

In conclusion

The total length of paper needed to cover all the tables is

X=288.5ft

For more information on this visit

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Answer:

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