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densk [106]
3 years ago
8

The number of bacteria in a certain population increases according to a continuous exponential growth model, with a growth rate

parameter of 2.4% per hour. How many hours does it take for the size of the sample to double?
Physics
1 answer:
sveta [45]3 years ago
6 0

Answer:

It would take approximately 289 hours for the population to double

Explanation:

Recall the expression for the continuous exponential growth of a population:

N(t)=N_0\,e^{kt}

where N(t) measures the number of individuals, No is the original population, "k" is the percent rate of growth, and "t" is the time elapsed.

In our case, we don't know No (original population, but know that we want it to double in a certain elapsed "t". We also have in mind that the percent rate "k" would be expressed in mathematical form as: 0.0024 (mathematical form of the given percent growth rate).

So we need to solve for "t" in the following equation:

2\,N_0=N_0\,e^{0.0024\,t}\\\frac{2\,N_0}{N_0} =e^{0.0024\,t}\\2=e^{0.0024\,t}\\ln(2)=0.0024\,t\\t=\frac{ln(2)}{0.0024} \\t=288.811\,\, hours

Which can be rounded to about 289 hours

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391.12 * 10^{16} mass is required for activity of 315 ci

The activity of the radioactive substance is the number of disintegrations taking place in a radioactive sample.

Isotopes are the members of an element having the same number of protons but the different number of neutrons.

the activity of the radioactive sample is given as:

A = λ * N

where λ is the decay constant and N is the number of disintegrations.

activity is given as 315 ci

A = 315 * 3.7 * 10^{10}

   = 1165.5 *  10^{10} Bq

λ = 0.693 / t_{1/2}

   = 0.693 / 2.69 * 24 * 3600

   = 2.98 * 10^{-6} t^-1

Therefore, N = A /  λ

                      = 1165.5 *  10^{10} /  2.98 * 10^{-6}

                  N = 391.12 * 10^{16}

391.12 * 10^{16} mass is required for activity of 315 ci

For more information click on the link below:

brainly.com/question/25746629

#SPJ4

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