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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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-450 m/s

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p₀ = p

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v = -450 m/s

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3 years ago
What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?
viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> V_{E} due to a point charge is expressed as:

V_{E}=k\frac{q}{r}

Where:

k=9(10)^{9}\frac{Nm^{2}}{C^{2}} is the <u>electric constant</u>

q=1.6(10)^{-19}C is the <u>electric charge of the hydrogen nucleus</u>, which is positive

r=5.292(10)^{-11}m is the <u>distance</u>

Rewritting the equation with the known values:

V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}

Finally:

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5 0
3 years ago
Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. T
alex41 [277]

Answer:

30643 J

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

t = Time taken = 1 ns

c = Speed of light = 3\times 10^8\ m/s

E_0 = Maximum electric field strength = 1.52\times 10^{11}\ V/m

A = Area = 1\ mm^2

Magnitude of magnetic field is given by

B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T

Intensity is given by

I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2

Power, intensity and time have the relation

E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J

The energy it delivers is 30643 J

4 0
3 years ago
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