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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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Object A is 71 degrees and object B is 75 degrees how will thermal energy flow
Tasya [4]

Given :

Object A is 71 degrees and object B is 75 degrees .

To Find :

How will thermal energy flow.

Solution :

We know, by law of thermodynamics thermal energy will flow from higher temperature to lower temperature.

So, in the given question energy will flow from object B from object A.

Hence, this is the required solution.

3 0
2 years ago
A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like prot
Rzqust [24]

Answer:

595391.482946 m/s

3.21875\times 10^{6}

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

e = Charge of electron = 1.6\times 10^{-19}\ C

t = Time taken = 1 second

m = Mass of proton = 1.67\times 10^{-27}\ kg

Velocity of proton is given by

v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s

The speed of the proton is 595391.482946 m/s

Current is given by

I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C

Number of protons is

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons

The number of protons is 3.21875\times 10^{6}

3 0
3 years ago
Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac
Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h#The speed toward each other.

\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0
2 years ago
What is the kinetic energy of an automobile with a mass of 1252 kg traveling at a speed of 12 m/s?
torisob [31]
KE = 1/ 2 * 1252 * 144
 as  KE = 1/2 * m * v ^2
 = 90144 J



4 0
3 years ago
How do you find the capacitance in this?
Lostsunrise [7]

Answer:

Explanation:

parallel capacitances add directly

Series capacitances add by reciprocal of sum of reciprocals.

Ceq = [ C ] + [1 / (1/C + 1/C)] + [1 / (1/C + 1/C + 1/C)]

Ceq = [ C ] + [C / 2] + [C / 3]

Ceq = [ 6C/6 ] + [3C / 6] + [2C / 6]

Ceq = 11C/6

3 0
2 years ago
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