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3 years ago

Help me on this question? log x-log 9=1

2 answers:

6 0

$\log { x } -\log { 9 } =1\\ \\ \log { \left( \frac { x }{ 9 } \right) } =1\\ \\ \log _{ 10 }{ \left( \frac { x }{ 9 } \right) } =1\\ \\ { 10 }^{ 1 }=\frac { x }{ 9 } \\ \\ 9\cdot 10=x\\ \\ x=90$

This is because:

$\log _{ a }{ \left( \frac { x }{ p } \right) } \\ \\ =\log _{ a }{ \left( \frac { { a }^{ m } }{ { a }^{ n } } \right) } \\ \\ =\log _{ a }{ \left( { a }^{ \left( m-n \right) } \right) } \\ \\ =\left( m-n \right) \cdot \log _{ a }{ a } \\ \\ =m-n\\ \\ =\log _{ a }{ x } -\log _{ a }{ p } \\$

Alika [10]3 years ago

4 0

   log(x) - log(9) = 1

Subtracting the logs of numbers gives you the log of
the quotient of the numbers.

   Log(x) - log(9) is the log of (x/9).

So the equation says: log (x/9) = 1

Raise 10 to the power of each side: 10^(log of x/9) = 10^1

But 10^(log of x/9) is x/9, and 10^1 is 10.

So x/9 = 10

   x = 90

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A quantum system has three states, with energies 0, 1.6 × 10-21, and 1.6 × 10-21, in Joules. It is coupled to an environment wit

xenn [34]

To develop the problem it is necessary to apply two concepts, the first is related to the calculation of average data and the second is the Boltzmann distribution.

Boltzmann distribution is a probability distribution or probability measure that gives the probability that a system will be in a certain state as a function of that state's energy and the temperature of the system. It is given by

$z = \sum\limit_i e^{-\frac{\epsilon_i}{K_0T}}$

Where,

$\epsilon_i =$ energy of that state

k = Boltzmann's constant

T = Temperature

With our values we have that

T= 250K

$k = 1.381*10^{23} m^2 kg s^{-2} K^{-1}$

$\epsilon_1=0J$

$\epsilon_2=1.6*10^{-21}J$

$\epsilon_3=1.6*10^{-21}J$

To make the calculations easier we can assume that the temperature and Boltzmann constant can be summarized as

$\beta = \frac{1}{kT}$

$\beta = \frac{1}{(1.381*10^{23} m^2)(250)}$

$\beta = 2.9*10^{20}J$

Therefore the average energy would be,

$\bar{\epsilon} =\frac{\sum \epsilon_i e^{-\beta \epsilon_i}}{\sum e^{-\beta \epsilon_i}}$

Replacing with our values we have

$\bar{\epsilon} = \frac{0e^{-0}+1.6*10^{-21}*e^{-\Beta(1.6*10^{-21})}+1.6*10^{-2-1}*e^{-(2.9*10^{20})(1.6*10^{-21})}}{1+2e^{-2.9*10^{20}*1.6*10^{-21}}}$

$\bar{\epsilon} = 0.9*10^{-22}J$

Therefore the average internal energy is $\bar{\epsilon} = 0.9*10^{-22}J$

3 0

2 years ago

Describe the transformation of mechanical energy in a ball when it is thrown straight up and then comes back down

zhenek [66]

I believe that the mechanical energy would transform from starting out as kinetic, the reaching the top it would be potential, then go back to kinetic as it is falling back down.

I'm not 100% sure that this is right but if I had to take a guess this is what I would say.

6 0

2 years ago

As the temperature of an object increases, the wavelength of the brightest light emitted _______

alexira [117]

Answer:

option (B) decreases

Explanation:

According to the Wein's displacement law, the minimum wavelength of the radiated emission is inversely proportional to the absolute temperature of the body which emits radiation.

$\lambda_{m}\alpha \frac{1}{T}$