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eduard
3 years ago
15

Wolfrich lived in Portugal and Brazil for a total period of 141414 months in order to learn Portuguese. He learned an average of

130130130 new words per month when he lived in Portugal and an average of 150150150 new words per month when he lived in Brazil. In total, he learned 192019201920 new words. How long did Wolfrich live in Portugal, and how long did he live in Brazil?
Mathematics
1 answer:
WINSTONCH [101]3 years ago
4 0

Answer:

Wolfrich lived in Portugal 9 months and 5 months in Brazil.

Step-by-step explanation:

The question is wrong. This are the corrections for the numbers :

141414 months is actually 14 months.

130130130 new words per month is actually 130 new words per month.

150150150 new words per month is actually 150 new words per month.

192019201920 new words is actually 1920 new words.

In order to solve this exercise, we are going to make a system of equations defining the following variables :

X= Months he lived in Portugal.

Y= Months he lived in Brazil.

We know that Wolfrich lived in Portugal and Brazil for a total period of 14 months in order to learn Portuguese. Therefore, the first equation of our system of equations will be :

X+Y=14 (I)

He learned an average of 130 new words per month when he lived in Portugal and an average of 150 new words per month when he lived in Brazil. In total, he learned 1920 new words. Therefore, our second and final equation of our system of equations will be :

130X+150Y=1920 (II)

With (I) and (II) we form our linear equation system :

\left \{ {{X+Y=14} \atop {130X+150Y=1920}} \right.

From equation (I) we can write

X+Y=14

X=14-Y (III)

If we use (III) in (II) :

130(14-Y)+150Y=1920

1820-130Y+150Y=1920

20Y=100

Y=5

If we replace this information in (I) :

X+5=14

X=14-5

X=9

We find that Wolfrich lived 9 months in Portugal and 5 months in Brazil. One way to verifies the answer is to replace the values of X and Y that we found  in the equations (I) and (II) :

X+Y=14\\9+5=14\\14=14

And finally

130X+150Y=1920\\(130).(9)+(150).(5)=1920\\1170+750=1920\\1920=1920

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evaluate the fermi function for an energy KT above the fermi energy. find the temperature at which there is a 1% probability tha
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Complete Question

Evaluate the Fermi function for an energy kT above the Fermi energy. Find the temperature at which there is a 1% probability that a state, with an energy 0.5 eV above the Fermi energy, will be occupied by an electron.

Answer:

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b

The temperature is  T_k  =  1261 \  K

Step-by-step explanation:

From the question we are told that

   The energy considered is  E = 0.5 eV

Generally the Fermi  function is mathematically represented as

       F(E_o) =  \frac{1}{e^{\frac{[E_o - E_F]}{KT} } + 1 }

    Here K is the Boltzmann constant with value k = 1.380649 *10^{-23} J/K

            E_F  is the Fermi energy

            E_o  is the initial energy level which is mathematically represented as

     E_o = E_F + KT

So

     F(E_o) =  \frac{1}{e^{\frac{[[E_F + KT] - E_F]}{KT} } + 1}

=>   F(E_o) =  \frac{1}{e^{\frac{KT}{KT} } + 1}

=>   F(E_o) =  \frac{1}{e^{ 1 } + 1}

=>   F(E_o) =  0.2689

Generally the probability that a state, with an energy 0.5 eV above the Fermi energy, will be occupied by an electron is mathematically represented by the  Fermi  function as

     F(E_k) =  \frac{1}{e^{\frac{[E_k - E_F]}{KT_k} } + 1 }  = 0.01

HereE_k is that energy level that is  0.5 ev above the Fermi energy  E_k = 0.5 eV  + E_F

=>   F(E_k) =  \frac{1}{e^{\frac{[[0.50 eV + E_F] - E_F]}{KT_k} } + 1 }  = 0.01

=>   \frac{1}{e^{\frac{0.50 eV ]}{KT_k} } + 1 }  = 0.01

=>   1 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} } + 0.01

=>   0.99 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} }

=>   e^{\frac{0.50 eV ]}{KT_k} }  = 99

Taking natural  log of both sides

=>   \frac{0.50 eV }{KT_k} }  =4.5951

=>    0.50 eV   =4.5951 *  K *  T_k

Note eV is electron volt and the equivalence in Joule is     eV  =  1.60 *10^{-19} \  J

So

     0.50 * 1.60 *10^{-19 }   =4.5951 *  1.380649 *10^{-23} *  T_k

=>   T_k  =  1261 \  K

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3 years ago
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