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Natalka [10]
2 years ago
11

Six - two times twelve plus five

Mathematics
1 answer:
fomenos2 years ago
6 0

Answer:

Ok, so the equation would be 6-2*12+5. The answer to that is -13.

Step-by-Step Explanation:

You would use PEMDAS. First, you multiply 2 and 12 and that's 24. Now your equation is 6-24+5. Then you subtract 6 and 24 and that's -18. Now your equation is -18+5, and the answer to that equation is -13. I hope this helps you!!

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Group technology requires that:
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Group technology is a technique in manufacturing that involves bringing together parts with similar process of manufacturing or function into the same location so that they can all go through the same process. This follows a general saying that problems with similar groups could possibly be solved using a single or similar solution, which could help save time, resources and efforts.

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3 years ago
Pls help
Andru [333]

Answer:

Step-by-step explanation:

K=2J-3

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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
dybincka [34]

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:

a

The Fermi function for the energy KT is  F(E_o) =  0.2689

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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