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2 years ago

15

Alex is building a circular base for a fountain. The radius of the circle is 4ft. what is the area of the base

2 answers:

sesenic [268]2 years ago

6 0

Area=
$\pi \times r^{2}$
$\pi \times 4^{2}$
$\pi \times 16$

50.2654824574.....

lions [1.4K]2 years ago

3 0

My teacher taught us a song for this. It's kind of my favorite thing. It's called the circle song if you ever want to look it up.

$\pi r^{2}$ is the formula. Therefore:

$\pi 4^{2} = 16 \pi$

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for a project in her geometry class, amira uses a mirror on the ground to measure the height of her school’s football goalpost.

kari74 [83]

Answer:

6.14 m

Step-by-step explanation:

8 0

2 years ago

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

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

Here $E_k$ 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

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

=> $T_k = 1261 \ K$

7 0

2 years ago

Is black......blue or red???

Len [333]

Black is black.
Blue is blue.
Red is red.

6 0

3 years ago

Read 2 more answers

When the price is $ 40 per radio, a producer will supply 90 radios each month for sale. For each $4 increase in price the produc

SVEN [57.7K]

Price for a radio = $ 40 for supply 90 radios in a month.

Ratio of increase in price to the number of radios supplies in a month 4:8 that is 1:2.

New unit price given = $76.

Difference in unit price given in starting and New unit price = 76-40=$36

So, increase in unit price = $36.

Let us assume number of supplied increased = x radios for increase in unit price $36.

So, we can setup a perportion as,

1/2 = 36/x.

On cross multiplication, we get

1*x = 2*36

x= 72.

So, 72 is the increased number of radios in 90 radios.

That is 72+90 = 162 radios each month.

Therefore, 162 radios are supplied if their per unit price is $76.

5 0

2 years ago

Can someone help me with my extra credit? It would be greatly appreciated.

topjm [15]

Answer: I think the answer you need is g. tell me if its wrong though.

4 0

2 years ago