The waves shown below represent sound waves. Which of the waves would have the highest-pitched sound?

Calculate the energy (in eV/atom) for vacancy formation in some metal, M, given that the equilibrium number of vacancies at 296o

Schach [20]

Explanation:

The given data is as follows.

Temperature of metal = $296^{o}C$ = (296 + 273) K

= 569 K

Density of the metal = 8.85 $g/cm^{3}$ = $8.85 \times 10^{-6} g/m^{3}$ (as $1 cm^{3} = 10^{-6} m^{3}$ )

Atomic mass = 51.40 g/mol

Vacancies = $9.19 \times 10^{23} m^{-3}$

Formula to calculate the number of atomic sites is as follows.

n = $\frac{\rho \times N_{A}}{\text{atomic weight}}$

= $\frac{8.85 \times 10^{-6} \times 6.022 \times 10^{23}}{51.40 g/mol}$

= $1.036 \times 10^{17} atom/m^{3}$

Now, we will calculate the energy as follows.

E = $-KT \times ln (\frac{\text{no. of vacancies}}{\text{no. of atomic sites}})$

where, K = $8.62 \times 10^{-5}$

E = $-8.62 \times 10^{-5} \times 569 K \times ln (\frac{9.19 \times 10^{23}}{1.036 \times 10^{17} atom/m^{3}})$

= $78.46 eV/atom$

Therefore, we can conclude that energy (in eV/atom) for vacancy formation in given metal, M, is $78.46 eV/atom$ .

6 0

4 years ago

1. What was the potential energy of Truck 1 at the top of the hill? (3 points)

Gelneren [198K]

Answer:

incomplete data

Explanation:

no mass is given.

no vertical height is given

5 0

3 years ago

(1 point) A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 new

Nonamiya [84]

Answer:

x(t)=0.337sin((5.929t)

Explanation:

A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.

Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation

$m \frac{d^{2}x}{dt^{2}} +kx=0$

Definition of parameters

m=mass 3kg

k=force constant

e=extension ,m

ω =angular frequency

k=90/1.6=56.25N/m

ω^2=k/m= 56.25/1.6

ω^2=35.15625

ω=5.929

General solution will be

$x(t)=c1cos(ωt)+c2Sin(ωt)$

$x(t)=c1cos(5.929t)+c2Sin(5.929t)$

differentiating x(t)

dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)

when x(0)=0, gives c1=0

dx(t0)=2m/s gives c2=0.337

Therefore, the position of the mass after t seconds is

x(t)=0.337sin((5.929t)

6 0

3 years ago

What is the prmary source of energy inside of the earth

FromTheMoon [43]

The correct answer is the energy of the sun

8 0

2 years ago

Read 2 more answers

Each degree on the Kelvin scale equals how many degrees on the Celius scale

Mamont248 [21]

Answer:

0° C = 273.15 K

0 K = - 273.15° C

Explanation:

Hello there!

The Kelvin temperature scale refers to the absolute temperature scale. The division in the Kelvin scale is equal to a degree on the Celsius scale only it's difference is zero

The Celsius scale of zero (0°C) is the freezing point.

0 K is Absolute Zero.

•°• 0 K = - 273.15° C

0° C = 273.15 K

Furthermore, this is what I would like you to know about this scales;

The Kelvin scale is used for very low or very high temperatures where water is not involved.

Note: 

The boiling point of water is 100°C.

I hope this helps you to understand more on this scales. Have a nice studies.

8 0

2 years ago