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1 year ago

6

where again p is the phonon momentum, E is the photon energy and c is the speed of light. When you divide the photon energy foun

d in #6 by the photon momentum found in #4, do you get the speed of light? (If not, check your work for questions #4 through #6).

1 answer:

creativ13 [48]1 year ago

8 0

Answer:

Yes

Explanation:

p = momentum of photon

E = energy of photon

c = velocity of light

Units of p = kg m /s

Units of E = kg m^2 / s^2

Units of E / p = {kg m^2 / s^2} / {kg m /s} = m/s

It is the unit of speed, so by the division of energy to the momentum, we get the speed. yes it is correct.

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the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point

Andreyy89

Answer:

The pressure at this point is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

$P=1.81\ kW$

We need to calculate the pressure at this point

Using formula of pressure

$P=F\times v$

$P=A_{1}P_{1}\times v_{1}$

Here, $A_{1}v_{1}=Q_{1}$

Where, v = velocity

Put the value into the formula

$1.81 =P_{1}\times0.124\times\dfrac{1}{60}$

$P_{1}=\dfrac{1.81\times60}{0.124}$

$P_{1}=875.80\ kPa$

$P_{1}=0.875\ mPa$

Hence, The pressure at this point is 0.875 mPa

5 0

1 year ago

Estimate the change in the equilibrium melting point of copper caused by a change in pressure of 10 kbar. The molar volume of co

mr Goodwill [35]

Answer:

The change in the equilibrium melting point is 4.162 K.

Explanation:

Given that,

Pressure = 10 kbar

Molar volume of copper $V=8.0\times10^{-6}\ m^3$

Volume of liquid $V=7.6\times10^{-6}\ m^3$

Latent heat of fusion $L= 13.05 kJ$

Melting point =1085°C

We need to calculate the change temperature

Using Clapeyron equation

$\dfrac{\Delta P}{\Delta T}=\dfrac{\Delta H}{T\Delta V}$

Put the value into the formula

$\dfrac{1000\times10^{5}}{\Delta T}=\dfrac{13050}{(1085+273)\times(8.0-7.6)\times10^{-6}}$

$\Delta T=\dfrac{1000\times10^{-5}\times(1085+273)\times(8.0-7.6)\times10^{-6}}{13050}$

$\Delta T=4.162\ K$

Hence, The change in the equilibrium melting point is 4.162 K.

5 0

1 year ago

Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley

kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = $\frac{4m}{2} x^2 + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2$

T = $4mx^2 + 2my^2 -2mxy$

also the general potential energy can be expressed as

U = $-4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant$

The Lagrangian of the problem can now be setup as

$L =4mx^2 +2my^2 -2mxy +2mgy + constant$

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange equation = $4x-y =0\\-2y+x +g = 0$

solving the equations simultaneously

x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

8 0

1 year ago

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

2 years ago

for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha

VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

$R=\dfrac{u^2\ \sin2\theta}{g}$

Here,

u is the speed of launch of projectile

Here, $\theta=30^{\circ}$

We need to find the other launch angle when the projectile have the same range, such that,

$\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}$

$\sin(60)=\sin2\alpha$

$\dfrac{\sqrt3}{2}=\sin2\alpha$

$\alpha =30^{\circ}$

So, the other angle is 30 degrees. Hence, this is the required solution.

3 0

2 years ago