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3 years ago

Which of the following processes forms clouds A. Boiling B.Condensation C.Freezing D.Precipitation

Physics

2 answers:

love history [14]3 years ago

6 0

Answer:

Explanation:

Anna007 [38]3 years ago

5 0

Answer:

Explanation:

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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

Answer: The change in entropy of the given process is 1324.8 J/K

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

where,

$\Delta S$ = Entropy change

$C_{p,m}$ = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

$\Delta S$ = Entropy change

m = mass of ice

$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

For process 1:

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

For process 2:

We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

For process 3:

We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

For process 4:

We are given:

$m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K$

For process 5:

We are given:

$m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K$

Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

Total entropy change for the process = $\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Total entropy change for the process = $[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K$

Hence, the change in entropy of the given process is 1324.8 J/K

4 0

3 years ago

Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.

Marina86 [1]

Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5
Frequency f = 2 Hz

3 0

3 years ago

At room temperature, glass used in windows actually has some properties of a liquid. it has a very slow, viscous flow. (visco

tamaranim1 [39]

The fahrenheit temperature is 927965. It is calculated using the formula 515515 Degree Cx1.8+32=927965. The degree celcius and fahrenheit are two units two measure temperature. If the value is given in celcius it can be converted into fahrenheit using the above formula.

8 0

3 years ago

4. How much time does it take for a student running at a speed of 5 m/s to cover a distance of 2,000 m?

Monica [59]

Answer:

6 Minutes 40 Seconds or 400 Seconds

Explanation:

Time to cover a distance of 5m = 1 Second

Time to cover a distance of 2000m = 2000÷5

= 400 Seconds

After converting 400 Seconds into minutes it will become 6 minutes 40 seconds.

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6 0

3 years ago