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Firdavs [7]
4 years ago
7

If the electron e − and the positron e + both have rest energy 0.51 MeV, nd the minimum energy a particle of light γ (a photon)

can have in the reaction e + + e − = γ + γ. Why can't there be a reaction producing only one photon, e + + e − = γ? Assume the center-of-mass frame.
Physics
1 answer:
QveST [7]4 years ago
3 0

Answer:

For Total energy and momentum to be conserved, the minimum energy of the photons released is equal to twice the rest mass energy of an electron that is  2 \times 0.51 MeV = 1.02 MeV

The annihilation of electron -positron cannot produce a single photon. It is prohibited by the law of conservation of energy and momentum.

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Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci
Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})

h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}

2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

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2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0
4 years ago
Determine the magnitude and direction of the resultant velocity of 75.0 m/s. 25.0 east of north, and 100.0 m/s, 25.0 east of sou
aleksandr82 [10.1K]

Answer:

77.35 m / s

Ф = -17° from + X axis or 343° from + X axis

Explanation:

v1 = 75 m/s 25° east of north

v2 = 100 m/s  25° east of south

Write the velocities in vector form ,we get

\overrightarrow{v_{1}}=75\left ( Sin25\widehat{i} +Cos25\widehat{j}\right )=31.7\widehat{i}+67.97\widehat{j}

\overrightarrow{v_{1}}=100\left ( Sin25\widehat{i} -Cos25\widehat{j}\right )=42.26\widehat{i}-90.63\widehat{j}

Now add the velocity vectors to get the resultant of the velocities.

\overrightarrow{v}=\overrightarrow{v_{1}}+\overrightarrow{v_{2}}

\overrightarrow{v}=\left (31.7+42.26  \right )\widehat{i}+\left ( 67.97- 90.63 \right )\widehat{j}

\overrightarrow{v}=73.96\widehat{i}-22.66\widehat{j}

magnitude of resultant velocity is \sqrt{\left ( 73.96 \right )^{2}+\left ( -22.66 \right )^{2}}

  = 77.35 m / s

The direction is Ф from X axis

tan\phi =\frac{-22.66}{73.96}=-0.306

Ф = -17° from + X axis or 343° from + X axis

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