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olga nikolaevna [1]
2 years ago
14

The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, thoug

h the nature of the prediction was not fully understood until the experimental discovery. Today, it is well accepted that all fundamental particles have antiparticles.
Required:
Suppose that an electron and a positron collide head-on. Both have kinetic energy of 3.58 MeV and rest energy of 0.511 MeV. They produce two photons, which by conservation of momentum must have equal energy and move in opposite directions. What is the energy Ephoton of one of these photons?
Physics
1 answer:
Taya2010 [7]2 years ago
8 0

Answer:

Energy of Photon = 4.091 MeV

Explanation:

From the conservation of energy principle, we know that total energy of the system must remain conserved. So, the energy or particles before collision must be equal to the energy of photons after collision.

K.E OF electron + Rest Energy of electron + K.E of positron + Rest Energy of positron = 2(Energy of Photon)

where,

K.E OF electron = 3.58 MeV

Rest Energy of electron = 0.511 MeV

Rest Energy of positron = 0.511 MeV

K.E OF positron = 3.58 MeV

Energy of Photon = ?

Therefore,

3.58 MeV + 0.511 MeV + 3.58 MeV + 0.511 MeV = 2(Energy of Photon)

Energy of Photon = 8.182 MeV/2

<u>Energy of Photon = 4.091 MeV</u>

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Two resistors R1 = 3 Ω and R2 = 6 Ω are connected in parallel. What is the net resistance in the circuit?​
gtnhenbr [62]

Answer:

"2Ω" is the net resistance in the circuit.

Explanation:

The given resistors are:

R1 = 3Ω

R2 = 6Ω

The net resistance will be:

⇒  \frac{1}{R_{net}} =\frac{1}{R_1} +\frac{1}{R_2}

On substituting the values, we get

⇒  \frac{1}{R_{net}} =\frac{1}{3} +\frac{1}{6}

On taking L.C.M, we get

⇒  \frac{1}{R_{net}} =\frac{2+1}{6}

⇒  \frac{1}{R_{net}} =\frac{3}{6}

⇒  \frac{1}{R_{net}} =\frac{1}{2}

On applying cross-multiplication, we get

⇒ R_{net}=2 \Omega

3 0
2 years ago
If the speed of a wave is 1500m/sec and its frequency is 200 Hz, what is its wavelength
Ray Of Light [21]

Answer:

The wavelength of wave is 7.5 meter.

Given:

Speed of wave = 1500 \frac{m}{s}

Frequency of wave = 200 Hz

To find:

Wavelength of wave = ?

Formula used:

\lambda = \frac{v}{n}

Where \lambda = wavelength of the wave

v = speed of wave

n = frequency of wave

Solution:

Wavelength of wave is given by,

\lambda = \frac{v}{n}

Where \lambda = wavelength of the wave

v = speed of wave

n = frequency of wave

\lambda = \frac{1500}{200}

\lambda = 7.5 m

The wavelength of wave is 7.5 meter.

4 0
3 years ago
Read the scenario and solve these two problems.
Burka [1]

Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy K of an object is given by:

K=\frac{1}{2}mV^{2}

Where:

m=12000 kg is the mass of the train

V=30 m/s is the speed of the train

Solving the equation:

K=\frac{1}{2}(12000 kg)(30 m/s)^{2}

K=5400000 J This is the train's kinetic energy at its top speed

b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

E_{i}=E_{f}

K_{i}+P_{i}=K_{f}+P_{f}

Where:

K_{i}=5400000 J is the train's initial kinetic energy

P_{i}=0 J is the train's initial potential energy

K_{f}=0 J is the train's final kinetic energy

P_{f}=mgh is the train's final potential energy, where g=9.8 m/s^{2} is the acceleration due gravity and h is the height.

Rewriting the equation with the given values:

5400000 J=(12000 kg)(9.8 m/s^{2})h

Finding h:

h=45.918 m \approx 45.92 m

7 0
2 years ago
Read 2 more answers
A cubic sample of a new kind of artificial tissue is subject to an increase in pressure of 160 kPa which results in a reduction
grin007 [14]

Answer:

0.82 MPa

Explanation:

the change in pressure 'σ'=160kPa

K= σ/∈Ф_v => σ/3∈Ф_L

K= 160/(3 x 0.065)

K=820 kPA=0.82 MPa

Thus,the bulk modulus of the tissue 'K' is 0.82 MPa

3 0
2 years ago
Porque deja de funcionar una estufa electrica cuando la electricidad es excesiva??
Brums [2.3K]
<h3>Porque deja de funcionar una estufa electrica cuando la electricidad es excesiva?</h3>

  • Es normal que los quemadores superiores de una estufa eléctrica o una estufa se enciendan y apaguen en configuraciones distintas a Hi. El quemador se encenderá y apagará más de lo normal cuando se utilizan cacerolas que no son planas o que son del tamaño incorrecto para el quemador.

<h2>☆彡Hanna</h2>

#CarryOnLearning

7 0
2 years ago
Read 2 more answers
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