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>
Explanation:
Consider the kinematic equation,
where x is the distance traveled, v is the initial velocity, a is the acceleration and t is time. By plugging in known values and solving for x,
through simple algebra we get
where this is the distance traveled in meters.
Answer:
Since F = G * m1 * m2 / r^2
F = 6.67 * 19E-11 * 2.79 * 9.47 * 10E23 / (1.2 * 10^7)^2
F = 126 * 10E-2 N = 1.22 N
