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>
Thank you for posting your question here at brainly. But your question seems incomplete. I will assume you based the situation below:
<span>An electrons moves at 2.0x10^6 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.4x10^-2 T.
The </span> largest possible magnitude of the acceleration of the electron due to the magnetic field is <span>= 2.6 × 10 ¹⁶ m / s ²</span>
Answer:
1700 kg
Explanation:
Let’s use conservation of momentum
32.5 * 388 = 7.42 * mc
mc = 1699.46
mc = 1700 kg
a = 7.8 m/s^2
Explanation:
Let Fnet = net force = ma
m = mass of the skydiver
a = acceleration caused by Fnet
W = weight = mg
f(air) = frictional force due to air resistance
Fnet = W - f(air)
= (100 kg)(9.8 m/s^2) - (200 N)
= 780 N
Therefore, the acceleration of the skydiver due to Fnet is
a = Fnet/m
= (780 N)/(100 kg)
= 7.8 m/s^2
