A related type of beta decay actually decreases the atomic number of the nucleus when a proton becomes a neutron. Due to charge conservation, this type of beta decay involves the release of a charged particle called a “positron” that looks and acts like an electron but has a positive charge.
Answer:
Change in electric potential energy ∆E = 365.72 kJ
Explanation:
Electric potential energy can be defined mathematically as:
E = kq1q2/r ....1
k = coulomb's constant = 9.0×10^9 N m^2/C^2
q1 = charge 1 = -2.1C
q2 = charge 2 = -5.0C
∆r = change in distance between the charges
r1 = 420km = 420000m
r2 = 160km = 160000m
From equation 1
∆E = kq1q2 (1/r2 -1/r1) ......2
Substituting the given values
∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)
∆E = 94.5 × 10^9 (3.87 × 10^-6) J
∆E = 365.72 × 10^3 J
∆E = 365.72 kJ
Answer:
The forms of energy involved are
1. Kinetic energy
2. Potential energy
Explanation:
The system consists of a ball initially at rest. The ball is pulled down from its equilibrium position (this builds up its potential energy) and then released. The released ball oscillates due to a continuous transition between kinetic and potential energy.
