High energy waves have Gamma rays
Answer:
B) Within an atom, an electron can have only particular energies.
Explanation:
As we know that electrons have energy but apart from electrons we know that protons and neutrons inside the nucleus of atom will also have energy in them.
rest all the statements are true as we have
A) Electrons orbit the nucleus rather like planets orbiting the Sun.
TRUE, because electrons can move in stationary orbit around the nucleus
C) Electrons can jump between energy levels in an atom only if they receive or give up an amount of energy equal to the difference in energy between the energy levels.
Difference amount of energy is lost or absorbed by the electron in form of photons
D) An electron has a negative electrical charge.
Charge of an electron is given as 
E) Electrons have very little mass compared to protons or neutrons
Mass of an electron is given as

mass of proton or neutron

Answer:
EP = 49.05Joules (J)
Explanation:
The equation for Potential energy (EP) is
EP = m g h
We are given the values below (do convert them into SI units)
m = 0.0025kg
h = 2000m
g = 9.81m/
Substitute the values into the equation and solve for EP
EP = 0.0025 * 2000 * 9.81
EP = 49.05Joules (J)
Answer:
48.16 %
Explanation:
coefficient of restitution = 0.72
let the incoming speed be = u
let the outgoing speed be = v
kinetic energy = 0.5 x mass x 
- incoming kinetic energy = 0.5 x m x
- coefficient of restitution =

0.72 =
v = 0.72u
therefore the outgoing kinetic energy = 0.5 x m x 
outgoing kinetic energy = 0.5 x m x 
outgoing kinetic energy = 0.5184 (0.5 x m x
)
recall that 0.5 x m x
is our incoming kinetic energy, therefore
outgoing kinetic energy = 0.5184 x (incoming kinetic energy)
from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.
The energy lost would be 100 - 51.84 = 48.16 %