Answer:
According to the law of conservation of energy, energy cannot be created or destroyed, although it can be changed from one form to another. KE + PE = constant. A simple example involves a stationary car at the top of a hill. As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases. On the way back up the hill, the car converts kinetic energy to potential energy. In the absence of friction, the car should end up at the same height as it started.
This law had to be combined with the law of conservation of mass when it was determined that mass can be inter-converted with energy.
One can also imagine the energy transformation in a pendulum. When the ball is at the top of its swing, all of the pendulum’s energy is potential energy. When the ball is at the bottom of its swing, all of the pendulum’s energy is kinetic energy. The total energy of the ball stays the same but is continuously exchanged between kinetic and potential forms
Answer:
The observed wavelength on Earth from that hydrogen atom is
.
Explanation:
Given that,
The actual wavelength of the hydrogen atom, 
A hydrogen atom in a galaxy moving with a speed of, 
We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

is the observed wavelength

So, the observed wavelength on Earth from that hydrogen atom is
. Hence, this is the required solution.
Answer:
maybe its heat sorry if it's wrong
because if friction is not in the problem so we are making heat or thermal energy
The relationships can best be described as follows:
As frequency increases, wavelength decreases. <span>The greater the </span>energy<span>, the larger the frequency </span>and<span> the shorter (smaller) the </span>wavelength<span>. </span>
<span>a) wavelength vs. frequency = inversely proportional
b) wavelength vs. energy = inversely proportional
c) frequency vs. energy = directly proportional
Hope this answers the questions. Have a nice day. Feel free to ask more questions.</span>