The absolute simplest way to explain E=MC^2 is: a lot of energy from a little mass.
Hello!
For the explanation of this energy conservation exercise, where we'll use <u>energy conservation law</u>, let's see what this principle proposes.
How you should know, mechanical energy conserves in every point, that is to say mechanical energy is same in A point like B point. (Mechanical energy will be represented by "Me")
Once time we know that, let's take the 220 Joules momentum like A point, and when 55 Joules momentum like B point.
Then, let's use the <u>energy conservation principle:</u>
Me(A) = Me(B)
- We know Mechanical energy in A point, so just lets replace according to our data:
220 J = Me(B)
- In B point, we know kinetic energy, but <u>we dont know gravitational potential energy</u>, so lets descompose Mechanical energy, into kinetic energy and gravitational potential energy:
220 J = Ke + Gpe
- We know kinetic energy value, so lets replace it:
220 J = 55 J + Gpe
- Finally, just clean Gpe and resolve it:
Gpe = 220 J - 55 J = 165 J
Gravitational potential energy is of One hundred sixty five Joules <u>(165 J).</u>
║Sincerely, ChizuruChan║
Distance is defined as the space between two points in space.
<h3>What is distance?</h3>
Distance is defined as the space between two points in space.
The unit of distance is metres or kilometres.
Distance can be measured directly usind instruments such as metre rule and tape rule.
The formula for calculating distance is given below:
Distnce = velocity × time.
Therefore, distance is defined as the space between two points in space.
Learn more about about distance at: brainly.com/question/4931057
Answer:
both mass take equal time to reach at bottom
Explanation:
As both masses M 1 and M 2 are situated at same height so that the potential energy at top is transformed to kinetic energy at the bottom by applying principle of energy conservation.
PE = KE
Mgh = (1/2)(M)(V^2)
V^2 = 2gh
V = (2gh)^(1/2)
So velocity is independent of mass
For M1
(M_1)(g)(h) = (1/2)(M1)(V^2)
V^2 = 2gh
V = (2gh)^(1/2)
For M2
(M_2)(g)(h) = (1/2)(M2)(V^2)
V^2 = 2gh
V = (2gh)^(1/2)
therefoer speed at bottom too are independent of mass
So we can say both speed are independent of therir respective mass
hence both take equal time to reach at bottom
