Answer:
For a wave, the <em>HIGHER </em>the amplitude, the <em>MORE </em>energy the wave carries.
The best system to talk about would be a galaxy system. Energy does not enter the galaxy, but it does "recycle" its energy. For instance, when the life of a star comes to its end, it can go super nova. and all the energy from that star is then released back into the galaxy to form nebula's and then eventually into other stars. The energy inside of a galaxy can change frequently. It can be in the form of heat from a star, or it can change into gamma radiation from an explosion. Gases like helium and hydrogen come together and form a ball of gas creating the heat. Then the heat is dispersed leaving different types of radiation like gamma, ultra-violet, microwave, and infrared. Energy can leave a system by the local black holes. Black holes with shoot out Hawking radiation is when the black hole disperses its own energy out into space, also known as Black Hole Evaporation<span>. The energy from that black hole is then dispersed into the rest of the universe or possibly back into the galaxy from which it came from. </span>
Answer:
The exploitation of high-value natural resources—oil, gas, minerals, and timber—has often been a key factor in triggering, escalating, or sustaining violent conflicts around the globe. Competition over renewable resources such as land and water is on the rise, and environmental degradation, population growth, and climate change are compounding the challenges. Governments are therefore under increasing pressure to sustainably manage natural resources and resolve conflicts around their ownership, management, allocation, and control.
You could do something about Einstein's theory of energy... it seems like it would be a simple and easy project. E=mc^2
Answer:
a) v = 4.64 m / s
, b) t = 0.947 s
, c) t = 0.947 s
Explanation:
We will work on this exercise with vertical launch kinematics, let's start by calculating the height of the jumper in the SI system
y₀ = 5 ’(0.3048 m / 1’) + 7 ”(2.54 10-2 m / 1”) = 1.70 m
The distance they give is the height of the jump
y = 1.10 m
Let's use energy conservation
Starting point. On the floor
Em₀ = K = ½ m v²
Final point. Maximum height
Em_{f} = U = m g y
Em₀ = 
½ m v² = m g y
v = √2gy
Let's calculate
v = √(2 9.8 1.10)
v = 4.64 m / s
b) Air time is the time to go up plus the time to go down, which is the same
For maximum height the speed is zero
v = v₀ - g t₁
t₁ = v₀ / g
t₁ = 4.64 /9.8
t₁ = 0.4735 s
The total time is
t = 2 t₁
t = 2 0.4735
t = 0.947 s
c) if it takes a distance of 0.40 to reach speed, what is the acceleration, as it stands on the floor its initial speed is zero
v² = v₀² + 2 a x
a = v² / 2x
a = 4.64²/2 0.40
a = 26.9 m / s²
