He reorganized life into five kingdoms
Answer:
The Troposphere is the lowermost portion of Earth’s atmosphere. It is the densest layer of the atmosphere and contains approximately 75 percent of the mass of the atmosphere and almost all the water vapor and aerosol. The troposphere extends from the Earth’s surface up to the tropopause where the stratosphere begins.
Explanation:
Answer: I think Its the Height is 11.76 Meters (38.582677 Feet) between the bridge and the ground
Explanation: Supposing that where not counting air resistance in the equation, the equation
states that 1/2 multiplied by earths gravitational acceleration multiplied by the amount of time to reach the bottom: 2.4 seconds equals 11.76 meters of height between the bridge and the ground.
Answer:
a) The Energy added should be 484.438 MJ
b) The Kinetic Energy change is -484.438 MJ
c) The Potential Energy change is 968.907 MJ
Explanation:
Let 'm' be the mass of the satellite , 'M'(6×
be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×
N/m) be the universal constant of gravitation.
We know that the orbital velocity(v) for a satellite -
v=
[(R+h) is the distance of the satellite from the center of the earth ]
Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)
For initial conditions ,
h =
= 98 km = 98000 m
∴Initial Energy (
) =
m
+
Substituting v=
in the above equation and simplifying we get,
= 
Similarly for final condition,
h=
= 198km = 198000 m
∴Final Energy(
) = 
a) The energy that should be added should be the difference in the energy of initial and final states -
∴ ΔE =
- 
=
(
-
)
Substituting ,
M = 6 ×
kg
m = 1036 kg
G = 6.67 × 
R = 6400000 m
= 98000 m
= 198000 m
We get ,
ΔE = 484.438 MJ
b) Change in Kinetic Energy (ΔKE) =
m[
-
]
=
[
-
]
= -ΔE
= - 484.438 MJ
c) Change in Potential Energy (ΔPE) = GMm[
-
]
= 2ΔE
= 968.907 MJ
Answer:
Explanation:
Let initial extension in the spring= x₀
Force on the spring = F₀
Let spring constant = k
Fo = k x₀
Fn = 3k x₀
Fn /Fo = 3
PEs0 ( ORIGINAL) =1/2 k x₀²
PEsn ( NEW) =1/2 k (3x₀)²
PEsn / PEs0 = 9