calculate the momentum of 5.07 kg egg that is dropped from a roof and falls 2 seconds before hitting the ground.
1 answer:
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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