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:
D. creates radioactive waste.
Explanation:
Nuclear energy can create nuclear radioactive waste
The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .
Net force would be towards the right and back (opposite direction of motion) since it's slowing down (decelerating) and turning right.
Answer:
V₁ = 6 V
, V₂ = V₃ = 3 V
Explanation:
To solve this circuit we must remember that there are two fundamental types of construction in series and parallel.
* a serial circuit there is only one path for current
in this circuit the constant current in the entire circuit and the voltage is the sum of the voltage of each term
* Parallel circuit in this there are two or more paths for the current
in this circuit the voltage is constant and the east is divided between each branch
with these principles let's analyze the proposed circuit
The DC battery is in parallel with resistor R1 and the equivalent of the other branch,
as in a parallel circuit the voltage is constant
V₁ = 6 V
in the other branch (23) it forms a series construction, where the current is constant
6 = iR₂ + iR₃
as they indicate that each resistance has the same value
6 = 2 iR
V = V₂ = V₃ = 3 V