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:
4500 N
Explanation:
When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:
a = v^2/r
where v is the velocity of the body and r is the radius of the circumference:
Therefore, a body with mass m, will feel a force f:
f = m v^2/r
Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:
fs = μN = μmg
The car will not slide if f = fs, i.e.
fs = μmg = m v^2/r
That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration
fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N
Answer:
noble gases are basically a group of gases that are similar in their chemical compounds, theres six of them : helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and the radioactive radon (Rn).
~batmans wife dun dun dun.....
Answer:
Part(a): The frequency is 
.
Part(b): The speed of the wave is 
.
Explanation:
Given:
The distance between the crests of the wave, 
.
The time required for the wave to laps against the pier, 
The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is 
.
Also, the time required for the wave for each laps is the time period of oscillation and it is given by 
.
Part(a):
The relation between the frequency and time period is given by
Substituting the value of 
in equation (1), we have
Part(b):
The relation between the velocity of a wave to its frequency is given by
Substituting the value of 
and 
in equation (2), we have
