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:
3) Ep = 13243.5[J]
4) v = 17.15 [m/s]
Explanation:
3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.
m = mass = 90 [kg]
h = elevation = 15 [m]
Potential energy is defined as the product of mass by gravity by height.
![E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%3D90%2A9.81%2A15%5C%5CE_%7Bp%7D%3D13243.5%5BJ%5D)
This energy will be transformed into kinetic energy.
Ek = 13243.5 [J]
4) The velocity can be determined by defining the kinetic energy, as shown below.
![E_{k}=\frac{1}{2} *m*v^{2} \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%20%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2%2AE_%7Bk%7D%20%7D%7Bm%7D%20%7D%5C%5C%20v%3D%20%5Csqrt%7B%5Cfrac%7B2%2A13243.5%20%7D%7B90%7D%20%7D%5C%5Cv%3D17.15%5Bm%2Fs%5D)
The period is 1/5 second, and the frequency is 5 Hz.
Explanation:
Let's start by reviewing some definitions about waves:
- The period of a wave is the time taken for the wave to make one complete cycle. It is indicated with T and it is measured in seconds (s)
- The frequency of a wave is the number of complete cycles made by the wave in one second. It is indicated with f and it is measured in Hertz (Hz). It is equal to the reciprocal of the period:
where f is the frequency and T is the period.
In this problem, we have a wave that passes a given point five time per second. This means that the number of oscillations per second is five, and so its frequency is:
It follows than the period of the wave is:
Therefore, the correct statement is
The period is 1/5 second, and the frequency is 5 Hz.
Learn more about waves, frequency and period:
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#LearnwithBrainly
60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg
Now,new weight with g = 6m/s^2
=m×g' (here g' is new acceleration of the new planet)
= 10×6=60N
