The principal of uniformitarian states that
1 answer:
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Answer:
7.55 km/s
Explanation:
The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:
where
is the gravitational constant
is the mass of the telescope
is the mass of the Earth
is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)
v = ? is the orbital velocity of the Hubble telescope
Re-arranging the equation and substituting numbers, we find the orbital velocity:
Answer:
h = 3.3 m (Look at the explanation below, please)
Explanation:
This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy
Kinetic energy = m
Plug in the numbers = (4.0)(
)
Solve = 2(64) = 128 J
Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.
Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)
Kinetic energy = Potential Energy
128 J = 39.2h
h = 3.26 m
h= 3.3 m (because of significant figures)
