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:
No, they will not change.
Explanation:
Answer:
D...............................
Answer:
The magnification of an astronomical telescope is -30.83.
Explanation:
The expression for the magnification of an astronomical telescope is as follows;

Here, M is the magnification of an astronomical telescope,
is the focal length of the eyepiece lens and
is the focal length of the objective lens.
It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.
Put
and
in the above expression.

M=-30.83
Therefore, the magnification of an astronomical telescope is -30.83.