What is the magnification of an astronomical telescope whose objective lens has a focal length of 71 cm and whose eyepiece has a
1 answer:
To solve this problem it is necessary to apply the concepts related to optical magnification (is the process of enlarging the apparent size, not physical size, of something.). Specifically the angular magnification of an optical telescope is given by
Where,
Focal length of the objective lens in a refractor
Focal length of the eyepiece
Our values are given as
71cm
2.1cm
Replacing we have
Therefore the magnification of this astronomical telescope is -33.81
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a) Potential energy: 147 m [J]
The gravitational potential energy of an object is given by
where
m is its mass
is the acceleration of gravity
h is the height of the object above the ground
In this problem,
h = 15 m
We call 'm' the mass of the ball, since we don't know it
So, the potential energy of the ball at the top of the hill is
(J)
b) Velocity of the ball at the bottom of the hill: 17.1 m/s
According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:
where
v is the final velocity of the ball
We know from part a) that
U = 147 m
Substituting into the equation above,
And re-arranging for v, we find the velocity: