Types of energy transfer involve
Conduction, Convection, Evaporation,and Condensation <span />
-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.
-- The mass of the satellite makes no difference.
Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is
(3.95 x 10⁵m) + (4.2 x 10⁶m)
= (3.95 x 10⁵m) + (42 x 10⁵m)
= 45.95 x 10⁵ m
The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.
The bird completes a revolution every 2.0 hours,
so its speed in orbit is
(91.9 π x 10⁵ m) / 2 hr
= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)
= 0.04 x 10⁵ m/sec
= 4 x 10³ m/sec
(4 kilometers per second)
initial velocity (u)=0m/s
final velocity (v)=10m/s
time( t)=5s
acceleration (a)=v-u÷t
acceleration (a)=10-0÷5
acceleration (a)=10÷5
acceleration (a)=2
therefore acceleration (a)=2m/s
Answer:
The magnifying power of this telescope is (-60).
Explanation:
Given that,
The focal length of the objective lens of an astronomical telescope, 
The focal length of the eyepiece lens of an astronomical telescope, 
To find,
The magnifying power of this telescope.
Solution,
The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :
m = -60
So, the magnifying power of this telescope is 60. Therefore, this is the required solution.
