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
![M = -\frac{f_o}{f_e}](https://tex.z-dn.net/?f=M%20%3D%20-%5Cfrac%7Bf_o%7D%7Bf_e%7D)
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
![M = -\frac{f_o}{f_e}](https://tex.z-dn.net/?f=M%20%3D%20-%5Cfrac%7Bf_o%7D%7Bf_e%7D)
![M = -\frac{71}{2.1}](https://tex.z-dn.net/?f=M%20%3D%20-%5Cfrac%7B71%7D%7B2.1%7D)
![M = - 33.81](https://tex.z-dn.net/?f=M%20%3D%20-%2033.81)
Therefore the magnification of this astronomical telescope is -33.81
Your answer should be A, because every reaction has an equal opposite reaction
Answer:
Option B (1.51 m)
Explanation:
U 2 can help me by marking as brainliest........
Answer: 259.2 KJ
Explanation:
The formula calculate work don in a circuit is given by :-
, where Q is charge and V is the potential difference.
The formula to calculate charge in circuit :-
, where I is current and t is time.
Given : Current : ![I=3A](https://tex.z-dn.net/?f=I%3D3A)
Potential difference : ![V=12\ V](https://tex.z-dn.net/?f=V%3D12%5C%20V)
Time : ![t=2\ hr=2(3600)\text{ seconds}=7200\text{ seconds}](https://tex.z-dn.net/?f=t%3D2%5C%20hr%3D2%283600%29%5Ctext%7B%20seconds%7D%3D7200%5Ctext%7B%20seconds%7D)
Now, ![Q=3(7200)=21,600\ C](https://tex.z-dn.net/?f=Q%3D3%287200%29%3D21%2C600%5C%20C)
Then, ![W=(21600)(12)=259,200\text{ Joules}=259.2\text{ KJ}](https://tex.z-dn.net/?f=W%3D%2821600%29%2812%29%3D259%2C200%5Ctext%7B%20Joules%7D%3D259.2%5Ctext%7B%20KJ%7D)
Hence, the work done = 259.2 KJ