Answer:
Period of the signal.
Explanation:
So, this question is all about a concept in physics or astronomy which is called or known as Radiation Astronomy and Galactic Nuclei that are active. This concept talks most about Quasars; a powerful radiating object which derives its power from black holes.
When You take a look at Quasars, we get the to know that the more you think you can see, the more they move away from us.
Thus, when "You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the "PERIOD OF THE SIGNAL.
NB: not the amplitude but the period.
Answer:
The time constant is ![\tau = 1.265 s](https://tex.z-dn.net/?f=%5Ctau%20%20%3D%201.265%20s)
Explanation:
From the question we are told that
the time take to charge is ![t = 2.4 \ s](https://tex.z-dn.net/?f=t%20%3D%202.4%20%5C%20%20s)
The mathematically representation for voltage potential of a capacitor at different time is
![V = V_o - e^{-\frac{t}{\tau} }](https://tex.z-dn.net/?f=V%20%20%3D%20%20V_o%20%20-%20e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%20%7D)
Where
is the time constant
is the potential of the capacitor when it is full
So the capacitor potential will be 100% when it is full thus
100% = 1
and from the question we are told that the at the given time the potential of the capacitor is 85% = 0.85 of its final potential so
V = 0.85
Hence
![0.85 = 1 - e^{-\frac{2.4}{\tau } }](https://tex.z-dn.net/?f=0.85%20%3D%20%201%20-%20%20e%5E%7B-%5Cfrac%7B2.4%7D%7B%5Ctau%20%7D%20%7D)
![- {\frac{2.4}{\tau } } = ln0.15](https://tex.z-dn.net/?f=-%20%7B%5Cfrac%7B2.4%7D%7B%5Ctau%20%7D%20%7D%20%20%3D%20%20ln0.15)
![\tau = 1.265 s](https://tex.z-dn.net/?f=%5Ctau%20%20%3D%201.265%20s)
Answer:
80 ft/s
Explanation:
Use III equation of motion
V^2 = U^2 + 2g h
Here, U = 0, g = 32 ft/s^2, h = 100 ft
V^2 = 0 + 2 × 32 ×100
V^2 = 6400
V = 80 ft/s