Answer:
the final potential energy of this system is 3U0/10
Explanation:
We are given
charge at left end and another test charge at point p
Potential energy is given by = 
where k is electrostatics constant = 
Q1 = first charge , Q2= test charge
R= distance between charges
potential at point p
U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1
now the test charge moves to point R
using Pytahgoreou theorem
R(distance) = 
= 10
New Potential energy
U1 = kq1*q2 / 10
substituting kq1q2 = 3U0 from 1
U1 = 3U0/10
So this is the final potential energy of this system.
Answer:
passive voice
many messengers all over the world was sent by emperor Ashoka to preach Buddhism.
Which pen was liked by you?
Has your passport size photo been taken for the application form?
A beautiful bicycle was given to me on my birthday by my father.
The plants is being watered by the gardener.
Paul said that he will never leave you and he will always be with you.
Impulse = Force * time
Impulse = 500N *0.5 s =250 N*s
The part of a river that would have animals with muscular bodies and adaptations that let survive in turbulent water is in the transition zone, the mid-transition zone to be precise.
Water at the source zone possesses a lot of potential energy and as it flows from the upper reaches the potential energy is turned into kinetic energy when the course of the river begins to gradually level out and this translates into increase in velocity. By the time river water reaches the middle of the transition zone, most of the potential energy would have been turned into kinetic energy and thus water velocity would be quite high here.
Animals living here would develop muscles because of constantly fighting against the strong current to avoid being swept downstream.
Explanation:
Formula which holds true for a leans with radii 
and 
and index refraction n is given as follows.
![\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%28n%20-%201%29%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
Since, the lens is immersed in liquid with index of refraction 
. Therefore, focal length obeys the following.
![\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf_%7B1%7D%7D%20%3D%20%5Cfrac%7Bn%20-%20n_%7B1%7D%7D%7Bn_%7B1%7D%7D%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
![\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%28n%20-%201%29%7D%20%3D%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
and, 
or, 
= 32.4 cm
Using thin lens equation, we will find the focal length as follows.
Hence, image distance can be calculated as follows.
= 47.9 cm
Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.
