Answer:
Secondary voltage of transformer is 905.23 volt
Explanation:
It is given number of turns in primary of transformer 
Number of turns in secondary 
Input voltage equation of the transformer
Here 
For transformer we know that
Therefore secondary voltage of transformer is 905.23 volt
Answer: A car initially traveling at 60 km/h accelerates at a constant rate of 2.0 m/s2. A spaceship far from any star or planet accelerates uniformly from 72 m/s to 160 m/s .
Explanation: i hoped that helped you.
Because there's no such thing as "really" moving.
ALL motion is always relative to something.
Here's an example:
You're sitting in a comfy cushy seat, reading a book and listening
to your .mp3 player, and you're getting drowsy. It's so warm and
comfortable, your eyes are getting so heavy, finally the book slips
out of your hand, falls into your lap, and you are fast asleep.
-- Relative to you, the book is not moving at all.
-- Relative to the seat, you are not moving at all.
-- Relative to the wall and the window, the seat is not moving at all.
-- But your seat is in a passenger airliner. Relative to people on the
ground, you are moving past them at almost 500 miles per hour !
-- Relative to the center of the Earth, the people on the ground are moving
in a circle at more than 700 miles per hour.
-- Relative to the center of the Sun, the Earth and everything on it are moving
in a circle at about 66,700 miles per hour !
How fast are they REALLY moving ?
There's no such thing.
It all depends on what reference you're using.
The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is
1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,
The distance from the center of the square to one of the corners is 
The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.
2) 
![V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a](https://tex.z-dn.net/?f=V_b%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B2%5Ctimes10%5E%7B-6%7D%7D%7B0.05%5Csqrt2%7D%20%2B%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B-2%5Ctimes10%5E%7B-6%7D%7D%7B0.05%7D%5C%5CV_b%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B2%5Ctimes10%5E%7B-6%7D%7D%7B0.05%7D%20%28%5Cfrac%7B1%7D%7B%5Csqrt2%7D-1%29%5C%5CV_b%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%20%284%5Ctimes%2010%5E%7B-5%7D%29%28-0.29%29%5C%5CV_b%20%3D%20%28-%5Cfrac%7B2.9%5Ctimes10%5E%7B-6%7D%7D%7B%5Cpi%5Cepsilon_0%7D%29%5Btex%5D%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E3%29%20The%20work%20done%20on%20q3%20by%20q1%20and%20q2%20is%20equal%20to%20the%20difference%20between%20%20energies.%20This%20is%20the%20work-energy%20theorem.%20So%2C%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%20U_b%20-%20U_a)
