Solution :
We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :
If we put an electron on point P, then force on point e is :
![F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B-eKQx%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B3%2F2%7D%7D%3D%20%5Cfrac%7B-eKQx%7D%7Br%5E3%5B1%2B%5Cfrac%7Bx%5E2%7D%7Br%5E2%7D%5D%5E%7B3%2F2%7D%7D)
If r >> x , then 
Then, 
Compare, a = -ω²x
We get,
Answer:
Electromagnetic radiation is an electric and magnetic disturbance traveling through space at the speed of light (2.998 × 108 m/s). It contains neither mass nor charge but travels in packets of radiant energy called photons, or quanta.
Answer:
12.6 (3 sig. fig.)
Explanation:
(12^2 + 4^2)^1/2 = 12.6 (3 sig. fig.)
I think it is either D) or E)
But i am going to go with E)
Answer:
Physics
Explanation:
Explanation:
We can use the Theorem of Work (W) and Kinetic Energy (K):
W=ΔK=Kf−Ki
it basically tells us that the work done on our system will show up as change in Kinetic Energy:
We know that the initial Kinetic Energy, Ki=12mv2i, is zero (starting from rest) while the final will be equal to 352J; Work will be force time displacement. so we get:
F⋅d=Ff
45d=352
and so:
d=35245=7.8≈8m
