Explanation:
Area of ring 
Charge of on ring 
Charge on disk
![\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dd%20v%20%26%3D%5Cfrac%7Bk%20d%20q%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5C%26%3D2%20%5Cpi-k%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5Cv%281%29%20%26%3D2%20%5Cpi%20c%20k%20%5Cint_%7B0%7D%5E%7BR%7D%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5Ccdot_%7B2%20%5Cvarepsilon_%7B0%7D%7D%5E%7B2%7D%20R%20%5C%5C%26%3D2%20%5Cpi%20%5Csigma%20k%5B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%26%3D%5Cfrac%7B2%20%5Cpi%20%5Csigma%7D%7B4%20%5Cpi%20%5Cvarepsilon_%7B0%7D%7D%5B%5Csqrt%7Bz%5E%7B2%7D%2BR%5E%7B2%7D%7D-%2821%29%5D%20%5C%5C%26%3D%5Cfrac%7B%5Csigma%7D%7B2%7D%28%5Csqrt%7B2%5E%7B2%7D%2BR%5E%7B2%7D%7D-2%29%5Cend%7Baligned%7D)
Note: Refer the image attached
Answer:
Magnetic Field Strength
Explanation:
The effect of the splitting of the spectral lines of an atom into several of its components when a static magnetic field is present is referred to as Zeeman Effect.
The spectral lines here corresponds to the various discrete energy levels in the sun at which the emission and absorption of the different energy levels of the sun occurs.
The splitting of the spectral lines occurs due to the presence of the strong magnetic field on the surface of the sun.
Thus the Zeeman Effect is used to measure the magnetic field strength on the sun's surface.
It a kilogram of feathers because i learned it in science
Answer:
0.1 rev/s
Explanation:
M = mass of the merry go round = 200 kg
R = radius of merry go round = 6 m
= Moment of inertia of merry go round = (0.5) MR² = (0.5) (200) (6)² = 3600 kgm²
m = mass of the man = 100 kg
= Moment of inertia of merry go round when man sits on it at the edge = (0.5) MR² + mR² = (0.5) (200) (6)² + (100) (6)² = 7200 kgm²
= initial Angular speed of merry-go-round before man sit = 0.2 rev/s
= Angular speed of merry-go-round after man sit = ?
Using conservation of angular momentum
=
(3600) (0.2) = (7200)
= 0.1 rev/s
Answer:
i dont know i will get someone to help u in a minute
Explanation:
