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:
6010.457N
Explanation:
Centripetal acceleration = a= V²/R
At a radius of 3.6m and velocity of 16.12m/s,
Acceleration is
a = 16.12²/ 3.6 = 72.182 m/s²
Force = Mass (m) * Acceleration (a)
36 = m * 72.182
m = 36/72.182
At breaking point
Radius = 0.468 m and Velocity = 75.1 m/s
a = V²/R = 75.1²/0.468
a = 12051.3 m/s
F = Mass(m) * Acceleration (a)
F = m * 12051.3
m = F/ 12051.3
Settings the ratio of mass equal
m = m
=> 36/72.182 = F/12051.3
F = 12051.3 * 36/72.182
F = 6010.457N
To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.
Heat flow is obtained as follows:

Where,
F =View Factor
A = Cross sectional Area
Stefan-Boltzmann constant
T= Temperature
Our values are given as
D = 0.6m

The view factor between two coaxial parallel disks would be


Then the view factor between base to top surface of the cylinder becomes
. From the summation rule


Then the net rate of radiation heat transfer from the disks to the environment is calculated as





Therefore the rate heat radiation is 780.76W
A decrease in mass will decrease an objects weight because
weight = mass x gravitational constant
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