Answer:
That is not true all objects fall at the same speed excepts things like feathers or paper.
The way I do it is suddenly, in the same sort of way that magicians try to pull a table cloth off a table when there's things on the table cloth.The sudden approach acts as an impulse of force and starts to accelerate the roll. But, the piece (assuming it has perforations) is off the roll before the roll can move, due to inertia. Then the roll will acclerate, move, slow down and stop. However, in accelerating, the roll will unravel. The bigger the impulse the more it will unravel.+++++++++++++++++++++++++++++++++++++++If on the other hand, the piece of paper is held firmly, and the roll is pulled, then the impulse is presumably given to the paper and the hand whose inertia is a lot more than that of the roll. So, I think I'd actually go for choice c)+++++++++++++++++++++++++++++++++++++This assumes that the roll is free to rotate.I think that a similar idea is behind the design and use of a "ballistic galvanometer". The charge is passed through the galvanometer quickly, as a current pulse. Then the needle starts to deflect, and the deflection is arranged to depend on the total charge that has passed through in the time of the current pulse.
Answer:
x=0.53![x10^{-3} m](https://tex.z-dn.net/?f=x10%5E%7B-3%7D%20m)
Explanation:
Using Gauss law the field is uniform so
E=ζ/ε
Charge densities ⇒ζ=1.![x10x^{-6} \frac{C}{m^{2}}](https://tex.z-dn.net/?f=x10x%5E%7B-6%7D%20%5Cfrac%7BC%7D%7Bm%5E%7B2%7D%7D)
ε=8.85![x10^{-12} \frac{C^{2}}{n*m^{2}}](https://tex.z-dn.net/?f=x10%5E%7B-12%7D%20%5Cfrac%7BC%5E%7B2%7D%7D%7Bn%2Am%5E%7B2%7D%7D)
![E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1x10%5E%7B-6%7D%5Cfrac%7BC%7D%7Bm%5E%7B2%7D%7D%7D%7B8.85x%5E%7B-12%7D%5Cfrac%7BC%5E%7B2%7D%20%7D%7BN%2Am%5E%7B2%7D%7D%7D%20%5C%5CE%3D0.11299%20x10%5E%7B-6%7D%20%5Cfrac%7BN%7D%7BC%7D)
Force of charge is
![F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N](https://tex.z-dn.net/?f=F_%7Bq%7D%3Dq%2AE%5C%5CF_%7Bq%7D%3D1.6x10%5E%7B-19%7DC%2A0.11299x10%5E%7B6%7D%5Cfrac%7BN%7D%7BC%7D%20%5C%5CF_%7Bq%7D%3D1.807x10%5E%7B-14%7D%20N)
![F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s](https://tex.z-dn.net/?f=F_%7Bq%7D%3Dm%2Aa%5C%5Ca%3D%5Cfrac%7BF_%7Bq%7D%7D%7Bm%7D%3D%5Cfrac%7B1.807x10%5E%7B-13%7DN%7D%7B1.67x10%5E%7B-27%7D%7D%5C%5C%20a%3D1.082x0%5E%7B14%7D%20%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%5C%5C%20x%3D2.1cm%5Cfrac%7B1m%7D%7B100cm%7D%3D0.021m%20%5C%5Cv%3D6.7x10%5E%7B6%7D%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5C%20t%3D%5Cfrac%7B0.021m%7D%7B6.7x10%5E%7B6%7D%5Cfrac%7Bm%7D%7Bs%7D%7D%20%5C%5Ct%3D3.13x10%5E%7B-9%7Ds)
So finally knowing the acceleration and the time the distance can be find using equation of uniform motion
![x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2} \\x_{f}=0.53x^{-3}m](https://tex.z-dn.net/?f=x_%7Bf%7D%3Dx_%7Bo%7D%2B%5Cfrac%7B1%7D%7B2%7D%2Aa%2At%5E%7B2%7D%5C%5C%20x_%7Bo%7D%3D0%5C%5Cx_%7Bf%7D%3D%5Cfrac%7B1%7D%7B2%7D%20a%2At%5E%7B2%7D%3D%5Cfrac%7B1%7D%7B2%7D%2A1.082x10%5E%7B14%7D%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%2A%283.134x%5E%7B-9%7Ds%29%5E%7B2%7D%20%20%5C%5Cx_%7Bf%7D%3D0.53x%5E%7B-3%7Dm)
neutron star is formed when the core of a star, that was one ginormous, collapse on its own mass thus it loses it's volume exponentially (y=a^x) while the mass of the star decreases linearly (y=mx+c) thus the neutron star has mass of 10 to 29 sun in the radius of 10 km( for scale earth has 6731 km) thus it's density is super high!!!!!!!!!!!!!!!!!!!
hope it helps you! ;)