Answer:
The maximum electrical force is
.
Explanation:
Given that,
Speed of cyclotron = 1200 km/s
Initially the two protons are having kinetic energy given by
![\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
When they come to the closest distance the total kinetic energy is converts into potential energy given by
Using conservation of energy
![mv^2=\dfrac{kq^2}{r}](https://tex.z-dn.net/?f=mv%5E2%3D%5Cdfrac%7Bkq%5E2%7D%7Br%7D)
![r=\dfrac{kq^2}{mv^2}](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7Bkq%5E2%7D%7Bmv%5E2%7D)
Put the value into the formula
![r=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{1.67\times10^{-27}\times(1200\times10^{3})^2}](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7B8.99%5Ctimes10%5E%7B9%7D%5Ctimes%281.6%5Ctimes10%5E%7B-19%7D%29%5E2%7D%7B1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%281200%5Ctimes10%5E%7B3%7D%29%5E2%7D)
![r=9.57\times10^{-14}\ m](https://tex.z-dn.net/?f=r%3D9.57%5Ctimes10%5E%7B-14%7D%5C%20m)
We need to calculate the maximum electrical force
Using formula of force
![F=\dfrac{kq^2}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bkq%5E2%7D%7Br%5E2%7D)
![F=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{(9.57\times10^{-14})^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B8.99%5Ctimes10%5E%7B9%7D%5Ctimes%281.6%5Ctimes10%5E%7B-19%7D%29%5E2%7D%7B%289.57%5Ctimes10%5E%7B-14%7D%29%5E2%7D)
![F=2.512\times10^{-2}\ N](https://tex.z-dn.net/?f=F%3D2.512%5Ctimes10%5E%7B-2%7D%5C%20N)
Hence, The maximum electrical force is
.
Answer:
The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".
Explanation:
Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.
Let's look at the speed of the car
![F = m a\\\\a =\frac{F}{m}](https://tex.z-dn.net/?f=F%20%3D%20m%20a%5C%5C%5C%5Ca%20%3D%5Cfrac%7BF%7D%7Bm%7D)
We use movies and find lips
![\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t](https://tex.z-dn.net/?f=%5Cto%20v%20%3D%20v_0%20%2B%20a%20t%5C%5C%5C%5C%5Cto%20v%20%3D%20v_0%20%2B%20%28%5Cfrac%7BF%7D%7Bm%7D%29%20t)
The moment is defined by
![\to p = m v](https://tex.z-dn.net/?f=%5Cto%20p%20%3D%20m%20v)
The moment change
![\Delta p = m v - m v_0](https://tex.z-dn.net/?f=%5CDelta%20p%20%3D%20m%20v%20-%20m%20v_0)
Let's replace the speeds in this equation
![\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0 + F t - m v_0\\\\\Delta p = F t](https://tex.z-dn.net/?f=%5CDelta%20p%20%3D%20m%20%28v_0%20%2B%20%5Cfrac%7BF%7D%7Bm%20t%7D%29%20-%20m%20v_0%5C%5C%5C%5C%5CDelta%20p%20%3D%20m%20v_0%20%20%2B%20F%20t%20-%20m%20v_0%5C%5C%5C%5C%5CDelta%20p%20%3D%20F%20t)
They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.
Answer:
Umm that's a personal question. All u have to do is say when have u pushed your personal limits....... Ummm one for me is when i had to try out for a select soccer and that is past my comfort zone.
Explanation: