impulse = F × t
The greater the impulse exerted on something, the greater will be the change in momentum.
impulse = change in momentum
Ft = ∆(mv)
A=F/m
a=(3000000)/(20000)
a=15 m/s^2
To rearrange these for v and r we must use BEDMAS.
First, times both sides by r, getting Ar = v²r/r
You can cancel out r/r to get Ar = v²
To get v, square root both sides. √Ar = √v²
Cancel out the square root and the squared v = √Ar
To get r, go back to Ar = v², then divide both sides by A. rA/A = v²/A
Then cancel out A/A to get r. r = v²/A
The force between two charged particles is an electrostatic force. The intensity of this force is given by the following formula:
![F_c=k_c\frac{q_1q_2}{r^2}](https://tex.z-dn.net/?f=F_c%3Dk_c%5Cfrac%7Bq_1q_2%7D%7Br%5E2%7D)
Where q1 and q2 are charges r is the distance, and k_c is Coulomb's constant.
Coulomb's constant has the value of:
![k_c=8.9875\cdot 10^9\frac{Nm^2}{C^2}](https://tex.z-dn.net/?f=k_c%3D8.9875%5Ccdot%2010%5E9%5Cfrac%7BNm%5E2%7D%7BC%5E2%7D)
When we plug all the given values into the formula we get:
![F_c=8.9875\cdot 10^9\cdot\frac{-2\cdot3}{80^2}=-8425781 $N](https://tex.z-dn.net/?f=F_c%3D8.9875%5Ccdot%2010%5E9%5Ccdot%5Cfrac%7B-2%5Ccdot3%7D%7B80%5E2%7D%3D-8425781%20%24N)
Negative sign means that force is attractive.