The current problem can solved by using Faraday's law of induction which is actually one of the Maxwell's equations. It is stated as follows:
F=qE
Where
F = Non-coulomb force
q = 6.0 C
E = (dB/dt)*r^2*(2/R); In which dB/dt = 4 T; r = 1 cm = 0.01 m, R = 6.0 cm = 0.06 m, E = Newtons/Coulob
Substituting;
E = 4*0.01^2*(2/0.06) = 1/75
Therefore,
F = qE = 6*1/75 = 0.08 N
Can you write the formula for kinetic energy ?
Here, let me help you:
Kinetic energy = (1/2) (mass) (speed)² .
Look at the (speed) in the formula. It's squared.
So if the speed gets multiplied by (something),
the kinetic energy gets multiplied by (something)² .
If the speed starts out at 1 m/s, but gets multiplied by 4,
then the kinetic energy gets multiplied by (4)² = 16 .
by momentum conservation we can say
![P_i = P_f](https://tex.z-dn.net/?f=P_i%20%3D%20P_f)
since gun and bullet was initially at rest so initial momentum of both must be ZERO
![0 = m_1v_{1f} + m_2v_{2f}](https://tex.z-dn.net/?f=0%20%3D%20m_1v_%7B1f%7D%20%2B%20m_2v_%7B2f%7D)
![0 = 3* 10^{-3}* 3000 + 1 * v_{2f}](https://tex.z-dn.net/?f=0%20%3D%203%2A%2010%5E%7B-3%7D%2A%203000%20%2B%201%20%2A%20v_%7B2f%7D)
![0 = 9 + v_{2f}](https://tex.z-dn.net/?f=0%20%3D%209%20%2B%20v_%7B2f%7D)
![v_{2f} = - 9 m/s](https://tex.z-dn.net/?f=v_%7B2f%7D%20%3D%20-%209%20m%2Fs)
<em>so it will recoil back with speed 9 m/s</em>
Answer:
I believe it's a compound.
Force is equal to mass multiply by acceleration.
force is also equal to mass multiply by gravity
F=ma
F=mg