The problem is a straightforward application of Ampere's law:
(the magnetic field on the loop) x (loop length) = (permeability) x (total current inside the loop)
Try solving it first for only 1 wire with radius r1 and current I1 for r>r1.
I mean 3rd equation in your original post, ##Ft=\left(m_1+m_2\right)v##. Try going through the whole derivation from the beginning and see if you can get ##v##
Yes, one could use decimals, although it's better to convert to decimals at the very end not to lose precision, e.g. if I were to convert 1/6 as ≈0.1667 early I would lose some precision since 1/6 has infinite number of digits in decimal notation which would be lost.
There is a technique called Logarithmic differentiation, which is applied here. It sometimes leads to answer faster as in this case. Of course the answer should be same regardless of whether it is used or not.
On the last line, there are 130 and 60 decibels under the log function... Instead, according to definition, log function should be applied to intensities, and decibels produced as a result...
The key here is what the "relative" speed of the two particles means in this problem. After realizing that it is the speed of one particle in the system of reference where the other one is at rest, everything else is straightforward. This system is one which moves at the same speed as one of the...
From the definition, (ratios of) intensities should be "fed" to the log function, and decibels are what comes out. In your equations, the log function seems to be "fed" with the decibels...
Sure, try plugging first two equations into second, you can see that ##Ft## cancels on both sides, after that, 1 remains the left, and ##\left(\frac{1}{4}+\frac{1}{6}\right)v## on the right. Then, rearrange to get ##v##.