The least net force applied : Car 3(12 N)
<h3>Further explanation </h3>
Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object
∑F = m. a
Car 1 ⇒m=0.5 kg, a=36 m/s²
Car 2⇒m=0.8 kg, a=50 m/s²
Car 3⇒m=0.6, a=20 m/s²
Car 4⇒m=1, a=19~m/s²
The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.
<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.
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A3B2
bond is ionic
A is in group 2 (you can pick any like Ca)
B is in group 5 (like B)
the other question:
the reason is they are neutral gas and they already have 8 electrons except for He which is 2 and are completely stable so don't want to loose any electron vs Li and Na which have only 1 electron in the outer layer and are willing to loose that one to become stable.
