Using Ohm's law:
#V = IR#
#V# is voltage in #"V"#.
#I# is current in #"C/s"#
#R# is resistance in whatever units convert #"C/s"# to #"V"#.
#(cancel("C")/cancel("s"))((cancel("s")*"V")/cancel("C")) = "V"#
Thus, #Omega = ("s"*"V")/"C"#
Since #"V" = "J"/"C"# #=> Omega = ("J"*"s")/("C"^2)#
#"J"*"s"# are the units of action. For example, if you had something like:
#int_(t_1)^(t_2) pdt = 1/2 mv^2Deltat = KDeltat#
the units would be according to #1/2 mv^2Deltat#, or #"J"*"s"#, and it means the linear momentum of an object's motion over time.
Therefore, #("V"*"s")/"C"^2# means that for every coulomb that travels through a circuit, the charge's momentum over time is decelerated some amount due to the electrical "density" of the wire. This magnitude of deceleration is directly proportional to the magnitude of the resistance (higher resistance, more deceleration of a charge, and thus the voltage drops across a resistor).
