Distance from Pawtucket to Aleksandrovac
The shortest distance (air line) between Pawtucket and Aleksandrovac is 4,431.43 mi (7,131.70 km).
How far is Pawtucket from Aleksandrovac
Pawtucket is located in Rhode Island, United States within 41° 52' 27.84" N -72° 37' 32.52" W (41.8744, -71.3743) coordinates. The local time in Pawtucket is 04:52 (12.11.2025)
Aleksandrovac is located in Rasinska oblast, Serbia within 43° 27' 19.08" N 21° 3' 5.04" E (43.4553, 21.0514) coordinates. The local time in Aleksandrovac is 10:52 (12.11.2025)
The calculated flying distance from Aleksandrovac to Aleksandrovac is 4,431.43 miles which is equal to 7,131.70 km.
Pawtucket, Rhode Island, United States
Related Distances from Pawtucket
Aleksandrovac, Rasinska oblast, Serbia