Distance between Lubusz and Jacksonville

The distance between Lubusz (Poland) and Jacksonville (Florida, USA) is ca. 10.554,4 km (air line).

Lubusz

Koordinaten: 2°11′43″N 15°20′51″E
Latidute: 2.19527778
Longitude: 15.3475

Jacksonville

Latidute: 30.332184
Longitude: -81.655648

Distance between Lubusz and other cities and places worldwide

Distance between Lubusz and other major cities