Distance between Ceuta and Waterbury

The distance between Ceuta (Spain) and Waterbury (Connecticut, USA) is ca. 5.758,1 km (air line).

Ceuta

Latidute: 35.88767602
Longitude: -5.30955744

Waterbury

Koordinaten: 41°33′22″N 73°2′29″W
Latidute: 41.55611111
Longitude: -73.04138889

Distance between Ceuta and other cities and places worldwide

Distance between Ceuta and other major cities