Distance from Secaucus to La Puebla del Rio
The shortest distance (air line) between Secaucus and La Puebla del Rio is 3,565.47 mi (5,738.07 km).
How far is Secaucus from La Puebla del Rio
Secaucus is located in New Jersey, United States within 40° 46' 51.6" N -75° 56' 2.76" W (40.7810, -74.0659) coordinates. The local time in Secaucus is 07:17 (17.11.2025)
La Puebla del Rio is located in Sevilla, Spain within 37° 16' 0.12" N -7° 57' 0" W (37.2667, -6.0500) coordinates. The local time in La Puebla del Rio is 13:17 (17.11.2025)
The calculated flying distance from La Puebla del Rio to La Puebla del Rio is 3,565.47 miles which is equal to 5,738.07 km.
Secaucus, New Jersey, United States
Related Distances from Secaucus
La Puebla del Rio, Sevilla, Spain