Distance from Secaucus to Lyaskovets
The shortest distance (air line) between Secaucus and Lyaskovets is 4,792.91 mi (7,713.45 km).
How far is Secaucus from Lyaskovets
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 10:39 (25.02.2025)
Lyaskovets is located in Veliko Tarnovo, Bulgaria within 43° 6' 0" N 25° 43' 0.12" E (43.1000, 25.7167) coordinates. The local time in Lyaskovets is 17:39 (25.02.2025)
The calculated flying distance from Lyaskovets to Lyaskovets is 4,792.91 miles which is equal to 7,713.45 km.
Secaucus, New Jersey, United States
Related Distances from Secaucus
Lyaskovets, Veliko Tarnovo, Bulgaria