Distance from Aleksandrovskoye to North Massapequa
Distance from Aleksandrovskoye to North Massapequa is 8,625 kilometers. This air travel distance is equal to 5,359 miles.
The air travel (bird fly) shortest distance between Aleksandrovskoye and North Massapequa is 8,625 km= 5,359 miles.
If you travel with an airplane (which has average speed of 560 miles) from Aleksandrovskoye to North Massapequa, It takes 9.57 hours to arrive.
Aleksandrovskoye
Aleksandrovskoye is located in Russia.
GPS Coordinates (DMS) | 44° 42´ 51.0120'' N 43° 0´ 2.9880'' E |
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Latitude | 44.71417 |
Longitude | 43.00083 |
Altitude | 295 m |
Country | Russia |
Aleksandrovskoye Distances to Cities
Aleksandrovskoye | Distance |
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Distance from Aleksandrovskoye to Chesterfield | 9,668 km |
Distance from Aleksandrovskoye to Cerritos | 11,096 km |
Distance from Aleksandrovskoye to Bloomfield | 8,656 km |
Distance from Aleksandrovskoye to Hanford | 10,861 km |
Distance from Aleksandrovskoye to Rapid City | 9,618 km |
North Massapequa
North Massapequa is located in United States.
GPS Coordinates | 40° 42´ 3.3480'' N 73° 27´ 43.4520'' W |
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Latitude | 40.70093 |
Longitude | -73.46207 |
Altitude | 17 m |
Country | United States |