Suchy Dwór, Pomeranian Voivodeship
See also: Suchy Dwór, Lower Silesian Voivodeship
Suchy Dwór | |
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Village | |
Suchy Dwór | |
Coordinates: 54°34′32″N 18°27′53″E / 54.57556°N 18.46472°E | |
Country | Poland |
Voivodeship | Pomeranian |
County | Puck |
Gmina | Kosakowo |
Population | 888 |
Suchy Dwór [ˈsuxɨ ˈdvur] is a village in the administrative district of Gmina Kosakowo, within Puck County, Pomeranian Voivodeship, in northern Poland.[1] It lies approximately 2 kilometres (1 mi) south-west of Kosakowo, 15 km (9 mi) south of Puck, and 26 km (16 mi) north-west of the regional capital Gdańsk.
For details of the history of the region, see History of Pomerania.
The village has a population of 888.
References
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Coordinates: 54°34′32″N 18°27′53″E / 54.57556°N 18.46472°E
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