Distance from Jefferson City to San Benito
Distance from Jefferson City to San Benito is 1,472 kilometers. This air travel distance is equal to 915 miles.
The air travel (bird fly) shortest distance between Jefferson City and San Benito is 1,472 km= 915 miles.
If you travel with an airplane (which has average speed of 560 miles) from Jefferson City to San Benito, It takes 1.63 hours to arrive.
Jefferson City
Jefferson City is located in United States.
| GPS Coordinates (DMS) | 38° 34´ 36.1200'' N 92° 10´ 24.6720'' W |
|---|---|
| Latitude | 38.57670 |
| Longitude | -92.17352 |
| Altitude | 195 m |
| Country | United States |
Jefferson City Distances to Cities
| Jefferson City | Distance |
|---|---|
| Distance from Jefferson City to Sacramento | 2,544 km |
| Distance from Jefferson City to Boston | 1,834 km |
| Distance from Jefferson City to San Francisco | 2,640 km |
| Distance from Jefferson City to Los Angeles | 2,386 km |
| Distance from Jefferson City to Dallas | 767 km |
San Benito
San Benito is located in United States.
| GPS Coordinates | 26° 7´ 57.2880'' N 97° 37´ 51.9600'' W |
|---|---|
| Latitude | 26.13258 |
| Longitude | -97.63110 |
| Altitude | 16 m |
| Country | United States |
San Benito Distances to Cities
| San Benito | Distance |
|---|---|
| Distance from San Benito to Weslaco | 36 km |
| Distance from San Benito to Phoenix | 1,612 km |
| Distance from San Benito to Bellevue Wa | 3,210 km |
| Distance from San Benito to Bel Air North | 2,474 km |
| Distance from San Benito to Arlington Heights | 1,978 km |