Air Canada AC4735 flight
Air Canada domestic flight AC4735 serves route Newark to Jacksonville. The flight departs daily. Provided details are valid for the flight departing on 19th April, 2024.
Flight AC4735: Newark - Jacksonville
Departure
| Departure day | 19th April (Friday) |
| Departure time | 07:55 (7:55 am) |
| Departure terminal | Terminal «C» |
| Airport name | Newark Liberty |
| Airport IATA code | EWR |
| City | Newark |
| State / Province | NJ |
| Country | United States |
Arrival
| Arrival day | 19th April (Friday) |
| Arrival time | 10:23 (10:23 am) |
| Arrival terminal | N/A |
| Airport name | Jacksonville |
| Airport IATA code | JAX |
| City | Jacksonville |
| State / Province | FL |
| Country | United States |
General
| Flight number | AC4735 |
| Airline name | Air Canada |
| IATA carrier code | AC |
| Flight type | Domestic |
Operational
| Airplane type | 7M8 |
| Arrival day | Same day |
| Flight duration | 2h 28m |
| Distance | 1321 km / 820 mi. |
| Operated by | United Airlines flight UA1207 |
See also
All non-stop flights from Newark to Jacksonville Return flights (Jacksonville to Newark)Flight departure days
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