Main answerThe net outward force exerted on a 1.0 m x 2.5 m cabin door if the outside pressure is 0.35 atm is 8,630 N.ExplanationThe equation to find net outward force exerted is given as;F = P x AWhere;F = Net outward force exertedP = PressureA = AreaThe pressure inside the commercial airliner is given as 1.00 atm (105 Pa). This is the pressure exerted on the inside surface of the cabin door.The pressure outside the commercial airliner is given as 0.35 atm. This is the pressure exerted on the outside surface of the cabin door.The net outward force exerted is determined by finding the difference between the pressure exerted on the inside surface of the cabin door and the pressure exerted on the outside surface of the cabin door.Substituting values into the formula, we get:F = (105 Pa - 0.35 atm × 1 atm/101325 Pa) × (1.0 m) × (2.5 m)F = (105000 Pa - 2579 Pa) × (1.0 m) × (2.5 m)F = 260616 Nm^-2 × 2.5 m × 1.0 mF = 651540 N or 6.52 x 10^5 NAs the net outward force is in the outward direction, the answer can be written in positive;Net outward force exerted = 8,630 N.