Question

The aqueous humor in a person’s eye is exerting a force of 0.300 N on the 1.10-cm2 area of the cornea. What pressure is this in mm Hg?

(a) 2.27 mm Hg
(b) 0.030 mm Hg
(c) 27.3 mm Hg
(d) 0.0027 mm Hg

Answer 1

The pressure exerted by the aqueous humor on the cornea is approximately 20.48 mm Hg. This value falls within the normal range for pressures in the eye.

Step-by-step explanation:

The pressure exerted by the aqueous humor on the cornea can be calculated using the formula Pressure = Force / Area. In this case, the force is 0.300 N and the area is 1.10 cm² (which can be converted to m² by dividing by 10000). Plugging these values into the formula, we get: Pressure = 0.300 N / (1.10 cm² / 10000) = 2730.3 N/m². To convert this pressure to mm Hg, we can use the conversion factor 1 N/m² = 0.0075 mm Hg. Multiplying the pressure by this conversion factor, we get a value of approximately 20.48 mm Hg.

Therefore, the pressure exerted by the aqueous humor on the cornea is approximately 20.48 mm Hg.

As for the second part of the question, the normal range for pressures in the eye is 12-24 mm Hg. Since the calculated pressure of 20.48 mm Hg falls within this range, it can be considered to be within the normal range for pressures in the eye.