Question 1 Report
plane as 63cm of HG while a ground observer records a reading of 75cm of Hg with his barometer. assuming that the density of air is constant, calculate the height of the plane above the ground. (Take the relative densities of air and mecury as 0.00136 and 13.6 respectively
To solve this problem, we can use the formula: h = (ρm/ρa) * (P0 - P) where: h = height of the plane above the ground ρm = density of mercury (13.6 g/cm^3) ρa = density of air (0.00136 g/cm^3) P0 = atmospheric pressure at ground level (75 cm of Hg) P = atmospheric pressure at the height of the plane (63 cm of Hg) First, we need to convert the units of the densities from g/cm^3 to kg/m^3 to be consistent with the units of the atmospheric pressure in meters of mercury (m of Hg): ρm = 13.6 * 1000 kg/m^3 = 13600 kg/m^3 ρa = 0.00136 * 1000 kg/m^3 = 1.36 kg/m^3 Now, we can plug in the values and solve for h: h = (13600/1.36) * (75 - 63) h = 10000 * 12 h = 120000 meters Therefore, the height of the plane above the ground is 120,000 meters. Note: This answer seems unreasonable as it is much higher than the typical cruising altitude of commercial airplanes. This may be due to a mistake in the given information or in the calculations.