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Apr 3, 2024
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Lockheed U-2 Author Affiliation Course Instructor Due Date
Question 1 The length of the airplane is 63 ft., which is equivalent to 19.20 m (Wikipedia contributors, 2024). Question 2 The diameter of the balloon can be estimated by the shadow cast by the U-2 aircraft. The method works based on: e As sunlight reaches the earth. the rays are nearly parallel due to the distance between earth and the sun. e Therefore, as the shadow falls on the balloon, the line which connects the shadow to the aircraft is nearly perpendicular to the airplane. Its long dimension intersects the balloon. e So, the line is used as a valuable stick to estimate the diameter of the balloon because it represents the actual distance from the aircraft to the shadow. e Measuring the length of the canopy helps to estimate the diameter of the balloon. The explanation can be actualized to determine the diameter of the balloon, which is equal to the airplane length of 19.20 m Question 3 To determine the volume of the balloon, the formula applied to calculate the volume of a sphere =§ 7r3 is used. The assumption is based on the information provided about the balloon being nearly spherical. The radius is half of the diameter = 19.20/2 = 9.6 m. V=1 = 2m(9.6)% = 3705.97 m. Question 4 Inspecting the units on the right-hand side of equation H: H = KT/mg (Wikipedia contributors, 2023)
Where K is Boltzmann constant = 1.380649 X 10™2%m?kg/s?K T is the temperature in Kelvin, K m is mass in kg g is the acceleration due to gravity in m/s? In unit form, m?.kg. K.s? ~ s2.K.kg.m Simplifying the equation through cancellation, m remains as the units for H, which is meters. Question 5 Nitrogen molecule has two nitrogen atoms. Therefore, the mass = 14x2=28. The atomic mass unit of nitrogen is converted to kg. One mass atomic unit =1.66 X 10727, For 28 =1.66 x 107?7 X 28 = 4.648 x 1072°. H=KT/mg (1.380649x1072%)x275 H=—rm——— 4.648X10726x9.81 =83268.52m = 83.27km Question 6 Calculating the atmospheric pressure at Balloon altitude: P=Po.e 2/ The scale height obtained in question 5 H of 83268.52m and an altitude, z of 18,000m are used. The surface pressure is 1013 hpa. P=1013.25 X e 18000/8326852 _g16 27 hpa. Question 7 Finding the number of molecules at the obtained atmospheric pressure: Using the ideal gas law: P = nKT.
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Therefore, n = P/KT. n =81627pa/(1.380649 X 1072%) X 275 = 2.15 x 102% molecules /volume. Question 8 The mass displaced is calculated using the density of air at that altitude and the volume of the balloon to find molecules displaced. The mass is then gotten from there. m=n. V where nis 2.15 x 1025 molecules /volume, and V is 3705.97 m3. m = 2.15 x 102°%3705.97 = 7.968 % 1022 molecules Mass = (7.968 X 1028) x (4.648 x 1072°) =3703.53 kg. Question 9 To find the lift in newtons, Newton's second law, F = mg is used. F=3703.53 x 9.81 =36331.59 N. Question 10 The capacity of the balloon to lift a given load diminishes with altitude because as the balloon ascends, the atmospheric pressure decreases. Thus, the air density decreases according to Archimedes' Principle of diminishing buoyancy force (Poso, 2018). It explains the reason, at higher altitudes, the balloon has a lower lifting force. Party balloons do not rise to the fringes of outer space because as they rise, they eventually reach an altitude where atmospheric pressure is too low for them to get sufficient lift.
References Poso, F. J. (2018). Buoyancy. The Archimedes principle. GRIN Verlag. Wikipedia contributors. (2023, December 12). Scale height. Wikipedia. https://en.wikipedia.org/wiki/Scale_height Wikipedia contributors. (2024, March 12). Lockheed U-2. Wikipedia. https://en.wikipedia.org/wiki/Lockheed_U-2
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НА
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