Tuesday, October 15, 2019

PHYSICS (SOLUTIONS) Essay Example | Topics and Well Written Essays - 750 words

PHYSICS (SOLUTIONS) - Essay Example 100Pa). Here, it is essential that the pressure decreases exponentially, i.e. without local minimums and maximums (with the exception of atmospheric fluctuations); so, our criterion mb is valid. Roughly, from the graph attached we can localize km, say, km. e) Let us analyze the pressure curve . It is necessary to note that values of altitude are in log-scale. Nevertheless, even in linear scale dependence is non linear; see ('Atmospheric pressure' 2008) and (Ahren 2000). This can be explained by 'exponential atmosphere' model. This is very rough model for a column of gas extending to a great height; see details in (Feynman et al 1964). Such gas column is supposed to be at thermal equilibrium without any disturbances, so in the model. Model gives an obvious relationship where is the mass of the gas molecule (supposed to be constant in the model), is the acceleration due to gravity, and is the total number of the gas molecules in the unit section of a gas column. Solution of this relationship is ; here . So, the pressure exponentially decreases with increasing of altitude. Such decreasing depends upon the mass of gas molecule: the pressure of lightweight gases (e.g. hydrogen) decreases more slowly with altitude then the pressure of heavy -weight gases (e.g. oxygen). Here, and are weights of a ball... oxygen). Task 2 This task can be solved using Archimedes' principle. Let us make schematic illustration for a helium balloon and all the forces applied to it: Here, and are weights of a balloon and helium inside it, is the net force of buoyancy (for both balloon and helium inside). In accordance with Archimedes' principle, or the law of upthrust, a balloon is buoyed up by a force equal to the weight of the displaced air. A balloon is in state of rest when resulting force equals to zero, i.e. when . a) A balloon is assumed to be a sphere with a diameter of 20cm, so m. Hence, its volume is , or m3, or roughly litres (because 1 litre equals to m3). b) Let us estimate the mass of helium in the balloon. Let us suppose that helium is an ideal gas at and atmosphere. The ideal gas law gives volume occupied by a mole of an ideal gas: litres; here is universal gas constant (Feynman et al 1964). Therefore, the balloon contains moles of helium. The relative atomic mass of helium is 4, so a mole of helium weights 4 grams. Hence, helium in the balloon weights approx. grams. c) Using the graph attached for the task 1, namely dependence , let us define where km. Roughly, mb, say, mb or atmosphere. Let us estimate density of the air at the altitude km. For exponential model of atmosphere we have ; see (Feynman et al 1964). The ideal gas law or allows us to make estimations in terms of air density ; here grams per mole is the mean molar mass of air. For isothermal atmosphere () , so . Then, , so . Let us compute , the density of air at sea level: grams per litre. Let us estimate , the density of the air at the altitude km: or grams per litre. d) It is necessary to note that almost all numeric

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