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Assignment 3 Life Sciences Math Problem Essay Paper (Math Problem Sample)

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NATS 1780 A 6.0 Y (Summer 2020) - Assignment 3 Due: No later than 11:00 pm EDT on August 12, 2020 via Moodle (Late Penalty: 25% per day - including weekends; 100% deduction after solutions are posted.) Instructions: ● You are expected to provide answers for every question. You are encouraged to show all of your work so that marks can be awarded for partially correct answers. ● Although you are encouraged to collaborate with your classmates, each of you is expected to submit a separate and distinct assignment. 1. For stations above Mean Sea Level (MSL), the pressure at the point of observation (pOBS ) can be corrected to its MSL equivalent pressure (pMSL ) through use of the Hydrostatic Equation, namely: 1 pMSL = pOBS + ρgzOBS . (Eq’n 1) In this equation, zOBS is the height of the observation point relative to MSL, ρ is the density of air at the point of observation and g is the acceleration due to gravity. a. Obtain a screenshot of EMOS similar to Figure 1, and include it in your submission. For the same date and time, obtain a pressure measurement from a weather service for a station close to EMOS; also include this screenshot in your submission. [4 marks, 4 mark deduction for not submitting two screenshots] b. Based on your EMOS screenshot (Question 1(a)), state the height (in m above MSL) of the EMOS meteorological observation station. [1 mark] c. Assuming that ρ is 1.2041 kg/m3 , g is 9.81 m/s 2 , and pOBS is your EMOS station air pressure in your figure (Question 1(a)), calculate pMSL from the equation above (i.e., Eq’n 1). [Recall: 1 hPa ≡ 100 Pa and 1 Pa = 1 kg / (m s 2 ). Note that 1.2041 kg/m3 is the value for the density of air at STP - namely at 20 °C and 1013.25 hPa.] [4 marks] d. Compare and contrast the result of your calculation above (Question 1(c)) with the EMOS-reported value for MSL air pressure in your EMOS screenshot (Question 1(a)). [2 marks] 1The Hydrostatic Equation was discussed during the “Air in Motion” lecture. This equation is also discussed in the course textbook Ahrens, Jackson & Jackson (1st Canadian Edition), pg. 250 and Appendix A-3. © L. I. Lumb - Sharing prohibited. Violators subject to legal and/or academic consequences. 1 e. Assuming that the reason for this difference is entirely due to density, calculate the density at EMOS using the Ideal Gas Law, namely: ρOBS = pOBS / RTOBS (Eq’n 2) Note that pOBS is the same EMOS station air pressure in the above figure, TOBS the EMOS air temperature from the above figure converted to K, and finally that R is a constant with the value 287.058 J/(kg K). [Note that the Joule, a unit of energy, is expressible as a (N m).] [4 marks] f. Re-calculate pMSL from Eq’n 1 above using the ‘improved’ value for density determined in Question 1(e). [4 marks] g. Compare and contrast the result of your calculation above (Question 1(f)) with the EMOS-reported value for MSL air pressure in your EMOS screenshot (Question 1(a)). [2 marks] h. Based on the comparisons (questions 1(d) and 1(g)) state the significance of pressure and temperature-adjusted density in the calculation of pMSL . [2 marks] i. Compare and contrast your calculated, density-corrected EMOS based value for MSL air pressure (Question 1(f)) with that obtained from a weather service for a station close to EMOS (Question 1(a)). [2 marks] Figure 1. Recent EMOS observables captured as indicated. This is sample data; you are expected to obtain current data as detailed in Question 1(a). You will need to include a screenshot like this one with your submission. © L. I. Lumb - Sharing prohibited. Violators subject to legal and/or academic consequences. 2 © L. I. Lumb - Sharing prohibited. Violators subject to legal and/or academic consequences. 3 2. Regarding precipitation. [40 marks] a. What determines the terminal velocity of a hydrometeor? [2 marks] b. How might each of the following affect terminal velocity: i. Updrafts. [1 mark] ii. Downdrafts. [1 mark] iii. Buoyancy. [2 marks] iv. Phase. [2 marks] v. Shape. [2 marks] c. Complete Table 1 by determining the terminal velocity of the hydrometeors based on their radii. Using this terminal velocity, and assuming a fall distance of 1.250 km, calculate the fall time and state whether the hydrometeor is cloud or rain. [12 marks] d. Determine the volume of the 397 μm radius hydrometeor in m3 . Assume it is spherical in shape, and comprised of pure water. [3 marks] e. Suppose 1,233,456 of these 397 μm hydrometeors accumulate over a period of 3 minutes. i. Determine the total volume of accumulated water in m3 . [2 marks] ii. Determine the height of the water, h, in mm that occupies a rain gauge of diameter 3 cm. Use h = n V / π R 2 to calculate the water height for a rain gauge of radius, R; here n represents the number of hydrometeors of volume, V (i.e., n V, your answer to 1(e)(i)). [4 marks] iii. What assumption is implied regarding the hydrometeors in the rain-gauge height formula of 1(e)(ii)? [1 mark] iv. Determine the corresponding hourly rain rate in mm/hr. [2 marks] v. Based on the intensity determined in 1(e)(iv), classify this as stratiform or cumuliform precipitation. Justify your answer. [2 marks] vi. How would the calculation of height in the rain gauge be affected if the precipitation type was snow versus rain? [2 marks] vii. EMOS measures precipitation. Yet, it is stated that: “precip only works when temperature > 0”. Why do you think this is the case? [2 marks] Radius (μm) Terminal Velocity (m/s) Fall Time Cloud or Rain 63 397 Table 1. Cloud versus rain classification on the basis of terminal-velocity derived fall times. *** THE END *** © L. I. Lumb - Sharing prohibited. Violators subject to legal and/or academic consequences. 4

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ASSIGNMENT 3
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ASSIGNMENT 3
Question: 1 (a)
Question: 1 (b)
From the figure above, the height of the EMOS is 196 m, therefore, z is also equal to 196 m.
Question: 1 (c)
pMSL = pOBS + ρgzOBS, 1 Pa = 1 kg / (ms2)
= (996.0 hPa) + (1.2041 kg/m 2) (196m)
= (996.0 hPa) + 2315 kg/ (ms2)
= (996.0 hPa) + 2315 Pa
= 996.0 hPa + (2315 Pa) (1 hPa/100 Pa)
= 996.0 hPa + 23.2 hPa
= 1019.2 hPa
Question: 1 (d)
The reported value for the MSL air pressure by EMOS was 1021.3 hP and my calculated MSL air pressure was 1019.2 hPa. These two values are very similar but they are approximately 2.1 hPa apart (Moser & Welzl, 2013). The error happens because we have to assume the standard temperature and pressure conditions (STP) the EMOS meteorological station is located in an area with a different temperature and pressure. Hence EMOS is not at STP (Baker, 2005).
Question: 1 (e)
P = 996.0 hPa / 287*270.25K
= 99600 Pa / 287*270.25
= 99600 / 77561.75
= 1.2841 kg/m^3

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