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Chrome File Edit View History Bookmarks People Window Help 4)) 100% E Sat 11:10 PM a E JJ5521 x Chegg.com -> https://www.webassign.net/web/Student/Assignment-Responses/last?dep=1 16542732 ABP ABS Apps A&M; Organizations TCC Miscellaneous PT HotSchedules - Onl Reading 9. 2/3 points | Previous Answers OSColPhys1 7.7.041 My Notes Ask Your Teacher (a) How long would it take a 1.50 × 105 kg airplane with engines that produce a maximum of 100 MW of power to reach a speed of 250 m/s and an altitude of 12.0 km if air resistance were negligible? 223 (b) If it actually takes 875 s, what is the power applied, in megawatts? 25.5 (c) Given this power, what is the average force of air resistance if the airplane takes 1200 s? (Hint: You must find the distance the plane travels in 1200 s assuming constant acceleration.) 49.4 MW Additional Materials Reading Submit Answer Save Progress Another Version 10. 2/2 points | Previous Answers OSColPhys1 7.8.047 My Notes Ask Your Teacher (a) What is the efficiency of an out-of-condition professor who does 2.00 x 10 J of useful work while metabolizing 510 kcal of food energy? 9.33 8

Explanation / Answer

a) KE = 0.5*mv² = 0.5* 1.5x105kg * (250m/s)² =4.6875 GJ
PE = mgh = 1.5x105kg * 9.8m/s² * 12000m = 17.64 GJ
total work done = 22.3275 MJ
time = work / power = 22.3275GJ / 100MW = 223 s

b) power = work / time = 22.3275GJ / 875s = 25.51MW

c) total work = 25.5MW * 1200s = 30.6GJ
minus the KE and PE leaves friction work

30.6GJ -22.3275= 8.28 GJ
a = v / t = 250m/s / 1200s = 0.208 m/s²
s = 0.5at² = 0.5 * 0.208m/s² * (1200s)² = 1.5 x105m
friction force F = work / distance = 8.28GJ / 1.5 x105m = 5.52 kN

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