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gavmur [86]
3 years ago
9

a) CALCULATE the acceleration of a 300,000 kg jumbo jet, just before takeoff when the thrust of each of its 4 engines is 36,000

N. If the jumbo jet take off speed is 184 mph, b) CONVERT 184 mph to m/s and c) DETERMINE how many seconds would it take for the plane to reach this speed? Show your work and round your answer to the nearest hundredth
Physics
1 answer:
Zanzabum3 years ago
5 0

Answer:

a) 0.12m/s^2

b)0.0511m/s

c)0.4258 seconds

Explanation:

a)

acceleration=  \frac{36000}{300000}

= 0.12m/s^2

b) 184m/h

184/3600 = 0.0511m/s

c)

time =  \frac{0.0511}{0.12}

Time = 0.4258 seconds

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Answer:

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Temperature at the left side surface is T₁ =  50°C

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Heat conduction process through wall is equal to the heat convection process so

Q_{conduction} = Q_{convection}

Expression for the heat conduction process is

Q_{conduction} = \frac{K(T_1 - T)}{L}

Expression for the heat convection process is

Q_{convection} = h(T_2 - T)

Substitute the expressions of conduction and convection in equation above

Q_{conduction} = Q_{convection}

\frac{K(T_1 - T_2)}{L} = h(T_2 - T)

Substitute the values in above equation

\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC

Now heat flux through the wall can be calculated as

q_{flux} = Q_{conduction} \\\\q_{flux}  = \frac{K(T_1 - T_2)}{L}\\\\q_{flux}  = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

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