Verified answer

A Boeing 747 Jumbo Jet of mass 3.70×105kg takes 40s to accelerate from rest to a take-off velocity of 290km/hr?

a) What is the take-off velocity ? Give your answer in m s-1

V t-o = 290 km/h = 290*1000/3600 = 80,56 m/sec

b)What is the kinetic energy of the aircraft when it is travelling at its take-off velocity ?Give your answer in Joules

Ek = 1/m*V^2 = 1.85*10^5*8.056^2*10^2 = 1,20*10^9 joule

c)What is the average power generated by the engines during take-off ?Give your answer in Watts

Pav = Ek/t = 1,20*10^9/40 = 3.00*10^7 watt (30 Mwatt)

d)If a typical household uses electrical energy at an average rate of 1100W , how many houses could be powered at the rate calculated in Part C ?Round your answer to 2 significant figures.

3.00*10^7/1.1*10^3 = 2.727*10^4

use correct units

F= Ma

ke = 1/2mv^2

power =fF* V and E/ T

notice at V= 0 power is 0 , all energy goes into exhaust

but F = change in momentum of exhaust /time