[tex]v = 2.45×10^3\:\text{m/s}[/tex]
Explanation:
Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as
[tex]F = \dfrac{d{p}}{d{t}}[/tex] (1)
Assuming that the velocity remains constant then
[tex]F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}[/tex]
Solving for [tex]v,[/tex] we get
[tex]v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)[/tex]
Before we plug in the given values, we need to convert them first to their appropriate units:
The thrust F is
[tex]F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}[/tex]
The exhaust rate dm/dt is
[tex]\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}[/tex]
[tex]\;\;\;\;\;= 1.36×10^4\:\text{kg/s}[/tex]
Therefore, the velocity at which the exhaust gases exit the engines is
[tex]v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}[/tex]
[tex]\;\;\;= 2.45×10^3\:\text{m/s}[/tex]
