Answer:
840 m/s
Explanation:
Given that,
In the first second the rocket ejects 1/160 of its mass as exhaust gas and has an acceleration of 14.0 m/s².
We need to find the speed of the exhaust gas relative to the rocket.
The thrust of rocket is given by :
[tex]T=v_{gas}\dfrac{dm}{dt}\\\\ma=v_{gas}\dfrac{dm}{dt}\\\\v_{gas}=\dfrac{ma}{\dfrac{dm}{dt}}\\\\v_{gas}=\dfrac{14m}{\dfrac{1}{60}m}\\\\v_{gas}=840\ m/s[/tex]
So, the speed of the exhaust gas relative to the rocket is 840 m/s.
The speed ( Vgas) of the exhaust gas relative to rocket is : 840 m/s
Given data :
In first round Rocket ejects 1/60 of mass as exhaust gas
Acceleration of rocket ( a ) = 14.0 m/s²
Determine the speed of the exhaust gas relative to rocket
We will apply the equation for Rocket thrust
T = Vgas * [tex]\frac{dm}{dt}[/tex]
where : T = ma
∴ Vgas = ma / [tex]\frac{dm}{dt}[/tex]
= 14 m / [tex]\frac{1}{60}[/tex] m
therefore V gas = 840 m/s
Hence we can conclude that the speed ( Vgas) of the exhaust gas relative to rocket is : 840 m/s
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