Question

. A jet airplane travelling at the speed of
500 km h–1 ejects the burnt gases at the speed
of 1100 km h–1 , relative to the jet airplane.
Find the speed of the burnt gases w.r.t. a
stationary observer on earth.

Answer 1

Answer:

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Explanation:

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Answer 2

Question:-

A jet airplane travelling at the speed of 400 km\h ejects gases at a speed of 800km\h the speed of the gas relative to rocket is or (speed of latter by seen by a observer in the ground.)

Explanation:

Given,

$\Large\vec{v}_{j}\:= \: 400\:Km\:h^{-1}$

$\Large\vec{v}_{cj}\:=\: -1200 \:Km\:h^{-1}h$

( Note:- here velocity of cumbustion or gas in negative direction)

To Find :-

$\vec{v}_{c}\:=\:?$

Solution:-

$\vec{v} _{cj} =\vec{v}_{c} - \vec{v}_{j} \\ - 1200 =\vec {v} _{c} - 400 \\ - 1200 + 400 = \vec{v} _{c} \\ ➜\vec{v} _{c} = - 800$

Note:-

$\vec{v}_{cj}$ is Velocity of cumbustion with respect to jet.

$\vec{v}_{c}$ is Velocity of cumbustion respect to the ground which we are finding.

$\vec{v}_{j}$ is Velocity of jet.

Velocity of cumbustion in 800 Km\h in opposite direction in plane.

The speed of gas is 800km\hr or latter

Or,

The speed observe by the observer is 800 Km\h.

Here, All statements are correct.