Question

PHYSICS
O
Q 19:-
A jet airplane travelling at the speed of 500
km h-1 ejects its products of combustion at
the speed of 1500 km h-1 relative to the jet
plane. The speed of the products of
combustion with respect to an observer on the
ground is​

Answer 1

Answer:

1000

Explanation:

please mark as brain list

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.