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Eddi Din [679]
3 years ago
6

As a rocket ascends, its acceleration increases even though the net force on it stays constant. why? (assume a traveling distanc

e small enough that the thrust, acceleration due to gravity and atmosphere do not change. the faster a rocket moves the more acceleration is imparted to it from a given force. as a rocket ascends it's momentum increases with it's speed. the greater the momentum of the rocket, the greater the acceleration imparted to it from a given force. as the rocket ascends, fuel is burned at a faster rate resulting in a larger acceleration. the rocket's mass decreases as its fuel is consumed. the same net force acting on a smaller mass results in a larger acceleration.
Physics
2 answers:
topjm [15]3 years ago
5 0
<span>the rocket's mass decreases as its fuel is consumed. the same net force acting on a smaller mass results in a larger acceleration</span>
bogdanovich [222]3 years ago
5 0

Answer:

Explanation:

A rocket is provided thrust and acceleration through the fuel which is filled within it, burning of fuel at faster rate provide larger acceleration to the rocket. However, as the fuel get consumed the mass of rocket also decreases and with the reduction of it's mass it's momentum and speed increases thereby, it ascends with larger acceleration despite the force of gravity acting upon it in opposite direction.

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\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

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\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

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