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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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Dimension.

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2 years ago
What is the equivalent resistance between points A and C if R1=1430, R2=1350, R3=1100, R4=1350, and R5=1150.
Marianna [84]

R1 + R4 = 1430 + 1350 = 2780 = R14    series combination of R1 & R4

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The circuit has been reduced to 3 resistors in parallel

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7 0
3 years ago
The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso
nalin [4]

Explanation:

It s given that,

Mass of a planet, M=3.7\times 10^{24}\ kg

Radius of a planet, R=9.2\times 10^{6}\ m

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

g=\dfrac{GM}{R^2}

g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}

g=2.91\ m/s^2

(2) The escape velocity is given by :

v=\sqrt{\dfrac{2GM}{R}}

v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}

v = 7324.61 m/s

Hence, this is the required solution.

3 0
3 years ago
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